Optical Communications Pune University MCQs
Optical Communications Pune University MCQs
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Ray Theory Transmission”.
1. Who proposed the idea of transmission of light via dielectric waveguide structure?
a) Christian Huygens
b) Karpon and Bockham
c) Hondros and debye
d) Albert Einstein
Answer: c
Explanation: It was in the beginning of 20 th century where Hondros and debye theoretical and experimental study demonstrated that information can be transferred as a form of light through a dielectric waveguide.
2. Who proposed the use of clad waveguide structure?
a) Edward Appleton
b) Schriever
c) Kao and Hockham
d) James Maxwell
Answer: c
Explanation: The invention of clad waveguide structure raised the eyebrows of the scientists. The proposals by Kao and Hockham proved beneficial leading in utilization of optical fibre as a communication medium.
3. Which law gives the relationship between refractive index of the dielectric?
a) Law of reflection
b) Law of refraction
c) Millman’s Law
d) Huygen’s Law
Answer: b
Explanation: Snell’s Law of refraction states that the angle of incidence Ø1 and refraction Ø2 are related to each other and to refractive index of the dielectrics.
It is given by n1sinØ1 = n2sinØ2
where n1 and n2 are the refractive indices of two mediums. Ø1 and Ø2 are angles of incidence and refraction.
4. The light sources used in fibre optics communication are ____________
a) LED’s and Lasers
b) Phototransistors
c) Xenon lights
d) Incandescent
Answer: a
Explanation: LED’s and Lasers are the light sources used in optical communication. During the working process of optical signals they are both supposed to be switched on and of rapidly and accurately enough to transmit the signal. Also they transmit light further with fewer errors.
5. The ________ ray passes through the axis of the fiber core.
a) Reflected
b) Refracted
c) Meridional
d) Shew
Answer: c
Explanation: When a light ray is passed through a perfect optical fiber, any discontinuities at the core cladding interface would result in refraction rather than total internal reflection. Such light ray passes through the axis of fiber core and is called as meridional ray. This principle is used while stating the fundamental transmission properties of optical fiber.
6. Light incident on fibers of angles________the acceptance angle do not propagate into the fiber.
a) Less than
b) Greater than
c) Equal to
d) Less than and equal to
Answer: b
Explanation: Acceptance angle is the maximum angle at which light may enter into the fiber in order to be propagated. Hence the light incident on the fiber is less than the acceptance angle, the light will propagate in the fiber and will be lost by radiation.
7. What is the numerical aperture of the fiber if the angle of acceptance is 16 degree?
a) 0.50
b) 0.36
c) 0.20
d) 0.27
Answer: d
Explanation: The numerical aperture of a fiber is related to the angle of acceptance as follows:
NA = sin Ѳa
Where NA = numerical aperture
Ѳ = acceptance angle.
8. The ratio of speed of light in air to the speed of light in another medium is called as _________
a) Speed factor
b) Dielectric constant
c) Reflection index
d) Refraction index
Answer: d
Explanation: When a ray travels from one medium to another, the ray incident from a light source is called as incident ray. In passing through, the speed varies. The ratio of the speed of incident and the refracted ray in different medium is called refractive index.
9. When a ray of light enters one medium from another medium, which quality will not change?
a) Direction
b) Frequency
c) Speed
d) Wavelength
Answer: b
Explanation: The electric and the magnetic field have to remain continuous at the refractive index boundary. If the frequency is changed, the light at the boundary would change its phase and the fields won’t match. In order to match the field, frequency won’t change.
This set of Optical Communications Interview Questions and Answers focuses on “Electromagnetic Mode Theory for Optical Propagation”.
1. Which equations are best suited for the study of electromagnetic wave propagation?
a) Maxwell’s equations
b) Allen-Cahn equations
c) Avrami equations
d) Boltzmann’s equations
Answer: a
Explanation: Electromagnetic mode theory finds its basis in electromagnetic waves. Electromagnetic waves are always represented in terms of electric field E, magnetic field H, electric flux density D and magnetic flux density B. These set of equations are provided by Maxwell’s equations.
2. When λ is the optical wavelength in vacuum, k is given by k=2Π/λ. What does k stand for in the above equation?
a) Phase propagation constant
b) Dielectric constant
c) Boltzmann’s constant
d) Free-space constant
Answer: a
Explanation: In the above equation, k = 2Π/λ, also termed as wave equation, k gives us the direction of propagation and also the rate of change of phase with distance. Hence it is termed as phase propagation constant.
3. Constructive interference occur when total phase change after two successive reflections at upper and lower interfaces is equal to?
a) 2Πm
b) Πm
c) Πm/4
d) Πm/6
Answer: a
Explanation: The component of phase waves which is in x direction is reflected at the interference between the higher and lower refractive index media. It is assumed that such an interference forms a lowest order standing wave, where electric field is maximum at the center of the guide, decaying towards zero.
4. When light is described as an electromagnetic wave, it consists of a periodically varying electric E and magnetic field H which are oriented at an angle?
a) 90 degree to each other
b) Less than 90 degree
c) Greater than 90 degree
d) 180 degree apart
Answer: a
Explanation: In case of electromagnetic wave which occur only in presence of both electric and magnetic field, a particular change in magnetic field will result in a proportional change in electric field and vice versa. These changes result in formation of electromagnetic waves and for electromagnetic waves to occur both fields should be perpendicular to each other in direction of wave travelling.
5. A monochromatic wave propagates along a waveguide in z direction. These points of constant phase travel in constant phase travel at a phase velocity V p is given by?
a) V p =ω/β
b) V p =ω/c
c) V p =C/N
d) V p =mass/acceleration
Answer: a
Explanation: Velocity is a function of displacement. Phase velocity V p is a measure of angular velocity.
6. Which is the most important velocity in the study of transmission characteristics of optical fiber?
a) Phase velocity
b) Group velocity
c) Normalized velocity
d) Average velocity
Answer: b
Explanation: Group velocity is much important in relation to transmission characteristics of optical fiber. This is because the optical wave propagates in groups or form of packets of light.
7. What is refraction?
a) Bending of light waves
b) Reflection of light waves
c) Diffusion of light waves
d) Refraction of light waves
Answer: a
Explanation: Unlike reflection, refraction involves penetration of a light wave from one medium to another. While penetrating, as it passes through another medium it gets deviated at some angle.
8. The phenomenon which occurs when an incident wave strikes an interface at an angle greater than the critical angle with respect to the normal to the surface is called as ____________
a) Refraction
b) Partial internal reflection
c) Total internal reflection
d) Limiting case of refraction
Answer: c
Explanation: Total internal reflection takes place when the light wave is in the more dense medium and approaching towards the less dense medium. Also, the angle of incidence is greater than the critical angle. Critical angle is an angle beyond which no propagation takes place in an optical fiber.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Cylindrical Fiber”.
1. A multimode step index fiber has a normalized frequency of 72. Estimate the number of guided modes.
a) 2846
b) 2592
c) 2432
d) 2136
Answer: b
Explanation: A step-index fiber has a constant refractive index core. The number of guided modes in a step-index fiber are given by M = /2. Here M denotes the number of modes and V denotes normalized frequency.
2. A graded-index fiber has a core with parabolic refractive index profile of diameter of 30μm, NA=0.2, λ=1μm. Estimate the normalised frequency.
a) 19.32
b) 18.84
c) 16.28
d) 17.12
Answer: b
Explanation: Normalized frequency for a graded index fiber is given by V = 2Πa/λ. Substituting and calculating the values, we get option 18.84. Here, V denotes normalized frequency and NA = numerical aperture.
3. A step-index fiber has core refractive index 1.46 and radius 4.5μm. Find the cutoff wavelength to exhibit single mode operation. Use relative index difference as 0.25%.
a) 1.326μm
b) 0.124μm
c) 1.214μm
d) 0.123μm
Answer: c
Explanation: The cutoff wavelength is the wavelength beyond which no single mode operation takes place. On solving λc = 2Πan1 \(\sqrt{2∆}\)/V, we get option c. Here, V=2.405, n1 = refractive index of core, a=radius of core.
4. A single-mode step-index fiber or multimode step-index fiber allows propagation of only one transverse electromagnetic wave.
a) True
b) False
Answer: a
Explanation: Single mode step index fiber is also called as mono-mode step index fiber. As the name suggests, only one mode is transmitted and hence it has the distinct advantage of low intermodal dispersion.
5. One of the given statements is true for intermodal dispersion. Choose the right one.
a) Low in single mode and considerable in multimode fiber
b) Low in both single mode and multimode fiber
c) High in both single mode and multimode fiber
d) High in single mode and low in multimode fiber
Answer: a
Explanation: Single mode propagates only one wave or only one mode is transmitted. Therefore, intermodal dispersion is low in single mode. In multimode fibers, higher dispersion may occur due to varying group velocities of propagating modes.
6. For lower bandwidth applications ______________
a) Single mode fiber is advantageous
b) Photonic crystal fibers are advantageous
c) Coaxial cables are advantageous
d) Multimode fiber is advantageous
Answer: d
Explanation: In multimode fibers, intermodal dispersion occurs. The group velocities often differ which gradually restricts maximum bandwidth attainability in multimode fibers.
7. Most of the optical power is carried out in core region than in cladding.
a) True
b) False
Answer: a
Explanation: In an ideal multimode fiber, there is no mode coupling. The optical power launched into a particular mode remains in that mode itself. The majority of these modes are mostly confined to fiber core only.
8. Meridional rays in graded index fibers follow ____________
a) Straight path along the axis
b) Curved path along the axis
c) Path where rays changes angles at core-cladding interface
d) Helical path
Answer: b
Explanation: Meridional rays pass through axis of the core. Due to the varying refractive index at the core, the path of rays is in curved form.
9. What is the unit of normalized frequency?
a) Hertz
b) Meter/sec
c) Coulombs
d) It is a dimensionless quantity
Answer: d
Explanation: Normalized frequency of optical fiber is the frequency which exists at cut-off condition. There is no propagation and attenuation above cut-off. It is directly proportional to numerical aperture which is a dimensionless quantity; hence itself is a dimensionless quantity.
10. Skew rays follow a ___________
a) Hyperbolic path along the axis
b) Parabolic path along the axis
c) Helical path
d) Path where rays changes angles at core-cladding interface
Answer: c
Explanation: The ray which does not pass through the fiber axis is termed as skew ray. Unlike Meridional rays, skew rays are more in number which makes them follow a round path called as helical path.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Single-Mode Fibers”.
1. An optical fiber has core-index of 1.480 and a cladding index of 1.478. What should be the core size for single mode operation at 1310nm?
a) 7.31μm
b) 8.71μm
c) 5.26μm
d) 6.50μm
Answer: d
Explanation: Normalized frequency V<=2.405 is the value at which the lowest order Bessel function J=0. Core size optical-communication-questions-answers-single-mode-fibers-q1 .
2. An optical fiber has a core radius 2μm and a numerical aperture of 0.1. Will this fiber operate at single mode at 600 nm?
a) Yes
b) No
Answer: a
Explanation: V= 2πa.NA/λ. Calculating this equation, we get the value of V. V is the normalised frequency and should be below 2.405 in order to operate the fiber at single mode. Here, V=2.094, is less than 2.405. Thus, this optical fiber exhibit single mode operation.
3. What is needed to predict the performance characteristics of single mode fibers?
a) The intermodal delay effect
b) Geometric distribution of light in a propagating mode
c) Fractional power flow in the cladding of fiber
d) Normalized frequency
Answer: b
Explanation: A mode field diameter is a fundamental parameter of single mode fibers. It tells us about the geometric distribution of light. MFD is analogous to core diameter in multimode fibers, except in single mode fibers not all the light that propagates is carried in the core.
4. Which equation is used to calculate MFD?
a) Maxwell’s equations
b) Peterman equations
c) Allen Cahn equations
d) Boltzmann’s equations
Answer: b
Explanation: Mode field diameter is an important parameter for single mode fibers because it is used to predict fiber properties such as splice loss, bending loss. The standard technique is to first measure the far-field intensity distribution and then calculating mode field diameter using Peterman equations.
5. A single mode fiber has mode field diameter 10.2μm and V=2.20. What is the core diameter of this fiber?
a) 11.1μm
b) 13.2μm
c) 7.6μm
d) 10.1μm
Answer: d
Explanation: For a single mode fiber, MFD=2w0. Here, core radius optical-communication-questions-answers-single-mode-fibers-q5 Solving this equation, we get a=5.05μm. Core-diameter=2a=10.1μm.
6. The difference between the modes’ refractive indices is called as ___________
a) Polarization
b) Cutoff
c) Fiber birefringence
d) Fiber splicing
Answer: c
Explanation: There are two propagation modes in single mode fibers. These two modes are similar but their polarization planes are orthogonal. In actual fibers, there are imperfections such as variations in refractive index profiles. These modes propagate with different phase velocities and their difference is given by B f =n y – n x . Here, n y and n x are refractive indices of two modes.
7. A single mode fiber has a beat length of 4cm at 1200nm. What is birefringence?
a) 2*10 -5
b) 1.2*10 -5
c) 3*10 -5
d) 2
Answer: c
Explanation: B f =n y – n x = λ/L p . Here, λ=wavelength and L p = beat length. Solving this equation, we will get the answer.
8. How many propagation modes are present in single mode fibers?
a) One
b) Two
c) Three
d) Five
Answer: b
Explanation: For a given optical fiber, the number of modes depends on the dimensions of the cable and the variations of the indices of refraction of both core and cladding across the cross section. Thus, for a single mode fiber, there are two independent, degenerate propagation modes with their polarization planes orthogonal.
9. Numerical aperture is constant in case of step index fiber.
a) True
b) False
Answer: a
Explanation: Numerical aperture is a measure of acceptance angle of a fiber. It also gives the light gathering capacity of the fiber. For a single mode fiber, core is of constant refractive index. There is no variation with respect to core. Thus, Numerical aperture is constant for single mode fibers.
10. Plastic fibers are less widely used than glass fibers.
a) True
b) False
Answer: a
Explanation: The majority of the fibers are made up of glass consisting of silica. Plastic fibers are used for short distance transmissions unlike glass fibers which can also be used for long haul applications. Also, plastic fibers have higher attenuation than glass fibers.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Photonic Crystal Fibers & Attenuation”.
1. Photonic crystal fibers also called as ___________
a) Conventional fibers
b) Dotted fibers
c) Stripped fibers
d) Holey fibers
Answer: d
Explanation: Photonic crystal fibers contain a fine array of air holes running longitudinally down the fiber cladding. The microstructure within the fiber is highly periodic.
2. Conventional optical fibers has more transmission losses than photonic crystal fibers.
a) True
b) False
Answer: a
Explanation: Conventional optical fibers have several hundreds of losses in transmission. Photonic crystal fibers have resulted in reduction in overall transmission losses.
3. Losses in photonic crystal fibers are reduced to a level of ___________
a) 0.1dB/km
b) 0.2dB/km
c) 0.3dB/km
d) 0.4dB/km
Answer: c
Explanation: Conventional fibers have losses of several hundred decibels per km. The invention of photonic crystal tubes has reduced the losses by hundreds of decibels.
4. The high index contrast enables the PCF core to be reduced from around 8 μmin conventional fiber to ___________
a) Less than 1μm
b) More than 5μm
c) More than 3μm
d) More than 2μm
Answer: a
Explanation: PCF’s have a wider range of optical properties in comparison with standard fibers. The lesser the core, more is the intensity of light in the core and enhances the non-linear effects.
5. The periodic arrangement of cladding air holes in photonic band gap fibers provides for the formation of a photonic band gap in the ___________
a) H-plane of fiber
b) E-plane of fiber
c) E-H-plane of fiber
d) Transverse plane of fiber
Answer: d
Explanation: Photonic band gap fibers are a class of micro structured fiber in which periodic arrangement of air holes is required. As a PBG fiber exhibits a 2-dimensional band gap, than the wavelengths within this band gap cannot propagate perpendicular to the fiber axis.
6. In index-guided photonic crystal fiber structure, the dark areas are air holes. What does white areas suggests?
a) Air
b) Silica
c) Water
d) Plasma
Answer: d
Explanation: Index-guided photonic crystal fibers have greater index contrast because the cladding contains air-holes having refractive index 1. Both index guided and conventional fibers arises from the manner in which guided mode interacts with the cladding region.
7. Which is the unit of measurement of attenuation in optical fibers?
a) km
b) dB
c) dB/km
d) Coulomb’s
Answer: c
Explanation: Attenuation is also referred to as transmission loss. Channel attenuation helped to determine the maximum transmission distance prior to signal restoration. Attenuation is usually expressed in logarithmic unit of decibel. It is given by
α dB L = 10 log 10 P i / P o
Where α dB = signal attenuation per unit length
P i & P o = Input and output power.
8. The optical fiber incurs a loss in signal power as light travels down the fiber which is called as ___________
a) Scattering
b) Attenuation
c) Absorption
d) Refraction
Answer: b
Explanation: When the light is passed through the fiber, it travels a large amount of distance before it starts fading. It needs restoration in the path. This loss or fading is called as Attenuation.
9. If the input power 100μW is launched into 6 km of fiber, the mean optical power at the fiber output is 2μW. What is the overall signal attenuation through the fiber assuming there are no connectors or splices?
a) 15.23dB
b) 16.98dB
c) 17.12dB
d) 16.62dB
Answer: b
Explanation: Signal attenuation is usually expressed in decibels. It is given by
Signal attenuation=10 log 10 P i / P o
Where, P i & P o = Input and output power.
10. A device that reduces the intensity of light in optical fiber communications is ___________
a) compressor
b) Optical attenuator
c) Barometer
d) Reducer
Answer: b
Explanation: A compressor compresses the signal before transmission. It does not affect the intensity of light. Optical attenuator is a device that affects the intensity of light and incurs a loss in transmission.
11. A decibel may be defined as the ratio of input and output optical power for a particular optical wavelength.
a) True
b) False
Answer: a
Explanation: Signal attenuation refers to the loss in transmission and it needs a logarithmic unit to express. Decibel is mainly used for comparing two power levels. It has the advantage that the operations of multiplication and division reduce to addition and subtraction.
12. When the input and output power in an optical fiber is 120μW & 3μW respectively and the length of the fiber is 8 km. What is the signal attenuation per km for the fiber?
a) 3dB/km
b) 2dB/km
c) 1dB/km
d) 4dB/km
Answer: b
Explanation: Signal attenuation per unit length is given by
α dB L = 10 log 10 P i / P o
α dB L = 16 dB
α dB = 16 dB/L = 2dB/km.
This set of Optical Communications Questions and Answers for Freshers focuses on “Material Absorption & Fiber Bend Losses In Silicon Glass Fibers”.
1. Which of the following statements best explain the concept of material absorption?
a) A loss mechanism related to the material composition and fabrication of fiber
b) A transmission loss for optical fibers
c) Results in attenuation of transmitted light
d) Causes of transfer of optical power
Answer: a
Explanation: Material absorption is a loss mechanism that results in dissipation of transmitted optical power as heat in a waveguide. It can be caused by impurities or interaction with other components of the core.
2. How many mechanisms are there which causes absorption?
a) One
b) Three
c) Two
d) Four
Answer: b
Explanation: Absorption is a loss mechanism. It may be intrinsic, extrinsic and also caused by atomic defects.
3. Absorption losses due to atomic defects mainly include ___________
a) Radiation
b) Missing molecules, oxygen defects in glass
c) Impurities in fiber material
d) Interaction with other components of core
Answer: b
Explanation: Atomic defects are imperfections in the atomic structure of fiber material. Atomic structure includes nucleus, molecules, protons etc. Atomic defects thus contribute towards loss of molecules, oxygen, etc.
4. The effects of intrinsic absorption can be minimized by ___________
a) Ionization
b) Radiation
c) Suitable choice of core and cladding components
d) Melting
Answer: c
Explanation: Intrinsic absorption is caused by interaction of light with one or more components of the glass i.e. core. Thus, if the compositions of core and cladding are chosen suitably, this effect can be minimized.
5. Which of the following is not a metallic impurity found in glass in extrinsic absorption?
a) Fe 2+
b) Fe 3+
c) Cu
d) Si
Answer: d
Explanation: In the optical fibers, prepared by melting techniques, extrinsic absorption can be observed. It is caused from transition metal element impurities. In all these options, Si is a constituent of glass and it cannot be considered as an impurity to glass itself.
6. Optical fibers suffer radiation losses at bends or curves on their paths.
a) True
b) False
Answer: a
Explanation: Optical fibers suffer radiation losses due to the energy in the bend or curves exceeding the velocity of light in the cladding. Hence, guiding mechanism is inhibited, which in turn causes light energy to be radiated from the fiber.
7. In the given equation, state what αr suggests?
a) Radius of curvature
b) Refractive index difference
c) Radiation attenuation coefficients
d) Constant of proportionality
Answer: c
Explanation: Above equation represents the fiber loss. This loss is seen at bends and curves as the fibers suffer radiation losses at curves. These radiation losses are represented by a radiation attenuation coefficient (α r ).
8. A multimode fiber has refractive indices n1 = 1.15, n2 = 1.11 and an operating wavelength of 0.7μm. Find the radius of curvature?
a) 8.60μ m
b) 9.30μ m
c) 9.1μ m
d) 10.2μ m
Answer: b
Explanation: The radius of curvature of the fiber bend of a multimode fiber is given by
optical-communication-questions-answers-material-absorption-fiber-bend-losses-silicon-glass-fibers-q8
Where, Rc = radius of curvature
n1, n2 = refractive indices
λ = wavelength.
9. A single mode fiber has refractive indices n1=1.50, n2 = 2.23, core diameter of 8μ m , wavelength = 1.5μ m cutoff wavelength = 1.214μ m . Find the radius of curvature?
a) 12 mm
b) 20 mm
c) 34 mm
d) 36 mm
Answer: c
Explanation: The radius of curvature of the fiber bend of a single mode fiber is given by-
optical-communication-questions-answers-material-absorption-fiber-bend-losses-silicon-glass-fibers-q9
Where R = radius of curvature,
n1, n2 = refractive indices,
λ c = cutoff wavelength,
λ = operating wavelength.
10. How the potential macro bending losses can be reduced in case of multimode fiber?
a) By designing fibers with large relative refractive index differences
b) By maintaining direction of propagation
c) By reducing the bend
d) By operating at larger wavelengths
Answer: a
Explanation: In the case of multimode fibers, radius of curvature is directly proportional to core refractive index and operating wavelength. In order to reduce the macro bending losses, the operative wavelength must be small and fibers must have large relative refractive index difference. Losses are inversely proportional to refractive index differences.
11. Sharp bends or micro bends causes significant losses in fiber.
a) True
b) False
Answer: a
Explanation: Sharp bends usually have a radius of curvature almost near to the critical radius. The fibers with the radius near to the critical radius cause significant losses and hence they are avoided.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Linear & Non-Linear Scattering Losses”.
1. Rayleigh scattering and Mie scattering are the types of _____________
a) Linear scattering losses
b) Non-linear scattering losses
c) Fiber bends losses
d) Splicing losses
Answer: a
Explanation: Rayleigh scattering and Mie scattering both result from non-ideal physical properties of the fiber. These losses may be impossible to eradicate. Linear scattering mechanisms cause the transfer of optical power contained within one propagating mode to be transferred linearly into a different mode.
2. Dominant intrinsic loss mechanism in low absorption window between ultraviolet and infrared absorption tails is ___________
a) Mie scattering
b) Rayleigh scattering
c) Stimulated Raman scattering
d) Stimulated Brillouin scattering
Answer: b
Explanation: Rayleigh scattering results from non-ideal physical properties of fiber. It is a type of linear scattering loss and is difficult or impossible to eradicate. Hence, it is termed as dominant intrinsic mechanism.
3. Rayleigh scattering can be reduced by operating at smallest possible wavelengths.
a) True
b) False
Answer: b
Explanation: Rayleigh scattering results from inhomogeneity of a random nature occurring on a small level compared with the wavelength of light. The Rayleigh scattering is inversely proportional to the wavelength. Thus, as wavelength scattering reduces.
4. The scattering resulting from fiber imperfections like core-cladding RI differences, diameter fluctuations, strains, and bubbles is?
a) Rayleigh scattering
b) Mie scattering
c) Stimulated Brillouin scattering
d) Stimulated Raman scattering
Answer: b
Explanation: Linear scattering also occurs at inhomogeneity which are comparable in size with the guided wavelength. These results from non-perfect cylindrical structures of the waveguide and hence caused by fiber imperfections.
5. Mie scattering has in-homogeneities mainly in ___________
a) Forward direction
b) Backward direction
c) All direction
d) Core-cladding interface
Answer: a
Explanation: In Mie scattering, the scattering in-homogeneities size is greater thanλ/10. Also, the scattered intensity has an angular dependence which is very large. The in-homogeneities are mainly in the direction of guided wavelength i.e. in forward direction.
6. The in-homogeneities in Mie scattering can be reduced by coating of a fiber.
a) True
b) False
Answer: a
Explanation: Mie scattering is a type of linear scattering loss. It results from fluctuations in diameter, differences in core-cladding refractive index, and differences along the fiber length. Therefore, such in-homogeneities can be reduced by controlled extrusion and coating of the fiber.
7. Raman and Brillouin scattering are usually observed at ___________
a) Low optical power densities
b) Medium optical power densities
c) High optical power densities
d) Threshold power densities
Answer: c
Explanation: Raman and Brillouin scattering mechanism are non-linear. They provide optical gain but with a shift in frequency, thus contributing to attenuation for light transmission at a particular wavelength. They can be seen at high optical power densities.
8. The phonon is a quantum of an elastic wave in a crystal lattice.
a) True
b) False
Answer: a
Explanation: A phonon is an elastic arrangement of atoms or molecules in condensed matter. This matter maybe solids or liquids. A phonon is a discrete unit of vibrational mechanical energy given by hf joules;
Where h = Planck’s constant
f = frequency.
9. A single-mode optical fiber has an attenuation of 0.3dB/km when operating at wavelength of 1.1μm. The fiber core diameter is 4μm and bandwidth is 500 MHz. Find threshold optical power for stimulated Brillouin scattering.
a) 11.20 mw
b) 12.77 mw
c) 13.08 mw
d) 12.12 mw
Answer: b
Explanation: The threshold optical power stimulated Brillouin scattering is given by-
P B = 4.4*10 -3 d 2 λ 2 α dB v
Where, P B = threshold optical power
d = diameter of core
λ = wavelength
α dB = attenuation.
10. 0.4 dB/km, 1.4μm, 6μm, 550MHz. Find threshold optical power for stimulated Raman scattering.
a) 1.98 W
b) 1.20 W
c) 1.18 W
d) 0.96 W
Answer: c
Explanation: The threshold optical power stimulated Raman scattering is given by-
P R = 5.9*10 -2 d 2 λα dB
Where, P R = optical power for Raman scattering
d = diameter of core
λ = wavelength
α dB = attenuation.
11. Stimulated Brillouin scattering is mainly a ___________
a) Forward process
b) Backward process
c) Upward process
d) Downward process
Answer: b
Explanation: The incident photon in Stimulated Brillouin scattering reduces a phonon of acoustic frequency as well as scattered photon. This produces an optical frequency shift which varies with the scattering angle. This frequency shift is max. in backward direction reducing to zero in forward direction making Stimulated Brillouin scattering a backward process.
12. High frequency optical phonon is generated in stimulated Raman scattering.
a) False
b) True
Answer: b
Explanation: An acoustic proton is generated in Stimulated Brillouin scattering. Raman scattering may have an optical power threshold higher than Stimulated Brillouin scattering.
13. Stimulated Raman scattering occur in ___________
a) Forward direction
b) Backward direction
c) Upward direction
d) Forward and backward direction
Answer: d
Explanation: Stimulated Raman scattering is similar to Stimulated Brillouin scattering except that a high frequency phonon is generated in Stimulated Raman scattering. Stimulated Raman scattering can occur in forward and backward direction as it has optical power threshold higher than Stimulated Brillouin scattering.
14. Stimulated Raman scattering may have an optical power threshold of may be three orders of magnitude ___________
a) Lower than Brillouin threshold
b) Higher than Brillouin threshold
c) Same as Brillouin threshold
d) Higher than Rayleigh threshold
Answer: b
Explanation: Stimulated Raman scattering involves generation of high- frequency phonon. Stimulated Brillouin scattering on the other hand, involves the generation of an acoustic phonon in a scattering process.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Dispersion – Chromatic Dispersion “.
1. What is dispersion in optical fiber communication?
a) Compression of light pulses
b) Broadening of transmitted light pulses along the channel
c) Overlapping of light pulses on compression
d) Absorption of light pulses
Answer: b
Explanation: Dispersion of transmitted optical signal causes distortion of analog as well as digital transmission. When the optical signal travels along the channel, the dispersion mechanism causes broadening of light pulses and thus in turn overlaps with their neighboring pulses.
2. What does ISI stand for in optical fiber communication?
a) Invisible size interference
b) Infrared size interference
c) Inter-symbol interference
d) Inter-shape interference
Answer: c
Explanation: Dispersion causes the light pulses to broaden and overlap with other light pulses. This overlapping creates an interference which is termed as inter-symbol interference.
3. For no overlapping of light pulses down on an optical fiber link, the digital bit rate BT must be ___________
a) Less than the reciprocal of broadened pulse duration
b) More than the reciprocal of broadened pulse duration
c) Same as that of than the reciprocal of broadened pulse duration
d) Negligible
Answer: a
Explanation: The digital bit rate and pulse duration are always inversely proportional to each other.
B T < = \(\frac{1}{2}\) Γ
Where B T = bit rate
2Γ = duration of pulse.
4. The maximum bit rate that may be obtained on an optical fiber link is 1/3Γ.
a) True
b) False
Answer: b
Explanation: The digital bit rate is function of signal attenuation on a link and signal to noise ratio. For the restriction of interference, the bit rate should be always equal to or less than 1/2Γ.
5. 3dB optical bandwidth is always ___________ the 3dB electrical bandwidth.
a) Smaller than
b) Larger than
c) Negligible than
d) Equal to
Answer: b
Explanation: Optical bandwidth is half of the maximum data rate. For non-return:0 , bandwidth is same as bit rate. The bandwidth B for metallic conductors is defined by electrical 3dB points. Optical communication uses electrical circuitry where signal power has dropped to half its value due to modulated portion of modulated signal.
6. A multimode graded index fiber exhibits a total pulse broadening of 0.15μsover a distance of 16 km. Estimate the maximum possible bandwidth, assuming no intersymbol interference.
a) 4.6 MHz
b) 3.9 MHz
c) 3.3 MHz
d) 4.2 MHz
Answer: c
Explanation: The maximum possible bandwidth is equivalent to the maximum possible bitrate. The maximum bit rate assuming no inter-symbol interference is given by
B T = \(\frac{1}{2}\) Γ
Where B T = bandwidth.
7. What is pulse dispersion per unit length if for a graded index fiber, 0.1μs pulse broadening is seen over a distance of 13 km?
a) 6.12ns/km
b) 7.69ns/km
c) 10.29ns/km
d) 8.23ns/km
Answer: b
Explanation: The dispersion mechanism causes broadening of light pulses. The pulse dispersion per unit length is obtained by dividing total dispersion of total length of fiber.
Dispersion = 0.1*10 -6 /13 = 7.69 ns/km.
8. Chromatic dispersion is also called as intermodal dispersion.
a) True
b) False
Answer: b
Explanation: Intermodal delay is a result of each mode having a different group velocity at a single frequency. The intermodal delay helps us to know about the information carrying capacity of the fiber.
9. Chromatic dispersion is also called as intermodal dispersion.
a) True
b) False
Answer: b
Explanation: Intermodal delay, the name only suggests, includes many modes. On the other hand chromatic dispersion is pulse spreading that takes place within a single mode. Chromatic dispersion is also called as intermodal dispersion.
10. The optical source used in a fiber is an injection laser with a relative spectral width σ λ /λ of 0.0011 at a wavelength of 0.70μm. Estimate the RMS spectral width.
a) 1.2 nm
b) 1.3 nm
c) 0.77 nm
d) 0.98 nm
Answer: c
Explanation: The relative spectral width σ λ /λ= 0.01 is given. The rms spectral width can be calculated as follows:
σ λ /λ = 0.0011
σ λ = 0.0011λ
= 0.0011*0.70*10 -6
= 0.77 nm.
11. In waveguide dispersion, refractive index is independent of ______________
a) Bit rate
b) Index difference
c) Velocity of medium
d) Wavelength
Answer: d
Explanation: In material dispersion, refractive index is a function of optical wavelength. It varies as a function of wavelength. In wavelength dispersion, group delay is expressed in terms of normalized propagation constant instead of wavelength.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Intermodal Dispersion”.
1. Intermodal dispersion occurring in a large amount in multimode step index fiber results in ____________
a) Propagation of the fiber
b) Propagating through the fiber
c) Pulse broadening at output
d) Attenuation of waves
Answer: c
Explanation: Pulse broadening due to intermodal dispersion is caused due to difference in propagation delay between different modes in the multimode fiber. As different modes travel with different group velocities, the pulse width at output depends on transmission time of all modes. This creates difference in overall dispersion which results in pulse broadening.
2. After Total Internal Reflection the Meridional ray __________
a) Makes an angle equal to acceptance angle with the axial ray
b) Makes an angle equal to critical angle with the axial ray
c) Travels parallel equal to critical angle with the axial ray
d) Makes an angle equal to critical angle with the axial ray
Answer: d
Explanation: The Meridional ray travels along the axis of the fiber. When the ray is incident, makes an angle equal to acceptance angle and thus it propagates through the fiber. As the propagating ray gets refracted from the boundary, it makes an angle with the normal.
3. Consider a single mode fiber having core refractive index n1= 1.5. The fiber length is 12m. Find the time taken by the axial ray to travel along the fiber.
a) 1.00μsec
b) 0.06μsec
c) 0.90μsec
d) 0.30μsec
Answer: b
Explanation: The time taken by the axial ray to travel along the fiber gives the minimum delay time
T min = Ln1/c
Where L = length of the fiber
n1 = Refractive index of core
c = velocity of light in vacuum.
4. A 4 km optical link consists of multimode step index fiber with core refractive index of 1.3 and a relative refractive index difference of 1%. Find the delay difference between the slowest and fastest modes at the fiber output.
a) 0.173 μsec
b) 0.152 μsec
c) 0.96 μsec
d) 0.121 μsec
Answer: a
Explanation: The delay difference is given by
δTs = Ln1/c
Where δTs = delay difference
n1 = core refractive index
Δ = Relative refractive index difference
c = velocity of light in vacuum.
5. A multimode step-index fiber has a core refractive index of 1.5 and relative refractive index difference of 1%. The length of the optical link is 6 km. Estimate the RMS pulse broadening due to intermodal dispersion on the link.
a) 92.6 ns
b) 86.7 ns
c) 69.3 ns
d) 68.32 ns
Answer: b
Explanation: The RMS pulse broadening due to intermodal dispersion is obtained by the equation is given below:
σs = Ln1Δ/2√3c
Where σs = RMS pulse broadening
L = length of optical link
C = velocity of light in vacuum
n1 = core refractive index.
6. The differential attenuation of modes reduces intermodal pulse broadening on a multimode optical link.
a) True
b) False
Answer: a
Explanation: Intermodal dispersion may be reduced by propagation mechanisms. The differential attenuation of various modes is due to the greater field penetration of the higher order modes into the cladding of waveguide. These slower modes exhibit larger losses at any core-cladding irregularities.
7. The index profile of a core of multimode graded index fiber is given by?
a) N = n1 [1 – 2Δ(r 2 /a) 2 ] 1/2 ; r<a
b) N = n1 [3 – 2Δ(r 2 /a) 2 ] 1/2 ; r<a
c) N = n1 [5 – 2Δ(r 2 /a) 2 ] 1/2 ; r>a
d) N = n1 [1 – 2Δ(r 2 /a) 2 ] 1/2 ; r<a
Answer: d
Explanation: In multimode graded index fibers, many rays can propagate simultaneously. The Meridional rays follow sinusoidal trajectories of different path length which results from index grading.
8. Intermodal dispersion in multimode fibers is minimized with the use of step-index fibers.
a) True
b) False
Answer: b
Explanation: As multimode graded index fibers show substantial bandwidth improvement over multimode step index fibers. So, inter-modal dispersion in multimode fiber is minimized with the use of multimode graded index fibers.
9. Estimate RMS pulse broadening per km due to intermodal dispersion for multimode step index fiber where length of fiber is 4 km and pulse broadening per km is 80.6 ns.
a) 18.23ns/km
b) 20.15ns/km
c) 26.93ns/km
d) 10.23ns/km
Answer: b
Explanation:
The RMS pulse broadening per km due to intermodal dispersion for multimode step index fiber is given by
(σ s /L = 80.6/4 = 20.15
Where L = length of fiber
σ s = pulse broadening.
10. Practical pulse broadening value for graded index fiber lies in the range of __________
a) 0.9 to 1.2 ns/km
b) 0.2 to 1 ns/km
c) 0.23 to 5 ns/km
d) 0.45 to 8 ns/km
Answer: b
Explanation: As all optical fiber sources have a finite spectral width, the profile shape must be altered to compensate for this dispersion mechanism. The minimum overall dispersion for graded index fiber is also limited by other intermodal dispersion mechanism. Thus pulse broadening values lie within range of 0.2 to 1 ns/km.
11. The modal noise occurs when uncorrected source frequency is?
a) δf>>1/δT
b) δf=1/δT
c) δf<<1/δT
d) Negligible
Answer: a
Explanation: Modal noise is dependent on change in frequency. Frequency is inversely proportional to time. The patterns are formed by interference of modes from a coherent source when coherence time of source is greater than intermodal dispersion time δT within fiber.
12. Disturbance along the fiber such as vibrations, discontinuities, connectors, splices, source/detectors coupling result in __________
a) Modal noise
b) Inter-symbol interference
c) Infrared interference
d) Pulse broadening
Answer: a
Explanation: Disturbance along the fiber cause fluctuations in specific pattern. These speckle patterns have characteristics time longer than resolution time of detector and is known as modal noise.
13. The modal noise can be reduced by __________
a) Decreasing width of signal longitudinal mode
b) Increasing coherence time
c) Decreasing number of longitudinal modes
d) Using fiber with large numerical aperture
Answer: d
Explanation: Disturbances along fiber cause fluctuations in speckle patterns. Fibers with large numerical apertures support the transmission of large number of modes giving greater number of speckle, thereby reducing modal noise.
14. Digital transmission is more likely to be affected by modal noise.
a) True
b) False
Answer: b
Explanation: Analog transmission is more affected by modal noise due to higher optical power levels which is required at receiver when quantum noise effects are considered. So it is important to look into design considerations.
This set of Optical Communications Interview Questions and Answers for freshers focuses on “Overall Fiber Dispersion & Modified Single Mode Fibers”.
1. A multimode step index fiber has source of RMS spectral width of 60nm and dispersion parameter for fiber is 150psnm -1 km -1 . Estimate rms pulse broadening due to material dispersion.
a) 12.5ns km -1
b) 9.6ns km -1
c) 9.0ns km -1
d) 10.2ns km -1
Answer: c
Explanation: The RMS pulse broadening per km due to material dispersion is given by
σ m = σ λ LM
= 60*1* 150pskm -1
= 9.0nskm -1
Where σ λ = rms spectral width
L = length of fiber
M = dispersion parameter.
2. A multimode fiber has RMS pulse broadening per km of 12ns/km and 28ns/km due to material dispersion and intermodal dispersion resp. Find the total RMS pulse broadening.
a) 30.46ns/km
b) 31.23ns/km
c) 28.12ns/km
d) 26.10ns/km
Answer: a
Explanation: The overall dispersion in multimode fibers comprises both chromatic and intermodal terms. The total RMS pulse broadening σ T is given by
optical-communication-questions-answers-overall-fiber-dispersion-modified single-mode-fibers-q2
Where σm = RMS pulse broadening due to material dispersion
σ i = RMS pulse broadening due to intermodal dispersion.
3. Γ g = dβ / C*dk. What is β in the given equation?
a) Attenuation constant
b) Propagation constant
c) Boltzmann’s constant
d) Free-space
Answer: b
Explanation: Above given equation is an equation of transit time or a group delay for a light pulse. This light pulse is propagating along a unit length of a single mode fiber.
4. Most of the power in an optical fiber is transmitted in fiber cladding.
a) True
b) False
Answer: b
Explanation: Most of the power in optical fiber is transmitted in fiber core. This is because in multimode fibers, majority of modes propagating in the core area are far from cutoff. Hence more power is transmitted.
5. A single mode fiber has a zero dispersion wavelength of 1.21μm and a dispersion slope of 0.08 psnm -2 km -1 . What is the total first order dispersion at wavelength 1.26μm.
a) -2.8psnm -1 km -1
b) -3.76psnm -1 km -1
c) -1.2psnm -1 km -1
d) 2.4psnm -1 km -1
Answer: b
Explanation: The total first order dispersion for fiber at two wavelength is obtained by
D T = λS 0 /4 [1-(λ 0 /λ) 4 ]
= (1260*0.08*10 -12 )/4 * (1-[1550/1260] 4 )
= -3.76psnm -1 km -1
Where
λ 0 = zero dispersion wavelength
λ = wavelength
S 0 = dispersion slope
D T = total first order dispersion.
6. The dispersion due to material, waveguide and profile are -2.8nm -1 km -1 , 20.1nm -1 km -1 and 23.2nm -1 km -1 respectively. Find the total first order dispersion?
a) 36.2psnm -1 km -1
b) 38.12psnm -1 km -1
c) 40.5psnm -1 km -1
d) 20.9psnm -1 km -1
Answer: c
Explanation: The total dispersion is given by
D T = D M + D W + D P (psnm -1 km -1 )
Where
D W = waveguide dispersion
D M = Material dispersion
D P = profile dispersion.
7. Dispersion-shifted single mode fibers are created by __________
a) Increasing fiber core diameter and decreasing fractional index difference
b) Decreasing fiber core diameter and decreasing fractional index difference
c) Decreasing fiber core diameter and increasing fractional index difference
d) Increasing fiber core diameter and increasing fractional index difference
Answer: c
Explanation: It is possible to modify the dispersion characteristics of single mode fibers by tailoring of some fiber parameters. These fiber parameters include core diameter and relative index difference.
8. An alternative modification of the dispersion characteristics of single mode fibers involves achievement of low dispersion gap over the low-loss wavelength region between __________
a) 0.2 and 0.9μm
b) 0.1 and 0.2μm
c) 1.3 and 1.6μm
d) 2 and 3μm
Answer: c
Explanation: Dispersion characteristics can be altered by changing fiber parameters and wavelength. The achievement of low dispersion gap over the region 1.3 and 1.6μm modifies the dispersion characteristics of single mode fibers.
9. The fibers which relax the spectral requirements for optical sources and allow flexible wavelength division multiplying are known as __________
a) Dispersion-flattened single mode fiber
b) Dispersion-enhanced single mode fiber
c) Dispersion-compressed single mode fiber
d) Dispersion-standardized single mode fiber
Answer: a
Explanation: The dispersion-flattened single mode fibers are obtained by fabricating multilayer index profiles with increased waveguide dispersion. This is tailored to provide overall dispersion say 2psnm -1 km -1 over the wavelength range 1.3 to 1.6μm.
10. For suitable power confinement of fundamental mode, the normalized frequency v should be maintained in the range 1.5 to 2.4μm and the fractional index difference must be linearly increased as a square function while the core diameter is linearly reduced to keep v constant. This confinement is achieved by?
a) Increasing level of silica doping in fiber core
b) Increasing level of germanium doping in fiber core
c) Decreasing level of silica germanium in fiber core
d) Decreasing level of silica doping in fiber core
Answer: b
Explanation: The tailoring of fiber parameters provides suitable power confinement. These parameters may be diameter, index-difference, frequency etc. The doping level of germanium contributes to the tailoring of fiber parameters; which in turn provides suitable power confinement.
11. Any amount of stress occurring at the core-cladding interface would be reduced by grading the material composition.
a) True
b) False
Answer: a
Explanation: A problem arises with that of simple step index approach to dispersion shifting is high. The fibers produced exhibit high dopant-dependent losses at operating wavelengths. These losses are caused by induced-stress in the region of core-cladding interface. This can be reduced by grading the material composition of the fiber.
12. The variant of non-zero-dispersion-shifted fiber is called as __________
a) Dispersion flattened fiber
b) Zero-dispersion fiber
c) Positive-dispersion fiber
d) Negative-dispersion fiber
Answer: d
Explanation: The dispersion profile for non-zero dispersion shifted fiber is referred to as bandwidth non-zero-dispersion-shifted fiber. It was introduced to provide wavelength division multiplexed applications to be extended into the s-band. The variant of non-zero-dispersion-shifted fiber can also be referred to as dispersion compensating fiber.
13. Non-zero-dispersion-shifted fiber was introduced in the year 2000.
a) True
b) False
Answer: b
Explanation: Non-zero-dispersion-shifted fiber was introduced in mid-1990s to provide wavelength division multiplexing applications. In the year 2000, the dispersion profile for non-zero-dispersion-shifted fiber was introduced.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Polarization”.
1. For many applications that involve optical fiber transmission, an intensity modulation optical source is not required.
a) True
b) False
Answer: b
Explanation: In many optical fibers transmission, the cylindrical fibers used generally do not maintain polarization state of light input source not more than a few meters. So for this reason, optical sources intensity modulation is required.
2. The optical source used for detection of optical signal is ____________
a) IR sensors
b) Photodiodes
c) Zener diodes
d) Transistors
Answer: b
Explanation: Optical signal is generally detected by photodiodes because photodiode is generally insensitive to optical polarization or phase of light with the fiber.
3. An optical fiber behaves as a birefringence medium due to differences in ___________
a) Effective R-I and core geometry
b) Core-cladding symmetry
c) Transmission/propagation time of waves
d) Refractive indices of glass and silica
Answer: a
Explanation: In an optical fiber with ideal optically circulatory symmetric core, both polarization modes propagate with same velocities. These fibers have variations in internal and external stress; fiber bending and so exhibit some birefringence.
4. The beat length in a single mode optical fiber is 8 cm when light from a laser with a peak wavelength 0.6μm is launched into it. Estimate the modal birefringence.
a) 1×10 -5
b) 3.5×10 -5
c) 2×10 -5
d) 4×10 -5
Answer: a
Explanation: Modal birefringence can be obtained by-
B F = λ/L B = 0.8×10 -6 /0.08
= 1×10 -5
Where
λ = peak wavelength
L B = beat length.
5. Beat length of a single mode optical fiber is 0.6cm. Calculate the difference between propagation constants for the orthogonal modes.
a) 69.8
b) 99.86
c) 73.2
d) 104.66
Answer: d
Explanation: The difference between the propagation constant for two orthogonal modes can be obtained by:
β x – β y = 2Π/L B = 2×3.14/0.06
= 104.66
Where
β x & β y are propagation constants for slow & fast modes resp.
L B = beat length.
6. A polarization maintaining fiber operates at a wavelength 1.2μm and have a modal birefringence of 1.8*10 -3 . Calculate the period of perturbation.
a) 0.7 seconds
b) 0.6 seconds
c) 0.23 seconds
d) 0.5 seconds
Answer: b
Explanation: The period of perturbation is given by-
T = λ/B F Where λ is operating wavelength, B F = Birefringence, T = period of perturbation.
7. When two components are equally excited at the fiber input, then for polarization maintaining fibers δΓ g should be around ___________
a) 1.5ns/km
b) 1 ns/km
c) 1.2ns/km
d) 2ns/km
Answer: b
Explanation: The differential group delay δΓ g is related to polarization mode dispersion of fiber. This linear relationship to fiber length however applies only to short fiber-lengths in which birefringence are uniform.
8. Polarization modal noise can _________ the performance of communication system.
a) Degrade
b) Improve
c) Reduce
d) Attenuate
Answer: a
Explanation: Polarization modal noise is generally of larger amplitude than modal noise. It is obtained within multimode fibers and so it degrades the performance of the communication system and prevents transmission of analog signals.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Non-Linear Effects”.
1. The nonlinear effects in optical fibers are large.
a) True
b) False
Answer: b
Explanation: The nonlinear effect arises from the interactions between light waves and the material transmitting them and thus affects the optical signals. The nonlinear effects are usually small in optical fibers. They have power levels of up to few milliWatts.
2. How many categories of nonlinear effects are seen in optical fibers?
a) One
b) Two
c) Three
d) Four
Answer: b
Explanation: The nonlinear effects are separated on the basis of their characteristics. There are two such categories; one is scattering effect and the other is Kerr effect.
3. Which of the following is not related to Kerr effects?
a) Self-phase modulation
b) Cross-phase modulation
c) Four-wave mixing
d) Stimulated Raman Scattering
Answer: d
Explanation: Stimulated Raman Scattering is related to scattering. The other effects include modulation and mixing which are parts of Kerr effect.
4. Linear scattering effects are _______ in nature.
a) Elastic
b) Non-Elastic
c) Mechanical
d) Electrical
Answer: a
Explanation: Linear scattering effects are elastic because the scattered wave frequency is equal to incident wave frequency. Nonlinear scattering effects are purely inelastic.
5. Which thing is more dominant in making a fiber function as a bidirectional optical amplifier?
a) Core material
b) Pump source
c) Cladding material
d) Diameter of fiber
Answer: b
Explanation: Brillouin gain is always greater than Raman gain. It exists for light propagation in opposite direction to the pump source. Also Brillouin frequency shifts and gain bandwidth are much smaller than Raman. Raman amplification occurs for light propagating in either direction. Thus, pump source is more important in making a fiber function as bidirectional optical amplifier.
6. _________ semiconductor laser sources generally have broader bandwidths.
a) Injection
b) Pulsed
c) Solid-state
d) Silicon hybrid
Answer: b
Explanation: Pulsed semiconductor lasers have broader bandwidths. Therefore, these sources prove to be inefficient pump sources. They prove inefficient especially for narrow gain spectrum.
7. Nonlinear effects which are defined by the intensity – dependent refractive index of the fiber are called as ________
a) Scattering effects
b) Kerr effects
c) Raman effects
d) Tomlinson effects
Answer: b
Explanation: Kerr effects are nonlinear effects. Nonlinear effects are divided into scattering and Kerr effects. Scattering effects include scattering of phonon whereas Kerr effects include intensity refractive index parameters.
8. Self-phase modulation causes modifications to the pulse spectrum.
a) True
b) False
Answer: a
Explanation: Kerr effect results in different transmission phase for the peak of the pulse compared with leading and trailing edges. Self-phase modulation can broaden the frequency spectrum of the pulse as the time varying phase creates a time varying frequency.
9. Self-phase modulation can be used for _____________
a) Enhancing the core diameter
b) Wavelength shifting
c) Decreasing the attenuation
d) Reducing the losses in the fiber
Answer: b
Explanation: Self phase modulation is related to phase change. It imposes a positive frequency sweep on the pulse which in turn enables wavelength or frequency shifting.
10. The beating between light at different frequencies or wavelengths in multichannel fiber transmission causes ________
a) Attenuation
b) Amplitude modulation of channels
c) Phase modulation of channels
d) Loss in transmission
Answer: c
Explanation: Phase modulation is related to frequency and wavelength shifting. In multichannel fiber transmission, phase modulation causes generation of modulation sidebands at new frequencies. This phenomenon is called as four-wave mixing.
11. What is different in case of cross-phase modulation from self-phase modulation?
a) Overlapping but same pulses
b) Overlapping but distinguishable pulses
c) Non-overlapping and same pulses
d) Non-overlapping but distinguishable pulses
Answer: b
Explanation: In cross phase modulation, variation in intensity of one pulse width modulates the refractive index of the fiber which causes phase modulation of the overlapping phases. In self-phase modulation, this phase modulation broadens the pulse spectrum.
12. When three wave components co-propagate at angular frequency w1, w2, w3, then a new wave is generated at frequency w4, which is given by?
a) w4 = w1 – w2 – w3
b) w4 = w1 + w2 + w3
c) w4 = w1 + w2 – w3
d) w4 = w1 – w2 + w3
Answer: c
Explanation: This type of frequency mixing is called as four-wave mixing. This frequency combination is problematic for multichannel optical communication as they become phase matched if the channel wavelengths are near to zero dispersion wavelengths.
13. _____________ results from a case of nonlinear dispersion compensation in which the nonlinear dispersion compensation in which the nonlinear chirp caused by self-phase modulation balances, postpones, the temporal broadening induced by group velocity delay.
a) Four wave mixing
b) Phase modulation
c) Soliton propagation
d) Raman scattering
Answer: c
Explanation: Soliton propagation is a nonlinear dispersion phenomenon. It limits the propagation distance that can be achieved when acting independently. It balances broadening of light pulse.
This set of Optical Communications Questions and Answers for Experienced people focuses on “Preparation of Optical Fibers – Liquid Phase Techniques”.
1. What is a fundamental necessity in the fabrication of fibers for light transmission?
a) Same refractive index for both core and cladding
b) Pump source
c) Material composition of fiber
d) Variation of refractive index inside the optical fiber
Answer: d
Explanation: For fabrication of fibers, two different transparent materials to light over a wavelength range of 0.8 to 1.7μm are required. Fiber should exhibit low attenuation, absorption and scattering losses. The variation of refractive indices in a fiber is a necessity for fiber fabrication.
2. Which materials are unsuitable for the fabrication of graded index fiber?
a) Glass-like-materials
b) Mono-crystalline structures
c) Amorphous material
d) Silica based material
Answer: b
Explanation: In case of graded index fiber, it is essential that the refractive index of the material is varied by suitable doping with another compatible material. These two materials should have mutual solubility over a wide range of concentration. This is achieved only in glass-like-materials.
3. How many different categories are available for the methods of preparing optical glasses?
a) 1
b) 2
c) 3
d) 4
Answer: b
Explanation: The methods of preparing optical glasses are divided into two categories. One is the conventional glass refining technique and other is vapor-phase-deposition method.
4. What is the first stage in liquid-phase-technique?
a) Preparation of ultra-pure material powders
b) Melting of materials
c) Decomposition
d) Crystallization
Answer: a
Explanation: In liquid-phase-technique melting, the first stage includes the preparation of ultra-pure material powders. These are usually oxides or carbonates which decomposes during glass melting.
5. Which processes are involved in the purification stage in liquid-phase-technique?
a) Filtration, Co-precipitation, Re-crystallization
b) Decomposition, Filtration, Drying
c) Doping, Drying, Decomposition
d) Filtration, Drying, Doping
Answer: a
Explanation: The compounds such as oxides and carbonates are formed during the glass melting. The purification accounts for a large proportion of material cost. These compounds are commercially available. The purification involves filtration, co-precipitation, re-crystallization and drying.
6. At what temperature range, does the melting of multi components glass systems takes place?
a) 100-300 degree Celsius
b) 600-800 degree Celsius
c) 900-1300 degree Celsius
d) 1500-1800 degree Celsius
Answer: c
Explanation: The glass materials in the powdered form and have relatively low melting point. Thus, the glass materials are melted at relatively low temperatures in the range of 900-1300 degrees Celsius.
7. Fiber drawing using preform was useful for the production of graded index fibers.
a) True
b) False
Answer: b
Explanation: A technique for producing fine optical fiber waveguides is to make a preform using the rod in the tube process. This technique was useful for the production of step-index fibers with large core diameters. In this technique, achievement of low attenuation is not critical as there is a danger of including bubbles at the core-cladding interface.
8. The minute perturbations and impurities in the fiber drawing process using preform technique can result in very high losses of _____________
a) Between 500 and 1000 dB/km
b) Between 100 and 300 dB/km
c) Between 1200 and 1600 dB/km
d) More than 2000 dB/km
Answer: a
Explanation: The minute perturbations and impurities in the fiber include formations of bubbles and involvement of particulate matter. The losses due to such impurities can be between 500 and 1000 dB/km.
9. The liquid-phase melting technique is used for the production of fibers ___________
a) With a core diameter of 50μm
b) With a core diameter less than 100μm
c) With a core diameter more than 200μm
d) With a core diameter of 100μm
Answer: c
Explanation: The multicomponent glass fibers prepared continuously by liquid-phase melting technique have losses in the range of 5 and 20 dB/km at a wavelength of 0.85μm. This method is thus used for preparation of fibers with a large core diameter. Also this technique is used for the continuous production of fibers.
10. Graded index fibers produced by liquid-phase melting techniques are less dispersive than step-index fibers.
a) True
b) False
Answer: a
Explanation: Liquid-phase melting technique does not offer optimum parabolic profile fibers. This parabolic profile yields minimum pulse dispersion. Graded index fibers prepared using liquid-phase melting techniques are less dispersive but do not have the bandwidth-length products of optimum profile fibers.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Vapor – Phase Deposition Techniques”.
1. Which of the following is not a technique for fabrication of glass fibers?
a) Vapor phase oxidation method
b) Direct melt method
c) Lave ring method
d) Chemical vapor deposition technique
Answer: c
Explanation: Lave ring method refers to the deposition of a crystalline layer on a substrate. All the other methods, except lave ring method, refer to optical fiber fabrication.
2. _____________ technique is method of preparing extremely pure optical glasses.
a) Liquid phase
b) Radio frequency induction
c) Optical attenuation
d) Vapor Phase Deposition
Answer: d
Explanation: Vapor Phase Deposition techniques are used to prepare silica-rich glasses. These glasses exhibit highest transparency and optimal optical properties.
3. Which of the following materials is not used as a starting material in vapor-phase deposition technique?
a) SiCl 4
b) GeCl 4
c) O 2
d) B 2 O 3
Answer: d
Explanation: In vapor-phase deposition technique, starting materials are volatile organic compounds. These materials are distilled to reduce the concentration of transition metal impurities. B 2 O 3 is used as a dopant.
4. P 2 O 5 is used as a _____________
a) Dopant
b) Starting material
c) Cladding glass
d) Core glass
Answer: a
Explanation: P 2 O 5 is a non silica material. Dopants are formed from non silica materials so that refractive index modification is achieved. Other dopants include Ti O 2 , Ge O 2 , etc.
5. How many types of vapor-phase deposition techniques are present?
a) One
b) Two
c) Three
d) Four
Answer: b
Explanation: Vapor-phase deposition techniques are divided into two types. The two types are flame hydrolysis and chemical vapor deposition . Further, these two types are subdivided into two more sections.
6. ___________ uses flame hydrolysis stems from work on soot processes which were used to prepare the fiber with losses below 20 dB/km.
a) Outside vapor phase oxidation
b) Chemical vapor deposition
c) Liquid phase melting
d) Crystallization
Answer: a
Explanation: Outside vapor phase oxidation is a type of vapor flame hydrolysis. It was originally developed by Hyde. In this process, the glass composition is deposited from a ‘soot’ generated by hydrolyzing the halide vapors in an oxygen-hydrogen flame.
7. Complete the given reaction.
SiCl4 + 2H2O → SiO2 + ______
a) 2HCl
b) 4HCl
c) 2Cl 2
d) 4Cl 2
Answer: b
Explanation: SiCl 4 is a starting material used in vapour-phase deposition technique. Dopants are added to the starting material in presence of heat to give glass compound. In the above reaction SiO 2 along with 4HCl is obtained.
8. In modified chemical vapor deposition, vapor phase reactant such as _________ pass through a hot zone.
a) Halide and oxygen
b) Halide and hydrogen
c) Halide and silica
d) Hydroxides and oxygen
Answer: a
Explanation: Halide and oxygen are passed through the hot zone during chemical vapor deposition. Glass particles formed during this travel are deposited on the walls of silica tube which are moved back and forth allowing the particles to deposit layer by layer.
9. _________ is the stimulation of oxide formation by means of non-isothermal plasma maintained at low pressure in a microwave cavity surrounding the tube.
a) Outside Vapor Phase Oxidation
b) Vapor Axial Deposition
c) Modified Chemical Vapor Deposition
d) Plasma-activated Chemical Vapor Deposition
Answer: d
Explanation: PCVD method was first developed by Cuppers and Koenig’s. It involves a plasma-induced chemical vapor deposition inside a silica tube. It is different from MCVD process as it involves stimulation of oxide formation by means of non-isothermal plasma.
10. Only graded index fibers are made with the help of vapor-phase deposition techniques.
a) True
b) False
Answer: b
Explanation: Vapor phase deposition techniques are used for preparation of both step-index and graded index fibers. These techniques provide fibers with low attenuation losses. Also, it gives similar performance for the fabrication of both single mode and multimode fibers.
11. Modified Chemical Vapor Deposition process is also called as an inside vapor phase oxidation technique.
a) True
b) False
Answer: a
Explanation: MCVD process was developed by Bell Telephone Laboratories and Southampton University, UK. It is called as inside vapor phase oxidation as it takes place inside the silica tube at the temperatures between 1400 and 1600 degrees Celsius.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Optical Fibers”.
1. Multimode step index fiber has ___________
a) Large core diameter & large numerical aperture
b) Large core diameter and small numerical aperture
c) Small core diameter and large numerical aperture
d) Small core diameter & small numerical aperture
Answer: a
Explanation: Multimode step-index fiber has large core diameter and large numerical aperture. These parameters provides efficient coupling to inherent light sources such as LED’s.
2. A typically structured glass multimode step index fiber shows as variation of attenuation in range of ___________
a) 1.2 to 90 dB km -1 at wavelength 0.69μm
b) 3.2 to 30 dB km -1 at wavelength 0.59μm
c) 2.6 to 50 dB km -1 at wavelength 0.85μm
d) 1.6 to 60 dB km -1 at wavelength 0.90μm
Answer: c
Explanation: A multimode step index fibers show an attenuation variation in range of 2.6 to 50dBkm -1 . The wide variation in attenuation is due to the large differences both within and between the two overall preparation methods i.e. melting and deposition.
3. Multimode step index fiber has a large core diameter of range is ___________
a) 100 to 300 μm
b) 100 to 300 nm
c) 200 to 500 μm
d) 200 to 500 nm
Answer: a
Explanation: A multimode step index fiber has a core diameter range of 100 to 300μm. This is to facilitate efficient coupling to inherent light sources.
4. Multimode step index fibers have a bandwidth of ___________
a) 2 to 30 MHz km
b) 6 to 50 MHz km
c) 10 to 40 MHz km
d) 8 to 40 MHz km
Answer: b
Explanation: Multimode step index fibers have a bandwidth of 6 to 50 MHz km. These fibers with this bandwidth are best suited for short -haul, limited bandwidth and relatively low-cost application.
5. Multimode graded index fibers are manufactured from materials with ___________
a) Lower purity
b) Higher purity than multimode step index fibers.
c) No impurity
d) Impurity as same as multimode step index fibers.
Answer: b
Explanation: Multimode graded index fibers have higher purity than multimode step index fiber. To reduce fiber losses, these fibers have more impurity.
6. The performance characteristics of multimode graded index fibers are ___________
a) Better than multimode step index fibers
b) Same as multimode step index fibers
c) Lesser than multimode step index fibers
d) Negligible
Answer: a
Explanation: Multimode graded index fibers use a constant grading factor. Performance characteristics of multimode graded index fibers are better than those of multimode step index fibers due to index graded and lower attenuation.
7. Multimode graded index fibers have overall buffer jackets same as multimode step index fibers but have core diameters ___________
a) Larger than multimode step index fibers
b) Smaller than multimode step index fibers
c) Same as that of multimode step index fibers
d) Smaller than single mode step index fibers
Answer: b
Explanation: Multimode graded index fibers have smaller core diameter than multimode step index fibers. A small core diameter helps the fiber gain greater rigidity to resist bending.
8. Multimode graded index fibers with wavelength of 0.85μm have numerical aperture of 0.29 have core/cladding diameter of ___________
a) 62.5 μm/125 μm
b) 100 μm/140 μm
c) 85 μm/125 μm
d) 50 μm/125μm
Answer: b
Explanation: Multimode graded index fibers with numerical aperture 0.29 having a core/cladding diameter of 100μm/140μm. They provide high coupling frequency LED’s at a wavelength of 0.85 μm and have low cost. They are also used for short distance application.
9. Multimode graded index fibers use incoherent source only.
a) True
b) False
Answer: b
Explanation: Multimode graded index fibers are used for short haul and medium to high bandwidth applications. Small haul applications require LEDs and low accuracy lasers. Thus either incoherent or incoherent sources like LED’s or injection laser diode are used.
10. In single mode fibers, which is the most beneficial index profile?
a) Step index
b) Graded index
c) Step and graded index
d) Coaxial cable
Answer: b
Explanation: In single mode fibers, graded index profile is more beneficial as compared to step index. This is because graded index profile provides dispersion-modified-single mode fibers.
11. The fibers mostly not used nowadays for optical fiber communication system are ___________
a) Single mode fibers
b) Multimode step fibers
c) Coaxial cables
d) Multimode graded index fibers
Answer: a
Explanation: Single mode fibers are used to produce polarization maintaining fibers which make them expensive. Also the alternative to them are multimode fibers which are complex but accurate. So, single-mode fibers are not generally utilized in optical fiber communication.
12. Single mode fibers allow single mode propagation; the cladding diameter must be at least ___________
a) Twice the core diameter
b) Thrice the core diameter
c) Five times the core diameter
d) Ten times the core diameter
Answer: d
Explanation: The cladding diameter in single mode fiber must be ten times the core diameter. Larger ratios contribute to accurate propagation of light. These dimension ratios must be there so as to avoid losses from the vanishing fields.
13. A fiber which is referred as non-dispersive shifted fiber is?
a) Coaxial cables
b) Standard single mode fibers
c) Standard multimode fibers
d) Non zero dispersion shifted fibers
Answer: b
Explanation: A standard single mode fiber having step index profile is known as non-dispersion shifted fiber. As these fibers have a zero dispersion wavelength of 1.31μm and so are preferred for single-wavelength transmission in O-band.
14. Standard single mode fibers are utilized mainly for operation in ___________
a) C-band
b) L-band
c) O-band
d) C-band and L-band
Answer: c
Explanation: SSMFs are utilized for operation in O-band only. It shows high dispersion in the range of 16 to 20ps/nm/km in C-band and L-band. So SSMFs are used in O-band.
15. Fiber mostly suited in single-wavelength transmission in O-band is?
a) Low-water-peak non dispersion-shifted fibers
b) Standard single mode fibers
c) Low minimized fibers
d) Non-zero-dispersion-shifted fibers
Answer: b
Explanation: Standard single mode fibers with a step index profile are called non dispersion shifted fiber and it is particularly used for single wavelength transmission in O-band and as if has a zero-dispersion wavelength at 1.31μm.
This set of Optical Communications Multiple Choice Questions & Answers focuses on ” Optical Fiber Cables”.
1. When optical fibers are to be installed in a working environment, the most important parameter to be considered is?
a) Transmission property of the fiber
b) Mechanical property of the fiber
c) Core cladding ratio of the fiber
d) Numerical aperture of the fiber
Answer: b
Explanation: Nowadays, optical fibers are used alternatively to electric transmission lines. They are installed safely and maintained in all environments including underground areas. This requires mechanical strengthening of fibers in order to ensure proper transmission.
2. It is not important to cover these optical fibers required for transmission.
a) True
b) False
Answer: b
Explanation: Unprotected optical fibers have number of losses regarding its strength and durability. Bare glass fibers are brittle and have small cross-section area that makes them highly susceptible to damages while handling and maintenance. Thus, to improve tensile strength, optical fibers should be covered by surrounding them with number of protective layers.
3. Optical fibers for communication use are mostly fabricated from ___________
a) Plastic
b) Silica or multicomponent glass
c) Ceramics
d) Copper
Answer: b
Explanation: Silica or a compound of glass are brittle and have almost perfect elasticity until reaching their breaking point. Strength of these materials is high. Thus, optical fibers are fabricated from these materials.
4. An Si-O bond with a Young’s modulus of 9*10 10 Nm -1 have an elliptical crack of depth 7nm. The surface energy is 2.29 J. Estimate fracture stress for silica fiber.
a) 4.32*10 9 Nm -1
b) 6.32*10 9 Nm -1
c) 5.2*10 9 Nm -1
d) 3*10 9 Nm -1
Answer: a
Explanation: For an elliptical crack, the fracture stress is given by-
S f = (2Eγ p /πC) 1/2
Where S f = fracture stress
γ p = surface energy
C = depth of crack.
5. Calculate percentage strain at break for a Si-O bond with a fracture strength of 3.52*10 10 Nm -1 and Young’s modulus of 9 *10 9 Nm -1 .
a) 3.1 %
b) 2.8 %
c) 4.5 %
d) 3.9 %
Answer: d
Explanation: Young’s modulus is given by-
E = Stress/Strain
To calculate strain from the above formula, we have to divide stress by Young’s modulus. Therefore, Strain = Stress/E.
6. Stress corrosion must be considered while designing and testing optical fiber cables.
a) True
b) False
Answer: a
Explanation: Stress corrosion means growth of flaws due to stress and water. This occurs as a result of molecular bonds at the tip of crack being attacked by water. Hence, it is important to have a protection against water to avoid stress corrosion.
7. Which statistics are used for calculations of strengths of optical fibers?
a) Edwin statistics
b) Newton statistics
c) Wei-bull statistics
d) Gamma statistics
Answer: c
Explanation: Calculations of strengths are conducted using Wei-bull statistics in case of optical fibers. It describes the strength behavior of a system that is dependent on the weakest link of the system. The Wei-bull statistics gives the probability of failure of the optical fiber at a given strength.
8. What does n denotes in the equation given below, if v c is the crack velocity; A is the constant for the fiber material and KI is the strength intensity factor?
vc = AKIn
a) Refractive index
b) Stress corrosion susceptibility
c) Strain
d) Young’s modulus
Answer: b
Explanation: The above equation allows estimation of the time to failure of a fiber under stress corrosion conditions. The constant n is called as stress corrosion susceptibility. It is typically in the range of 15 to 50 for a glass.
This set of Optical Communications Interview Questions and Answers for Experienced people focuses on “Stability of the Fiber Transmission Characteristics”.
1. ____________ results from small lateral forces exerted on the fiber during the cabling process.
a) Attenuation
b) Micro-bending
c) Dispersion
d) Stimulated Emission
Answer: b
Explanation: Optical fibers must be designed so that the transmission characteristics of the fiber are maintained after the cabling process. The main problem which occurs in the cabling process is the meandering of the axis of the fiber core on a microscopic scale within the cable form. This phenomenon is called as micro-bending.
2. Microscopic meandering of the fiber core axis that is micro-bending is caused due to ___________
a) Environmental effects
b) Rough edges of the fiber
c) Large diameter of core
d) Polarization
Answer: a
Explanation: Micro-bending can be generated at any stage during manufacturing process, cable installation process or during service. This is mainly due to environmental effects, mainly varying temperatures causing differential expansion or contraction.
3. How many forms of modal power distribution are considered?
a) One
b) Two
c) Three
d) Four
Answer: b
Explanation: Two forms of modal power distribution are considered. The first form is seen when a fiber is excited by a diffuse Lambertian source, and is called as fully filled mode distribution. The second form occurs when, due to mode coupling and attenuation, the distribution of optical power becomes invariant with the distance of propagation along the fiber, and is called as steady-state mode distribution.
4. What does micro-bending losses depend on _____________
a) Core material
b) Refractive index
c) Diameter
d) Mode and wavelength
Answer: d
Explanation: Micro-bending losses cause differential expansion or contraction. These losses are mode dependent. The number of modes is a function inverse to the wavelength of the transmitted light and thus micro-bending losses are wavelength dependent.
5. The fiber should be________________ to avoid deterioration of the optical transmission characteristics resulting from mode-coupling-induced micro-bending.
a) Free from irregular external pressure
b) Coupled with plastic
c) Large in diameter
d) Smooth and in a steady state
Answer: a
Explanation: Micro-bending losses results from environmental effects such as temperature variation. The irregular external pressure deteriorates the quality of transmission through the fiber. Thus, controlled coating and cabling of the fiber is essential in order to reduce the cabled fiber attenuation.
6. The diffusion of hydrogen into optical fiber affects the ______________
a) Transmission of optical light in the fiber
b) Spectral attenuation characteristics of the fiber
c) Core of the fiber
d) Cladding of the fiber
Answer: b
Explanation: The hydrogen absorption by an optical fiber increases optical fiber losses. It forms absorption peaks where the hydrogen diffuses into interstitial spaces in the glass. At high temperatures, these losses can increase and reduced if the hydrogen source is removed.
7. __________ can induce a considerable amount of attenuation in optical fibers.
a) Micro-bending
b) Dispersion
c) Diffusion of hydrogen
d) Radiation Exposure
Answer: d
Explanation: The optical transmission characteristics of the fiber cables can be degraded by exposure to nuclear radiation. The nature of this attenuation depends upon fiber structures, optical intensity, wavelength, etc. The radiation-induced attenuation comprises both permanent and temporary components which makes the exposure irreversible and reversible respectively.
8. The radiation-induced attenuation can be reduced through photo-bleaching.
a) True
b) False
Answer: a
Explanation: Photo-bleaching can be exploited to study the diffusion of molecules. It is used to remove the radiation exposure by quenching auto-fluorescence. It helps to increase signal-to-noise ratio of the fiber and thus reduces attenuation.
9. The losses due to hydrogen absorption and reaction with fiber deposits can be temporary.
a) True
b) False
Answer: b
Explanation: Hydrogen absorption occurs in two mechanisms. First phenomenon affects silica-based glass fibers whereas the second one occurs when hydrogen reacts with the fiber deposits to give P-OH, Ge-OH absorption. These losses are permanent.
10. The losses caused due to hydrogen absorption mechanisms are in the range of ___________
a) 20 dB/km to 25 dB/km
b) 10 dB/km to 15 dB/km
c) 25 dB/km to 50 dB/km
d) 0 dB/km to 5 dB/km
Answer: c
Explanation: The diffusion of hydrogen into optical fiber leads to an increase in optical fiber losses, causing damage to spectral loss characteristics. This phenomenon gets vibrant at higher temperatures. The losses caused due to such absorption are greater than 25 dB/km.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Cable Design”.
1. The cable must be designed such that the strain on the fiber in the cable does not exceed __________
a) 0.002%
b) 0.01%
c) 0.2%
d) 0.160%
Answer: c
Explanation: The constraints included in cable design are stability, protection, strength and jointing of the fibers. The fiber cable does not get affected if the strain exerted on it is below 0.2%. Although, it is suggested that the permanent strain on the fiber should be less than 0.1%.
2. How many categories exists in case of cable design?
a) Two
b) Three
c) One
d) Four
Answer: b
Explanation: Cable design is separated into three categories. They are fiber buffering, cable structural and strength and cable sheath and water barrier. After successfully going through these tests, an optical cable is designed.
3. How many types of buffer jackets are used in fiber buffering?
a) Three
b) One
c) Two
d) Four
Answer: a
Explanation: The buffer jacket is designed to protect the fiber from micro-bending losses. There are three types of buffer jackets used in fiber buffering. They are tight buffer jackets, loose tube buffer jackets and filled loose tube buffer jacket.
4. Loose tube buffer jackets exhibits a low resistance to movement of the fiber.
a) True
b) False
Answer: a
Explanation: Loose tube buffering is achieved by using a hard, smooth, flexible material in the form of extruded tube. The buffer tube is smooth from inside. Thus, it exhibits a low resistance to movement of the fiber. Also, it can be easily stripped for jointing or fiber termination.
5. An inclusion of one or more structural members in an optical fiber so as to serve as a cable core foundation around which the buffer fibers may be wrapped is called _____________
a) Attenuation
b) Splicing
c) Buffering
d) Stranding
Answer: d
Explanation: Optical fiber is made structurally stronger by adding one or more strength members. The core fiber is trapped with buffered fibers or they are slotted in the core foundation. This approach is called as stranding.
6. Which of the following is not a strength member used in optical cable?
a) Steel wire
b) Germanium
c) Aramid yarns
d) Glass elements
Answer: b
Explanation: Strength members or tensile members are added to the fiber to make it stronger and durable. These members include solid steel wire, dielectric aramid yarns , glass elements etc. Germanium is not a structural or strength member.
7. When the stranding approach consists of individual elements than the cable is termed as _____________
a) Optical unit cable
b) Coaxial cable
c) Layer cable
d) Bare glass cable
Answer: c
Explanation: The stranding approach consists of a fiber core foundation around which the buffered fibers are wrapped. The cable elements are stranded in one, two or several layers around the central structural member. When the stranding is composed of individual elements, then the cable is termed as layer cable. If the cable core consists of stranding elements each of which comprises a unit of stranding elements, then it is termed as optical unit cable.
8. The primary function of the structural member is load bearing.
a) True
b) False
Answer: b
Explanation: The primary function of the structural member is not load bearing. It’s function is to provide suitable accommodation for the fiber ribbons within the cables. These fiber ribbons lie in the helical grooves or slots formed in the surface of the structural members.
9. What is the Young’s modulus of Kevlar, an aromatic polyester?
a) 9 ×10 10 Nm -2
b) 10 ×10 10 Nm -2
c) 12 ×10 10 Nm -2
d) 13 ×10 10 Nm -2
Answer: d
Explanation: Kevlar is used as a strength member in an optical fiber. The Young’s modulus of Kevlar is very high which gives it strength to weight ratio advantage four times that of steel. Kevlar is coated with extruded plastic to provide a smooth surface which in turn prevents micro-bending losses.
10. The cable is normally covered with an outer plastic sheath to reduce _______________
a) Abrasion
b) Armor
c) Friction
d) Dispersion
Answer: a
Explanation: Abrasion is the process of scraping or wearing something away. If the cable is not coated with plastic sheath, it gives rise to effects such as abrasion and crushing. The most common plastic sheath material used in covering a cable is polyethylene .
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Fiber Alignment and Joint Loss”.
1. A measure of amount of optical fiber emitted from source that can be coupled into a fiber is termed as ______________
a) Radiance
b) Angular power distribution
c) Coupling efficiency
d) Power-launching
Answer: c
Explanation: Coupling efficiency depends upon the type of fiber attached to the source which should consider the parameters such as numerical aperture, core size, R.I. profile, radiance, core-cladding index difference. All these parameters relate to the performance of the fibers determined by power coupled into the fiber to power emitted by the source. This is called coupling efficiency ηwhich is given by
η = P F /P s
Where P F = power coupled into the fiber
P s = power emitted by the source.
2. The ratio r = / indicates ____________
a) Fresnel reflection
b) Reflection coefficient
c) Refraction coefficient
d) Angular power distribution coefficient
Answer: b
Explanation: The ratio, r = / is known as Reflection coefficient. It relates the amplitude of the reflected ray to the amplitude of the incident wave.
3. A GaAs optical source having a refractive index of 3.2 is coupled to a silica fiber having a refractive index of 1.42. Determine Fresnel reflection at interface in terms of percentage.
a) 13.4%
b) 17.4%
c) 17.6%
d) 14.8%
Answer: d
Explanation: If the fiber end and the source are in close physical contact, the reflection is given by
r = /) 2
Multiplying r by 100, we get the value of r in terms of percentage.
4. A particular GaAs fiber has a Fresnel reflection magnitude of 17.6% i.e. 0.176. Find the power loss between the source and the fiber?
a) 0.86 dB
b) 0.78 dB
c) 0.84 dB
d) 0.83 dB
Answer: c
Explanation: The optical losses in decibels at the joint is given by
Loss = -10log 10
Where L = loss due to Fresnel reflection
R = magnitude of Fresnel reflection.
5. Two joined step index fibers are perfectly aligned. What is the coupling loss of numerical aperture are NA R = 0.26 for emitting fiber?
a) -0.828 dB
b) -0.010 dB
c) -0.32 dB
d) 0.32 dB
Answer: b
Explanation: Coupling loss for two joined step index fibers is given by
L F = -10 log (NA R /NA E ) 2
Where L F = coupling loss
NA R = Numerical aperture of receiving fiber
NA E = Numerical aperture of emitting fiber.
6. Two joined graded index fibers that are perfectly aligned have refractive indices α R = 1.93 for receiving fiber α E = 2.15 for emitting fiber. Calculate the coupling loss.
a) 0.23 dB
b) 0.16 dB
c) 0.82 dB
d) 0.76 dB
Answer: a
Explanation: Coupling loss for two joined and perfectly aligned graded index fiber is given by
L F = -10log 10 α R (α E +2)/α E (α R +2)
Where L F = Coupling loss
α R = refractive index of receiving fiber
α E = refractive index of emitting fiber.
7. How many types of misalignments occur when joining compatible fiber?
a) One
b) Two
c) Five
d) Three
Answer: d
Explanation: There are three layers of fiber misalignments and they are: Longitudinal, lateral and angular misalignments.
8. Losses caused by factors such as core-cladding diameter, numerical aperture, relative refractive index differences, different refractive index profiles, fiber faults are known as ____________
a) Intrinsic joint losses
b) Extrinsic losses
c) Insertion losses
d) Coupling losses
Answer: a
Explanation: There are inherent connection problems while joining fibers. These connection problem cause different losses in the fibers and are called as Intrinsic joint losses.
9. A step index fiber has a coupling efficiency of 0.906 with uniform illumination of all propagation modes. Find the insertion loss due to lateral misalignment?
a) 0.95 dB
b) 0.40 dB
c) 0.42 dB
d) 0.62 dB
Answer: c
Explanation: The insertion loss due to lateral misalignment is given by
Loss 10t = -10log 10t η 10t
Where, Loss 10t = insertion loss due to lateral misalignment
η 10t = Coupling efficiency.
10. A graded index fiber has a parabolic refractive index profile and core diameter of 42μm. Estimate an insertion loss due to a 2 μm lateral misalignment when there is index matching and assuming there is uniform illumination of all guided modes only.
a) 0.180
b) 0.106
c) 0.280
d) 0.080
Answer: d
Explanation: The misalignment loss is given by
L t = 0.85
Where y = lateral misalignment
a = core radius.
11. Determine coupling efficiency if the misalignment loss in a graded index fiber is 0.102.
a) 0.136
b) 0.898
c) 0.982
d) 0.684
Answer: b
Explanation: If the misalignment loss is known, the coupling efficiency is defined by
η = 1-L t
Where η = coupling efficiency
L t = misalignment loss.
12. In a single mode fiber, the losses due to lateral offset and angular misalignment are given by 0.20 dB and 0.46 dB respectively. Find the total insertion loss.
a) 0.66 dB
b) 0.26 dB
c) 0.38 dB
d) 0.40 dB
Answer: a
Explanation: The total insertion loss in a single mode fiber is given by
T T = T L + T a
Where, T T = total insertion loss
T L = lateral offset loss
T a = Angular misalignment loss.
13. The intrinsic loss through a multimode fiber joint is independent of direction of propagation.
a) True
b) False
Answer: b
Explanation: Intrinsic loss is defined as the summation of lateral offset loss and angular misalignment loss. In case of multimode fibers, the intrinsic loss is dependent on the refractive index gradient. The intrinsic loss through a single mode fiber joint is independent of direction of propagation.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Fiber Splices”.
1. A permanent joint formed between two different optical fibers in the field is known as a ____________
a) Fiber splice
b) Fiber connector
c) Fiber attenuator
d) Fiber dispersion
Answer: a
Explanation: The jointing of two individual fibers is called as fiber splicing. It is used to establish long-haul optical fiber links by joining two small length fibers.
2. How many types of fiber splices are available?
a) One
b) Two
c) Three
d) Four
Answer: b
Explanation: Splices are divided into two types depending upon the splicing technique used. These are fusion splicing and mechanical splicing.
3. The insertion losses of the fiber splices are much less than the Fresnel reflection loss at a butted fiber joint.
a) True
b) False
Answer: a
Explanation: The Fresnel reflection loss is usually more because there is no large step change in refractive index with the fusion splice as it forms a continuous fiber connection. Also, some method of index matching tends to be utilized with mechanical splices.
4. What is the main requirement with the fibers that are intended for splicing?
a) Smooth and oval end faces
b) Smooth and square end faces
c) Rough edge faces
d) Large core diameter
Answer: b
Explanation: A curved mandrel is used which cleaves the fiber to achieve end preparation. The edges must be smooth and have square face at the end for splicing purpose.
5. In score and break process, which of the following is not used as a cutting tool?
a) Diamond
b) Sapphire
c) Tungsten carbide
d) Copper
Answer: d
Explanation: The score and break process is also called as scribe and break. It involves the scribing of the fiber surface under tension with a cutting tool. Copper is not used as a cutting tool.
6. The heating of the two prepared fiber ends to their fusing point with the application of required axial pressure between the two optical fibers is called as ____________
a) Mechanical splicing
b) Fusion splicing
c) Melting
d) Diffusion
Answer: b
Explanation: Fusion splicing is also called as welding. It refers to the welding of two fiber ends. It is essential for fusion splicing that the fiber ends are adequately positioned and aligned in order to achieve good continuity of the transmission medium at the junction point.
7. Which of the following is not used as a flame heating source in fusion splicing?
a) Microprocessor torches
b) Ox hydric burners
c) Electric arc
d) Gas burner
Answer: d
Explanation: Micro-plasma torches uses argon and hydrogen and alcohol vapor. The most widely used heating source is an electric arc. Thus, gas burner is not used in fusion splicing.
8. The rounding of the fiber ends with a low energy discharge before pressing the fibers together and fusing with a stronger arc is called as ____________
a) Pre-fusion
b) Diffusion
c) Crystallization
d) Alignment
Answer: a
Explanation: Pre-fusion involves rounding of fiber ends. It removes the requirement for fiber end preparation which has a distinct advantage in the field environment. It is utilized with multimode fibers giving average splice losses of 0.09dB.
9. _____________ is caused by surface tension effects between the two fiber ends during fusing.
a) Pre-fusion
b) Diffusion
c) Self-alignment
d) Splicing
Answer: c
Explanation: The two fiber ends are close but not aligned before fusion. During fusion, the surface tension affects the fiber ends to get aligned. After fusion, they are aligned in such a way that a transmission medium can get a good continuity.
10. Average insertion losses as low as _________ have been obtained with multimode graded index and single-mode fibers using ceramic capillaries.
a) 0.1 dB
b) 0.5 dB
c) 0.02 dB
d) 0.3 dB
Answer: a
Explanation: Mechanical techniques for splicing involve the use of an accurately produced rigid tube in which fiber ends are permanently bonded. It utilizes a ceramic capillary in which an epoxy resin is injected through a transverse bore to provide mechanical sealing and index matching. This technique which uses ceramic capillaries provides insertion losses as low as 0.1dB.
11. _____________ are formed by sandwiching the butted fiber ends between a V-groove glass substrate and a flat glass retainer plate.
a) Springroove splices
b) V-groove splices
c) Elastic splices
d) Fusion splices
Answer: b
Explanation: In V-groove splices, a V-groove glass substrate is used with a flat glass plate. The name V-groove suggests that the fiber ends are spliced in a V-shape.
These splices provide losses as low as 0.01dB.
12. Mean splice insertion losses of 0.05 dB are obtained using multimode graded index fibers with the Springroove splice.
a) True
b) False
Answer: a
Explanation: Springroove utilizes a bracket containing two cylindrical pins which act as alignment guide for two fiber ends. An elastic element is used to press the fibers into a groove. The assembly is secured with a drop of epoxy resin. It provides a loss of 0.05 dB and has found a practical use in Italy.
13. Alignment accuracy of the order ___________ is obtained using the three glass rod alignment sleeve.
a) 0.23 μm
b) 0.15 μm
c) 0.05 μm
d) 0.01 μm
Answer: c
Explanation: Alignment accuracies as high as 0.05 μmare necessary to obtain low losses. The mode-field diameter for single-mode fiber is in the range 8 to 10μm. The three glass rod alignment provides higher accuracies than rotary splice sleeve.
14. In case of multiple fusion, splice losses using an electric arc fusion device with multimode graded index fiber range from ____________
a) 0.01 to 0.04 dB
b) 0.19 to 0.25 dB
c) 0.12 to 0.15 dB
d) 0.04 to 0.12 dB
Answer: d
Explanation: In multiple fusions, an electric arc fusing device allows splicing of 12 fibers simultaneously. It takes a tool time of 6 minutes, which requires only 30 seconds per splice. The splice losses for single mode fiber are of 0.04 dB as maximum whereas for graded index fibers, losses are up to 0.12dB.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Fiber Connectors”.
1. Demountable fiber connectors are more difficult to achieve than optical fiber splices.
a) True
b) False
Answer: a
Explanation: Fiber connectors must maintain tolerance requirements similar to splices in order to couple light efficiently between the fibers. Also, fiber connectors must accomplish this in a removable fashion. The connector design must allow repeated connection and disconnection without any problems of fiber alignment.
2. What is the use of an index-matching material in the connector between the two jointed fibers?
a) To decrease the light transmission through the connection
b) To increase the light transmission through the connection
c) To induce losses in the fiber
d) To make a fiber dispersive
Answer: b
Explanation: The index-matching material used might be epoxy resin. It increases the light transmission through the connection while keeping dust and dirt from between the fibers. It also provides optimum optical coupling.
3. How many categories of fiber connectors exist?
a) One
b) Three
c) Two
d) Four
Answer: c
Explanation: Fiber connectors are separated into two broad categories. They are butt-jointed connectors and expanded beam connectors. Butt-jointed connectors rely upon alignment of the two fiber ends butted to each other whereas expanded beam connectors uses interposed optics at the joint.
4. The basic ferrule connector is also called as _____________
a) Groove connector
b) Beam connector
c) Multimode connector
d) Concentric sleeve connector
Answer: d
Explanation: The basic ferrule connector is the simplest connector. The ferrules are placed in an alignment sleeve within the connector. The alignment sleeve is concentric which allows the fiber ends to be butt-jointed.
5. What is the use of watch jewel in cylindrical ferrule connector?
a) To obtain the diameter and tolerance requirements of the ferrule
b) For polishing purposes
c) Cleaving the fiber
d) To disperse a fiber
Answer: a
Explanation: Ferrule connectors have a watch jewel in the ferrule end face. It is used instead of drilling of the metallic ferrule end face which takes time. It is used to obtain close diameter and tolerance requirements of the ferrule end face whole easily.
6. The concentricity errors between the fiber core and the outside diameter of the jeweled ferrule are in the range of ___________ with multimode step-index fibers.
a) 1 to 3μm
b) 2 to 6μm
c) 7 to 10μm
d) 12 to 20μm
Answer: b
Explanation: The fiber alignment accuracy of the basic ferrule connector is dependent on the ferrule hole into which the fiber is inserted. The concentricity errors in the range of 2 to 6μm gives insertion losses in the range 1 to 2dB with multimode step index fibers.
7. The typical average losses for multimode graded index fiber and single mode fiber with the precision ceramic ferrule connector are _____________ respectively.
a) 0.3 and 0.5 dB
b) 0.2 and 0.3 dB
c) 0.1 and 0.2 dB
d) 0.4 and 0.7 dB
Answer: b
Explanation: Unlike metal and plastic components, the ceramic ferrule material is harder than the optical fiber. Thus, it is unaffected by grinding and polishing process. This factor enables to provide the low-loss connectors which have low losses as low as 0.2 and 0.3 dB in case of optical fibers.
8. Bi-conical ferrule connectors are less advantageous than cylindrical ferrule connectors.
a) FalseStat
b) True
Answer: a
Explanation: Cylindrical and bi-conical ferrule connectors are assembled in housings to form a multi-fiber configuration. The force needed to insert multiple cylindrical ferrules can be large when multiple ferrules are involved. The multiple bi-conical ferrule connectors are more advantageous as they require less insertion force.
9. In connectors, the fiber ends are separated by some gap. This gap ranges from ____________
a) 0.040 to 0.045 mm
b) 0.025 to 0.10 mm
c) 0.12 to 0.16 mm
d) 0.030 to 0.2mm
Answer: b
Explanation: In connectors, gaps are introduced to prevent them from rubbing against each other and becoming damaged during connector fixing/engagement. The gap ranges from 0.025 to 0.10 mm so as to reduce the losses below 8dB for a particular diameter fiber say 50μm.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Expanded Beam Connectors”.
1. What is the use of interposed optics in expanded beam connectors?
a) To achieve lateral alignment less critical than a butt-joined fiber connector
b) To make a fiber loss free
c) To make a fiber dispersive
d) For index-matching
Answer: a
Explanation: Expanded beam connector utilize interposed optics at the joint in order to expand the beam from transmitting fiber end before reducing it to a size compatible with the receiving fiber end. It helps to achieve lateral alignment less critical than a butt-jointed connector. Also, the longitudinal separation is critical in expanded beam connectors.
2. The expanded beam connectors use ____________ for beam expansion and reduction.
a) Square micro-lens
b) Oval micro-lens
c) Spherical micro-lens
d) Rectangular micro-lens
Answer: c
Explanation: Expanded beam connectors use the principle of transmission of digital data to the receiver. It uses spherical micro-lens to first expand the beam from the transmitting end and reduces the beam at the receiving end.
3. Lens-coupled expanded beam connectors exhibit average losses of _________ in case of single mode and graded index fibers.
a) 0.3 dB
b) 0.7 dB
c) 0.2 dB
d) 1.5 dB
Answer: b
Explanation: Lens-coupled expanded beam connectors use spherical micro-lenses. The average losses are in the range of 1dB. With the antireflection coating on the lenses, the losses are reduced to 0.7 dB in case of single mode fibers.
4. Sapphire ball lens expanded beam design is successful than spherical lens coupled design.
a) True
b) False
Answer: a
Explanation: Spherical lens coupled design exhibits losses in the range 0.7 dB to 1dB. Sapphire ball lens expanded beam design achieved successful single mode fiber connection with losses as low as 0.4dB.
5. The fiber is positioned at the ________ of the lens in order to obtain a collimated beam and to minimize lens-to-lens longitudinal misalignment effects.
a) Aperture
b) Focal length
c) Curve
d) Exterior circumference
Answer: b
Explanation: The expanded beam connector also uses a molded spherical lens. A lens alignment sleeve is used to minimize the effects of angular misalignment. The fiber is positioned at the focal length of the lens to achieve losses as low as 0.7dB.
6. ___________ exhibits a parabolic refractive index profile with a maximum at the axis similar to graded index fiber.
a) Lens coupled design
b) Sapphire ball lens
c) Spherical micro-lens
d) GRIN-rod lens
Answer: d
Explanation: GRIN-rod lens geometry has a parabolic refractive index profile. It facilitates efficient beam expansion and collimation within expanded beam connectors. It finds its applications in fiber couplers and source-to-fiber coupling.
7. The GRIN-rod lens can produce a collimated output beam with a divergent angle αof between _____________ from a light source situated on, or near to, the opposite lens face.
a) 1 to 5 degrees
b) 9 to 16 degrees
c) 4 to 8 degrees
d) 25 to 50 degrees
Answer: a
Explanation: GRIN-rod lens comprises of a cylindrical glass rod typically 0.2 to 2 mm in diameter. It exhibits a parabolic refractive index profile. It facilitates efficient beam expansion and collimation with an angle in the range 1 to 5 degrees.
8. In the given equation, if r is the radial distance, n is the refractive index; what does z stands for?
dr2/dz2 = (1/n) (d n/dr)
a) Focal length
b) Distance along the optical axis
c) Axial angle
d) Diameter
Answer: b
Explanation: The above equation is known as paraxial ray equation which governs the ray propagation through the GRIN-rod lens. GRIN-rod lens geometry is parabolic in nature. Thus z is the distance along the optical axis of a parabolic profile.
9. The majority of the GRIN-rod lenses have diameters in the range of ____________
a) 2 to 2.5 mm
b) 3 to 4 mm
c) 0.1 to 0.4 mm
d) 0.5 to 2 mm
Answer: d
Explanation: The GRIN-rod lenses performance directly depends on the radial distance. The diameters in the range of 0.5 to 2 mm may be employed with either single mode or multimode fibers. They are available with numerical apertures of 0.37, 0.46 and 0.6.
10. Which of the following factors does not cause divergence of the collimated beam from a GRIN-rod lens?
a) Lens cut length
b) Size of fiber core
c) Refractive index profile
d) Chromatic aberration
Answer: c
Explanation: Various factors contribute to the divergence of the collimated beam from a GRIN-rod lens. Error in lens cut length, finite size of the fiber core and chromatic aberration are the factors that cause divergence.
11. GRIN-rod lens connectors have loss characteristics that are independent of the modal power distribution in the fiber.
a) True
b) False
Answer: a
Explanation: GRIN-rod lens geometry is analogous to butt-jointed multimode fiber connectors. The loss characteristics of butt-jointed connectors are dependent on modal power distribution in the fiber.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Fiber Couplers”.
1. When considering source-to-fiber coupling efficiencies, the ________ is an important parameter than total output power.
a) Numerical aperture
b) Radiance of an optical source
c) Coupling efficiency
d) Angular power distribution
Answer: b
Explanation: Radiance is the optical power radiated into a unit solid angle per unit emitting surface area. Since this optical power is dependent on radiance, radiance is much important factor than optical power.
2. It is a device that distributes light from a main fiber into one or more branch fibers.
a) Optical fiber coupler
b) Optical fiber splice
c) Optical fiber connector
d) Optical isolator
Answer: a
Explanation: Nowadays, requirements to divide combined optical signals for applications are increasing. Optical fiber coupler is one such device that is used for dividing and combining optical signals. It is generally used in LANs, computer networks etc.
3. Optical fiber couplers are also called as ________________
a) Isolators
b) Circulators
c) Directional couplers
d) Attenuators
Answer: c
Explanation: Optical fiber couplers are passive devices. The power transfer in couplers takes place either through the fiber core cross-section by butt jointing the fibers or by using some form of imaging optics between the fibers. It distributes light from one fiber to many fibers and hence it is also called as a directional coupler.
4. How many types of multiport optical fiber couplers are available at present?
a) Two
b) One
c) Four
d) Three
Answer: d
Explanation: Multiport optical fiber couplers are subdivided into three types. These are three and four port couplers, star couplers and wavelength division multiplexing couplers. These couplers distribute light among the branch fibers with no scattering loss.
5. The optical power coupled from one fiber to another is limited by ____________
a) Numerical apertures of fibers
b) Varying refractive index of fibers
c) Angular power distribution at source
d) Number of modes propagating in each fiber
Answer: d
Explanation: When two fibers are coupled to each other, the optical power is limited by number of modes propagating in each fiber. For example, when a fiber propagating with 500 modes is connected to a fiber that propagates only 400 modes, then at maximum, only 80% of power is coupled into the other fiber.
6. ________ couplers combine the different wavelength optical signal onto the fiber or separate the different wavelength optical signal output from the fiber.
a) 3-port
b) 2*2-star
c) WDM
d) Directional
Answer: c
Explanation: WDM coupler is abbreviated as wavelength division multiplexing coupler. It is a category of multiport optical fiber couplers. It is designed to permit a number of different peak wavelength optical signals to be transmitted in parallel on a single fiber.
7. How many fabrication techniques are used for 3 port fiber couplers?
a) One
b) Two
c) Three
d) Four
Answer: b
Explanation: There are two fabrication techniques available for three port couplers. First is a lateral offset method which relies on the overlapping of the fiber end faces and the other is the semi-transparent mirror method. Using these techniques, three port couplers with both multimode and single-mode fibers can be fabricated.
8. Which is the most common method for manufacturing couplers?
a) Wavelength division multiplexing
b) Lateral offset method
c) Semitransparent mirror method
d) Fused bi-conical taper technique
Answer: d
Explanation: The FBT technique is basic and simple. In this technique, the fibers are generally twisted together and then spot fused under tension such that the fused section is elongated to form a bi-conical taper structure. A three port coupler can be obtained by removing one of the input fibers.
9. Couplers insertion loss is same as that of excess loss.
a) True
b) False
Answer: b
Explanation: Excess loss is defined as the ratio of input power to output power. The insertion loss is defined as the loss obtained for a particular port-to-port optical path. Thus, the insertion loss and excess loss are different in nature.
10. A four-port multimode fiber FBT coupler has 50 μW optical power launched into port 1. The measured output power at ports 2,3 and 4 are 0.003, 23.0 and 24.5 μW respectively. Determine the excess loss.
a) 0.22 dB
b) 0.33 dB
c) 0.45 dB
d) 0.12 dB
Answer: a
Explanation: Excess loss is a ratio of power input to power output of the fiber and it is given by Excess loss = 10log 10 P 1 /(P 3 +P 4 )
WhereP 1 , P 3 , P 4 = output power at ports 1,3 and 4 resp.
11. A four-port FBT coupler has 60μW optical power launched into port one. The output powers at ports 2, 3, 4 are 0.0025, 18, and 22 μW respectively. Find the split ratio?
a) 42%
b) 46%
c) 52%
d) 45%
Answer: d
Explanation: Split ratio indicates the percentage division of optical power between the outputs ports. It is given by
Split ratio = [P 3 /(P 3 +P 4 )]*100%
Where P 3 and P 4 are output powers at ports 3 and 4 respectively.
12. How many manufacturing methods are used for producing multimode fiber star couplers?
a) Two
b) One
c) Three
d) Five
Answer: a
Explanation: The manufacturing methods of star couplers are mixer-rod technique and FBT technique. In the mixer-rod method, a thin platelet of glass is employed, which mixes light from one fiber, dividing it among the outgoing fibers. FBT method involves twisting, heating and pulling of fiber.
13. Calculate the splitting loss if a 30×30 port multimode fiber star coupler has 1 mW of optical power launched into an input port.
a) 13 dB
b) 15 dB
c) 14.77 dB
d) 16.02 dB
Answer: c
Explanation: The splitting loss is related to the number of output ports N of a coupler. It is given by-
Splitting loss = 10log 10 N .
14. A _____________ coupler comprises a number of cascaded stages, each incorporating three or four-port FBT couplers to obtain a multiport output.
a) Star
b) Ladder
c) WDM
d) Three-port
Answer: a
Explanation: A star coupler can be realized by constructing a ladder coupler. It consists of many cascaded stages. If a three-port coupler is used, then a ladder coupler does not form symmetrical star coupler. It is a useful device to achieve a multiport output with low insertion loss.
15. A number of three-port single-mode fiber couplers are used in the fabrication of a ladder coupler with 16 output ports. The three-port couplers each have an excess loss of 0.2 dB along with a splice loss of 0.1 dB at the interconnection of each stage. Determine the excess loss.
a) 1.9 dB
b) 1.4 dB
c) 0.9 dB
d) 1.1 dB
Answer: d
Explanation: The number of stages M within the ladder design is given by 2 M =16. Hence M=4.
Thus, excess loss is given by-
Excess loss = +
Where number of splices = 3 .
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Optical Isolators and Circulators”.
1. An FBG is developed within a fiber core having a refractive index of 1.30. Find the grating period for it to reflect an optical signal with a wavelength of 1.33μm.
a) 0.51 μm
b) 0.58 μm
c) 0.61 μm
d) 0.49 μm
Answer: a
Explanation: The grating period is denoted by Λ. It is given by-
Λ = λ B / 2n
Where λ B = wavelength
n = refractive index.
2. It is a passive device which allows the flow of optical signal power in only one direction and preventing reflections in the backward direction.
a) Fiber slice
b) Optical fiber connector
c) Optical isolator
d) Optical coupler
Answer: c
Explanation: Ideally, an optical isolator transmits the signal power in the desired forward direction. Material imperfections in the isolator medium generate backward reflections. Optical isolators can be implemented by using FBG.
3. Which feature of an optical isolator makes it attractive to use with optical amplifier?
a) Low loss
b) Wavelength blocking
c) Low refractive index
d) Attenuation
Answer: b
Explanation: Optical isolators are made using FBGs. Since FBGs are wavelength dependent, the optical isolators can be designed to allow or block the optical signal at particular wavelength. The wavelength blocking feature makes the optical isolator a very attractive device for use with optical amplifier in order to protect them from backward reflections.
4. Magneto-optic devices can be used to function as isolators.
a) True
b) False
Answer: a
Explanation: Magneto-optic devices use the principle of Faraday rotation. It relates the TM mode characteristics and polarization state of an optical signal with its direction of propagation. The rotation of polarization plane is proportional to the intensity of component of magnetic field in the direction of optical signal. Therefore, it is possible to block and divert an optical signal using magnetic properties which is a function of an isolator.
5. How many implementation methods are available for optical isolators?
a) One
b) Four
c) Two
d) Three
Answer: d
Explanation: Optical isolators can be implemented using three techniques. These are as follows:
-By using FBGs
-By using magnetic oxide materials
-By using semiconductor optical amplifiers .
6. A device which is made of isolators and follows a closed loop path is called as a ____________
a) Circulator
b) Gyrator
c) Attenuator
d) Connector
Answer: a
Explanation: Isolator can be connected together to form multiport devices. A circulator is formed from isolators connected together to form a closed circular path. In circulator, the signal continues to travel in closed loop and does not get discarded unlike isolator.
7. The commercially available circulators exhibit insertion losses around ________________
a) 2 dB
b) 0.7 dB
c) 0.2 dB
d) 1 dB
Answer: d
Explanation: A number of isolators can be used to implement a circulator. However, as the number of ports increases, the device complexity increases. Hence, three-or four-port circulators are used for optical interconnection with insertion losses around 1 dB and high isolation in the range of 40-50dB.
8. A combination of a FBG and optical isolators can be used to produce non-blocking optical wavelength division add/draw multiplexers.
a) True
b) False
Answer: b
Explanation: Optical wavelength divisions add/draw multiplexers can be produced by a combination of a FBG and a circulator. Non-blocking NXM optical wavelengths divisions add/draw multiplexer is produced where N and M denotes the number of wavelength channels and add/drop channels.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Optical Sources : Laser Basics”.
1. A device which converts electrical energy in the form of a current into optical energy is called as ___________
a) Optical source
b) Optical coupler
c) Optical isolator
d) Circulator
Answer: a
Explanation: An Optical source is an active component in an optical fiber communication system. It converts electrical energy into optical energy and allows the light output to be efficiently coupled into the Optical fiber.
2. How many types of sources of optical light are available?
a) One
b) Two
c) Three
d) Four
Answer: c
Explanation: Three main types of optical light sources are available. These are wideband sources, monochromatic incoherent sources. Ideally the optical source should be linear.
3. The frequency of the absorbed or emitted radiation is related to difference in energy E between the higher energy state E 2 and the lower energy state E 1 . State what h stands for in the given equation?
E = E2 - E1 = hf
a) Gravitation constant
b) Planck’s constant
c) Permittivity
d) Attenuation constant
Answer: b
Explanation: In the given equation, difference in the energy E is directly proportional to the absorbed frequency where h is used as a constant and is called as Planck’s constant. The value of h is measured in Joules/sec & is given by-
h = 6.626×10 -34 Js.
4. The radiation emission process can occur in __________ ways.
a) Two
b) Three
c) Four
d) One
Answer: a
Explanation: The emission process can occur in two ways. First is by spontaneous emission in which the atom returns to the lower energy state in a random manner. Second is by stimulated emission where the energy of a photon is equal to the energy difference and it interacts with the atom in the upper state causing it to return to the lower state along with the creation of a new photon.
5. Which process gives the laser its special properties as an optical source?
a) Dispersion
b) Stimulated absorption
c) Spontaneous emission
d) Stimulated emission
Answer: d
Explanation: In Stimulated emission, the photon produced is of the same energy to the one which cause it. Hence, the light associated with stimulated photon is in phase and has same polarization. Therefore, in contrast to spontaneous emission, coherent radiation is obtained. The coherent radiation phenomenon in laser provides amplification thereby making laser a better optical source than LED.
6. An incandescent lamp is operating at a temperature of 1000K at an operating frequency of 5.2×10 14 Hz. Calculate the ratio of stimulated emission rate to spontaneous emission rate.
a) 3×10 -13
b) 1.47×10 -11
c) 2×10 -12
d) 1.5×10 -13
Answer: b
Explanation: The ratio of the stimulated emission rate to the spontaneous emission rate is given by-
Stimulated emission rate/ Spontaneous emission rate = 1/exp -1.
7. The lower energy level contains more atoms than upper level under the conditions of ________________
a) Isothermal packaging
b) Population inversion
c) Thermal equilibrium
d) Pumping
Answer: c
Explanation: Under the conditions of thermal equilibrium, the lower energy level contains more atoms than the upper level. To achieve optical amplification, it is required to create a non-equilibrium distribution such that the population of upper energy level is more than the lower energy level. This process of excitation of atoms into the upper level is achieved by using an external energy source and is called as pumping.
8. __________________ in the laser occurs when photon colliding with an excited atom causes the stimulated emission of a second photon.
a) Light amplification
b) Attenuation
c) Dispersion
d) Population inversion
Answer: a
Explanation: Laser emits coherent radiation of one or more discrete wavelength. Lasers produce coherent light through a process called stimulated emission. Light amplification is obtained through stimulated emission. Continuation of this process creates avalanche multiplication.
9. A ruby laser has a crystal of length 3 cm with a refractive index of 1.60, wavelength 0.43 μm. Determine the number of longitudinal modes.
a) 1×10 2
b) 3×10 6
c) 2.9×10 5
d) 2.2×10 5
Answer: d
Explanation: The number of longitudinal modes is given by-
q = 2nL/λ
Where
q = Number of longitudinal modes
n = Refractive index
L = Length of the crystal
λ = Peak emission wavelength.
10. A semiconductor laser crystal of length 5 cm, refractive index 1.8 is used as an optical source. Determine the frequency separation of the modes.
a) 2.8 GHz
b) 1.2 GHz
c) 1.6 GHz
d) 2 GHz
Answer: c
Explanation: The modes of laser are separated by a frequency internal δf and this separation is given by-
δf = c/2nL
Where
c = velocity of light
n = Refractive index
L = Length of the crystal.
11. Doppler broadening is a homogeneous broadening mechanism.
a) True
b) False
Answer: b
Explanation: Doppler broadening is a inhomogeneous broadening mechanism. In this broadening, the individual groups of atoms have different apparent resonance frequencies. Atomic collisions usually provide homogeneous broadening as each atom in collection has same resonant frequency and spectral spread.
12. An injection laser has active cavity losses of 25 cm -1 and the reflectivity of each laser facet is 30%. Determine the laser gain coefficient for the cavity it has a length of 500μm.
a) 46 cm -1
b) 51 cm -1
c) 50 cm -1
d) 49.07 cm -1
Answer: d
Explanation: The laser gain coefficient is equivalent to the threshold gain per unit length and is given by –
g th = α + 1/L ln
Where
α = active cavity loss
L = Length of the cavity
r = reflectivity.
13. Longitudinal modes contribute only a single spot of light to the laser output.
a) True
b) False
Answer: a
Explanation: Laser emission includes the longitudinal modes and transverse modes. Transverse modes give rise to a pattern of spots at the output. Longitudinal modes give only a spot of light to the output.
14. Considering the values given below, calculate the mode separation in terms of free space wavelength for a laser.
a) 1.4×10 -11
b) 1.6×10 -12
c) 1×10 -12
d) 6×10 -11
Answer: b
Explanation: The mode separation in terms of free space wavelength is given by-
δλ = λ 2 /c δf
Where
δf = frequency separation
λ = wavelength
c = velocity of light.
This set of Optical Communications test focuses on “Optical Emission From Semiconductors”.
1. A perfect semiconductor crystal containing no impurities or lattice defects is called as __________
a) Intrinsic semiconductor
b) Extrinsic semiconductor
c) Excitation
d) Valence electron
Answer: a
Explanation: An intrinsic semiconductor is usually un-doped. It is a pure semiconductor. The number of charge carriers is determined by the semiconductor material properties and not by the impurities.
2. The energy-level occupation for a semiconductor in thermal equilibrium is described by the __________
a) Boltzmann distribution function
b) Probability distribution function
c) Fermi-Dirac distribution function
d) Cumulative distribution function
Answer: c
Explanation: For a semiconductor in thermal equilibrium, the probability P that an electron gains sufficient thermal energy at an absolute temperature so as to occupy a particular energy level E, is given by the Fermi-Dirac distribution. It is given by-
P = 1/(1+exp(E-E F /KT))
Where K = Boltzmann constant, T = absolute temperature, E F = Fermi energy level.
3. What is done to create an extrinsic semiconductor?
a) Refractive index is decreased
b) Doping the material with impurities
c) Increase the band-gap of the material
d) Stimulated emission
Answer: b
Explanation: An intrinsic semiconductor is a pure semiconductor. An extrinsic semiconductor is obtained by doping the material with impurity atoms. These impurity atoms create either free electrons or holes. Thus, extrinsic semiconductor is a doped semiconductor.
4. The majority of the carriers in a p-type semiconductor are __________
a) Holes
b) Electrons
c) Photons
d) Neutrons
Answer: a
Explanation: The impurities can be either donor impurities or acceptor impurities. When acceptor impurities are added, the excited electrons are raised from the valence band to the acceptor impurity levels leaving positive charge carriers in the valence band. Thus, p-type semiconductor is formed in which majority of the carriers are positive i.e. holes.
5. _________________ is used when the optical emission results from the application of electric field.
a) Radiation
b) Efficiency
c) Electro-luminescence
d) Magnetron oscillator
Answer: c
Explanation: Electro-luminescence is encouraged by selecting an appropriate semiconductor material. Direct band-gap semiconductors are used for this purpose. In band-to-band recombination, the energy is released with the creation of photon. This emission of light is known as electroluminescence.
6. In the given equation, what does p stands for?
p = 2πhk
a) Permittivity
b) Probability
c) Holes
d) Crystal momentum
Answer: d
Explanation: The given equation is a relation of crystal momentum and wave vector. In the given equation, h is the Planck’s constant, k is the wave vector and p is the crystal momentum.
7. The recombination in indirect band-gap semiconductors is slow.
a) True
b) False
Answer: a
Explanation: In an indirect band-gap semiconductor, the maximum and minimum energies occur at different values of crystal momentum. However, three-particle recombination process is far less probable than the two-particle process exhibited by direct band-gap semiconductors. Hence, the recombination in an indirect band-gap semiconductor is relatively slow.
8. Calculate the radioactive minority carrier lifetime in gallium arsenide when the minority carriers are electrons injected into a p-type semiconductor region which has a hole concentration of 10 18 cm -3 . The recombination coefficient for gallium arsenide is 7.21*10 -10 cm 3 s -1 .
a) 2ns
b) 1.39ns
c) 1.56ns
d) 2.12ms
Answer: b
Explanation: The radioactive minority carrier lifetime ςrconsidering the p-type region is given by-
ς r = [B r N] -1 where Br = Recombination coefficient in cm 3 s -1 and N = carrier concentration in n-region.
9. Which impurity is added to gallium phosphide to make it an efficient light emitter?
a) Silicon
b) Hydrogen
c) Nitrogen
d) Phosphorus
Answer: c
Explanation: An indirect band-gap semiconductor may be made into an electro-luminescent material by the addition of impurity centers which will convert it into a direct band-gap material. The introduction of nitrogen as an impurity into gallium phosphide makes it an effective emitter of light. Such conversion is only achieved in materials where the direct and indirect band-gaps have a small energy difference.
10. Population inversion is obtained at a p-n junction by __________
a) Heavy doping of p-type material
b) Heavy doping of n-type material
c) Light doping of p-type material
d) Heavy doping of both p-type and n-type material
Answer: d
Explanation: Population inversion at p-n junction is obtained by heavy doping of both p-type and n-type material. Heavy p-type doping with acceptor impurities causes a lowering of the Fermi-level between the filled and empty states into the valence band. Similarly n-type doping causes Fermi-level to enter the conduction band of the material.
11. A GaAs injection laser has a threshold current density of 2.5*10 3 Acm -2 and length and width of the cavity is 240μm and 110μm respectively. Find the threshold current for the device.
a) 663 mA
b) 660 mA
c) 664 mA
d) 712 mA
Answer: b
Explanation: The threshold current is denoted by I th . It is given by-
I th = J th * area of the optical cavity
Where J th = threshold current density
Area of the cavity = length and width.
12. A GaAs injection laser with an optical cavity has refractive index of 3.6. Calculate the reflectivity for normal incidence of the plane wave on the GaAs-air interface.
a) 0.61
b) 0.12
c) 0.32
d) 0.48
Answer: c
Explanation: The reflectivity for normal incidence of the plane wave on the GaAs-air interface is given by-
r = /) 2 where r=reflectivity and n=refractive index.
13. A homo-junction is an interface between two adjoining single-crystal semiconductors with different band-gap energies.
a) True
b) False
Answer: b
Explanation: The photo-emissive properties of a single p-n junction fabricated from a single-crystal semiconductor material are called as homo-junction. A hetero-junction is an interface between two single-crystal semiconductors with different band-gap energies. The devices which are fabricated with hetero-junctions are said to have hetero-structure.
14. How many types of hetero-junctions are available?
a) Two
b) One
c) Three
d) Four
Answer: a
Explanation: Hetero-junctions are classified into an isotype and an-isotype. The isotype hetero-junctions are also called as n-n or p-p junction. The an-isotype hetero-junctions are called as p-n junction with large band-gap energies.
15. The ______________ system is best developed and is used for fabricating both lasers and LEDs for the shorter wavelength region.
a) InP
b) GaSb
c) GaAs/GaSb
d) GaAs/Alga AS DH
Answer: d
Explanation: For DH device fabrication, materials such as GaAs, Alga AS are used. The band-gap in this material may be tailored to span the entire wavelength band by changing the AlGa composition. Thus, GaAs/ Alga As DH system is used for fabrication of lasers and LEDs for shorter wavelength region .
This set of Optical Communications Multiple Choice Questions & Answers focuses on “The Semiconductor Injection Laser”.
1. Stimulated emission by recombination of injected carriers is encouraged in __________
a) Semiconductor injection laser
b) Gas laser
c) Chemist laser
d) Dye laser
Answer: a
Explanation: Stimulated emission by use of optical cavity in crystal structure is used in semiconductor injection diodes. This provides the feedback of protons which gives injection laser many important advantages over other sources like LED’s.
2. In semiconductor injection laser, narrow line bandwidth is of the order?
a) 1 nm or less
b) 4 nm
c) 5 nm
d) 3 nm
Answer: a
Explanation: A narrow line bandwidth of order 1 nm or less is used. This narrow bandwidth is useful in minimizing the effects of material dispersion.
3. Injection laser have a high threshold current density of __________
a) 10 4 Acm -2 and more
b) 10 2 Acm -2
c) 10 -2 Acm -2
d) 10 -3 Acm -2
Answer: a
Explanation: Injection laser have a high threshold current density of 10 4 Acm -2 due to lack of matter and in-efficient light sources. These high current densities are largely utilized in pulse mode in order to minimize junction and thus avert damage.
4. η T is known as slope quantum efficiency.
a) True
b) False
Answer: b
Explanation: η D is known as slope quantum efficiency. It gives a measure rate of optical output power change with current and thus it determines slope of output characteristics in the region. So, η T is referred to as slope quantum efficiency.
5. The total efficiency of an injection laser with GaAs active region is 12%. The applied voltage is 3.6 V and band gap energy for GaAs is 2.34 eV. Determine external power efficiency.
a) 7.8 %
b) 10 %
c) 12 %
d) 6 %
Answer: a
Explanation: The total external power efficiency is defined as
η = η T *100
= 0.12 *100
= 7.8 %.
6. In a DH laser, the sides of cavity are formed by _______________
a) Cutting the edges of device
b) Roughening the edges of device
c) Softening the edges of device
d) Covering the sides with ceramics
Answer: b
Explanation: In a DH laser, the sides of cavity are formed by the roughening edges of the device. This is done so as to reduce the unwanted emission in these directions and limit the number of horizontal transversal modes.
7. A particular laser structure is designed so that the active region extends the edges of devices.
a) True
b) False
Answer: a
Explanation: Laser structures are particularly designed so that the active region does not extend beyond the edges. This is done to reduce problems like difficult heat sinking, lasing from multiple filament in wide active areas, unsuitable light output geometry for efficient coupling and also to reduce required threshold current.
8. Gain guided laser structure are __________
a) Chemical laser
b) Gas laser
c) DH injection laser
d) Quantum well laser
Answer: c
Explanation: DH injection lasers are known as gain guided laser structure. This is because the optical mode distribution along the junction plane is decided by optical gain.
9. Laser modes are generally separated by few __________
a) Tenths of micrometer
b) Tenths of nanometer
c) Tenths of Pico-meter
d) Tenths of millimeter
Answer: b
Explanation: The spacing in between modes is about a few tenths of nanometer. The spacing of the modes depends on optical cavity length where each one corresponds to an integral number of lengths.
10. The spectral width of emission from the single mode device is __________
a) Smaller than broadened transition line-width
b) Larger than broadened transition line-width
c) Equal the broadened transition line-width
d) Cannot be determined
Answer: a
Explanation: Single mode device has a smaller spectral width as compared to that of broadening transition line-width. This is because for a single-mode operation, the laser optical output must have only a single longitudinal and single transverse mode.
11. Single longitudinal mode operation is obtained by __________
a) Eliminating all transverse mode
b) Eliminating all longitudinal modes
c) Increasing the length of cavity
d) Reducing the length of cavity
Answer: d
Explanation: Single longitudinal mode operation is obtained by reducing the length L of cavity. Length must be reduced until the frequency separation of adjacent modes is given in the equation
δf = c/2nL is larger than gain curve. Then only single mode falling in transition line width will oscillate in laser cavity.
12. A correct DH structure will restrict the vertical width of waveguide region is?
a) 0.5μm.
b) 0.69 μm
c) 0.65 μm
d) Less than 0.4 μm
Answer: d
Explanation: The vertical width DH structure should be less than 0.4μm. This allows only fundamental transverse mode and removes any interference of higher order transverse modes on emitted longitudinal waves.
13. The external power efficiency of an injection laser with a GaAs is 13% having band gap energy of 1.64 eV. Determine external power efficiency.
a) 0.198
b) 0.283
c) 0.366
d) 0.467
Answer: a
Explanation: The external power efficiency of an injection laser is given by
η ep = η T *100
η T = η ep /100
= 13/100
= 0.198.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Some Injection Laser Structures”.
1. In multimode injection lasers, the construction of current flow to the strip is obtained in structure by __________
a) Covering the strip with ceramic
b) Intrinsic doping
c) Implantation outside strip region with protons
d)Implantation outside strip region with electrons
Answer: c
Explanation: The current flow is realized by implanting the region outside strip with protons. This implantation makes the laser highly resistive and gives superior thermal properties due to absence of silicon dioxide layer.
2. What is the strip width of injection laser?
a) 12 μm
b) 11.5 μm
c) Less than 10 μm
d) 15 μm
Answer: c
Explanation: A strip width less than or equal to 10 μm is usually preferred in injection lasers. This width range provides the lasers highly efficient coupling into multimode fibers as comapred to single mode fibers.
3. Some refractive index variation is introduced into lateral structure of laser.
a) True
b) False
Answer: a
Explanation: Gain guided lasers possess several undesirable characteristics, nonlinearities in light output versus current characteristics, high threshold current, low differential quantum efficiency, movement of optical a;ong junction plane. This problems can be reduced by introducing refractive index variations into lateral structure of lasers so that optical mode is determined along the junction plane.
4. Buried hetero-junction device is a type of _____________ laser where the active volume is buried in a material of wider band-gap and lower refractive index.
a) Gas lasers.
b) Gain guided lasers.
c) Weak index guiding lasers.
d) Strong index guiding lasers.
Answer: d
Explanation: In strong index guiding lasers, a uniformly thick, planar active waveguide is achieved by lateral variations in confinement layer thickness or refractive index. In Buried hetero-junction devices, strong index guiding along junction plane introduces transverse mode control in injection lasers.
5. In Buried hetero-junction lasers, the optical field is confined within __________
a) Transverse direction
b) Lateral direction
c) Outside the strip
d) Both transverse and lateral direction
Answer: d
Explanation: Optical field is strongly confined in both transverse and lateral direction. This provides strong index guiding of optical mode along with good carrier confinement.
6. A double-channel planar buried hetero-structure has a planar active region, the confinement material is?
a) Alga AS
b) InGaAsP
c) GaAs
d) SiO 2
Answer: b
Explanation: The planar active region made up of InGaAsP can be seen in double-channel planar buried hetero-structure . This material confinement provides a very high power operation with CW output power up to 40 mW in longer wavelength region.
7. Problems resulting from parasitic capacitances can be overcome __________
a) Through regrowth of semi-insulating material
b) By using oxide material
c) By using a planar InGaAsP active region
d) By using a AlGaAs active region
Answer: a
Explanation: The use of reverse-biased current confinement layers introduces parasitic capacitances which reduces high speed modulation of BH lasers. This problem can be reduced by regrowth of semi-insulating material or deposition of dielectric material. This causes increase in modulation speeds of 20 GHz.
8. Quantum well lasers are also known as __________
a) BH lasers
b) DH lasers
c) Chemical lasers
d) Gain-guided lasers
Answer: b
Explanation: DH lasers are known as Quantum well lasers. The carrier motion normal to active layer is restricted in these devices. This results in quantization of kinetic energy into discrete energy levels for carriers moving in that direction. This phenomenon is similar to quantum mechanical problem of one dimensional potential well which is seen in DH lasers.
9. Quantum well lasers are providing high inherent advantage over __________
a) Chemical lasers
b) Gas lasers
c) Conventional DH devices
d) BH device
Answer: c
Explanation: Quantum well lasers exhibit high incoherent advantage over conventional DH lasers. In Quantum well laser structures, the thin active layer results in drastic changes in electronic and optical properties over conventional DH laser structures. This changes are due to quantized nature of discrete energy levels with step-like density and also allow high gain and low carrier density.
10. Strip geometry of a device or laser is important.
a) True
b) False
Answer: a
Explanation: Near fluid intensity distribution corresponding to single optical output power level in plane of junction can be seen in GaAs or AlGaAs lasers. This distribution is in lateral direction and is determined by the nature of lateral waveguide. The single intensity maximum shows the fundamental lateral mode is dominant.
11. Better confinement of optical mode is obtained in __________
a) Multi Quantum well lasers
b) Single Quantum well lasers
c) Gain guided lasers
d) BH lasers
Answer: a
Explanation: As compared to all lasers including single quantum well lasers, multi-Quantum well lasers are having better confinement of optical mode. This results in a lower threshold current density for these devices.
12. Multi-quantum devices have superior characteristics over __________
a) BH lasers
b) DH lasers
c) Gain guided lasers
d) Single-quantum-well devices
Answer: b
Explanation: Lower threshold currents, narrower bandwidths, high modulation speeds, lower frequency chirps and less temperature dependence are parameters determining characteristics of a particular laser. All the above parameters make multi-quantum devices superior over DH lasers.
13. Dot-in-well device is also known as __________
a) DH lasers
b) BH lasers
c) QD lasers
d) Gain guided lasers
Answer: c
Explanation: Quantum well lasers are devices in which device contains a single discrete atomic structure or Quantum-dot. These are elements that contain electron tiny droplets which forms a quantum well structure.
14. A BH can have anything from a single electron to several electrons.
a) True
b) False
Answer: b
Explanation: Quantum-dot lasers are fabricated using semiconductor crystalline materials. They have a particular dimension ranging from nm to few microns. The size, shape of these structures and number of electrons they contain are precisely controlled.
15. QD lasers have a very low threshold current densities of range __________
a) 0.5 to 5 A cm -2
b) 2 to 10 A cm -2
c) 10 to 30 A cm -2
d) 6 to 20 A cm -2
Answer: d
Explanation: Low-threshold current density between 6 to 20 A cm -2 is obtained with InAs/InGaAs QD lasers which emit at a wavelength of 1.3 μm and 1.5 μm Such low values of threshold current densities make these lasers possible to create stacked or cascaded QD structures. These structures provide high optical gain for short-cavity transmitters and vertical cavity surface-emitting lasers.
This set of Optical Communications Multiple Choice Questions & Answers focuses on ” Single Frequency Injection Lasers”.
1. __________________ may be improved through the use of frequency-selective feedback so that the cavity loss is different for various longitudinal modes.
a) Frequency selectivity
b) Longitudinal mode selectivity
c) Electrical feedback
d) Dissipated power
Answer: b
Explanation: Improved longitudinal mode selectivity can be achieved using structures which gives loss discrimination between the desired and all the unwanted modes. Thus, mode discrimination can be seen. To allow for stable mode operation, the use of frequency-selective feedback approach is undertaken.
2. Device which apply the frequency-selective feedback technique to provide single longitudinal operation are referred to as ________________
a) DSM lasers
b) Nd: YAG lasers
c) Glass fiber lasers
d) QD lasers
Answer: a
Explanation: DSM lasers are also known as single frequency lasers. Such devices provide single longitudinal mode operation hence called as dynamic single mode lasers. These lasers reduce fiber intra-modal dispersion within high speed systems.
3. Which of the following does not provide single frequency operation?
a) Short cavity resonator
b) DSM lasers
c) Coupled cavity resonator
d) Fabry-Perot resonator
Answer: d
Explanation: DSM lasers, short cavity resonators, coupled cavity resonators employ frequency selective feedback approach and provide single mode operation. However, the Fabry-Perot resonator allows several longitudinal modes to exist within the gain spectrum of the device.
4. A method for increasing the longitudinal mode discrimination of an injection laser which is commonly used?
a) Decreasing refractive index
b) Increasing the refractive index
c) Increasing cavity length
d) Shortening of cavity length
Answer: d
Explanation: The longitudinal mode discrimination of an injection laser is indirectly proportional to the cavity length. Thus, as the cavity length is shortened, the mode discrimination will get increase. If the cavity length is reduced from 250 to 25 units, the mode spacing is increased from 1 to 10 nm.
5. Conventional cleaved mirror structures are difficult to fabricate with the cavity lengths below __________
a) 200 μm and greater than 150 μm
b) 100 μm and greater than 50 μm
c) 50 μm
d) 150 μm
Answer: c
Explanation: cleaved laser mirrors are used in Fabry-Perot resonator which does not give result for shorter cavity lengths. These lengths may vary from 20 μm to 50μm. Hence micro-cleaved or etched resonator is used for shorter cavity lens.
6. In the given equation, corrugation period is given by lλ b /2N e . If λb is the Bragg wavelength, then what does ‘l’ stand for?
a) Length of cavity
b) Limitation index
c) Integer order of grating
d) Refractive index
Answer: c
Explanation: The period of corrugation is given by
Period of corrugation = lλ b /2N e
Where, λ b = Bragg wavelength
L = integer order of grating.
7. The first order grating provide the strongest coupling within the device.
a) True
b) False
Answer: a
Explanation: The period of corrugation is given by lλ b /2N e includes order of grating. The second grating provide larger spatial period and thus helps in fabrication. If the order of grating is 1, then the device is coupled at high level.
8. The semiconductor lasers employing the distributed feedback mechanism are classified in _________________ categories.
a) One
b) Two
c) Three
d) Four
Answer: b
Explanation: Considering the device operation, semiconductor lasers are classified into two broad categories referred to as distributed feedback laser and distributed Bragg reflector laser. In the DFB laser, optical grating is applied over the entire active region, whereas in DBR lasers, the grating is etched only near the cavity ends.
9. DBF-BH lasers exhibit low threshold currents in the range of ________________
a) 40 to 50 mA
b) 21 to 30 mA
c) 2 to 5 mA
d) 10 to 20 mA
Answer: d
Explanation: DFB lasers are used to provide single frequency semiconductor optical sources. DFB-BH lasers, developed in laboratories exhibit high modulation speeds, output power but low threshold currents in the range of 10 to 20 mA.
10. Fabry-Perot devices with BH geometries high modulation speeds than DFB-BH lasers.
a) True
b) False
Answer: b
Explanation: DFB-BH lasers exhibit low threshold currents but high output power and modulation speeds. DFB-BH laser is fabricated by etching or grating. Fabry-perot devices provide modulation speeds of M-bits per seconds whereas, DFB-BH lasers provides modulation speeds of G-bits/sec.
11. The InGaAsP/InP double channel planar DFB-BH laser with a quarter wavelength shifted first order grating provides a single frequency operation and incorporates a phase shift of ______________
a) π/2 Radians
b) 2π Radians
c) π Radians
d) 3π/2 radians
Answer: a
Explanation: The performance of DFB laser is improved by modifying a grating, which in turn introduces an optical phase shift. The phase shift depends on the wavelength used. A quarter wavelength shifted first order grating incorporates the phase shift of π/2 in the corrugation at the center of laser cavity.
12. The narrow line-width obtained under the CW operation for quarter wavelength shifted DFB laser is ________________
a) 2 MHz
b) 10 MHz
c) 3 MHz
d) 1 MHz
Answer: c
Explanation: A quarter wavelength shifted DFB laser provides a large gain difference between the central mode and side modes. It provides improved dynamic single mode stability. Narrow line-width of around 3 MHz can be obtained under CW operation.
13. Line-width narrowing is achieved in DFB lasers by a strategy referred as _______________
a) Noise partition
b) Grating
c) Tuning
d) Bragg wavelength detuning
Answer: d
Explanation: Line-width narrowing is achieved in DFB lasers by detuning the lasing wavelength towards the shorter wavelength side of gain peak. It increases the differential gain between the central mode and nearest side mode. This strategy is called as Bragg wavelength detuning.
14. _________________ is a technique used to render the non-conducting material around the active cavity by producing permanent defects in the implanted area.
a) Dispersion
b) Ion de-plantation
c) Ion implantation
d) Attenuation
Answer: c
Explanation: Ion implantation approach concentrates the injection current in active region. Current confinement is realized by ion implantation. Ions are implanted into a selective area of a semiconducting material to make it non-conducting.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Injection Laser Characteristics”.
1. The threshold temperature coefficient for InGaAsP devices is in the range of __________
a) 10-40 K
b) 40-75 K
c) 120-190 K
d) 150-190 K
Answer: b
Explanation: The threshold temperature coefficient for InGaAsP devices is in between 40 and 75 K. This range shows higher temperature sensitivity due to intrinsic physical properties of InGaAsP material system, Auger recombination, inter-valence band absorption, carrier leakage effects over hetero-junctions.
2. The process where the energy released during the recombination of an electron-hole event getting transferred to another carrier is known as __________
a) Inter-valence bond absorption
b) Auger recombination
c) Carrier leakage effects
d) Exothermic actions
Answer: b
Explanation: Auger recombination is a process where energy is released. This energy is released during the recombination of electron-hole and this released energy is transferred to another electron-hole event. During this process, when a carrier is excited to a higher energy level, it loses its excessive amount of energy by emitting a phonon in order to maintain thermal equilibrium. It consists of number of different processes each process involving three particles .
3. Auger recombination can be reduced by using __________
a) Strained MQW structure
b) Strained SQW structure
c) Gain-guided strained structure
d) Strained Quantum dots lasers
Answer: a
Explanation: Auger recombination is a process where energy is released during recombination of electron-hole event is transferred to another event. This loss mechanism can be reduced by using strained by using MCQ laser structure. Strain can be either compressive or tensile, modifying the valence band energy levels of material and therefore can be used to increase energy.
4. High strain in strained MCQ structure should be incorporated.
a) True
b) False
Answer: b
Explanation: Strain is introduced in thin layers of quantum wells by making small differences in lattice constants. High strain should be avoided because it causes damage in these thin-quantum layers. Also carrier leakage adds at high temperatures since it represents processes that prevent carrier from recombination thus reducing device efficiency.
5. The parameter that prevents carrier from recombination is __________
a) Auger recombination
b) Inter-valence band absorption
c) Carrier leakage
d) Low temperature sensitivity
Answer: c
Explanation: Carrier leakage is the parameter that prevents carriers from recombination. At high temperatures, carrier leakage represents all those processes preventing carriers from recombination. It therefore increases the lasing threshold and thus reduces device efficiency.
6. Determine the threshold current density for an AlGaAs injection laser with T 0 =180k at 30°C.
a) 6.24
b) 9.06
c) 3.08
d) 5.09
Answer: d
Explanation: The threshold current density for a laser is given by-
I th = exp(T/T 0 )
For AlGaAs device,
I th =exp(T/T 0 ) = exp = 5.09.
7. The phenomenon occurring when the electron and photon population within the structure comes into equilibrium is known as __________
a) Auger recombination
b) Inter-valence band absorption
c) Carrier leakage
d) Relaxation oscillations
Answer: d
Explanation: Phenomenon occurring when the electron and photon population within the structure comes into equilibrium is known as Relaxation oscillations. The application of a current state to device resulting in a switch delay which is followed by high frequency damped oscillations.
8. When a current pulse reaches a laser having parasitic capacitance after the initial delay time, that pulse will __________
a) Have no effect
b) Will get vanished
c) Becomes narrower
d) Gets broader
Answer: d
Explanation: The pulse will be broadened when it will reach a laser with parasitic capacitance after initial time delay. This is because when a current pulse reaches the laser, the parasitic capacitance of laser provides a source of current over the period when there is high photon density. As electron density is repetitively built up and reduced quickly, there will be several pulses at laser output as photon density will be high resulting in relaxation oscillations.
9. Reducing delay time and ____________ are of high importance for lasers.
a) Auger recombination
b) Inter-valence band absorption
c) Carrier leakage effects
d) Relaxation oscillations
Answer: d
Explanation: For lasers generally a switch-on delay time may last for 0.5ns and relaxation oscillations behind twice that period. This behavior can produce serious deterioration in shape of laser pulse at a data rate of 100Mbits. So time delay and Relaxation oscillations are highly desirable for lasers.
10. Dynamic line-width broadening under the direct modulation of injection current is known as __________
a) Auger recombination
b) Inter-valence band absorption
c) Carrier leakage effects
d) Frequency Chirping
Answer: d
Explanation: Frequency Chirping is a phenomenon which is due to Dynamic line-width broadening under direct modulation of a single longitudinal mode. Semiconductor laser cause a dynamic shifting of peak wavelength emitted from device. Strong coupling between the free carrier density and refractive index of device present in semiconductor structure results in gain-induced variations which also causes Frequency Chirping.
11. A particular characteristic or parameter that occurs during analog transmission of injection lasers is?
a) Noise
b) Mode hopping
c) Carrier leakage effects
d) Frequency Chirping
Answer: a
Explanation: During analog transmission, noise behavior of device is main thing that affects the operation of injection laser. This noise may be due to instabilities in kinks in light output versus current characteristics, reflection of light back to device and mode partition noise.
12. Intensity of output from semiconductor injection lasers leading to optical intensity noise is due to __________
a) Fluctuations in amplitude
b) Mode hopping
c) Carrier leakage effects
d) Frequency Chirping
Answer: a
Explanation: Fluctuations in the laser output or intensity of laser output leads to optical intensity noise. These fluctuations are generally caused by temperature variations and spontaneous emission in the laser output. This randomness in fluctuations creates a noise source known as relative intensity noise .
13. In multimode lasers the optical feedback from unnecessary external reflections affecting stability of frequency and intensity is?
a) Remains unaffected
b) Increased gradually
c) Reduced
d) Gets totally vanished
Answer: c
Explanation: The effect due to unwanted external reflections in multimode laser is reduced. This is because the reflections are spread along any fiber modes so they are weakly coupled back into laser mode.
14. Reduction in the number of modes in multimode fiber increases the mode partition noise.
a) False
b) True
Answer: a
Explanation: Mode partition noise is a result of laser spectral fluctuations and so a reduce in number of modes results in low pulse-width spreading thus providing low values of intermodal dispersion in the fiber. And so, the mode partition noise is decreased in multimode fiber due to reducing the number of modes.
15. The behavior of laser occurring when current is increased above threshold particularly is?
a) Mode hopping
b) Auger recombination
c) Frequency chirping
d) Noise
Answer: a
Explanation: Mode hopping results in the hopping of modes to a higher wavelength. This mode hopping occurs in all injection lasers and is due to increase in temperature. Mode hopping is not a continuous function of drive current but occurs above 1 to 2 mA. Mode hopping alters characteristics of laser and results in kinks in characteristics of single mode device.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Non – Semiconductor Lasers”.
1. ____________________ lasers are presently the major laser source for optical fiber communications.
a) Semiconductor
b) Non-Semiconductor
c) Injection
d) Solid-state
Answer: c
Explanation: Injection laser coupling using discrete lasers have proved to fruitful. Such lenses provide for relaxation for an alignment tolerances normally required for fiber coupling. Certain non-semiconductor sources are making its lace in the optical fiber communication. At slowly present, injection lasers are mostly used as laser sources.
2. In Nd: YAG lasers, the maximum doping levels of neodymium is ____________
a) 0.5 %
b) 1.5 %
c) 1.8 %
d) 2 %
Answer: b
Explanation: The Nd: YAG laser structure is formed by doping of yttrium- aluminum -garnet with neodymium. The energy levels for lasing transition and pumping are provided by neodymium ions. The maximum doping level of neodymium in YAG is around 1.5 %.
3. Which of the following is not a property of Nd: YAG laser that enables its use as an optical fiber communication source?
a) Single mode operation
b) Narrow line-width
c) Long lifetime
d) Semiconductors and integrated circuits
Answer: d
Explanation: Nd: YAG laser is a non-semiconductor laser. It does not include the use of semiconductors and thus cannot take advantage of well-developed technology associated with integrated circuits. Single mode operation, narrow line-width, lifetime are the properties that are useful for optical communication.
4. The Nd: YAG laser has a narrow line-width which is ________________
a) < 0.01 nm
b) > 0.01 nm
c) > 1 mm
d) > 1.6 mm
Answer: a
Explanation: The Nd: YAG laser has several properties which make it an active optical source. One of such properties is its narrow line-width. It is less than 0.01 nm which is useful for reducing dispersion of optical links.
5. The strongest pumping bands is a four level system of Nd: YAG laser at wavelength of range_________________
a) 0.25 and 0.56 nm
b) 0.75 and 0.81 nm
c) 0.12 and 0.23 nm
d) 1 and 2 nm
Answer: b
Explanation: The Nd: YAG laser is a four level system. It consists of number of pumping bands and fluorescent transitions. The strongest pumping bands are the wavelengths of 0.75μm and 0.81μm. and gives lasing transition at 1.064μm and 1.32μm. Single mode emission is usually obtained at these wavelengths.
6. The Nd: YAG laser is costlier than earth-doped glass fiber laser.
a) True
b) False
Answer: a
Explanation: The most important requirement of the Nd: YAG laser is pumping and modulation. These two requirements tend to give a cost disadvantage in comparison with earth-doped glass fiber laser. Also it is easier and less expensive to fabricate glass fiber in earth-doped laser.
7. It is a resonant cavity formed by two parallel reflecting mirrors separated by a mirror separated by a medium such as air or gas is?
a) Optical cavity
b) Wheatstone’s bridge
c) Oscillator
d) Fabry-perot resonator
Answer: d
Explanation: Resonant cavity is formed between two mirrors where fiber core doped with earth ions is placed. This cavity is 250-500 μm long and 5 to 15 μm wide. A Fabry-perot resonator oscillates at resonant frequency for which there is high gain.
8. In a three level system, the threshold power decreases inversely with the length of the fiber gain medium.
a) True
b) False
Answer: b
Explanation: If the imperfection losses are low then in a four level system the threshold power decreases inversely with the length of the fiber gain medium. A three level consists of an optimum length. This optimum length gives the minimum threshold power which is independent of the value of imperfection losses.
9. Which of the following co-dopant is not employed by neodymium and erbium doped silica fiber lasers?
a) Phosphorus pent oxide
b) Germania
c) Nitrogen
d) Alumina
Answer: c
Explanation: Silica based glass fibers are proved to be the best host material till date. These silica fibers are doped with neodymium and erbium. These dopants include co-dopants such as phosphorus pent-oxide, germanium and alumina.
10. Dopants levels in glass fiber lasers are generally ___________
a) Low
b) High
c) Same as that of GRIN rod lens laser
d) Same as that of semiconductor laser
Answer: a
Explanation: Dopant levels are low in glass fibers . This is because of increasing in concentration quenching which increases with the doping level. It may cause the reduction in the population of the upper lasing level as well as crystallization within the glass matrix.
11. _______________ fibers include addition of lead fluoride to the core glass in order to raise the relative refractive index.
a) Solid-state
b) GaAs
c) Semiconductor
d) ZBLANP
Answer: d
Explanation: Up-conversion pumping of laser material is used to convert an infrared laser output to a visible laser output. ZBLANP is host material on which laser action at all wavelengths can be obtained by pumping. The relative refractive index is increased by addition of lead fluoride which makes it a very interesting host material.
12. The lasing output of the basic Fabry-perot cavity fiber is restricted to between ____________
a) 1 and 2 nm
b) 5 and 10 nm
c) 3 and 6 nm
d) 15 and 30 nm
Answer: b
Explanation: the gain spectrum of rare earth ions may be seen over a wavelength range of 50 nm. The lasing output will thus be narrow unless the dielectric on the mirror is arranged. Such a narrow line-width is not used for a broadband optical source.
13. In Fabry-perot laser, the lower threshold is obtained by ___________
a) Increasing the refractive index
b) Decreasing the refractive index
c) Reducing the slope efficiency
d) Increasing the slope efficiency
Answer: c
Explanation: The finesse of Fabry-perot cavity provides a measure of its filtering properties. When the finesse is high the splitting ratio is low thus lowering the laser threshold in an optical cavity without mirror. In Fabry-perot laser, mirrors are present and thus lower threshold is obtained by reducing the slope efficiency.
14. When did the non-semiconductor laser developed?
a) 1892
b) 1946
c) 1985
d) 1993
Answer: c
Explanation: Non-semiconductor sources are crystalline and glass wave-guiding structures. They are doped with rare earth ions and are good optical sources. The development of these sources started in the year 1985. Example: Nd: YAG laser.
15. Y 3 A l5 O 12 is a molecular formula for _____________
a) Ytterbium aluminate
b) Yttrium oxide
c) Ytterbium oxy-aluminate
d) Yttrium-aluminum garnet
Answer: d
Explanation: The atomic number of Yttrium is 39. It is the base element of Yttrium-aluminum garnet. Y 3 A l5 O 12 , doped with rare earth ion neodymium to form Nd: YAG laser structure.
This set of Optical Communications Quiz focuses on “Narrow – Linewidth and Wavelength – Tunable Lasers”.
1. Which of these factors are critical in affecting the system performance in the case of coherent optical fiber transmission?
a) Laser line-width and stability
b) Refractive index and index difference
c) Core cladding diameter
d) Frequency
Answer: a
Explanation: The system employing intensity modulation does not consider line-width and stability as the factors of utmost importance. In coherent optical source transmission, laser line-width and stability are critical factors. These factors affect the system performance and are in the range of 0.5-1 Megahertz.
2. _______________ occurs as a result of the change in lasing frequency with gain.
a) Frequency multiplication
b) Dispersion
c) Attenuation
d) Line-width broadening
Answer: d
Explanation: Line-width broadening is a fundamental consequence of spontaneous emission process. It is related to the fluctuations in the phase of the optical fields. These phase fluctuations are due to the phase noises associated with the spontaneous emission process.
3. Laser cavity length can be extended by ___________
a) Increasing the refractive index
b) Reducing frequency
c) Introduction of external feedback
d) Using GRIN-rod lenses
Answer: c
Explanation: the lasers having long external cavity are referred to as LEC lasers. The extension of the laser cavity length by introduction of external feedback can be achieved by using an external cavity with a wavelength dispersive element.
4. What is the purpose of wavelength dispersive element is LEC lasers?
a) Wavelength selectivity
b) Reduction of line-width
c) Frequency multiplication
d) Avalanche multiplication
Answer: a
Explanation: A wavelength dispersive element is a part of the laser cavity. It is required because the long resonator structure has very closely spaced longitudinal modes which necessitates additional wavelength selectivity.
5. An effective method to reduce the line-width is to make the cavity longer.
a) True
b) False
Answer: a
Explanation: As the laser power increases, the device line-width decreases. The output power f laser cannot be mode arbitrarily large. Thus, the line-width is reduced by making the cavity longer. Longer cavity also enables increased wavelength selectivity.
6. Which devices are used to modulate the external cavity in order to achieve the higher switching speeds?
a) Electromagnetic
b) Acousto-optic
c) Dispersive
d) Lead
Answer: b
Explanation: The devices are tuned mechanically to extend the cavity of laser. The disadvantage of using mechanically tuned devices is low. Thus, electro-optic devices are used to modulate the external cavity in order to achieve higher switching speeds.
7. How many techniques are used to tune monolithic integrated devices ?
a) Five
b) One
c) Two
d) Three
Answer: c
Explanation: There are two techniques which can be employed to tune monolithic integrated devices. In the first method, the mode selectivity of a coupled cavity structure is used. Other method is used to a refractive index change in the device cavity provided by application of an electric field.
8. _________________ laser can be produced when a coupler section is introduced between the amplifier and phase sections of a structure.
a) SG-DBR
b) GCSR
c) Y 4-shifted
d) DSM
Answer: b
Explanation: DBR lasers are capable of wavelength tuning. Grating assisted co-directional coupler with sampled reflector . Laser is capable of a tuning range greater than 40 nm. It consists of a co-directional coupler between the amplifier and the phase section.
9. The rare-earth-doped fiber lasers have spectral line-width in the range of _________________
a) 0.1 to 1 nm
b) 1.2 to 1.5 nm
c) 6 to 10 nm
d) 2 to 2.3 nm
Answer: a
Explanation: The rare-earth-doped fiber lasers have spectral line-width in the range of 0.1 to 1 nm. These line-widths are too long for high speed transmission is possible in this range.
10. The lasing line-width of Fox-smith resonator is ____________________
a) Less than 1 MHz
b) 1 MHz
c) 2 MHz
d) Greater than 3 MHz
Answer: a
Explanation: Fox-smith resonator employs a fused coupled fabricated from erbium-doped fiber. Narrower spectral line-width can be obtained using a resonator. It provides favorable line-widths than semiconductor laser.
11. What is the widest tuning range obtained in optical fiber laser structure?
a) 60 nm
b) 80 nm
c) More than 100 nm
d) 100 nm
Answer: c
Explanation: A tuning range greater than 100 nm by using an erbium-doped photonic crystal fiber. A wider tuning range greater than 100 nm is obtained at wavelength 1.55 nm.
12. How many techniques can be used to increase the injection cavity length?
a) One
b) Two
c) Three
d) Four
Answer: b
Explanation: Two techniques can be used to increase the injection laser cavity length. These are using laser chips and by extending a cavity with a passive medium such as air, glass etc.
13. The mechanism which results from a refractive index change in the passive waveguide layer is called as ___________
a) Absorption
b) Spontaneous emission
c) Monolithic inversion
d) Bragg wavelength control
Answer: d
Explanation: A wider wavelength tuning length is obtained by separating the Bragg region in the passive waveguide and by introducing a phase region within a waveguide control mechanism provides phase control. It takes place by some changes in a passive waveguide layer.
14. How many sections are included in a sampling grating distributed Bragg-reflector laser ?
a) Four
b) Five
c) Three
d) Two
Answer: b
Explanation: In SG-DBR laser, five sections are longitudinally integrated together on a semiconductor substrate. These five sections include two diffraction Bragg grating sections, a gain, a phase and an amplifier section.
15. Fiber based lasers provide diffraction-limited power at higher levels than solid-state laser.
a) True
b) False
Answer: a
Explanation: In fiber lasers, the active gain medium is doped with rare earth elements. These lasers have active regions several kilometers long and thus provide high optical gain. Solid-state lasers, on the other hand, provide diffraction limited power at lower levels.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Mid Infrared and Far Infrared Lasers”.
1. The parameters having a major role in determining threshold current of efficiency of injection laser are ___________
a) Angle recombination and optical losses
b) Frequency chirping
c) Relaxation oscillation
d) Mode hopping
Answer: a
Explanation: Optical losses due to free carrier absorption are more because of their dependence on square of the wavelength. Also irradiative recombination through Auger recombination contributes to it. Both these effects cause more problems in md-infrared wavelengths and so are of much importance art high temperature due to high concentration of free carriers. They also limit maximum operating temperatures.
2. Auger current is mostly ___________________ for material with band gap providing longer wavelength emission.
a) Unaffected
b) Lesser
c) Larger
d) Vanishes
Answer: c
Explanation: The total current required for injection laser threshold is more than that provided to radioactive recombination as Auger current is added. This current depends on electronic band structure of material and often consists of different Auger transitions. So it is larger for materials with band gaps providing longer wavelength emission.
3. Injection lasers operating in smaller wavelengths are subjected to increased carrier losses.
a) True
b) False
Answer: b
Explanation: Injection lasers operating in longer wavelengths are subjected to increased carrier losses as compared to devices operating up to 1.6μm. This is from nonradiative recombination through Auger interaction. This recombination energy is dissipated as thermal energy to other free carriers. If band gap of semiconductor is increased, occurrence of these events gets increased.
4. Devices based on quaternary PbSnSeTe and their ternary compounds, emit at wavelength?
a) Between 3-4 μm
b) Longer than 4 μm
c) Between 3.5 to 4.2 μm
d) Between 2 to 3 μm
Answer: b
Explanation: Quaternary devices emit at wavelength longer than 4μm. Auger effects are less in these alloys which provide lower current thresholds and higher maximum operating temperature.
5. Replacing Sn with Eu, Cd or Ge in some _________________ the band gap.
a) Remove the band gap
b) Does not affect
c) Decreases
d) Increases
Answer: d
Explanation: When in a particular alloy laser for example PbSnSeTe, if Sn is replaced with Eu, Cd or Ge, there is an increase in band gap. This increase in band gap provides the laser to operate in shorter wavelength.
.
6. Lasing obtained in __________ when 191 mW of pump light at a wavelength of 0.477 μm is launched into laser.
a) Ternary PbSnSeTe alloy laser
b) Quaternary PbSnSeTe alloy laser
c) Doped Fluoro-zirconate fiber
d) Ternary PbEuTe alloy laser
Answer: c
Explanation: When Fluoro-zirconate fiber lasers are doped with Erbium helium or thulium, there are emission at 2-3 μm wavelength range. But lasing was obtained in this doped Fluorozirconate fiber at a wavelength of 0.477μm.
7. The thulium doped fiber laser when pumped with alexandrite laser output at 0.786 μm, the laser emits at ___________
a) 0.6 μm
b) 0.8 μm
c) 2.3 μm
d) 1.2μm
Answer: c
Explanation: The thulium system emits at 2.3 μmwhen subjected to alexandrite laser at 0.786 μm. this system is four levels in which the pump band is upper lasing level at 2.3μm.
8. The diode-cladding-pumped Erbium praseodymium-doped fluoride device operates at wavelength.
a) Around 3 μm
b) 4 μm
c) 2.6 μm
d) 1.04 μm
Answer: a
Explanation: The diode-cladding-pumped Erbium praseodymium-doped fluoride device operates at a wavelength of 3 μm. This laser is capable of producing a very high output power of about 1W or more. It consists of double clad fluoride fiber.
9. A technique based on inter-sub band transition is known as ___________
a) Auger recombination
b) Frequency chirping
c) Inter-valence band absorption
d) Quantum cascading
Answer: d
Explanation: The quantum cascaded laser is a layered semiconductor device having a series of coupled quantum wells grown on GaAs or Imp substrate. This principle of QC lasers provides emission of an optical signal around full wavelength range. Quantum mechanical band structure determines the emitted wavelength.
10. In a QC laser, a same electron can emit number of photons.
a) True
b) False
Answer: a
Explanation: The QC laser operates by pumping a energy level and then using the energy in a controlled manner. This gives some energy each time over several steps. And since a QC laser structure includes a series of energy levels the same electron emits a number of photons while cascading down through each energy level.
11. The phenomenon resulting in the electrons to jump from one state to another each time emitting of photon is known as ___________
a) Inter-valence band absorption
b) Mode hopping
c) Quantum cascading
d) Quantum confinement
Answer: d
Explanation: In Quantum confinement, charge carriers are trapped in a small area and this occurs in quantum wells at nanometer scale. When the quantum layer size raises to a size comparable to emission wavelength, the electron motion becomes perpendicular to plane of layer. Due to this, the electrons jump from one state to another each time from one state to another.
12. A QC laser is sometimes referred as ___________
a) Unipolar laser
b) Bipolar laser
c) Gain guided laser
d) Non semiconductor laser
Answer: a
Explanation: A QC laser utilizes only n-type of charge carriers. Their operation is entirely based on electrons and holes play no part in this, so they are known as unipolar lasers.
13. In QC lasers, it is possible to obtain different output signal wavelengths. This can be achieved by ___________
a) Inter-valence band absorption
b) Mode hopping
c) Quantum cascading
d) Selecting layers of different thickness
Answer: d
Explanation: In QC laser, electrons emit energy. This energy emitted at this stage determines wavelength of radiation and it depends only on thickness of the layer. Thus output signal wavelength is dependent on thickness of lasers.
14. QC lasers ______________ the performance characteristics.
a) Have negligible effects
b) Does not affects
c) Improves
d) Degrades
Answer: c
Explanation: QC lasers are based on inter sub band transition techniques. They have ability of carrying large amount of currents. A single electron is enough to generate number of photons. Thus, provides an increase in output signal power which is greater than thousands at same wavelength due to large number of cascaded stages.
15. An MQW cascaded laser is more advantageous because of ___________
a) Mode hopping
b) Auger recombination
c) Control over layers of material
d) Properties of material
Answer: c
Explanation: In MQW cascaded layers, cascading creates number of injector/collector and active region in single stage. Each region contains a single quantum wells. Such structures permit maximum injection/collection of current and thereby produce a large number of photons. This formation of any injector/collector and active regions is achieved through precise control of several hundreds of layers of the material, where each layer should only be few nanometers thick.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “LED Power and Efficiency”.
1. The absence of _______________ in LEDs limits the internal quantum efficiency.
a) Proper semiconductor
b) Adequate power supply
c) Optical amplification through stimulated emission
d) Optical amplification through spontaneous emission
Answer: c
Explanation: The ratio of generated electrons to the electrons injected is quantum efficiency. It is greatly affected if there is no optical amplification through stimulated emission. Spontaneous emission allows ron-radiative recombination in the structure due to crystalline imperfections and impurities.
2. The excess density of electrons Δnand holes Δpin an LED is ____________
a) Equal
b) Δpmore than Δn
c) Δn more than Δp
d) Does not affects the LED
Answer: a
Explanation: The excess density of electrons ΔnandΔp is equal. The charge neutrality is maintained within the structure due to injected carriers that are created and recombined in pairs. The power generated internally by an LED is determined by taking into considering the excess electrons and holes in p- and n-type material respectively.
3. The hole concentration in extrinsic materials is _________ electron concentration.
a) much greater than
b) lesser than
c) equal to
d) negligible difference with
Answer: a
Explanation: In extrinsic materials, one carrier type will be highly concentrated than the other type. Hence in p-type region, hole concentration is greater than electron concentration in context of extrinsic material. This excess minority carrier density decays with time.
4. The carrier recombination lifetime becomes majority or injected carrier lifetime.
a) True
b) False
Answer: b
Explanation: The initial injected excess electron density and τrepresents the total carrier recombination time. In most cases, Δnis a small fraction of majority carriers and contains all minority carriers. So in these cases, carrier recombination lifetime becomes minority injected carrier lifetime τ i .
5. In a junction diode, an equilibrium condition occurs when ____________
a) Δngreater than Δp
b) Δnsmaller than Δp
c) Constant current flow
d) Optical amplification through stimulated emission
Answer: c
Explanation: The total rate at which carriers are generated in sum of externally supplied and thermal generation rates. When there is a constant current flow in this case, an equilibrium occurs in junction diode.
6. Determine the total carrier recombination lifetime of a double heterojunction LED where the radioactive and nonradioactive recombination lifetime of minority carriers in active region are 70 ns and 100 ns respectively.
a) 41.17 ns
b) 35 ns
c) 40 ns
d) 37.5 ns
Answer: a
Explanation: The total carrier recombination lifetime is given by
τ = τ r τ nr /τ r +τ nr = 70× 100/70 + 100 ns = 41.17 ns
Where
τ r = radiative recombination lifetime of minority carriers
τ nr = nonradioactive recombination lifetime of minority carriers.
7. Determine the internal quantum efficiency generated within a device when it has a radiative recombination lifetime of 80 ns and total carrier recombination lifetime of 40 ns.
a) 20 %
b) 80 %
c) 30 %
d) 40 %
Answer: b
Explanation: The internal quantum efficiency of device is given by
η int = τ/τ r = 40/80 ×100 = 80%
Where
τ = total carrier recombination lifetime
τ r = radiative recombination lifetime.
8. Compute power internally generated within a double-heterojunction LED if it has internal quantum efficiency of 64.5 % and drive current of 40 mA with a peak emission wavelength of 0.82 μm.
a) 0.09
b) 0.039
c) 0.04
d) 0.06
Answer: b
Explanation: The power internally generated within device i.e. double-heterojunction LED can be computed by
P int = η int hci/eλ = 0.645×6.626×10 -34 ×3×10 8 ×40×10 -3 / 1.602×10 -19 × 0.82 × 10 -6
= 0.039 W
Where
η int = internal quantum efficiency
h = Planck’s constant
c = velocity of light
i = drive current
e = electron charge
λ = wavelength.
9. The Lambertian intensity distribution __________ the external power efficiency by some percent.
a) Reduces
b) Does not affects
c) Increases
d) Have a negligible effect
Answer: a
Explanation: In Lambertian intensity distribution, the maximum intensity I 0 is perpendicular to the planar surface but is reduced on the sides in proportion to the cosine of θ i.e. viewing angle as apparent area varies with this angle. This reduces the external power efficiency. This is because most of the light is tapped by total internal refraction when radiated at greater than the critical angle for crystal air interface.
10. A planar LED fabricated from GaAs has a refractive index of 2.5. Compute the optical power emitted when transmission factor is 0.68.
a) 3.4 %
b) 1.23 %
c) 2.72 %
d) 3.62 %
Answer: c
Explanation: The optical power emitted is given by
P e = P int F n 2 /4n x 2 = P int 2 ) = 0.0272 P int .
Hence power emitted is only 2.72 % of optional power emitted internally.
Where,
F n 2 = transmission factor
nx = refractive index.
11. A planar LED is fabricated from GaAs is having a optical power emitted is 0.018% of optical power generated internally which is 0.018% of optical power generated internally which is 0.6 P. Determine external power efficiency.
a) 0.18%
b) 0.32%
c) 0.65%
d) 0.9%
Answer: d
Explanation: Optical power generated externally is given by
η cp = (0.018P int /2P int )*100
Where,
P int = power emitted
η cp = external power efficiency.
12. For a GaAs LED, the coupling efficiency is 0.05. Compute the optical loss in decibels.
a) 12.3 dB
b) 14 dB
c) 13.01 dB
d) 14.6 dB
Answer: c
Explanation: The optical loss in decibels is given by-
Loss = -10log 10 η c
Where,
η c = coupling efficiency.
13. In a GaAs LED, compute the loss relative to internally generated optical power in the fiber when there is small air gap between LED and fiber core. (Fiber coupled = 5.5 * 10 -4 P int )
a) 34 dB
b) 32.59 dB
c) 42 dB
d) 33.1 dB
Answer: b
Explanation: The loss in decibels relative to P int is given by-
Loss = -10log 10 P c /P int
Where,
P c = 5.5 * 10 -4 P int .
14. Determine coupling efficiency into the fiber when GaAs LED is in close proximity to fiber core having numerical aperture of 0.3.
a) 0.9
b) 0.3
c) 0.6
d) 0.12
Answer: a
Explanation: The coupling efficiency is given by
η c = 2 = 2 = 0.9.
15. If a particular optical power is coupled from an incoherent LED into a low-NA fiber, the device must exhibit very high radiance.
a) True
b) False
Answer: a
Explanation: Device must have very high radiance specially in graded index fiber where Lambertian coupling efficiency with same NA is about half that of step-index fibers. This high radiance is obtained when direct bandgap semiconductors are fabricated with DH structure driven at high current densities.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “LED Structures”.
1. The amount of radiance in planer type of LED structures is ____________
a) Low
b) High
c) Zero
d) Negligible
Answer: a
Explanation: Planer LEDs are fabricated using liquid or vapor phase epitaxial processes. Here p-type is diffused into n-type substrate which creates junction. Forward current flow through junction provides Lambertian spontaneous emission. Thus, device emits light from all surfaces. However a limited amount of light escapes the structure due to total internal reflection thus providing low radiance.
2. In optical fiber communication _____________ major types of LED structures are used.
a) 2
b) 4
c) 6
d) 3
Answer: c
Explanation: Optical fiber communication involves the use of 6 different major LED structure. These are the surface emitter, edge emitter, the super luminescent, the resonant cavity LED, planar LEDs and Dome LEDs.
3. As compared to planar LED structure, Dome LEDs have ______________ External power efficiency ___________ effective emission area and _____________ radiance.
a) Greater, lesser, reduced
b) Higher, greater, reduced
c) Higher, lesser, increased
d) Greater, greater, increased
Answer: b
Explanation: In Dome LEDs, the diameter of dome is selected so as to maximum the internal emission reaching surface within critical angle of GaAs. Thus, dome LEDs have high external power efficiency. The geometry of Dome LEDs is such that dome is much larger than active recombination area, so it has greater emission era and reduced of radiance.
4. The techniques by Burros and Dawson in reference to homo structure device is to use an etched well in GaAs structure.
a) True
b) False
Answer: a
Explanation: Burros and Dawson provided a technique to restrict emission to small active region within device thus providing high radiance. Etched well in a GaAs substrate is used to prevent heavy absorption of emitted region and physically accommodating the fiber. These structures provide low thermal impedance allowing high current densities of high radiance.
5. In surface emitter LEDs, more advantage can be obtained by using ____________
a) BH structures
b) QC structures
c) DH structures
d) Gain-guided structure
Answer: c
Explanation: DH structures provide high efficiency from electrical and optical confinement. Along with efficiency, they provide less absorption of emitted radiation.
6. Internal absorption in DH surface emitter Burros type LEDs is ____________
a) Cannot be determined
b) Negligible
c) High
d) Very low
Answer: d
Explanation: The larger band gap confining layers and the reflection coefficient at the back crystal space is high in DH surface emitter Burros type LEDs. This provides good forward radiance. Thus these structure LEDs have very less internal absorption.
7. DH surface emitter generally give ____________
a) More coupled optical power
b) Less coupled optical power
c) Low current densities
d) Low radiance emission into-fiber
Answer: a
Explanation: The optical power coupled into a fiber depends on distance, alignment between emission area and fiber, SLED emission pattern and medium between emitting area and fiber. All these parameters if considered, reduces refractive index mismatch and increases external power efficiency thus providing more coupled optical power.
8. A DH surface emitter LED has an emission area diameter of 60μm. Determine emission area of source.
a) 1.534*10 -6
b) 5.423*10 -3
c) 3.564*10 -2
d) 2.826*10 -9
Answer: d
Explanation: The emission area A of source is given by
A = π(30*10 -6 ) 2 = 2.826*10 -9 cm 2 .
9. Estimate optical power coupled into fiber of DH SLED having emission area of 1.96*10 -5 , radiance of 40 W/rcm 2 , numerical aperture of 0.2 and Fresnel reflection coefficient of 0.03 at index matched fiber surface.
a) 5.459*10 -5
b) 1.784*10 -3
c) 3.478*10 2
d) 9.551*10 -5
Answer: d
Explanation: The optical power coupler in the step index fiber of SLED is given by
Pc = π A R D 2
= 3.14 *1.96*10 -5 *40* 2
= 9.551*10 -5 W.
10. In a multimode fiber, much of light coupled in the fiber from an LED is ____________
a) Increased
b) Reduced
c) Lost
d) Unaffected
Answer: c
Explanation: Optical power from an incoherent source is initially coupled into large angle rays falling within acceptance angle of fiber but have more energy than Meridional rays. Energy from these rays goes into the cladding and thus may be lost.
11. Determine the overall power conversion efficiency of lens coupled SLED having forward current of 20 mA and forward voltage of 2 V with 170 μWof optical power launched into multimode step index fiber.
a) 1.256*10 -5
b) 4.417*10 2
c) 4.25*10 -3
d) 2.14*10 -3
Answer: c
Explanation: The overall power conversion efficiency is determined by
η pc = P c /P = 170*10 -6 /20*10 -3 *2
= 4.25*10 -3 .
12. The overall power conversion efficiency of electrical lens coupled LED is 0.8% and power applied 0.0375 V. Determine optical power launched into fiber.
a) 0.03
b) 0.05
c) 0.3
d) 0.01
Answer: a
Explanation: Optical power launched can be computed by
η pc = P c /P
Pc = η p c * P
= 0.8 * 0.0375
= 0.03.
13. Mesa structured SLEDs are used ____________
a) To reduce radiance
b) To increase radiance
c) To reduce current spreading
d) To increase current spreading
Answer: c
Explanation: The planar structures of Burros-type LED allow lateral current spreading specially for contact diameters less than 25 μm.This results in reduced current density and effective emission area greater than contact area. This technique to reduce current spreading in very small devices is Mesa structured SLEDs.
14. The InGaAsP is emitting LEDs are realized in terms of restricted are ____________
a) Length strip geometry
b) Radiance
c) Current spreading
d) Coupled optical power
Answer: a
Explanation: The short striped structure of these LEDs around 100 μmimproves the external efficiency of LEDs by reducing internal absorption of carriers. These are also called truncated strip E-LEDs.
15. The active layer of E-LED is heavily doped with ____________
a) Zn
b) Eu
c) Cu
d) Sn
Answer: a
Explanation: Zn doping reduces the minority carrier lifetime. Thus this improves the device modulation bandwidth hence active layer is doped in Zn in E-LEDs.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “LED Characteristics”.
1. Intrinsically _________________ are a very linear device.
a) Injection lasers
b) DH lasers
c) Gain-guided
d) LEDs
Answer: d
Explanation: The ideal light output power against current characteristics for an LED linear. This tends to be more suitable for analog transmission where several constraints are put in linearity of optical source.
2. Linearizing circuit techniques are used for LEDs.
a) True
b) False
Answer: a
Explanation: In practice, LEDs exhibit nonlinearities depending on configuration used. Thus, to allow its used in high quality analog transmission system and to ensure linear performance of device, linearizing circuit techniques is used.
3. The internal quantum efficiency of LEDs decreasing _______________ with ________________ temperature.
a) Exponentially, decreasing
b) Exponentially, increasing
c) Linearly, increasing
d) Linearly, decreasing
Answer: b
Explanation: The light emitted from LEDs decreases. This is due to increase in p-n junction temperature. Thus, this results in exponentially decreasing internal quantum efficiency with temperature increment.
4. To utilize _____________________ of SLDs at elevated temperatures, the use of thermoelectric coolers is important.
a) Low-internal efficiency
b) High-internal efficiency
c) High-power potential
d) Low-power potential
Answer: c
Explanation: The output characteristics of SLDs are typically of nonlinear in nature. This is observed with a knee becoming apparent at an operating temperature around 20 degree c. Thus, to utilize high-power potential of these devices at elevated temperature, thermoelectric coolers are necessarily used.
5. For particular materials with smaller bandgap energies operating in _____________ wavelength, the linewidth tends to ______________
a) 2.1 to 2.75 μm, increase
b) 1.1 to 1.7 μm, increase
c) 2.1 to 3.6 μm, decrease
d) 3.5 to 6 μm, decrease
Answer: b
Explanation: For materials with smaller bandgap, linewidth increases to 50 to 160 nm. This increases in band gap is due to increased doping levels and formation of bandtail states.
6. The active layer composition must be adjusted if a particular center wavelength is desired.
a) True
b) False
Answer: a
Explanation:There is a difference in output spectra between surface and edge emitting LEDs when devices have generally heavily doped and lightly doped active layers by reduction in doping.
7. In optical fiber communication, the electrical signal dropping to half its constant value due to modulated portion of optical signal corresponds to _______
a) 6 dB
b) 3 dB
c) 4 dB
d) 5 dB
Answer: b
Explanation: Modulation bandwidth in optical communication is often defined in electrical/optical terms. So when considering electrical circuitry in optical fiber system, electrical 3 dB point or frequency at which output electrical power is reduced by 3 dB bandwidth with respect to input electrical power.
8. The optical 3 dB point occurs when currents ratio is equal to _____________
a) \
\
\
\(\frac{3}{4}\)
Answer: c
Explanation: In optical regime, the bandwidth is defined by frequency at which output current has dropped to ½ output input current system.
9. The optical bandwidth is _____________ the electrical bandwidth.
a) Smaller
b) Greater
c) Same as
d) Zero with respect to
Answer: b
Explanation: The difference between optical and electrical bandwidth In terms of frequency depends on the shape of the frequency response of the system. If the system response is assumed to be Gaussian, then optical bandwidth is a factor of √2 greater than electrical bandwidth.
10. When a constant d.c. drive current is applied to device, the optical o/p power is 320 μm. Determine optical o/p power when device is modulated at frequency 30 MHz with minority carrier recombination lifetime of LED i.e. 5ns.
a) 4.49*10 -12
b) 6.84*10 -9
c) 1.29*10 -6
d) 2.29*10 -4
Answer: d
Explanation: The output o/p at 30 MHz is
Pc = Pdc/ 2 ) 1/2
= 320*10 -6 /(1+(2π*30*10 -6 *5*10 -9 ) 2 ) 1/2
= 2.29*10 -4 W.
11. The optical power at 20 MHz is 246.2 μW. Determine dc drive current applied to device with carrier recombination lifetime for LED of 6ns.
a) 3.48*10 -4
b) 6.42*10 -9
c) 1.48*10 -3
d) 9.48*10 -12
Answer: a
Explanation: The optical output power at 20 MHz is
Pe = Pdc/ 2 ) 1/2
246.2*10 -6 = Pdc/(1+(2π*20*10 -6 *5*10 -9 ) 2 ) 1/2
Pdc = 3.48*10 -4 .
12. Determine the 3 dB electrical bandwidth at 3 dB optical bandwidth Bopt of 56.2 MHz.
a) 50.14
b) 28.1
c) 47.6
d) 61.96
Answer: b
Explanation: The 3dB electrical bandwidth is given by
B = Bopt/ √2
= 56.2/2
= 28.1 MHz.
13. The 3 dB electrical bandwidth B is 42 MHz. Determine 3dB optical bandwidth Bopt.
a) 45.18
b) 59.39
c) 78.17
d) 94.14
Answer: b
Explanation: The 3dB electrical bandwidth is
B = Bopt/√2
Bopt = B*√2
= 42*√2
= 59.39 MHz.
14. Determine degradation rate βrif constant junction temperature is 17 degree celsius.
a) 7.79*10 -11
b) 7.91*10 -11
c) 6.86*10 -11
d) 5.86*10 -11
Answer: a
Explanation: The degradation rate βris determined by
β r = β 0 exp
= 1.89*10 7 exp (-1*1.602*10 -19 /1.38*10 -23 *290)
= 7.79*10 -11 h -1 .
15. Determine CW operating lifetime for LED with βrt = -0.58 and degradation rate βr = 7.86*10 -11 h -1 .
a) 32.12
b) 42
c) 22.72
d) 23.223
Answer: c
Explanation: The CW operating lifetime is given by
t = Ln 0.58/7.86*10 -11
= 22.72h -1 .
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Device Types”.
1. ____________ converts the received optical signal into an electrical signal.
a) Detector
b) Attenuator
c) Laser
d) LED
Answer: a
Explanation: A detector is an essential component of an optical fiber communication system. It dictates the overall system performance. Its function is to convert optical signal into an electrical signal. This electrical signal is then amplified before further processing.
2. The first generation systems of optical fiber communication have wavelengths between ___________
a) 0.2 and 0.3 μm
b) 0.4 and 0.6 μm
c) 0.8 and 0.9 μm
d) 0.1 and 0.2 μm
Answer: c
Explanation: The first generation systems operated at a bit-rate of 45 Mbps with repeater spacing of 10 km. It operates at wavelengths between 0.8 and 0.9μm. These wavelengths are compatible with AlGaAs laser and LEDs.
3. The quantum efficiency of an optical detector should be high.
a) True
b) False
Answer: a
Explanation: The detector must satisfy stringent requirements for performance and compatibility. The photo detector thus produces a maximum electrical signal for a given amount of optical power; i.e. the quantum efficiency should be high.
4. Which of the following does not explain the requirements of an optical detector?
a) High quantum efficiency
b) Low bias voltages
c) Small size
d) Low fidelity
Answer: d
Explanation: The size of the detector must be small for efficient coupling to the fiber. Also, ideally, the detector should not require excessive bias voltages and currents. The fidelity and quantum efficiency should be high.
5. How many device types are available for optical detection and radiation?
a) One
b) Two
c) Three
d) Four
Answer: b
Explanation: Two types of devices are used for optical detection and radiation. These are external photoemission and internal photoemission devices. External photoemission devices are too bulky and require high voltages for operation. Internal devices provide good performance and compatibility.
6. The ___________ process takes place in both extrinsic and intrinsic semiconductors.
a) Avalanche multiplication
b) External photoemission
c) Internal photoemission
d) Dispersion
Answer: c
Explanation: During intrinsic absorption, the received photons excite electrons from the valence band and towards the conduction band in the semiconductor. Extrinsic absorption involves impurity centers created with the material. Generally, intrinsic absorption is preferred for internal photoemission.
7. ____________ are widely used in first generation systems of optical fiber communication.
a) p-n diodes
b) 4-alloys
c) 3-alloys
d) Silicon photodiodes
Answer: d
Explanation: The first generation systems operates at wavelengths 0.8 and 0.9 μm. Silicon photodiodes have high sensitivity over the 0.8-0.9 μm wavelength band with adequate speed, long term stability. Hence, silicon photodiodes are widely used in first generation systems.
8. Silicon has indirect band gap energy of __________________
a) 1.2 eV
b) 2 eV
c) 1.14 eV
d) 1.9 eV
Answer: c
Explanation: Silicon’s indirect band gap energy of 1.14 eV gives a loss in response above 1.09μm. To avoid this, narrower bandgap materials are used. Hence, silicon’s usefulness is limited to first generation systems and not for second and third generation systems.
9. Which of the following detector is fabricated from semiconductor alloys?
a) Photoconductive detector
b) p-i-n detector
c) Photodiodes
d) Photoemission detectors
Answer: a
Explanation: The detectors fabricated from semiconductor alloys can be used for longer wavelengths. Photoconductive detector and hetero-junction transistor have found favor as a potential detector over a wavelength range of 1.1 to 1.6μm.
10. Silicon photodiodes provide high shunt conductance.
a) True
b) False
Answer: b
Explanation: Semiconductor photodiodes provide best solution for detection in optical fiber communications. Silicon photodiodes have high sensitivity, negligible shunt conductance and low dark current.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Optical Detection Principles”.
1. P-n photodiode is forward biased.
a) True
b) False
Answer: b
Explanation: p-n photodiode includes p and n regions. The electric field developed across the p-n junction sweeps holes and electrons to p and n regions respectively. P-n photodiode is thus reverse biased due to reverse leakage current.
2. The depletion region must be ____________ to allow a large fraction of the incident light to be absorbed in the device.
a) Thick
b) Thin
c) Long
d) Inactive
Answer: a
Explanation: In p-n photodiode, intrinsic conditions are created in the depletion region. The depletion region must be thick in order to achieve maximum carrier pair generation. Also, its width must be limited to enhance the speed of operation of the p-n photodiode.
3. The process of excitation of an electron from valence band to conduction band leaves an empty hole in the valence band and is called as ____________
a) Detection
b) Absorption
c) Degeneration of an electron-hole pair
d) Regeneration of an electron-hole pair
Answer: d
Explanation: A photon is incident in the depletion region of a device has an energy greater than or equal to the band gap energy of the fabricating material. This will cause excitation of an electron from valence to the conduction band. This creates an empty hole in valence band which is referred to as photo-generation of an electron-hole pair.
4. __________________ always leads to the generation of a hole and an electron.
a) Repulsion
b) Dispersion
c) Absorption
d) Attenuation
Answer: c
Explanation: Absorption affects the electron and excites it to some other level say conduction band. This is called as photo-generation as absorption always leads to the generation of hole and electron. This does not mean that both contribute to the electronic transport.
5. The electron hole pairs generated in a photodiode are separated by the ____________
a) Magnetic field
b) Electric field
c) Static field
d) Depletion region
Answer: b
Explanation: Electric field separates the electron-hole pairs in a photodiode. The electric field distribution is determined by an internal and an external field component. A reverse bias voltage is usually applied to the p-n photodiode.
6. Electric field in the depletion region should be high.
a) True
b) False
Answer: a
Explanation: The electric field in the depletion region is always kept high in order to extract all photogenerated carriers. Only the extracted electron hole pairs contribute to the overall photocurrent.
7. The photocurrent of an optical detector should be __________
a) Less
b) More
c) Linear
d) Non-linear
Answer: c
Explanation: A linear relationship must exist between the intensity of the incident light and the photocurrent. This makes the photodiode free of noise. It increases system performance.
8. How many types of optical detectors are available?
a) One
b) Four
c) Two
d) Three
Answer: d
Explanation: Three types of optical detectors are available. These are diodes, photoconductors and photo-transistors. Diodes include p-n photodiodes, p-i-n diodes, avalanche photodiodes and schottky diodes.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Absorption”.
1. The absorption of photons in a photodiode is dependent on __________
a) Absorption Coefficient α 0
b) Properties of material
c) Charge carrier at junction
d) Amount of light
Answer: a
Explanation: Absorption in a photodiode is for producing carrier pans. Thus, photocurrent is dependent on absorption coefficient α 0of the light in semiconductor used to fabricate device.
2. The photocurrent in a photodiode is directly proportional to absorption coefficient.
a) True
b) False
Answer: a
Explanation: The absorption of photons produces carrier pairs. Thus, photocurrent is dependent on absorption coefficient and is given by
I = P o e/hf(1-exp (-α rd ))
Where r = Fresnel coefficient
D = width of absorption region.
3. The absorption coefficient of semiconductor materials is strongly dependent on __________
a) Properties of material
b) Wavelength
c) Amount of light
d) Amplitude
Answer: b
Explanation: In some common semiconductors, there is a variation in absorption curves for materials. It is found that they are each suitable for different wavelength and related applications. This is due to difference in band gap energies. Thus absorption coefficient depends on wavelength.
4. Direct absorption requires assistance of photon.
a) True
b) False
Answer: b
Explanation: Indirect absorption requires photon assistance resulting in conversation of energy and momentum. This makes transition probability less likely for indirect absorption than direct absorption where no photon is included.
5. In optical fiber communication, the only weakly absorbing material over wavelength band required is?
a) GaAs
b) Silicon
c) GaSb
d) Germanium
Answer: c
Explanation: The transition over wavelength band in silicon is due to indirect absorption mechanism. This makes silicon weakly absorbent over particular wavelength band.
6. The threshold for indirect absorption occurs at wavelength __________
a) 3.01 μm
b) 2.09 μm
c) 0.92 μm
d) 1.09 μm
Answer: d
Explanation: The band gap for silicon is 4.10 eV corresponding to threshold of 0.30 μm in ultraviolet. Thus it’s outside wavelength range is the one which is required.
7. The semiconductor material for which the lowest energy absorption takes place is?
a) GaAs
b) Silicon
c) GaSb
d) Germanium
Answer: d
Explanation: Germanium absorption is by indirect optical transition. The threshold for direct absorption is at 1.53μm. Below this, germanium becomes strongly absorbing to corresponding link.
8. The wavelength range of interest for Germanium is __________
a) 0.8 to 1.6 μm
b) 0.3 to 0.9 μm
c) 0.4 to 0.8 μm
d) 0.9 to 1.8 μm
Answer: a
Explanation: Germanium is used in fabrication of detectors over the whole wavelength range i.e. first and second generation 0.8 to 1.6 μm while specially taking into consideration that indirect absorption will occur up to a threshold of 1.85 μm.
9. A photodiode should be chosen with a ________________ less than photon energy.
a) Direct absorption
b) Band gap energy
c) Wavelength range
d) Absorption coefficient
Answer: d
Explanation: A photodiode selection must be made by choosing that diode having band gap energy less than photon energy corresponding to longest operating wavelength. This provides high absorption coefficient which ensures a good response and limits the thermally generated carriers to obtain low dark current with no incident light.
10. ________________ photodiodes have large dark currents.
a) GaAs
b) Silicon
c) GaSb
d) Germanium
Answer: c
Explanation: Germanium photodiodes provide narrow band gaps as compared to other semiconductor materials. This is main disadvantage with use of germanium photodiodes at shorter wavelength and thus they have large dark current.
11. For fabrication of semiconductor photodiodes, there is a drawback while considering _________________
a) GaAs
b) Silicon
c) GaSb
d) Germanium
Answer: d
Explanation: Due to drawback with germanium to be used as fabricating material, there
is an increased investigation of direct band gap III and V alloys for longer wavelength region.
12. _________________ materials are potentially superior to germanium.
a) GaAs
b) Silicon
c) GaSb
d) III – V alloys
Answer: d
Explanation: The band gap energies for III – V alloys materials can be tailored to required wavelength. This can be achieved by changing relative concentration of their constituents which results in low dark currents. Thus, III – V alloys are superior potentially to germanium.
13. ____________ alloys such as InGaAsP and GaAsSb deposited on InP and GaSb substrate.
a) Ternary
b) Quaternary
c) Gain-guided
d) III – V alloys
Answer: a
Explanation: Ternary alloys are used to fabricate photodiodes for longer wavelength band. Thus, these alloys such as InGaAsP and GaAsSb are deposited on InP and GaSb substrates.
14. _________________ alloys can be fabricated in hetero-junction structures.
a) InGaSb
b) III – V alloys
c) InGaAsP
d) GaAsSb
Answer: b
Explanation: III – V alloys enhances the high speed operations of hetero-junction structures. Thus these structures can be fabricated with III-V alloys.
15. The alloys lattice matched to InP responds to wavelengths up to 1.7μm is?
a) InAsSb
b) III – V alloys
c) InGaSb
d) InGaAs
Answer: d
Explanation: Although there were difficulties in growth of IOnGaAs alloys, the problems are now reduced. These alloys lattice matched to InP responding to wavelength around 1.7 μmare widely utilized for fabrication of photodiodes operating around 1.7μm.
This set of Optical Communications Mcqs focuses on “Quantum Efficiency , Responsivity and Long – Wavelength Cut-Off”.
1. The fraction of incident photons generated by photodiode of electrons generated collected at detector is known as ___________________
a) Quantum efficiency
b) Absorption coefficient
c) Responsivity
d) Anger recombination
Answer: a
Explanation: Efficiency of a particular device is obtained by ratio of input given to that of output obtained. Thus, similarly, in photodiode, input i.e. incident photon and output generated electrons and their ratio is quantum efficiency.
2. In photo detectors, energy of incident photons must be ________________ band gap energy.
a) Lesser than
b) Greater than
c) Same as
d) Negligible
Answer: b
Explanation: While considering intrinsic absorption process, the energy of incident photon must be greater than band gap energy of material fabricating photo detector.
3. GaAs has band gap energy of 1.93 eV at 300 K. Determine wavelength above which material will cease to operate.
a) 2.431*10 -5
b) 6.424*10 -7
c) 6.023*10 3
d) 7.234*10 -7
Answer: b
Explanation: The long wavelength cutoff is given by
λ c = hc/Eg = 6.6268*10 -34 *2.998*10 8 /1.93*1.602*10 -19
= 6.424*10 -7 μm.
4. The long cutoff wavelength of GaAs is 0.923 μm. Determine bandgap energy.
a) 1.478*10 -7
b) 4.265*10 -14
c) 2.784*10 -9
d) 2.152*10 -19
Answer: d
Explanation: Long wavelength cutoff of photo detector is given by
λ c = hc/Eg
Eg = hc/λ c = 6.6268*10 -34 *2.998*10 8 /0.923*10 -6
= 2.152*10 -19 eV.
5. Quantum efficiency is a function of photon wavelength.
a) True
b) False
Answer: a
Explanation: Quantum efficiency is less than unity as all of incident photons are not absorbed to create electrons holes pairs. For example quantum efficiency of 60% is equivalent to 60% of electrons collected per 100 photons. Thus efficiency is a function of photon wavelength and must be determined at a particular wavelength.
6. Determine quantum efficiency if incident photons on photodiodes is 4*10 11 and electrons collected at terminals is 1.5*10 11 ?
a) 50%
b) 37.5%
c) 25%
d) 30%
Answer: b
Explanation: Quantum efficiency is given by
Quantum Efficiency = No. of electrons collected/No. of incident photons
= 1.5*10 11 /4*10 11
= 0.375 * 100
= 37.5%.
7. A photodiode has quantum efficiency of 45% and incident photons are 3*10 11 . Determine electrons collected at terminals of device.
a) 2.456*10 9
b) 1.35*10 11
c) 5.245*10 -7
d) 4.21*10 -3
Answer: b
Explanation: Quantum efficiency is given by
Quantum efficiency = No. of electrons collected/No. of incident photons
Electrons collected = Quantum efficiency * number of incident photons
= 45/100 * 3*10 11
= 1.35*10 11 .
8. The quantum efficiency of photodiode is 40% with wavelength of 0.90*10 -6 . Determine the responsivity of photodiodes.
a) 0.20
b) 0.52
c) 0.29
d) 0.55
Answer: c
Explanation: Responsivity of photodiodes is given by
R = ηe λ/hc
= 0.4*1.602*10 -19 * 0.90*10 -6 /6.626*10 -34 * 3*10 8
= 0.29 AW -1 .
9. The Responsivity of photodiode is 0.294 AW -1 at wavelength of 0.90 μm. Determine quantum efficiency.
a) 0.405
b) 0.914
c) 0.654
d) 0.249
Answer: a
Explanation: Responsivity of photodiode is
R = ηe λ/hc
η = RXhc/eλ
= 0.294*6.626*10 -34 *3*10 8 / 1.602*10 -19 *0.90*10 8
= 0.405 AW -1 .
10. Determine wavelength of photodiode having quantum efficiency of 40% and Responsivity of 0.304 AW -1 .
a) 0.87 μm
b) 0.91 μm
c) 0.88 μm
d) 0.94 μm
Answer: d
Explanation: The Responsivity of photodiode is
R = ηe λ/hc
λ = Rhc/ηe
= 0.304*6.626*10 -34 *3*10 8 /0.4*1.602*10 -19
= 0.94 μm.
11. Determine wavelength at which photodiode is operating if energy of photons is 1.9*10 -19 J?
a) 2.33
b) 1.48
c) 1.04
d) 3.91
Answer: c
Explanation: To determine wavelength,
λ = hc/t
= 6.626*10 -34 *3*10 8 /1.9*10 -19
= 1.04 μm.
12. Determine the energy of photons incident on a photodiode if it operates at a wavelength of 1.36 μm.
a) 1.22*10 -34 J
b) 1.46*10 -19 J
c) 6.45*10 -34 J
d) 3.12*10 9 J
Answer: b
Explanation: The wavelength of photodiode is given by
λ = hc/t
E = hc/λ
= 6.626*10 -34 *3*10 8 /1.36*10 -6
= 1.46*10 -19 J.
13. Determine Responsivity of photodiode having o/p power of 3.55 μm and photo current of 2.9 μm.
a) 0.451
b) 0.367
c) 0.982
d) 0.816
Answer: d
Explanation: The Responsivity of photodiode is
R = Ip/Po
= 2.9*10 -6 /3.55*10 -6
= 0.816 A/W.
14. Determine incident optical power on a photodiode if it has photocurrent of 2.1 μA and responsivity of 0.55 A/W.
a) 4.15
b) 1.75
c) 3.81
d) 8.47
Answer: c
Explanation: The Responsivity of photodiode is
R = Ip/Po
Po = Ip/R
= 2.1*10 -6 /0.55
= 3.81 μm.
15. If a photodiode requires incident optical power of 0.70 A/W. Determine photocurrent.
a) 1.482
b) 2.457
c) 4.124
d) 3.199
Answer: b
Explanation: The Responsivity of photodiode is given by
R = Ip/Po
Ip = R*Po
= 0.70*3.51*10 -6
= 2.457μm.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Semiconductor Photodiodes Without Internal Gain”.
1. The width of depletion region is dependent on ___________ of semiconductor.
a) Doping concentrations for applied reverse bias
b) Doping concentrations for applied forward bias
c) Properties of material
d) Amount of current provided
Answer: a
Explanation: The depletion region is formed by immobile positively and immobile negatively charged donor and acceptor atoms in n- and p-type respectively. When carriers are swept towards majority side under electric field, lower the doping, wider the depletion region.
2. Electron-hole pairs are generated in ___________
a) Depletion region
b) Diffusion region
c) Depletion region
d) P-type region
Answer: c
Explanation: Photons are absorbed in both depletion and diffusion regions. The position and width of absorption region depends on incident photons energy. The absorption region may extend throughout device in weakly absorption of photons. Thus carriers are generated in both regions.
3. The diffusion process is _____________ as compared with drift.
a) Very fast
b) Very slow
c) Negligible
d) Better
Answer: b
Explanation: None.
4. Determine drift time for carrier across depletion region for photodiode having intrinsic region width of 30μm and electron drift velocity of 10 5 ms -1 .
a) 1×10 -10 Seconds
b) 2×10 -10 Seconds
c) 3×10 -10 Seconds
d) 4×10 -10 Seconds
Answer: c
Explanation: The drift time is given by
t drift = w/v d = 30×10 -6 /1×10 -10 = 3×10 -10 seconds.
5. Determine intrinsic region width for a photodiode having drift time of 4×10 -10 s and electron velocity of 2×10 -10 ms -1 .
a) 3×10 -5 M
b) 8×10 -5 M
c) 5×10 -5 M
d) 7×10 -5 M
Answer: b
Explanation: The drift time is given by
t drift = w/v d
4×10 -10 = w/2×10 5
= 4×10 -10 ×2×10 5
= 8×10 -5 m.
6. Determine velocity of electron if drift time is 2×10 -10 s and intrinsic region width of 25×10 -6 μm.
a) 12.5×10 4
b) 11.5×10 4
c) 14.5×10 4
d) 13.5×10 4
Answer: a
Explanation: The drift time is given by
t drift = w/v d
v d = 25×10 -6 /2×10 -10 = 12.5×10 4 ms -1 .
7. Compute junction capacitance for a p-i-n photodiode if it has area of 0.69×10 -6 m 2 , permittivity of 10.5×10 -13 Fcm -1 and width of 30μm.
a) 3.043×10 -5
b) 2.415×10 -7
c) 4.641×10 -4
d) 3.708×10 -5
Answer: b
Explanation: The junction capacitance is given by,
C j = ε s A/w = 10.5×10 -13 ×0.69×10 -6 /30×10 -13
= 2.415×10 -7 F.
8. Determine the area where permittivity of material is 15.5×10 -15 Fcm -1 and width of 25×10 -6 and junction capacitance is 5pF.
a) 8.0645×10 -5
b) 5.456×10 -6
c) 3.0405×10 -2
d) 8.0645×10 -3
Answer: d
Explanation: The junction capacitance is given by,
C j = ε s A/ w = 5×10 -12 ×25×10 -6 /15.5×10 -15
= 8.0645×10 -3 m 2 .
9. Compute intrinsic region width of p-i-n photodiode having junction capacitance of 4pF and material permittivity of 16.5×10 -13 Fcm -1 and area of 0.55×10 -6 m 2 .
a) 7.45×10 -6
b) 2.26×10 -7
c) 4.64×10 -7
d) 5.65×10 -6
Answer: b
Explanation: The junction capacitance is given by,
C j = ε s A/ W
w = ε s A/C j
= 16.5×10 -13 × 0.55×10 -6 /4×10 -12
= 2.26×10 -7 .
10. Determine permittivity of p-i-n photodiode with junction capacitance of 5pF, area of 0.62×10 -6 m 2 and intrinsic region width of 28 μm.
a) 7.55×10 -12
b) 2.25×10 -10
c) 5×10 -9
d) 8.5×10 -12
Answer: b
Explanation: The junction capacitance is given by,
C j = ε s A/ W
ε s = C j w/A = 5×10 -12 ×28×10 -6 /0.62×10 -6
= 2.25×10 -10 Fcm -1 .
11. Determine response time of p-i-n photodiode if it has 3 dB bandwidth of 1.98×10 8 Hz.
a) 5.05×10 -6 sec
b) 5.05×10 -7 Sec
c) 5.05×10 -7 sec
d) 5.05×10 -8 Sec
Answer: c
Explanation: The maximum response time is
Maximum response time = 1/Bm = 1/1.98×10 8 = 5.05×10 -9 sec.
12. Compute maximum 3 dB bandwidth of p-i-n photodiode if it has a max response time of 5.8 ns.
a) 0.12 GHz
b) 0.14 GHz
c) 0.17 GHz
d) 0.13 GHz
Answer: c
Explanation: The maximum response time is
Maximum response time = 1/Bm
= 1/5.8×10 -9 = 0.17 GHz.
13. Determine maximum response time for a p-i-n photodiode having width of 28×10 -6 m and carrier velocity of 4×10 4 ms -1 .
a) 105.67 MHz
b) 180.43 MHz
c) 227.47 MHz
d) 250.65 MHz
Answer: c
Explanation: Maximum 3 dB bandwidth of photodiode is given by
Bm = V d /2ΠW = 4×10 -4 /2×3.14×28×10 -6 = 227.47 MHz.
14. Determine carrier velocity of a p-i-n photodiode where 3dB bandwidth is1.9×10 8 Hz and depletion region width of 24μm.
a) 93.43×10 -5
b) 29.55×10 -3
c) 41.56×10 -3
d) 65.3×10 -4
Answer: b
Explanation: Maximum 3 dB bandwidth of photodiode is given by
Bm = V d /2ΠW
V d = Bm × 2Π × W
= 1.98×10 8 ×2Π×24×10 -6
= 29.55×10 -3 .
15. Compute depletion region width of a p-i-n photodiode with 3dB bandwidth of 1.91×10 8 and carrier velocity of 2×10 4 ms -s .
a) 1.66×10 -5
b) 3.2×10 -3
c) 2×10 -5
d) 2.34×10 4
Answer: a
Explanation: Maximum 3 dB bandwidth of photodiode is given by
Bm = V d /2ΠW
W = V d /Bm2Π
= 2×10 -5 /1.91×10 8 ×2Π
= 1.66×10 -5 m.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Semiconductor Photodiodes With Internal Gain”.
1. ___________ has more sophisticated structure than p-i-n photodiode.
a) Avalanche photodiode
b) p-n junction diode
c) Zener diode
d) Varactor diode
Answer: a
Explanation: Avalanche photodiode is second major type of detector in optical communications. This diode is more sophisticated so as to create a much higher electric field region.
2. The phenomenon leading to avalanche breakdown in reverse-biased diodes is known as _______
a) Auger recombination
b) Mode hopping
c) Impact ionization
d) Extract ionization
Answer: c
Explanation: In depletion region, almost all photons are absorbed and carrier pairs are generated. So there comes a high field region where carriers acquire energy to excite new carrier pairs. This is impact ionization.
3. _______ is fully depleted by employing electric fields.
a) Avalanche photodiode
b) P-I-N diode
c) Varactor diode
d) P-n diode
Answer: a
Explanation: APD is fully depleted by electric fields more than 10 4 V/m. This causes all the drifting of carriers at saturated limited velocities.
4. At low gain, the transit time and RC effects ________
a) Are negligible
b) Are very less
c) Dominate
d) Reduce gradually
Answer: c
Explanation: Low gain causes the dominance of transit time and RC effects. This gives a definitive response time and thus device obtains constant bandwidth.
5. At high gain, avalanche buildup time ________
a) Is negligible
b) Very less
c) Increases gradually
d) Dominates
Answer: d
Explanation: High gain causes avalanche buildup time to dominate. Thus the bandwidth of device decreases as increase in gain.
6. Often __________ pulse shape is obtained from APD.
a) Negligible
b) Distorted
c) Asymmetric
d) Symmetric
Answer: c
Explanation: Asymmetric pulse shape is acquired from APD. This is due to relatively fast rise time as electrons are collected and fall time dictated by transit time of holes.
7. Fall times of 1 ns or more are common.
a) False
b) True
Answer: b
Explanation: The use of suitable materials and structures give rise times between 150 and 200 ps. Thus fall times of 1 ns or more are common which in turn limits the overall response of device.
8. Determine Responsivity of a silicon RAPD with 80% efficiency, 0.7μm wavelength.
a) 0.459
b) 0.7
c) 0.312
d) 0.42
Answer: a
Explanation: The Responsivity of a RAPD is given by-
R = ηeλ/hc A/w where, η=efficiency, λ = wavelength, h = Planck’s constant.
9. Compute wavelength of RAPD with 70% efficiency and Responsivity of 0.689 A/w.
a) 6μm
b) 7.21μm
c) 0.112μm
d) 3μm
Answer: c
Explanation: The wavelength can be found from the Responsivity formula given by-
R = ηeλ/hc. The unit of wavelength isμm.
10. Compute photocurrent of RAPD having optical power of 0.7 μw and responsivity of 0.689 A/W.
a) 0.23 μA
b) 0.489 μA
c) 0.123 μA
d) 9 μA
Answer: b
Explanation: The photocurrent is given byI P =P 0 R. Here I P = photocurrent, P 0 =Power, R = responsivity.
11. Determine optical power of RAPD with photocurrent of 0.396 μAand responsivity of 0.49 A/w.
a) 0.91 μW
b) 0.32 μW
c) 0.312 μW
d) 0.80 μW
Answer: d
Explanation: The photocurrent is given by I P = P 0 R. Here I P = photocurrent, P 0 = Power, R = responsivity.
P 0 = IP/R gives the optical power.
12. Determine the Responsivity of optical power of 0.4μW and photocurrent of 0.294 μA.
a) 0.735
b) 0.54
c) 0.56
d) 0.21
Answer: a
Explanation: The photocurrent is given by I P = P 0 R. Here I P = photocurrent, P 0 = Power, R = responsivity.
R = I P /P0 gives the responsivity.
13. Compute multiplication factor of RAPD with output current of 10 μAand photocurrent of 0.369μA.
a) 25.32
b) 27.100
c) 43
d) 22.2
Answer: b
Explanation: The multiplication factor of photodiode is given by-
M = I/I P where I = output current, I P = photocurrent.
14. Determine the output current of RAPD having multiplication factor of 39 and photocurrent of 0.469μA.
a) 17.21
b) 10.32
c) 12.21
d) 18.29
Answer: d
Explanation: The multiplication factor of photodiode is given by-
M = I/I P where I = output current, I P = photocurrent. I = M*I P gives the output current inμA.
15. Compute the photocurrent of RAPD having multiplication factor of 36.7 and output current of 7μA.
a) 0.01 μA
b) 0.07 μA
c) 0.54 μA
d) 0.9 μA
Answer: a
Explanation: The multiplication factor of photodiode is given by-
M = I/I P where I = output current, I P = photocurrent. I P = I/M Gives the output current inμA.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Mid Infrared and Far Infrared Photodiodes”.
1. In the development of photodiodes for mid-infrared and far-infrared transmission systems, lattice matching has been a problem when operating at wavelengths ____________
a) 1 µm
b) Greater than 2 µm
c) 2 µm
d) 0.5 µm
Answer: b
Explanation: Lattice matching for alloy materials is obtained at wavelengths above 2 µm. For example, a lattice-matched alloy material system was utilized in a p-i-n photodiode for high speed operation at wavelengths up to 2.3 µm.
2. What is generally used to accommodate a lattice mismatch?
a) Alloys
b) Attenuator
c) Graded buffer layer
d) APD array
Answer: c
Explanation: The use of indium alloy cause inherent problems of dislocation-induced junction leakage and low quantum efficiency. To avoid these problems, a compositionally graded buffer layer is used to accommodate a lattice mismatch.
3. HgCdTe material system is utilized to fabricate long-wavelength photodiodes.
a) True
b) False
Answer: a
Explanation: HgCdTe family alloys allow resonant characteristics via hole ionization. Its band gap energy variation enables optical detection to far-infrared. Thus, this material can be used for fabrication of long-wavelength photodiodes.
4. Avalanche photodiodes based on HgCdTe are used for ______________ in both the near and far infrared.
a) Dispersion
b) Dislocation
c) Ionization
d) Array applications
Answer: d
Explanation: Avalanche photodiodes based on HgCdTe are used for array applications. The materials of APDs based on HgCdTe possess uniform avalanche gain across an array. This variation in gain is variation in gain is lower in HgCdTe as compared with silicon.
5. The detection mechanism in ____________ relies on photo excitation of electrons from confined states in conduction band quantum wells.
a) p-i-n detector
b) Quantum-dot photo detector
c) p-n photodiode
d) Avalanche photodiodes
Answer: b
Explanation: Quantum-dot photo detector’s detection mechanism involves photo excitation of electrons. This process of photo excitation in photo detectors is similar to that in the Quantum-dot semiconductor optical amplifier. The dots-in-well in Quantum-dot detector is called as DWELL structure.
6. When determining performance of a photo detector ___________ is often used.
a) No. of incident photon
b) No. of electrons collected
c) Responsivity
d) Absorption coefficient
Answer: c
Explanation: The expression for quantum efficiency does not include photon energy. Thus for characterizing performance of photo detector, Responsivity is used.
7. The important parameter for exciting an electron with energy required from valence band to conduction band is?
a) Wavelength
b) Absorption coefficient
c) Responsivity
d) Band gap energy
Answer: a
Explanation: As wavelength of incident photon becomes longer, the photon energy is less than energy required to excite electron. Mostly parameters of photodiode are dependent on wavelength.
8. __________ is less than or unity for photo detectors.
a) Absorption coefficient
b) Band gap energy
c) Responsivity
d) Quantum efficiency
Answer: d
Explanation: Quantum efficiency determines the absorption coefficient of semiconductor material of photo detector. It is not all incident photons are absorbed to create electron-hole pairs. Thus quantum efficiency must be less than unity.
9. There must be improvement in __________ of an optical fiber communication system.
a) Detector
b) Responsivity
c) Absorption Coefficient
d) Band gap energy
Answer: a
Explanation: If proper and improved and highly efficient detector is utilized, it will then reduce the repeated stations. It will also lower down both capital investment and maintenance cost.
This set of Optical Communications online test focuses on “Phototransistors and Metal – Semiconductor – Metal Photodetectors”.
1. The _____________ is photosensitive to act as light gathering element.
a) Base-emitter junction
b) Base-collector junction
c) Collector-emitter junction
d) Base-collector junction and Base-emitter junction
Answer: a
Explanation: Base-collector junction is photosensitive in n-p-n phototransistor and act as light gathering element. This light absorbed affects the base current and gives multiplication of primary photocurrent in device.
2. A large secondary current _________________ in n-p-n InGaAs phototransistor is achieved.
a) Between base and collector
b) Between emitter and collector
c) Between base and emitter
d) Plasma
Answer: b
Explanation: The photo-generated holes are swept to the base. This increases the forward bias device. This generates secondary current between emitter and collector.
3. _______ emitter-base and collector-base junction capacitances is achieved by use of hetero-structure along with _________ base resistance.
a) Low, high
b) High, low
c) Low, low
d) High, negligible
Answer: c
Explanation: In hetero-structure, there is low doping level in emitter and collector which is coupled with heavy doping base. This is due low emitter-base and collector-base junction capacitance and low base resistance. This allows large current gain.
4. A ________ is created by hetero-junction at collector-base junction.
a) Potential barrier
b) Depletion region
c) Parasitic capacitance
d) Inductance
Answer: a
Explanation: Potential barrier is created at emitter-base junction by hetero-junction. This eliminates hole junction from base. This is achieved when junction is forward-biased and provides good emitter-base efficiency.
5. Phototransistors based on hetero-junction using _________ material are known as waveguide phototransistors.
a) InGaP
b) InGaAs
c) InGaAsP/ InAlAs
d) ErGaAs
Answer: c
Explanation: Phototransistor using InGaAsP/ InAlAs are known as waveguide phototransistors. They function as waveguide phototransistors. They function as high performance photo-detectors at 1.3 micro-meter wavelength. They utilize a passive waveguide layer under active transistor region.
6. A phototransistor has collector current of 18 mA, incident optical power of 128 μW with a wavelength of 1.24 μm. Determine an optical gain.
a) 1.407 *10 2
b) 19.407 *10 2
c) 2.407 *10 2
d) 3.407 *10 2
Answer: a
Explanation: The optical gain is given by-
G 0 =hcI c /λeP 0 , where h=Planck’s constant, I c =collector current, λ=wavelength, P 0 =incident optical power.
7. For a phototransistor having gain of 116.5, wavelength of 1.28 μm, optical power 123μW. Determine collector current.
a) 0.123 mA
b) 0.0149 mA
c) 1.23 mA
d) 0.54 mA
Answer: b
Explanation: The collector current is given by-
I c = G 0 λeP 0 / hc, where h=Planck’s constant, I c =collector current, λ=wavelength, P 0 =incident optical power.
8. The detection mechanism in the ____________ photo-detector includes inter sub-band transitions.
a) Dwell
b) Set
c) Avalanche
d) Futile
Answer: a
Explanation: The inter sub-band transitions are also known as type-2 transitions. It comprises of mini-bands within a single energy band, The detection mechanism in DWELL photo-detector includes inter sub-band transitions.
9. Which of the following is the difference between the n-p-n and conventional bipolar transistor?
a) Electric property
b) Magnetic property
c) Unconnected base
d) Emitter base efficiency
Answer: c
Explanation: The n-p-n bipolar transistor differs in the following ways: base is unconnected, base-collector junction is photosensitive as a light gathering element.
10. The n-p-n hetero-junction phototransistor is grown using ______________
a) Liquid-phase tranquilizers
b) Liquid-phase epistaxis
c) Solid substrate
d) Hetero poleax
Answer: b
Explanation: The technique LPE consists of a thin layer of n-type collector based on a p-type base layer. Liquid phase epistaxis is used in hetero-junction technology.
11. The _____________ at emitter-base junction gives good emitter base injection efficiency.
a) Homo-junction
b) Depletion layer
c) Holes
d) Hetero-junction
Answer: d
Explanation: The hetero-junction at the emitter-base junction effectively eliminates hole injection from the base when the junction is forward biased. This gives good emitter-base injection efficiency.
12. Waveguide phototransistors utilize a ___________ waveguide layer under the _________ transistor region.
a) Active, passive
b) Passive, active
c) Homo, hetero
d) Hetero, homo
Answer: b
Explanation: Waveguide phototransistors are based on hetero-junction structure. They function as high-performance photo-detectors and thus utilize a passive waveguide layer under the active transistor region.
13. What is the main benefit of the waveguide structure over conventional hetero-junction phototransistor?
a) High depletion region
b) Depletion width
c) Increased photocurrent, responsivity
d) Low gain
Answer: c
Explanation: Waveguide structure offers increased photocurrent. Photocurrent is directly proportional to the responsivity; thus in turn increases responsivity.
14. Waveguide structure provides high quantum efficiency.
a) True
b) False
Answer: b
Explanation: Responsivity and quantum efficiency follow a different path. They are indirectly proportional to each other. Thus, in waveguide structure, as the responsivity increases, quantum efficiency remains low.
15. Metal-semiconductor-metal photo-detectors are photoconductive detectors.
a) True
b) False
Answer: a
Explanation: MSM photo-detectors are the simplest of photo-detectors. It provides the simplest form of photo-detection within optical fiber communications and are photoconductive.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Noise”.
1. _____________ refers to any spurious or undesired disturbances that mask the received signal in a communication system.
a) Attenuation
b) Noise
c) Dispersion
d) Bandwidth
Answer: b
Explanation: Noise is an unwanted and undesirable quantity. It affects the received signal in a communication system. In optical fiber communication systems, noise is due to the spontaneous fluctuations rather than erratic disturbances.
2. How many types of noise are observed because of the spontaneous fluctuations in optical fiber communication systems?
a) One
b) Four
c) Two
d) Three
Answer: d
Explanation: There are three types of noise because of the spontaneous fluctuations in optical fiber communication systems. These are thermal noise, the dark current noise and quantum noise. These noise types are not caused by the electronic interference.
3. ______________ is caused due to thermal interaction between the free electrons and the vibrating ions in the conduction medium.
a) Thermal noise
b) Dark noise
c) Quantum noise
d) Gaussian noise
Answer: a
Explanation: Thermal noise is basically a spontaneous fluctuation caused due to thermal interaction of electrons and ions. It is especially prevalent in resistors at room temperature. Thermal noise is measured in the form of current and is called as thermal noise current.
4. A small leakage current still flows from the device terminals even if there is no optical power incident on the photo detector.
a) True
b) False
Answer: a
Explanation: A reverse leakage current that flows from the device terminals is called as dark current. This dark current contributes to the total system noise. This gives random fluctuations about the average particle flow of the photocurrent.
5. ___________ distribution provides the description the random statistics of light emitted in black body radiation.
a) Poisson
b) Cumulative
c) Probability
d) Bose-Einstein
Answer: d
Explanation: Incoherent light is emitted by independent atoms and therefore there is no phase relationship between the emitted photons. The property dictates an exponential intensity distribution which is identical to Bose-Einstein distribution.
6. The probability of zero pairs being generated when a light pulse is present is given by which of the following equation?
a) P = exp(-Z m )
b) P = exp (Z m )
c) P = x + x
d) P = P(-Z m )
Answer: a
Explanation: The probability of zero pairs being generated when a light pulse is present is given by equation –
P = exp(-Z m )
Where, P represents the system error probability p and Z m is variance of the probability distribution.
7. The minimum pulse energy needed to maintain a given bit-error-rate which any practical receiver must satisfy is known as ___________
a) Minimal energy
b) Quantum limit
c) Point of reversed
d) Binary signaling
Answer: b
Explanation: A perfect photo detector emits no electron-hole pairs in the absence of illumination. The error probability determines a standardized fundamental limit in digital optical communications. This limit is termed as quantum limit.
8. A digital optical fiber communication system requires a maximum bit-error-rate of 10 -9 . Find the average number of photons detected in a time period for a given BER.
a) 19.7
b) 21.2
c) 20.7
d) 26.2
Answer: c
Explanation: The probability of error is given by-
P = exp(-Z m )
Where, Z m = No. of photons
Here P = 10 -9 , therefore Z m is calculated from above relation.
9. For a given optical fiber communication system, P = 10 -9 , Z m = 20.7, f = 2.9×10 14 , η = 1. Find the minimum pulse energy or quantum limit.
a) 3.9×10 -18
b) 4.2×10 -18
c) 6.2×10 -14
d) 7.2×10 -14
Answer: a
Explanation: The minimum pulse energy or quantum limit is given by –
E min = Z m hf/η
Where, Z m = Number of photons
h = Planck’s constant
f = frequency
η = Quantum efficiency.
10. An analog optical fiber system operating at wavelength 1μmhas a post-detection bandwidth of 5MHz. Assuming an ideal detector and incident power of 198 nW, calculate the SNR (f = 2.99×10 14 Hz).
a) 46
b) 40
c) 50
d) 52
Answer: c
Explanation: The SNR is given by –
S/N = ηP 0 /2hfB
Where, η = 1
P 0 = incident power
h = Planck’s constant
B = Bandwidth.
11. The incident optical power required to achieve a desirable SNR is 168.2nW. What is the value of incident power in dBm?
a) -37.7 dBm
b) -37 dBm
c) – 34 dBm
d) -38.2 dBm
Answer: a
Explanation: Incident power in denoted by P 0 . It is given by –
P 0 = 10log 10 (P 0 )
Where P 0 = incident power in Watts/milliWatt.
12. In the equation given below, what does τstands for?
Zm = ηP0τ/hf
a) Velocity
b) Time
c) Reflection
d) Refractive index
Answer: b
Explanation: In the given equation, Z m is the variance of the probability distribution. The number of electrons generated in time τis equal to the average of the number of photons detected over this time period Z m . Hence, τ is the time and P 0 is the incident power, ηis the quantum efficiency and f is the frequency.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Receiver Noise”.
1. Which are the two main sources of noise in photodiodes without internal gain?
a) Gaussian noise and dark current noise
b) Internal noise and external noise
c) Dark current noise & Quantum noise
d) Gaussian noise and Quantum noise
Answer: c
Explanation: The two main sources of noise in photodiodes without internal gain are dark current noise and quantum noise. They are regarded as shot noise on the photocurrent. These noise are together called as analog quantum noise.
2. The dominating effect of thermal noise over the shot noise in photodiodes without internal gain can be observed in wideband systems operating in the range of ________
a) 0.4 to 0.5 μm
b) 0.8 to 0.9 μm
c) 0.3 to 0.4 μm
d) 0.7 to 0.79 μm
Answer: b
Explanation: When the photodiode is without internal avalanche gain, the detector load resistor and active elements’ thermal noise in the amplifier tends to dominate. It is seen in wideband systems operating in the 0.8 to 0.9 μmwavelength band. This is because the dark currents in the silicon diodes can be made very small.
3. A silicon p-i-n photodiode incorporated in an optical receiver has following parameters:
Quantum efficiency = 70%
Wavelength = 0.8 μm
Dark current = 3nA
Load resistance = 4 kΩ
Incident optical power = 150nW.
Bandwidth = 5 MHz
Compute the photocurrent in the device.
a) 67.7nA
b) 81.2nA
c) 68.35nA
d) 46.1nA
Answer: a
Explanation: The photocurrent is given by
I p = ηP 0 eλ/hc
Where η = Quantum efficiency
P 0 = Incident optical power
e = electron charge
λ = Wavelength
h = Planck’s constant
c = Velocity of light.
4. In a silicon p-i-n photodiode, if load resistance is 4 kΩ, temperature is 293 K, bandwidth is 4MHz, find the thermal noise in the load resistor.
a) 1.8 × 10 -16 A 2
b) 1.23 × 10 -17 A 2
c) 1.65 × 10 -16 A 2
d) 1.61 × 10 -17 A 2
Answer: d
Explanation: The thermal noise in the load resistor is given by –
i t 2 = 4KTB/R L
Where T = Temperature
B = Bandwidth
R L = Load resistance.
5. ________________ is a combination of shunt capacitances and resistances.
a) Attenuation
b) Shunt impedance
c) Shunt admittance
d) Thermal capacitance
Answer: c
Explanation: Admittance is a measure of how easily a circuit will allow a current to flow. It is the inverse of impedance and is measured in Siemens. It is a combination of shunt capacitances and resistances.
6. ______________ is used in the specification of optical detectors.
a) Noise equivalent power
b) Polarization
c) Sensitivity
d) Electron movement
Answer: a
Explanation: Noise equivalent power is defined as the amount of incident optical power per unit bandwidth required to produce an output power equal to detector output noise power.
Noise equivalent power is the value of incident power which gives an output SNR of unity.
7. A photodiode has a capacitance of 6 pF. Calculate the maximum load resistance which allows an 8MHz post detection bandwidth.
a) 3.9 kΩ
b) 3.46 kΩ
c) 3.12 kΩ
d) 3.32 kΩ
Answer: d
Explanation: The load resistance is given by-
R L = 1/2πC d B
Where
B = Post detection bandwidth
C d = Input capacitance
R L = Load resistance.
8. The internal gain mechanism in an APD is directly related to SNR. State whether the given statement is true or false.
a) True
b) False
Answer: a
Explanation: The internal gain mechanism in an APD increases the signal current into the amplifier. This improves the SNR because the load resistance and amplifier noise remains unaffected.
9. ____________ is dependent upon the detector material, the shape of the electric field profile within the device.
a) SNR
b) Excess avalanche noise factor
c) Noise gradient
d) Noise power
Answer: b
Explanation: Excess avalanche noise factor is represented as F . Its value depends upon the detector material, shape of electric field profile and holes and electrons inclusion. It is a function of multiplication factor.
10. For silicon APDs, the value of excess noise factor is between _________
a) 0.001 and 0.002
b) 0.5 and 0.7
c) 0.02 and 0.10
d) 1 and 2
Answer: c
Explanation: The excess noise factor is same as that of the multiplication factor. In case of holes, the smaller values of K produce high performance and therefore the performance is achieved when k is small. For silicon APDs, k = 0.02 to 0.10.
11. __________ determines a higher transmission rate related to the gain of the APD device.
a) Attenuation
b) Gain-bandwidth product
c) Dispersion mechanism
d) Ionization coefficient
Answer: b
Explanation: Gain-bandwidth product is defined as Gain multiplied by the bandwidth. Gain is a dimensionless quantity but the gain-bandwidth product is therefore measured in the units of frequency.
12. _________________ APDs are recognized for their high gain-bandwidth products.
a) GaAs
b) Alloy-made
c) Germanium
d) Silicon
Answer: d
Explanation: Silicon APDs possess a large asymmetry of electron and hole ionization coefficient. Thus, they possess high gain-bandwidth products. These APDs do not operate at high transmission rates.
13. APDs do not operate at signal wavelengths between 1.3 and 1.6μm.
a) True
b) False
Answer: a
Explanation: APDs having high gain-bandwidth products do not operate at signal wavelengths between 1.3 and 1.6 μm.Hence, these APDs are not prefered for use in receivers operating at high transmission rates.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Receiver Structures”.
1. How many circuits are present in an equivalent circuit for the digital optical fiber receiver?
a) Four
b) One
c) Three
d) Two
Answer: a
Explanation: A full equivalent circuit for the digital optical fiber receiver includes four circuits. These are the detector circuit, noise sources, and amplifier and equalizer circuit.
2. __________ compensates for distortion of the signal due to the combined transmitter, medium and receiver characteristics.
a) Amplification
b) Distortion
c) Equalization
d) Dispersion
Answer: c
Explanation: Equalization adjusts the balance between frequency components within an electronic signal. It compensates for distortion of the signal. The distortion may be due to the transmitter, receiver etc.
3. ____________ is also known as frequency-shaping filter.
a) Resonator
b) Amplifiers
c) Attenuator
d) Equalizer
Answer: d
Explanation: Equalizer, often called as frequency-shaping filter has a frequency response inverse to that of the overall system frequency response. In wideband systems, it boosts the high frequency components to correct the overall amplitude of the frequency response.
4. The phase frequency response of the system should be ____________ in order to minimize inter-symbol interference.
a) Non-Linear
b) Linear
c) More
d) Less
Answer: b
Explanation: An equalizer is used as frequency shaping filter. The phase frequency response of the system should be linear to acquire the desired spectral shape for digital systems. This, in turn, minimizes the inter-symbol interference.
5. Noise contributions from the sources should be minimized to maximize the receiver sensitivity.
a) True
b) False
Answer: a
Explanation: Noise sources include transmitter section, medium and the receiver section. As the noise increases, the sensitivity at the receiver section decreases. Thus, noise contributions should be minimized to maximize the receiver sensitivity.
6. How many amplifier configurations are frequently used in optional fiber communication receivers?
a) One
b) Two
c) Three
d) Four
Answer: c
Explanation: Three amplifier configurations are used in optical fiber communication receivers. These are voltage amplifiers, semiconductor optical amplifier and current amplifier. Voltage amplifier is the simplest and most common amplifier configuration.
7. How many receiver structures are used to obtain better receiver characteristics?
a) Two
b) One
c) Four
d) Three
Answer: d
Explanation: The various receiver structures are low-impedance front end, high-impedance front end and trans-impedance front-end. The noise in the trans-impedance amplifier will always exceed than the front end structure.
8. The high-impedance front-end amplifier provides a far greater bandwidth than the trans-impedance front-end.
a) True
b) False
Answer: a
Explanation: The noise in the trans-impedance amplifier exceeds that incurred by the high-impedance amplifier. Hence, the trans-impedance front-end provides a greater bandwidth without equalization than the high-impedance front end.
9. A high-impedance amplifier has an effective input resistance of 4MΩ. Find the maximum bandwidth that may be obtained without equalization if the total capacitance is 6 pF and total effective load resistance is 2MΩ.
a) 13.3 kHz
b) 14.2 kHz
c) 15.8 kHz
d) 13.9 kHz
Answer: a
Explanation: The maximum bandwidth obtained without equalization is given by –
B = 1/2ΠR TL C T
Where,
R TL = Total load resistance
C T = Total capacitance.
10. A high-input-impedance amplifier has following parameters . Find the mean square thermal noise current per unit bandwidth for the high-impedance configuration.
a) 8.9×10 -27 A 2 /Hz
b) 8.12×10 -27 A 2 /Hz
c) 8.29×10 -27 A 2 /Hz
d) 8.4×10 -27 A 2 /Hz
Answer: c
Explanation: the mean square thermal noise current per unit bandwidth for the high-impedance configuration is given by –
i T 2 = 4KT/R TL
Where, K = constant
T = Temperature
R TL = total effective load resistance.
11. The mean square thermal noise current in the trans-impedance configuration is _________ greater than that obtained with the high-input-impedance configuration.
a) 30
b) 20
c) 15
d) 10
Answer: b
Explanation: 13 dB noise penalties are incurred with the trans-impedance amplifier over that of the high-input-impedance configuration. It is the logarithmic function of the noise current value. However, the trans-impedance amplifiers can be optimized for noise performance.
12. The major advantage of the trans-impedance configuration over the high-impedance front end is ______________
a) Greater bandwidth
b) Less bandwidth
c) Greater dynamic range
d) Less dynamic range
Answer: c
Explanation: Greater dynamic range is a result of the different attenuation mechanism for the low-frequency components of the signal. This attenuation is obtained in the trans-impedance amplifier through the negative feedback and therefore the low frequency components are amplified by the closed loop. This increases the dynamic range.
13. The trans-impedance front end configuration operates as a __________ with negative feedback.
a) Current mode amplifier
b) Voltage amplifier
c) Attenuator
d) Resonator
Answer: a
Explanation: The trans-impedance configuration overcomes the drawbacks of the high-impedance front end. It utilizes a low-noise, high-input-impedance amplifier with negative feedback. It operates as a current mode amplifier where high impedance is reduced by negative feedback.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “FET Pre – Amplifiers”.
1. ____________ is the lowest noise amplifier device.
a) Silicon FET
b) Amplifier-A
c) Attenuator
d) Resonator-B
Answer: a
Explanation: FET operates by controlling the current flow with an electric field produced by an applied voltage on the gate of the device. Silicon FET is fabricated for low noise devices. It is the lowest noise amplifier device available.
2. FET device has extremely high input impedance greater than _________
a) 10 7 Ohms and less than 10 8
b) 10 6 Ohms and less than 10 7
c) 10 14 Ohms
d) 10 23 Ohms
Answer: c
Explanation: FET operation involves the applied voltage on the gate of the device. The gate draws virtually no current, except for leakage, giving the device extremely high input impedance.
3. The properties of a bipolar transistor are superior to the FET.
a) True
b) False
Answer: b
Explanation: bipolar transistor operates by controlling the current flow with an electric field produced with a base current. The properties of a bipolar transistor are limited by its high trans-conductance than the FET.
4. Bipolar transistor is more useful amplifying device than FET at frequencies _____________
a) Above 1000 MHz
b) Equal to 1 MHz
c) Below 25 MHz
d) Above 25 MHz
Answer: d
Explanation: In FETs, the current gain drops to values near unity at frequencies above 25MHz. The trans-conductance is fixed with decreasing input impedance. Therefore, bipolar transistor is more useful amplifying device at frequencies above 25MHz.
5. High-performance microwave FETs are fabricated from ___________
a) Silicon
b) Germanium
c) Gallium arsenide
d) Zinc
Answer: c
Explanation: Since the mid- 1970s, the development of high-performance microwave FETs found its way. These FETs are fabricated from gallium arsenide and are called as GaAs metal Schottky field effect transistors .
6. Gallium arsenide MESFETs are advantageous than Silicon FETs.
a) True
b) False
Answer: a
Explanation: Gallium arsenide MESFETs are Schottky barrier devices. They operate with both low noise and high gain at microwave frequencies . Silicon FETs cannot operate with wide bands.
7. The PIN-FET hybrid receivers are a combination of ______________
a) Hybrid resistances and capacitances
b) Pin photodiode and low noise amplifier
c) P-N photodiode and low noise amplifier
d) Attenuator and low noise amplifier
Answer: b
Explanation: The PIN-FET or p-i-n/FET receiver utilizes a p-i-n photodiode along with a low noise preamplifier . It is fabricated using thick-film integrated circuit technology. This hybrid integration reduces the stray capacitance to negligible levels.
8. PIN-FET hybrid receiver is designed for use at a transmission rate of _____________
a) 130 Mbits -1
b) 110 Mbits -1
c) 120 Mbits -1
d) 140 Mbits -1
Answer: d
Explanation: At 140 Mbits -1 , the performance of PIN-FET hybrid receiver is found to be comparable to germanium and alloy APD receivers. A digital equalizer is necessary as the high-impedance front end effectively integrates the signal at 140 Mbits -1 .
9. It is difficult to achieve higher transmission rates using conventional __________
a) Voltage amplifier
b) Waveguide Structures
c) PIN-FET or APD receivers
d) MESFET
Answer: c
Explanation: It is difficult to achieve higher transmission rates due to limitations in their gain bandwidth products. Also, the trade-off between the multiplication factor requirement and the bandwidth limits the performance of conventional receivers.
10. Which receiver can be fabricated using PIN-FET hybrid approach?
a) Trans-impedance front end receiver
b) Gallium arsenide receiver
c) High-impedance front-end
d) Low-impedance front-end
Answer: a
Explanation: Trans-impedance front-end receivers are fabricated using the PIN-FET hybrid approach. An example of such receivers consists of a GaAs MESFET and two complementary bipolar microwave transistors.
11. A silicon p-i-n photodiode utilized with the amplifier and the receiver is designed to accept data at a rate of ___________
a) 276Mbits -1
b) 274 Mbits -1
c) 278Mbits -1
d) 302Mbits -1
Answer: b
Explanation: A silicon p-i-n photodiode is used with the low-noise preamplifier. This preamplifier is based on a GaAs MESFET. Thus, a receiver using p-i-n photodiode accepts a data rate of 274 Mbits -1 giving a sensitivity around -35dB m .
12. What is usually required by FETs to optimize the figure of merit?
a) Attenuation of barrier
b) Matching with the depletion region
c) Dispersion of the gate region
d) Matching with the detector
Answer: d
Explanation: Total capacitance is given by C t = C d + C a . The figure of merit is optimized when C d =C a . This requires FETs to be matched with the detectors. This requires FETs to be matched with the detectors. This procedure is usually not welcomed by the device and is not permitted in current optical receiver design.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “High Performance Receivers”.
1. How many design considerations are considered while determining the receiver performance?
a) Three
b) Two
c) One
d) Four
Answer: a
Explanation: Three main considerations are utilized for determining the receiver performance. Noise performance is a major design consideration providing a limitation to the sensitivity. Other two considerations are bandwidth and dynamic range.
2. FET preamplifiers provide higher sensitivity than the Si-bipolar device.
a) True
b) False
Answer: a
Explanation: At low speeds, the FET preamplifiers provide higher sensitivity than the Si-bipolar device. It is apparent that below 10Mbits -1 the Si MOSFET preamplifier provides a lower noise performance than GaAs MESFET.
3. What is the abbreviation of HBT?
a) Homo-junction unipolar transistor
b) Homo-junction bipolar transistor
c) Hetero-junction bipolar transistor
d) Hetero-Bandwidth transcendence
Answer: c
Explanation: HBT is abbreviated as Hetero-junction bipolar transistor. It comprises a selectively doped hetero-junction FET. It is a high-speed, low-noise transistor device.
4. What type of receivers are used to provide wideband operation, low-noise operation?
a) APD optical receivers
b) Optoelectronic integrated circuits
c) MESFET receivers
d) Trans-impedance front-end receivers
Answer: b
Explanation: A strategy for the provision of wideband, low-noise receivers involves the use of p-i-n photodiode detector along with the monolithic integration of the device with semiconductor alloy FETs. It has an operating wavelength of 1.1 to 1.6 μmranges.
5. ___________ circuits extends the dynamic range of the receiver.
a) Monolithic
b) Trans-impedance
c) Automatic Error Control
d) Automatic Gain Control
Answer: d
Explanation: AGC circuit extends the dynamic range by diverting excess photocurrent away from the input of the receiver. The receiver dynamic range is an important performance parameter as it provides a measure of the difference between the sensitivity and its overload level.
6. The sensitivity of the low-impedance configuration is ____________
a) Good
b) Poor
c) Great
d) Same as that of high-impedance configuration
Answer: b
Explanation: A receiver saturation level is determined by the value of the photodiode bias resistor. The photodiode bias resistor valve is indirectly proportional to the sensitivity but is directly proportional in low impedance configuration. The low resistor value provides less sensitivity in the low-impedance configuration.
7. What is generally used to determine the receiver performance characteristics?
a) Noise
b) Resistor
c) Dynamic range & sensitivity characteristics
d) Impedance
Answer: c
Explanation: Dynamic range and sensitivity characteristics involve a graph of received power level and the value of feedback resistor. The high value of photodiode bias resistor in the high impedance front end causes high sensitivity and a narrow dynamic range. These factors prove useful for determining the performance characteristics of receiver.
8. The __________ technique eliminates the thermal noise associated with the feedback resistor in the trans-impedance front end design.
a) Compensation
b) Resonating impedance
c) Electromagnetic
d) Optical feedback
Answer: d
Explanation: The optical feedback strategy proves most useful at low transmission rate. The use of optically coupled feedback has demonstrated dynamic ranges of around 40 dB for p-i-n receivers operating at modest bit rates. It removes thermal noise associated with the feedback resistor.
9. The removal of the feedback resistor in the optical feedback technique allows reciever sensitivity of the order of _______________
a) -54 dB m at 2Mbit/sec
b) -12 dB m at 2Mbit/sec
c) -64 dB m at 2Mbit/sec
d) -72 dB m at 2Mbit/sec
Answer: c
Explanation: The removal of feedback resistor in the optical feedback technique allows low noise performance. Low noise performance, in turn, affects sensitivity. The receiver sensitivity gets high of the order of -64 dB m at 2Mbit/sec transmission rates.
10. The optical feedback technique is useful at low transmission rates.
a) True
b) False
Answer: a
Explanation: The optical feedback technique is useful at low transmission rates because in this case the feedback resistors employed are smaller than the optimum value for low-noise performance. This is done to maintain the resistor at a practical size of 1MΩ. Large values of feedback resistor limits the dynamic range.
11. How many types of optical amplifier technologies are available.
a) One
b) Three
c) Four
d) Two
Answer: d
Explanation: There are two basic optical amplifier technologies available. They are semiconductor optical amplifiers and fiber amplifiers. Both these devices are utilized in the pre-amplification role.
12. The optimum filter bandwidth is typically in the range ________________
a) 0.1 to 0.3 nm
b) 0.5 to 3 nm
c) 0.1 to 0.3 μm
d) 0.5 to 3 μm
Answer: b
Explanation: The optimum fiber bandwidth is determined by detector noise, transmission rate and the transmitter chirp characteristics. It is typically in the range of 0.5 to 3 nmas it depends upon the filter insertion loss.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Optical Amplifiers – Semiconductor Optical Amplifiers”.
1. For linear as well as in nonlinear mode _______________ are most important network elements.
a) Optical amplifier
b) Optical detector
c) A/D converter
d) D/A converters
Answer: a
Explanation: In single-mode fiber system, signal dispersion is very small, hence there is attenuation. These systems don’t require signal regeneration as optical amplification is sufficient so optical amplifier are most important.
2. The more advantages optical amplifier is ____________
a) Fiber amplifier
b) Semiconductor amplifier
c) Repeaters
d) Mode hooping amplifier
Answer: b
Explanation: Semiconductor optical amplifiers are having smaller size. They can be integrated to produce subsystems. Thus are more profitable than other optical amplifier.
3. ________________ cannot be used for wideband amplification.
a) Semiconductor optical amplifier
b) Erbium-doped fiber amplifier
c) Raman fiber amplifier
d) Brillouin fiber amplifier
Answer: d
Explanation: Brillouin fiber amplifiers provide a very narrow spectral bandwidth. These bandwidth can be around 50 MHz, hence cannot be employed for wideband amplification.
4. ____________ is used preferably for channel selection in a WDM system.
a) Semiconductor optical amplifier
b) Erbium-doped fiber amplifier
c) Raman fiber amplifier
d) Brillouin fiber amplifier
Answer: d
Explanation: Brillouin fiber provides amplification of a particular channel. This amplification can be done without boosting other channels besides that particular channel.
5. For used in single-mode fiber __________ are used preferably.
a) Semiconductor optical amplifier
b) Erbium-doped fiber amplifier
c) Raman fiber amplifier
d) Brillouin fiber amplifier
Answer: a
Explanation: Semiconductor optical amplifiers have low power consumption. There single mode structure makes them appropriate and suitable for used in single mode fiber.
6. Mostly ____________ are used in nonlinear applications.
a) Semiconductor optical amplifier
b) Erbium-doped fiber amplifier
c) Raman fiber amplifier
d) FPAs
Answer: d
Explanation: FPAs have a resonant nature. This can be combined with their high internal fields. They provide pulse shaping and bi-stable elements. Thus, are used widely in nonlinear application.
7. _______________ is superior as compared to _________________
a) TWA, FPA
b) FPA, TWA
c) EDFA, FPA
d) FPA, EDFA
Answer: a
Explanation: In TWA operating in single-pass amplification mode, the Fabry-Perot resonance is suppressed by facet reflectivity reduction. This affects in increasing of amplifier spectral bandwidth. This makes them less dependence of transmission characteristics on fluctuations in biased current, input signal polarization. Thus FPA are superior to TWA.
8. ______________ are operated at current beyond normal lasing threshold current, practically.
a) Semiconductor optical amplifier
b) Erbium-doped fiber amplifier
c) Raman fiber amplifier
d) Brillouin fiber amplifier
Answer: a
Explanation: The anti-reflection facet coatings affects in the form of increasing lasing current threshold. This causes SOAs to be operated at current beyond normal lasing threshold current.
9. An uncoated FPA has peak gain wavelength 1.8μm, mode spacing of 0.8nm, and long active region of 300 v. Determine RI of active medium.
a) 4.25×10 6
b) 3.75×10 7
c) 3.95×10 7
d) 4.25×10 9
Answer: b
Explanation: n=λ 2 /2δλL=1.8×10 -6 /2×0.8×10 -9 ×300×10 -6 =3.75×10 7 .
10. Determine the peak gain wavelength of uncoated FPA having mode spacing of 2nm,and 250μmlong active region and R.I of 3.78.
a)2.25×10 -4
b)4.53×10 -8
c)1.94×10 -6
d)4.25×10 9
Answer: c
Explanation: The peak gain wavelength is given by
λ 2 =n2δλL=3.78×2×2×10 -9 ×250×10 -6 =1.94×10 -6 m.
11. An SOA has net gain coefficient of 300, at a gain of 30dB. Determine length of SOA.
a) 0.32 m
b) 0.023 m
c) 0.245 m
d) 0.563 m
Answer: b
Explanation: The length of SOA is determined by
L = G s /10×g×log e = 30/10×300×0.434`= 0.023 m.
12. An SOA has length of 35.43×10 -3 m, at 30 dB gain. Determine net gain coefficient.
a) 5.124×10 -3
b) 1.12×10 -4
c) 5.125×10 -3
d) 2.15×10 -5
Answer: c
Explanation: The net gain coefficient of SOA is given by
g = L×10×log e /G s = 35.43×10 -3 ×10×0.434/30
=5.125×10 -3 .
13. An SOA has mode number of 2.6, spontaneous emission factor of 4, optical bandwidth of 1 THz. Determine noise power spectral density.
a) 1.33×10 -3
b) 5.13×10 12
c) 3.29×10 -6
d) 0.33×10 -9
Answer: a
Explanation: The noise power spectral density P ast is
P ast = mn sp (G s -1) hfb
= 2.6×4×6.63×10 -34 ×1.94×10 14 ×1×10 12
= 1.33×10 -3 W.
14. An SOA has noise power spectral density of 1.18mW, spontaneous emission factor of 4, optical bandwidth of 1.5 THz. Determine mode number.
a) 1.53 × 10 28
b) 6.14 × 10 12
c) 1.78 × 10 16
d) 4.12 × 10 -3
Answer: a
Explanation: The mode number is determined by
m = P ast /n sp hfB
= 1.18×10 -3 /4×6.63×10 -34 ×1.94×10 14 ×1.3×10 12
= 1.53 × 10 -34 .
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Fiber and Waveguide Amplifiers”.
1. The spectral dependence on gain is always constant.
a) True
b) False
Answer: b
Explanation: The spectral dependence on gain is mostly not constant. Thus the spectral bandwidth for erbium-doped silica fibers is restricted to around 300 GHz.
2. ESA ________ the pumping efficiency of device.
a) Increases
b) Does not affects
c) Reduces
d) Has negligible effect on
Answer: c
Explanation: In erbium fiber amplifier photons at pump wavelength promotes the electrons in upper lasing level into a high state of excitation. These electrons decay non-radiate to intermediate levels and then back to upper lasing level thereby reducing pumping efficiency.
3. Signal amplification is obtained in ____________
a) Erbium-doped fluoro-zir-carbonate fiber multimode
b) Rare-earth-doped fiber amplifiers
c) Raman fiber systems
d) Brillouin fiber amplifier
Answer: a
Explanation: To avoid excited state absorption . We should use different glass technology in place by using a 488 nmpump wavelength; erbium-doped multimode fluoro zir carbonate fiber provides gain at 1.525 μmwavelengths.
4. It is possible to construct a single rare-earth-doped fiber amplifier which provides amplification for all-bands.
a) True
b) False
Answer: b
Explanation: Each material has different absorption emission properties to absorb energy either in single or multi steps. Also it possesses property to emit light in one or more narrow spectral ranges. Thus we cannot construct a single earth-doped fiber for all bands.
5. _______________ is constructed using erbium-doped glass.
a) An erbium-based micro fiber amplifier
b) Rare-earth-doped fiber amplifiers
c) Raman fiber systems
d) Brillouin fiber amplifier
Answer: a
Explanation: As compared to other glass, erbium-based micro fiber amplifier is more advantageous. This amplifier provides high optical gain over just a few centimeters of fiber over many meters.
6. ____________ uses Er 3+ -doped erbium glass.
a) An erbium-based micro fiber amplifier
b) Rare-earth-doped fiber amplifiers
c) Raman fiber systems
d) Brillouin fiber amplifier
Answer: a
Explanation: The erbium-based micro fiber amplifier uses Er 3+ -doped erbium glass. It supports the doping constructions of erbium ions at high levels as compared to conventional glasses.
7. The most advantageous amplification is ____________
a) An erbium-based micro fiber amplifier
b) Rare-earth-doped fiber amplifiers
c) Raman fiber systems
d) Brillouin fiber amplifier
Answer: c
Explanation: As compared to all the amplifications, Raman amplification is more advantageous. It has self-phase matching between pump of signal together with broad gain bandwidth as compared to other nonlinear processes.
8. _________ is also known as lump Raman amplifiers.
a) An erbium-based micro fiber amplifier
b) Rare-earth-doped fiber amplifiers
c) Raman fiber systems
d) Discrete Raman amplifiers
Answer: d
Explanation: Discrete Raman Amplifiers are lumped elements. This lumped element is to be inserted in transmission line to provide gain.
9. _______________ extends the pump power into transmission line fiber.
a) An erbium-based micro fiber amplifier
b) Rare-earth-doped fiber amplifiers
c) Raman fiber systems
d) Distributed Raman amplification
Answer: d
Explanation: In Distributed Raman amplification, all pump power is confined to lumped element. And it is distributed when the amplification takes place among several kilometers.
10. _____________ are called hybrid Raman amplifier.
a) Lumped and distributed Raman Amplifiers
b) Rare-earth-doped fiber amplifiers
c) Raman fiber systems
d) Distributed Raman amplification
Answer: a
Explanation: Lumped and distributed Raman Amplifiers can be combined together to be used in wideband application. This combination increases overall amplified spectral bandwidth.
11. In ___________ the ASE contributes most of noise.
a) An erbium-based micro fiber amplifier
b) Rare-earth-doped fiber amplifiers
c) Raman fiber systems
d) Distributed Raman amplification
Answer: d
Explanation: ASE contributes most of noise in Raman Amplification. The common sources of noise include beating of signal with ASE, mixing, self-phase modulation and cross-plane modulation.
12. In _____________ Rayleigh scattering can be reduced.
a) An erbium-based micro fiber amplifier
b) Rare-earth-doped fiber amplifiers
c) Raman fiber systems
d) Distributed Raman amplification
Answer: d
Explanation: Rayleigh scattering adverse effects can be reduced in Raman Amplification. This can be done by employing two or more stages of amplification over single stage amplification over fiber.
13. Compute the fiber nonlinear coefficient of a parametric optical amplifier having parametric peak gain of 63.6 dB, signal power of 1.6W, length 520.
a) 2.78×10 -2 W -1 km -1
b) 9.61×10 -3 W -1 km -1
c) 3.25×10 -3 W -1 km -1
d) 5.61×10 -4 W -1 km -1
Answer: b
Explanation: The fiber nonlinear coefficient can be found by
γ = G p -log 10 /P pl ×L × 1/10log 10 2
= 63.6+6/1.6×1.6×520×1/8.7 = 9.61×10 -3 W -1 km -1 .
14. Compute signal power for parametric amplifier having length of 500, nonlinear gain coefficient 12.6×10 -3 W -1 km -1 and parametric peak gain of 63.9 dB.
a) 0.245 W
b) 0.012 W
c) 0.19 W
d) 0.342 W
Answer: b
Explanation: Signal power is given by
P p =G p -log 10 /γL× 1/10log 10 2 = 63.9+6/12.6×10 -3 ×1/ 8.7
= 0.012 W.
15. Compute the gain of parametric amplifier having signal power of 1.6W, length of 500, non-linear coefficient of 10.19 * 10 -3 W -1 km -1 .
a) 34.890
b) 19.15
c) 18.22
d) 16.11
Answer: c
Explanation: Quadratic gain is given by-
G p =10log 10 (γP pl L) 2
Where L=length of amplifier
P pl =signal power
γ=nonlinear coefficient.
This set of Optical Communications online quiz focuses on “Wavelength Conversion and Optical Regeneration”.
1. ___________________ is defined as a process by which the wavelength of the transmitted signal is changed without altering the data carried by the signal.
a) Wavelength conversion
b) Attenuation
c) Sigma management
d) Wavelength dispersion
Answer: a
Explanation: Wavelength conversion observes the changes in the length of the wave. It does not proportionate with the data carried by the signal or wave.
2. The device which is used to perform wavelength conversion is called as ___________
a) Attenuator
b) Wavelength Gyrator
c) Wavelength Circulator
d) Wavelength translator
Answer: d
Explanation: Wavelength translator changes the frequency of the wave and hence it is also called as frequency changer. It does not affect the data carried by the wave.
3. A wavelength converter is termed as _______ if the converted wavelength is longer than the original signal wavelength.
a) Down converter
b) Up converter
c) Attenuator
d) Shifter
Answer: b
Explanation: A wavelength converter is capable of receiving an incoming signal at any wavelength at the input port and produces output at the output port. A converter is termed as up converter when the output signal wavelength is longer than the original signal wavelength.
4. The ___________ converters cannot process different modulation formats.
a) Shifting
b) Optoelectronic wavelength
c) Opt-circular
d) Magnetic simulating
Answer: b
Explanation: In optoelectronic wavelength converters, the information contained in the intensity, frequency, phase of the signal is required to be reprocessed for the purpose of wavelength conversion. It does not process all the modulation formats.
5. The optical medium, in case of optical wavelength conversion is ___________
a) Depleted
b) Linear
c) Non-linear
d) Dispersive
Answer: c
Explanation: The implementation of optical wavelength conversion involves non-linearity of the optical medium. It can be either active or passive, each providing different nonlinear effects.
6. The process of imposing the nonlinear response of the medium onto the control signal is known as ______________ scheme.
a) Demodulation
b) Absorption
c) Cross-modulation
d) Repeater mixing
Answer: c
Explanation: The cross-modulation scheme involves changes produced due to the intensity variation of the intensity-modulated input signal. It takes place in the active cavity.
7. How many approaches are adopted by the cross-modulation scheme?
a) Four
b) Three
c) Two
d) Five
Answer: a
Explanation: Based on the properties of the nonlinear medium, the cross-modulation scheme can be divided into four main approaches. These are cross-gain modulation, cross-phase modulation, cross-absorption modulation, differential polarization modulation.
8. __________ wavelength converters make use of a passive optical medium to exploit non-linear effects.
a) Bipolar
b) Optoelectronic
c) Magnetic
d) Coherent
Answer: d
Explanation: The nonlinear effects include four-wave mixing and difference frequency generation. Coherent wavelength converters use a passive medium to extend the changes of nonlinear effects.
9. A _____________ wavelength converter utilizes the nonlinear properties of a semiconductor optical amplifier to perform the conversion process.
a) Cross-gain modulation
b) Cross-phase modulation
c) Cross-absorption modulation
d) Differential polarization modulation
Answer: a
Explanation: Cross-gain modulation wavelength converter is also called as XGM wavelength converter. It uses semiconductor optical amplifier along with its nonlinear properties for the conversion process.
10. The intensity modulated data on one signal wavelength is called as _______
a) Dispersed data
b) Pump signal
c) Probe signal
d) Frequency signal
Answer: b
Explanation: Pump signal is intensity modulated data. It produces variations in the carrier density within the SOA which provides inverted gain modulation in the SOA medium.
11. The probe signal is inverse to that of the pump signal.
a) True
b) False
Answer: a
Explanation: The gain modulations of the pump signal are imprinted onto the probe signal. Thus, the probe signal acquires the inverse copy of the pump signal, thereby contributing to the wavelength conversion with the pump signal.
12. In the XGM converter, the transfer function maintains the rectangular shape.
a) True
b) False
Answer: b
Explanation: By default, the ideal transfer function should be rectangular in shape. But it does not apply the same for XGM converter as the amplitude gradually decreases.
13. The speed of operation of XGM wavelength conversion is determined by the _______________ of the SOA.
a) Depletion level
b) Hole concentration
c) Carrier dynamics
d) Electron concentration
Answer: c
Explanation: The carrier dynamics deals with the interaction time between the input and the probe signal. On increasing the interaction time, the speed of operation of XGM wavelength conversion is increased.
14. ____________ is defined as the deviation in the emission frequency with respect to time when a laser is driven by a time-varying current source.
a) Intensity probe
b) Dispersion
c) Attenuation
d) Frequency chirp
Answer: d
Explanation: Frequency chirp occurs during the process of XGM and XPM. It is often termed as instantaneous frequency variation.
15. When frequency chirp shifts the optical frequency towards the shorter wavelength, it is known as ________
a) Red shift
b) Green shift
c) Yellow shift
d) Blue shift
Answer: d
Explanation: When frequency chirp shifts the optical frequency towards the shorter wavelength, it is known as blue shift. Similarly, when frequency chirp shifts the optical frequency towards the longer wavelength, it is known as red shift.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Integrated Optics and Photonics Technologies”.
1. Integrated technology for optical devices are developed within optical fiber communication.
a) True
b) False
Answer: a
Explanation: Integration of optical devices enable fabrication of the whole system onto a single chip. Integration of such devices has become a confluence of several optical terms.
2. When both active and passive devices are integrated on a single chip, in multilayered form, then these devices are known as _____________
a) IP devices
b) IO devices
c) Wavelength converters
d) Optical parametric amplifiers
Answer: a
Explanation: IP technology enables fabrication of subsystems and systems. This is all realized on a single substrate. The integration on a single chip is done in IP technology.
3. _________ is a further enhancement of ________
a) IP, IO
b) IO, IP
c) IO, wavelength converters
d) IP, wavelength converters
Answer: a
Explanation: IP seems to be a miniaturization process and integration of optical systems on a single chip. IO devices are formed when both active and passive elements are interconnected. Thus, IP is a developed version of IO.
4. Thin transparent dielectric layers on planar substrates are used in _________ and ______ devices.
a) Wavelength converters and amplification devices
b) IP and IO
c) IP and wavelength converters
d) IO and amplification devices
Answer: b
Explanation: IP and IO provide an alternative to conversion of optical signal back to electrical signal. Thin transparent dielectric layers act as optical waveguides to produce small-scale and miniature circuits.
5. __________ did not make significant contribution to earlier optical fiber systems.
a) IO
b) IP
c) Wavelength amplifiers
d) Couplers
Answer: a
Explanation: IO is based on single mode optical waveguides. Thus it is incompatible with multimode fiber systems. Thus, IO has less importance than IP.
6. Side or edge-emitting or conducting optical devices cannot be integrated on same substrate.
a) True
b) False
Answer: b
Explanation: In serial integration of device, different elements of optical chip can be interconnected in a consecutive manner. Thus, integration of side or edge emitting optical devices can be done on a single substrate.
7. Hybrid ________ integration demands _________ IP circuits to be produced on a single substrate.
a) IP, single-layered
b) IO, multilayered
c) IP, multilayered
d) IO, multilayered
Answer: c
Explanation: To gain control of optical signals, elements can be directly attached to IP circuit. Both active and passive devices should be on the same substrate. To make devices compatible with 3d structures of other IP/IO devices, hybrid IP integration demands multilayered IP circuits.
8. Using SOI integration technique __________ components can be coupled to IP devices.
a) Passive
b) Layered
c) Demounted
d) Active
Answer: d
Explanation: SOI is used to produce micro-waveguide bends and couplers thereby maintaining compatibility with silicon fabrication techniques. Thus, active components like optical sources, detectors can be coupled to other IP devices using SOI technique.
9. Who invented the IO technology?
a) Albert Einstein
b) Anderson
c) M.S Clarke
d) Robert
Answer: b
Explanation: The birth of IO can be traced back to the basic ideas outlined by Anderson in 1966. He suggested the micro-fabrication technology which in turn led to the term integrated optics in 1969.
10. Electronic circuits have a practical limitation on speed of operation at a frequency of around _________
a) 10 10 Hz
b) 10 12 Hz
c) 10 14 Hz
d) 10 11 Hz
Answer: a
Explanation: The speed of operation of electronic devices or circuits results from their use of metallic conductors to transport electronic charges and build up signals. It has a limitation to speed of operation of frequency around 10 10 Hz.
11. The use of light as an electromagnetic wave of high frequency provides high speed operation around ____________ times the conceivable employing electronic circuits.
a) 10 8 Hz
b) 10 5 Hz
c) 10 6 Hz
d) 10 4 Hz
Answer: d
Explanation: The use of light with its property as an electromagnetic wave offers the possibility of high speed operation. For this, the frequency should be high as 10 14 to 10 15 Hz.
12. How many layers are possessed by waveguide structures of silica-on-silicon?
a) Two
b) Three
c) Four
d) One
Answer: b
Explanation: The SOS is a part of IP technology. The waveguide structures provided by it comprises of three layers. They are buffer, the core and the cladding.
13. The ________________ is a versatile solution-based technique for making ceramic and glass materials.
a) SOL gel process
b) SSL gel process
c) SDL gel process
d) SAML gel process
Answer: a
Explanation: The SOL gel process involves the transition of system from a liquid to a gel. The SOL gel process along with SOS technique is used for the fabrication of ceramic fibers, film coatings and waveguide based optical amplifiers.
This set of Basic Optical Communications Questions and Answers focuses on “Planar Waveguides and Integrated Optical Devices”.
1. Optical fibre communications uses _______ dielectric waveguide structures for confining light.
a) Rectangular
b) Circular
c) Triangular
d) Planar
Answer: b
Explanation: The use of circular dielectric waveguide structures is universally acceptable. This has been a boon for optical fibre communications.
2. __________ waveguide is formed when the film is sandwiched by layers of different refractive index.
a) Planar
b) Circular
c) Asymmetric
d) Symmetric
Answer: c
Explanation: When the film is sandwiched between layers of same refractive index, symmetric waveguide is formed. Owing to the different refractive index, asymmetry is observed and hence asymmetric waveguide is formed.
3. When the dimensions of the guide are reduced, the number of ___________ also decreases.
a) Propagating nodes
b) Electrons
c) Holes
d) Volume of photons
Answer: a
Explanation: The dimensions of the guide are directly proportional to the number of propagating nodes. As the dimensions are reduced, the number of propagating nodes also decreases.
4. What does h ff stands for in the equation h ff = h+x+x2?
a) Frequency of layer
b) Diameter of curve
c) Effective guide layer thickness
d) Space propagation
Answer: c
Explanation: In the above equation, h is the height, x and x2 are the evanescent field penetration depths. h ff Denotes the effective guided layer thickness.
5. ___________ waveguides are plagued by high losses.
a) Circular
b) Planar
c) Depleted
d) Metal-clad
Answer: d
Explanation: All suitable waveguide materials are subject to limitations in the confinement. However, metal-clad waveguides are not so limited. Hence, they are plagued by high losses.
6. The planar waveguides may be fabricated from glasses and other isotropic materials such as ___________ and ______________
a) Octane and polymers
b) Carbon monoxide and diode
c) Fluorides and carbonates
d) Sulphur dioxide and polymers
Answer: d
Explanation: These materials are isotropic. However, their properties do not affect the fabrication of planar waveguides. Their properties cannot be controlled by external energy sources.
7. Which of the following devices are less widely used in the field of optical fibre communications?
a) Acousto-optic devices
b) Regenerators
c) Reflectors
d) Optical translators
Answer: a
Explanation: Acousto-optic devices are less widely used, mainly in the area of field deflection. Regenerators, reflectors form a base for the optical fibre communications.
8. Which of the following materials have refractive index near two?
a) GA As
b) Zinc
c) InP
d) AlSb
Answer: b
Explanation: Two basic groups are distinguished on the basis of the respective refractive indices near two and near three. GaAs, InP, AlSb have refractive indices near 3.
9. Passive devices are fabricated by __________ technique.
a) Fassbinder
b) High density integration
c) Radio-frequency sputtering
d) Lithium implantation
Answer: c
Explanation: Passive devices’ fabrication comes mainly from microelectronics industry. Radio frequency sputtering is used to deposit thin films of glass onto glass substrates.
10. Strip pattern in waveguide structures is obtained through ____________
a) Lithography
b) Cryptography
c) Depletion of holes
d) Implantation
Answer: a
Explanation: Field strength is an important aspect when it comes to strip patterns in waveguide structures. The electron and laser beam lithography is used to obtain stripe pattern in waveguide structures.
11. Propagation losses in slab and strip waveguides are smaller than the single mode fibre losses.
a) True
b) False
Answer: b
Explanation: The losses are in the range of 0.1 to 0.3 dB/cm. In case of slab and stripe waveguides, the losses are much higher whereas in case of single-mode fibres, they are much less.
12. A passive Y-junction beam splitter is fabricated from __________
a) GaAs
b) ZnS
c) AlbS
d) LiNbO 3
Answer: d
Explanation: A passive Y-junction splitter is used to combine signals from separate sources or to divide a signal into two or more channels. It is fabricated from the waveguide materials such as LiNbO 3 .
13. A passive Y-junction beam splitter is also used as a switch.
a) True
b) False
Answer: a
Explanation: A passive junction beam splitter finds application where equal power division of the incident beam is required. It can be used as a switch if it is fabricated from an electro-optic material.
14. The linear variation of refractive index with the electric field is known as the ________
a) Linear implantation
b) Ionization
c) Koppel effect
d) Pockels effect
Answer: d
Explanation: The change in refractive index is related by the applied field via the linear and quadratic electro-optic coefficients. The variation of R.I with the electric field is known as Pockels effect.
15. Planar waveguides are used to produce _______ coupler.
a) MMI
b) CMI
c) Frequency
d) Differential
Answer: a
Explanation: MMI couplers are abbreviated as Multimode interference couplers. These are similar to fused fibre couplers. These are easily produced by using planar waveguides.
This set of Optical Communications Question Bank focuses on “Optoelectronic Integration and Photonic Integrated Circuits”.
1. Monolithic integration for optical sources are confined to the use of __________ semiconductors.
a) Ⅲ-Ⅴ
b) Ⅱ-Ⅲ
c) Ⅰ-Ⅱ
d) Ⅶ-Ⅷ
Answer: a
Explanation: Ⅲ-Ⅴsemiconductor compounds are much useful. They possess both optical and electronic properties. These properties can be exploited to produce high performance devices.
2. Circuits fabricated from GaAs or AlGaAs operate in wavelength region of __________
a) 0.1 and 0.2 μm
b) 0.8 and 0.9 μm
c) 0.4 and 0.6 μm
d) 0.6 and 0.7 μm
Answer: b
Explanation: Circuits fabricated from GaAs use injection laser which is fabricated on GaAs with a MESFET. This is used to bias and modulate the laser.
3. The OEICs realization __________ as compared to the other developments in IO.
a) Scripted
b) Decreased
c) Lagged behind
d) Increased
Answer: c
Explanation: IO devices use dielectric materials such as lithium niobate. This lagging behind is caused by inherent difficulties in fabrication of OEICs even if Ⅲ-Ⅴ semiconductors are used.
4. Compositional and structural differences between photonic and electronic devices __________
a) Provide high efficiency
b) Provide low efficiency
c) Highly used
d) Create problems
Answer: d
Explanation: Compositional and structural differences cause epitaxial crystal growth, planarization for lithography, electrical interconnections. They also cause thermal and chemical stability of materials, electric matching and heat dissipation.
5. To avoid large chip __________ devices are used.
a) InGaAsP
b) InGa
c) GaAs
d) InGaAs
Answer: a
Explanation: To avoid large chip, InGaAsP devices are used with directly modulated semiconductor lasers. This gives good dynamic characteristics at 40 Gbit/s at 1.55 μmwavelength.
6. Devices operating at transmission rates greater than 40 Gb/s are _________
a) GaAs and InP
b) GaAs
c) InGa
d) InGaAs
Answer: a
Explanation: Optoelectronic integrated circuits are based on heterojunction bipolar transistor and electron mobility transistor use GaAs and InP. These are capable of operating at transmission rates higher than 40 Gb/s.
7. HEMT based __________ have a spot-size convertor with a photodiode.
a) p-n junction diode
b) p-i-n photoreceiver
c) IGBT
d) BJT
Answer: b
Explanation: P-I-N photoreceiver comprises of spot-size convertor with a photodiode. Spot-size convertor increases fiber alignment tolerances by one order of magnitude. This enables use of cleaved instead of lensed fiber.
8. P-I-N photoreceiver based on HEMT is integrated with _________ guiding layers.
a) GaAs and InP
b) GaAs
c) InGa
d) InGaAsP
Answer: d
Explanation: P-I-N photoreceiver is integrated with InGaAsP guiding layers. In this HEMT based technology, InGaAsP provides more confinement.
9. An optical power splitter integrated with optical waveguide amplifier is more useful.
a) True
b) False
Answer: a
Explanation: The aim of optical waveguide amplifier is to reduce the number of amplifiers in system. Alongwith, it also reaches maximum number of nodes.
10. The use of intelligent optical switches is necessary.
a) False
b) True
Answer: b
Explanation: Most applications of OEICs in optical networks require large switching capacity to support a large number of WDM channels. This also provides control of both optical signal wavelength and signal power.
11. The wafer scale replication technology uses ____________
a) SOL gel
b) GaAs
c) InGa
d) InGaAsP
Answer: a
Explanation: Replication technology employs hot embossing, molding and ultraviolet lithography. Ultraviolet curable SOL gel enables refractive and diffractive micro-optical elements to be replicated directly on glass substrates.
12. ___________ is useful for production of both planar micro-optical elements and stacked optical microsystems.
a) Wavelength amplifier
b) Wavelength convertor
c) Replication technology
d) Optical switching matrix
Answer: c
Explanation: SOL gel materials used in replication technology allows combination of replication with lithography. This leaves selected areas material-free for sawing and bunding.
13. Optical interconnection between optoelectronic device is achieved in _________
a) Wavelength amplifier
b) Wavelength convertor
c) Replication technology
d) Chip-to-chip interconnection
Answer: d
Explanation: The chip-to-chip interconnection of optical components have a vertical cavity surface-emitting laser. These are assembled in micro-trenches in which embedded electrodes are connected through passive junction of poliver waveguide on alignment pits.
14. Multilevel interconnections are incorporated in _______
a) PIC
b) AWG based coupler
c) Convertors
d) OEIC technologies
Answer: a
Explanation: PIC reduces the overall size of optical functions. This causes the interconnection of several modules growing on same substrate.
15. When there is M number of WDM channels present at N input ports, then the output port 1 produces a _________
a) CW signal
b) WDM signal
c) Amplified signal
d) Distorted signal
Answer: b
Explanation: The reconstituted spectrum of WDM signal at any output port consists of a different set of wavelength channels with at least one wavelength channel from each input port producing a WDM signal having wavelength signal from each of input ports.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Optical Bistability, Digital Optics and Optical Computation”.
1. ___________ provides a series of optical processing functions.
a) Wavelength convertors
b) Wavelength amplifiers
c) Detectors
d) Bi-stable optical devices
Answer: d
Explanation: Optical bi-stable devices include optical logic and memory elements, A-D convertors. Their response to light is nonlinear giving the basis of optical communication.
2. ___________ comprise of Fabry-Perot cavity.
a) Wavelength convertors
b) Wavelength amplifiers
c) Bi-stable optical devices
d) Detectors
Answer: c
Explanation: Fabry-Perot cavity consists of a material in which there are variations in refractive index with optical intensity. These variations are nonlinear giving rise to bistability.
3. The optical path length in nonlinear medium is integer number of ______ wavelength.
a) Half
b) Double
c) Three-fourth
d) Single
Answer: a
Explanation: Fabry-Perot cavity exhibits a sharp resonance to optical power passing into and through it. This is achieved when optical path length is integer number of half wavelength in nonlinear medium.
4. As compared to laser, the value of _________ in the cavity controls the optical transmission.
a) Amplification
b) Refractive index
c) Rectification
d) Reflection
Answer: b
Explanation: The refractive index value in the Fabry-Perot cavity controls the optical transmission. This provides high optical output on resonance and low optical output off resonance.
5. ___________ are able to latch between two distinct optical states.
a) Wavelength converters
b) Wavelength amplifiers
c) Detectors
d) Bistable optical devices
Answer: d
Explanation: The transfer characteristic for Bistable optical devices exhibit two state hysteresis resulting from turning in and out of resonance. So they can be latched between two states responding to external signal acting as flip-flop.
6. __________ can act as AND, OR, NOT gate.
a) Wavelength converters
b) Wavelength amplifiers
c) Detectors
d) Bistable optical devices
Answer: d
Explanation: BOD’s exhibit 2-state hysteresis. Thus they are able to latch between two operating states thereby providing logic functions.
7. _______ proves superior to _______
a) BOD’s, electronic devices
b) Electronic devices, BOD’s
c) BOD’s, convertors
d) Convertors, BOD’s
Answer: a
Explanation: There is also a thing of picosecond switching using only Pico-joules of energy. A BOD comprises of these switching properties. Thus, it proves superior to electronic devices.
8. ________ BOD’s provides optical feedback.
a) Extrinsic
b) Intrinsic
c) Detector
d) Bistable
Answer: b
Explanation: All optical or intrinsic devices which utilize a nonlinear optical medium between a pair of partially reflecting mirrors forming a nonlinear etalon in which feedback is provided optical.
9. ___________ devices employ artificial nonlinearity.
a) Extrinsic
b) Intrinsic
c) Hybrid
d) Bistable
Answer: c
Explanation: Hybrid devices have artificial nonlinearity in an electro-optic medium in the cavity. This produces variations in refractive index through electro-optic effect.
10. Hybrid devices have limited ________ speed.
a) Switching
b) Planar
c) Curvature
d) Electrical
Answer: a
Explanation: Hybrid BOD’s provides flexibility. But at the same time their switching speeds are limited by use of electrical feedback. These devices are interconnected to provide a more complex logic circuit.
11. _______ exhibit optical bistability.
a) Extrinsic lasers
b) Intrinsic lasers
c) Detectors
d) Semiconductor lasers
Answer: d
Explanation: Semiconductor lasers have optical bistability. This is due to nonlinearities in absorption, gain, dispersion, wave guiding and the selection of output polarization.
12. ___________ is fabricated with tandem electrode.
a) Full convertor
b) Semiconductor
c) Detector diode
d) Bistable laser diode
Answer: d
Explanation: Bistable laser diode is fabricated with tandem electrode. The tandem electrode provides two gain sections. Also it has a loss region between them.
13. Optical pulsing can be obtained using _________
a) BODs
b) WDM
c) Detector
d) Semiconductor
Answer: a
Explanation: BODs with a very narrow bi-stable loop can provide optical pulsing. This type of device can be used to shape, clean up and amplify a noisy input pulse.
14. A weak second beam is introduced in _________
a) BOD differential amplifier
b) WDM
c) Detector
d) Semiconductor laser
Answer: a
Explanation: A weak second beam in BOD differential amplifier is introduced into the nonlinear optical cavity. This is used to control the resonance and transmission of the main beam through effects of its own stored energy.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “The Optical Transmitter Circuit”.
1. _____________ must be operated in stimulated emission region.
a) Injection laser
b) LED’s
c) Detector
d) Receiver
Answer: a
Explanation: Injection laser is a threshold device. In stimulated emission region, continuous optical output power levels are in the range of 1 to 10mW.
2. Coherent radiation is relatively __________
a) Parabolic
b) Elliptic
c) Directional
d) Rectangular
Answer: c
Explanation: Most of the light output is coupled into optical fibre. This is because of the isotropic distribution of narrow-line width, coherent radiation is directional.
3. _____________ are capable of launching powers between 0.5 and several mW.
a) LED’s
b) Injection laser
c) Attenuator
d) Reflector
Answer: b
Explanation: Coupling efficiency up to 30% may be obtained by placing a fiber close to laser mirror. These can approach 90% with suitable lens and optical coupling arrangements. So they can launch 0.5 to several mW of optical power into fiber.
4. LED’s display good linearity.
a) True
b) False
Answer: a
Explanation: LED’s appear to be suited to analog transmission. This is because of its output which is directly proportional to the drive current.
5. Which behaviour may prove as a limitation for injection lasers and LED’s?
a) Isotropic
b) Radioactive
c) Thermal
d) Photosensitive
Answer: c
Explanation: The thermal behaviour of the injection lasers and the LED’s limits their operation within the optical transmitter. The main problem is caused by the variation of injection laser threshold current.
6. Optical output power from an LED is directly proportional to the device junction temperature.
a) False
b) True
Answer: b
Explanation: Output power is dependent on the junction temperature in case of LED’s. Most LED’s exhibit a decrease in the optical output power following an increase in junction temperature.
7. _____________ from the LED is dependent on the effective minority carrier lifetime in the semiconductor material.
a) Spontaneous emission
b) Stimulated emission
c) Absorption
d) Diffusion
Answer: a
Explanation: The speed of the response of the LED is dictated by the respective emission mechanism. Spontaneous emission is related to the carrier lifetime and hence dictating the speed of response.
8. The _________ of the LED is twice that of the effective minority carrier lifetime.
a) Dwell time
b) Reflection scatters
c) Sensitivity
d) Rise time
Answer: d
Explanation: The response of the optical fiber source is specified in terms of the rise time. This rise time is reciprocally related to the device frequency response.
9. The finite spectral width of the optical source causes ___________
a) Depletion
b) Frequency burst
c) Pulse broadening
d) Efficient reflection
Answer: c
Explanation: The finite spectral width causes pulse broadening due to material dispersion on an optical fiber communication link. This results in a limitation on the bandwidth-length product.
10. The coherent emission from an injection laser has a line width of ________
a) 2 nm
b) 3nm
c) 8 nm
d) 1nm
Answer: d
Explanation: An optical source such as an injection laser is a narrow line width device as compared to the LED. It has a narrow line width of 1 nm or less.
11. Extinction ratio is denoted by symbol __________
a) ε
b) σ
c) β
d) ρ
Answer: a
Explanation: Extinction ratio is defined as the ratio of the optical energy emitted in the 0 bit period to that emitted during the 1 bit period. It is denoted by ε.
12. The use of low impedance driving circuit may increase _____________
a) Noise
b) Width
c) Intensity
d) Switching speed
Answer: d
Explanation: Pulse shaping is usually required to increase the switching speed. However, increased switching speed may be obtained from an LED without a speed-up element by use of a low-impedance driving circuit.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “The Optical Receiver Circuit”.
1. ____________ limits receiver sensitivity.
a) Noise
b) Depletion layer
c) Avalanche
d) Current
Answer: a
Explanation: Receiver noise affects receiver sensitivity. It can dictate the overall system design. The noise can be temperature, environmental factor or due to components.
2. A ____________ performs the linear conversion of the received optical signal into an electric current.
a) Receiver
b) Converter
c) Detector
d) Reflector
Answer: c
Explanation: An optical signal is always fed to a detector. A detector is an optoelectronic converter which linearly converts the received optical signal into an electric current.
3. __________ are provided to reduce distortion and to provide a suitable signal shape for the filter.
a) Detector
b) Equalizer
c) Filters
d) Amplifier
Answer: b
Explanation: Optical detectors are linear devices. They do not introduce distortion themselves but other components may exhibit nonlinear behaviour. To compensate for distortion, an equalizer is provided in the receiver circuit.
4. A _________ maximizes the received signal-to-noise ratio in the receiver circuitry.
a) Filter
b) Equalizer
c) Detector
d) Reflector
Answer: a
Explanation: A filter reduces the noise bandwidth as well as inbounds noise levels. A filter maximizes the received signal-to-noise ratio while preserving the essential features of the signal. It also reduces ISI.
5. ________ can be operated in three connections.
a) Reflectors
b) Diodes
c) LED’s
d) FET’s
Answer: d
Explanation: FET’s or bipolar transistors are operated in three useful connections. These are the common emitter, the common base or gate, and the emitter or source follower.
6. How many structures of pre-amplifiers exist?
a) Two
b) Three
c) Four
d) One
Answer: b
Explanation: The basic structures of pre-amplifiers are observed in three forms. These are low-impedance, high-impedance and trans-impedance front end preamplifier structures.
7. What is the main factor contributing to the choice of the operational amplifier?
a) Gain
b) Impedance
c) Conductance
d) Gain-Bandwidth product
Answer: d
Explanation: A TTL interface stage is always used with the operational amplifier. A device that requires higher accuracy often tends to depend on gain-bandwidth product.
The choice of amplifier for receiver accuracy is dependant on gain-bandwidth product.
8. The multiplication factor for the APD varies with the device temperature.
a) True
b) False
Answer: a
Explanation: Optimum multiplication factor is required for smooth voltage variance. The multiplication factor for APD varies with the device temperature thus making provision of fine control for bias voltage.
9. How many categories of dynamic gain equalizers are available?
a) One
b) Two
c) Three
d) Four
Answer: b
Explanation: Dynamic gain equalizers are categorized into two types. These are single-channel and multichannel equalizers, thus providing operation using single or multiple wavelengths.
10. How many simultaneous channels can be provided in a band DGE?
a) Six
b) Two
c) Eight
d) Ten
Answer: c
Explanation: Generally, eight channels are provided simultaneously in a band DGE. These are for the attenuation purpose of channels along with gain equalization.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “System Design Considerations”.
1. __________ is the unique property of the glass fiber.
a) Transmission
b) Opaque property
c) Ductile
d) Malleable
Answer: a
Explanation: Glass fibres have a unique property as a transmission medium which enables their use in the communication. The major transmission characteristics are dispersion and attenuation.
2. __________ limits the maximum distance between the optical fiber transmitter and receiver.
a) Attenuation
b) Transmission
c) Equipment
d) Fiber length
Answer: a
Explanation: Attenuation along with dispersion and the conductor size are some of the factors that limit the maximum distance between the optical transmitter and the receiver. The associated constraints within the equipment also affect the distance.
3. The ___________ incorporates a line receiver in order to convert the optical signal into the electrical regime.
a) Attenuator
b) Transmitter
c) Repeater
d) Designator
Answer: c
Explanation: Repeaters are a mediator between transmitter and receiver. The weak signal is strengthened back by the repeaters on its path to the receiver.
4. A regenerative repeater is called as ____________
a) Repetitive repeater
b) Regenerator
c) Attenuator
d) Gyrator
Answer: b
Explanation: When digital transmission techniques are used, the repeater also regenerates the original digital signal in the electrical signal before it is retransmitted as an optical signal via a line transmitter.
5. The wavelength range of __________ will be fruitful for the operating wavelength of the system referring to the system performance.
a) 0.8 – 0.9 μm
b) 1.1 – 2 μm
c) 5.2 – 5.7 μm
d) 3.1 – 3.2 μm
Answer: a
Explanation: It is useful if the operating wavelength of the system is established to range of 0.8-0.9 μm. This will be dictated by the overall requirements for the system performance, cost, etc.
6. How many encoding schemes are used in optical fiber communication system design requirements?
a) Three
b) One
c) Two
d) Four
Answer: c
Explanation: Encoding schemes are used for digital transmission of data. These are bi-phase and delay modulation codes. They are also called as Manchester and Miller codes respectively.
7. In ________ the optical channel bandwidth is divided into non-overlapping frequency bands.
a) Time division multiplexing
b) Frequency division multiplexing
c) Code division multiplexing
d) De-multiplexing
Answer: b
Explanation: In FDM, the non-overlapping frequency bands are divided to the individual frequencies. These individual signals can be extracted from the combined FDM signal by electrical filtering at the receiver terminal.
8. A multiplexing technique which does not involve the application of several message signals onto a single fiber is called as _________
a) Time division multiplexing
b) Frequency division multiplexing
c) Code division multiplexing
d) Space division multiplexing
Answer: d
Explanation: In SDM, each signal channel is carried on a separate fiber within a fiber bundle or multi-fiber cable form. The cross coupling between channels is negligible.
9. Which of the following is not an optical fiber component?
a) Fiber
b) Connector
c) Circulator
d) Detector
Answer: c
Explanation: Circulator is a device used in electromagnetic theory. All others are optical components.
10. ________technique involves an increase in the number of components required.
a) Time division multiplexing
b) Space division multiplexing
c) Code division multiplexing
d) Frequency division multiplexing
Answer: b
Explanation: SDM involves good optical isolation due to the negligible cross coupling between channels. It uses separate fiber and thus requires more number of components.
11. Time division multiplexing is inverse to that of frequency division multiplexing.
a) True
b) False
Answer: a
Explanation: TDM involves distribution of channels in time slots whereas FDM involves bands that are run on different frequencies. Both of these techniques improve accuracy and reduce complexity.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Digital System Planning Considerations”.
1. Sampling rate for each speech channel on 32-channel PCM is 8 KHz each encoded into 8 bits. Determine number of bits in a frame.
a) 64
b) 128
c) 32
d) 256
Answer: d
Explanation: Number of bits in a frame can be calculated as follows:
Bits in a frame = No. of channels * Sampling rate for each channel.
2. Sampling rate for each speech channel on 32-channel PCM is 8 KHz each encoded into 8 bits. Determine the transmission rate for system with 256 bits in a frame.
a) 2.96 Mbits/s
b) 2.048 Mbits/s
c) 3.92 Mbits/s
d) 4 Mbits/s
Answer: b
Explanation: Transmission rate can be determined by-
Transmission rate = Sampling rate * No. of bits in a frame.
3. Sampling rate for each speech channel on 32-channel PCM is 8 KHz each encoded into 8 bits. Determine the bit duration with transmission rate of 2.048 M bits/s.
a) 388 ns
b) 490 ns
c) 488 ns
d) 540 ns
Answer: c
Explanation: Bit duration is the reciprocal of the transmission rate. Thus, it is given by-
Bit duration = 1/transmission rate.
4. The bit duration is 488 ns. Sampling rate for each channel on 32-channel PCM is 8 KHz encoded into 8 bits. Determine the time slot duration.
a) 3.2 μs
b) 3.1 μs
c) 7 μs
d) 3.9 μs
Answer: d
Explanation: Time slot duration is given by –
Time slot duration = Encoded bits * bit duration.
5. Sampling rate for each speech channel on 32-channel PCM is 8 KHz each encoded into 8 bits. Determine duration of frame with time slot duration of 3.9μs.
a) 125 μs
b) 130 μs
c) 132 μs
d) 133 μs
Answer: a
Explanation: Duration of a frame is determined by –
Duration of a frame = 32 * time slot duration.
6. Sampling rate for each speech channel on 32-channel PCM is 8 KHz each encoded into 8 bits. Determine the duration of multi-frame if duration of a frame is 125μs.
a) 2ms
b) 3ms
c) 4ms
d) 10ms
Answer: a
Explanation: Multi-frame duration can be determined by –
Multi-frame duration = 16 * Duration of a single frame.
7. Determine excess avalanche noise factor F if APD has multiplication factor of 100, carrier ionization rate of 0.02.
a) 3.99
b) 3.95
c) 4.3
d) 4
Answer: b
Explanation: Excess avalanche noise factor is computed by –
F = k*M + , where k is ionization rate and M is the multiplication factor.
8. Compute average number of photons incident at receiver in APD if quantum efficiency is 80%, F = 4, SNR = 144.
a) 866
b) 865
c) 864
d) 867
Answer: c
Explanation: Average number of photons arez m =[2β ς F]*[S/N*η]
Here, η = quantum efficiency, S/N = signal to noise ratio.
9. Determine incident optical power if z m =864, wavelength = 1μm.
a) -85 dBm
b) -80 dBm
c) -69.7 dBm
d) -60.7 dBm
Answer: d
Explanation: Incident optical power is P0=z m h c B T /2λ. Here z m =average number of photons, h c =Planck’s constant.
10. Determine wavelength of incident optical power if z m =864, incident optical power is -60.7 dB, B T =1 * 10 7 .
a) 1 μs
b) 2 μs
c) 3 μs
d) 4 μs
Answer: a
Explanation: Wavelength is determined by λ=z m h c B T /2P 0 . Here z m =average number of photons, h c =Planck’s constant, P 0 =incident optical power.
11. Determine total channel loss if connector loss at source and detector is 3.5 and 2.5 dB and attenuation of 5 dB/km.
a) 34 dB
b) 35 dB
c) 36 dB
d) 38 dB
Answer: a
Explanation: The total channel loss is C L =(α fc +α j )L + α cr . Here α cr =loss at detector and source combined, α fc = attenuation in dB/km.
12. Determine length of the fiber if attenuation is 5dB/km, splice loss is 2 dB/km, connector loss at source and detector is 3.5 and 2.5.
a) 5 km
b) 4 km
c) 3 km
d) 8 km
Answer: b
Explanation: Length of the fiber is L = C L /(α fc +α j ) – α cr . Here α cr = loss at detector and source combined, α fc = attenuation in dB/km.
13. Determine total RMS pulse broadening over 8 km if RMS pulse broadening is 0.6ns/km.
a) 3.6 ns
b) 4 ns
c) 4.8 ns
d) 3 ns
Answer: c
Explanation: Total RMS pulse broadening is given by –
σ T = σ*L Where σ = rms pulse broadening and L = length of the fiber.
14. Determine RMS pulse broadening over 8 km if total RMS pulse broadening is 5.8ns/km.
a) 0.2ns/km
b) 0.1ns/km
c) 0.4ns/km
d) 0.72ns/km
Answer: d
Explanation: RMS pulse broadening is given by –
σ = σ T /L where σ = rms pulse broadening and L = length of the fiber.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Analog Systems”.
1. Determine dispersion equalization penalty if total RMS pulse broadening is 4.8ns, B T is 25 Mbits/s.
a) 0.03 dB
b) 0.08 dB
c) 7 dB
d) 0.01 dB
Answer: a
Explanation: Dispersion equalization penalty is denoted by D L . It is given by-
D L = 2 (2σ T B T √2) 4 . Here σ T =RMS pulse broadening.
2. Determine RMS pulse broadening with mode coupling if pulse broadening is 0.6 over 8km.
a) 1.6ns
b) 1.7ns
c) 1.5ns
d) 1.4ns
Answer: b
Explanation: Total RMS pulse broadening with mode coupling is given by-
σ T = σ√L. Here σ T = RMS pulse broadening, L = length of the fiber.
3. Determine dispersion equalization penalty with mode coupling of 1.7ns if B T is 25 Mbits/s.
a) 4.8 * 10 4 dB
b) 4 * 10 4 dB
c) 4.2 * 10 4 dB
d) 3.8 * 10 4 dB
Answer: c
Explanation: Dispersion equalization penalty is denoted by D L . With mode coupling, it is given by-
D L =2 (2σ T B T √2) 4 . Here σ T =RMS pulse broadening.
4. Determine dispersion equalization penalty without mode coupling if B T is 150 Mbits/s and total rms pulse broadening is 4.8ns.
a) 34 dB
b) 33 dB
c) 76.12 dB
d) 34.38 dB
Answer: d
Explanation: Dispersion equalization penalty is denoted by D L . It is given by-
D L = 2 (2σ T B T √2) 4 . Here σ T = RMS pulse broadening, = without mode coupling.
5. Determine ratio of SNR of coaxial system to SNR of fiber system if peak output voltage is 5V, quantum efficiency of 70%, optical power is 1mW, wavelength of 0.85μm.
a) 1.04 * 10 4
b) 2.04 * 10 4
c) 3.04 * 10 4
d) 4.04 * 10 4
Answer: a
Explanation: Ratio of SNR of coaxial system to SNR of fiber system is given by-
Ratio = V 2 hc/2KTZ 0 ηP i λ. Here, η=quantum efficiency, P i = 0ptical power in mW, V=optical output voltage.
6. Determine the peak output voltage if efficiency is 70%, wavelength is 0.85μm and output power is 1mW.
a) 7V
b) 8V
c) 5V
d) 6V
Answer: b
Explanation: Peak output voltage is given by-
V 2 = (2KTZ 0 ηP i λ * Ratio)/hc. Here, η = quantum efficiency, P i =0ptical power in mW, V=optical output voltage.
7. Determine the efficiency of a coaxial cable system at 17 degree Celsius with peak output voltage 5V, 0.85 μm wavelength and SNR ratio of 1.04 * 10 4 .
a) 80%
b) 70%
c) 40%
d) 60%
Answer: b
Explanation: The efficiency of a coaxial cable system is η=V 2 hc/2KTZ 0 ηP i λ * Ratio. Hereη=quantum efficiency, P i = 0ptical power in mW, V=optical output voltage.
8. Determine the wavelength of a coaxial cable system operating at temperature 17 degree Celsius at output voltage of 5V, 100Ω impedance, optical power of 1mW, 70% quantum efficiency.
a) 0.39μm
b) 0.60μm
c) 0.85μm
d) 0.98μm
Answer: c
Explanation: The wavelength can be determined by –
λ = V 2 hc/2KTZ 0 ηP i * Ratio. Hereη=quantum efficiency, P i = 0ptical power in mW, V = optical output voltage.
9. Determine the impedance of a coaxial cable system operating at temperature 17 degree Celsius at output voltage of 5V, 0.85μmwavelength, optical power of 1mW, 70% quantum efficiency and SNR ratio of 1.04 * 10 4 .
a) 80Ω
b) 50Ω
c) 90Ω
d) 100Ω
Answer: d
Explanation: The impedance is given by-Z 0 =V 2 hc/2KTP i * Ratio. Hereη=quantum efficiency, P i = Optical power in mW, V=optical output voltage.
10. The 10-90% rise times for components used in D-IM analog optical link is given. . Link is of 5km. Determine the total rise time.
a) 62ns
b) 53ns
c) 50ns
d) 52ns
Answer: d
Explanation: Total rise time is given by-
T syst =1.1[T s 2 +T n 2 +T c 2 +T D 2 ] 1/2 . Here T s = rise time, T n = intermodal time, T c = Chromatic time.
11. The 10-90% rise times for components used in D-IM analog optical link is given. . Link is of 5km. It has an optical bandwidth of 6MHz. Determine maximum permitted system rise time.
a) 58.3ns
b) 54ns
c) 75ns
d) 43.54ns
Answer: a
Explanation: The maximum permitted system rise time is given by-
T syst = 0.35/B opt . Here, B opt =Optical Bandwidth.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Multiplexing Strategies”.
1. What is the full form of ETDM?
a) Electronic tube di-cyclic mechanism
b) Electrical time division multiplexing
c) Emphasis tier division mechanism
d) Electrical tube dielectric medium
Answer: b
Explanation: ETDM is the major baseband digital strategy. It allows for greater exploitation of available fiber bandwidth.
2. The practical limitations of the speed of electronic circuits have been pushed towards operational frequencies around ___________
a) 100 MHz
b) 120 MHz
c) 100GHz
d) 80 Hz
Answer: c
Explanation: The speed of the circuitry in the fiber optic communication plays an important role in its performance. It is pushed around 100 GHz frequency allowing for 100 Gbit/s feasibility.
3. A strategy used for increasing the bitrate of digital optical fiber systems beyond the bandwidth capabilities of the drive electronics is known as ___________
a) Optical time division multiplexing
b) Electrical time division multiplexing
c) Frequency division multiplexing
d) Code division multiplexing
Answer: a
Explanation: OTDM is favourable for long distance transmission of signal. It is designed to push the bitrate of the fiber systems beyond the bandwidth limits to gain performance.
4. ____________ semiconductor laser sources provide low duty cycle pulse streams for subsequent time multiplexing.
a) Diameter preferred
b) Mode locked
c) Divine
d) Depletion
Answer: b
Explanation: Mode locked semiconductor laser sources were used at the transmitter side. They provide effective distribution of time multiplexing providing low duty cycle pulse streams.
5. ______________ are the devices which are employed to eliminate the laser chirp.
a) Optical intensity modulators
b) Demodulators
c) Circulators
d) Optical Isolators
Answer: a
Explanation: Optical intensity modulators eliminate the laser chirp. This laser chirp may result in dispersion of the transmitted pulses as they propagate within the single mode fiber, thus limiting the achievable transmission distance.
6. _____________ provides operation at high transmission rate.
a) Optical intensity modulators
b) Demodulators
c) Circulators
d) Electro-absorption modulators
Answer: d
Explanation: Electro-absorption modulators are employed at the transmitter and receiver sections. They provide operation at high transmission rate and for field trial.
7. In __________ the microwave frequency are modulated with an optical carrier and transmitted using a single wavelength channel.
a) Subcarrier multiplexing
b) TDM
c) FDM
d) Code division multiplexing
Answer: a
Explanation: Optical Subcarrier multiplexing is transmitted using a single wavelength channel. It enables multiple broadband signals to be transmitted over single-mode fiber.
8. Which of the following techniques is easy to implement?
a) Amplitude shift keying
b) Phase shift keying
c) Frequency shift keying
d) SCM
Answer: c
Explanation: Frequency shift keying has an advantage of being simple to implement at the modulator as well as demodulator side. It is formed by up converting to a narrowband channel at high frequency employing frequency.
9. Which of the following is the disadvantage of SCM?
a) Source nonlinearity
b) Linearity
c) Distortion
d) Narrow bandwidth
Answer: a
Explanation: The problem associated with SCM is source nonlinearity. The distortion caused by this becomes noticeable when several subcarriers are transmitted from a single optical source.
10. In CATV, the signal must be received with a carrier to noise ratio of between __________
a) 90 and 100 dB
b) 10 and 30 dB
c) 60 and 70 dB
d) 45 and 55 dB
Answer: d
Explanation: The CATV multichannel spectrum tends to minimize the required bandwidth. The carrier to noise ratio must be between to avoid degradation of picture quality.
11. The IF signal can be input to a demodulator to recover the baseband signal.
a) True
b) False
Answer: a
Explanation: The IF signal is obtained through SCM at the receive terminals. The baseband video signal in a CATV is obtained through IF signal by using it with a demodulator input.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Application of Optical Amplifiers “.
1. Which of the following is not a drawback of regenerative repeater?
a) Cost
b) Bandwidth
c) Complexity
d) Long haul applications
Answer: d
Explanation: The regenerative repeaters are useful in long haul applications. However, such devices increase the cost and complexity of the optical communication system. It act as a bottleneck by restricting the system operational bandwidth.
2. The term flexibility, in terms of optical amplifiers means the ability of the transmitted signal to remain in the optical domain in a long haul link.
a) True
b) False
Answer: a
Explanation: Repeaters are usually used to maintain the transmitted signal in the optical domain. But, it has its own drawbacks. Thus, flexible systems which include optical amplifiers are used for such purpose.
3. How many configurations are available for employment of optical amplifiers?
a) Three
b) Four
c) Two
d) Five
Answer: a
Explanation: Optical amplifiers can be employed in three configurations. These are simplex mode, duplex mode, multi-amplifier configuration.
4. Repeaters are bidirectional.
a) True
b) False
Answer: b
Explanation: Repeaters are unidirectional. Optical amplifiers have the ability to operate simultaneously in both directions at the same carrier wavelength.
5. It is necessary to ____________ the optical carriers at different speeds to avoid signal interference.
a) Inculcate
b) Reduce
c) Intensity-modulate
d) Demodulate
Answer: c
Explanation: Optical amplifiers are bidirectional. They operate in both directions at the same carrier wavelength. In order to avoid interference, the optical carriers should be intensity modulated.
6. The _________________ increases the system reliability in the event of an individual amplifier failure.
a) Simplex configuration
b) Duplex configuration
c) Serial configuration
d) Parallel multi-amplifier configuration
Answer: d
Explanation: The optical amplifiers with spectral bandwidths in the range 50 to 100 nm allow amplifiers to be more reliable than repeaters. The parallel multi-amplifier configuration increases system reliability and relaxes the linearity.
7. Which of the following is not an application of optical amplifier?
a) Power amplifier
b) In-line repeater amplifier
c) Demodulator
d) Preamplifier
Answer: c
Explanation: Optical amplifiers have a wide variety of applications in the transmitter as well as receiver side. It is used as the power amplifier in the transmitter side and as preamplifier at the receiver side.
8. _________ reconstitutes a transmitted digital optical signal.
a) Repeaters
b) Optical amplifiers
c) Modulators
d) Circulators
Answer: a
Explanation: Optical amplifiers simply act as gain blocks on an optical fiber link. However, in contrast, the regenerative repeaters reconstitute a transmitted digital optical signal.
9. _____________ are transparent to any type of signal modulation.
a) Repeaters
b) Optical amplifiers
c) Modulators
d) Circulators
Answer: b
Explanation: The main benefit of acting as a gain block for optical amplifier is that it can be transparent to modulation bandwidth. However, both the noise and signal distortions are continuously amplified.
10. _________________ imposes serious limitations on the system performance.
a) Fiber attenuation
b) Fiber modulation
c) Fiber demodulation
d) Fiber dispersion
Answer: d
Explanation: The fiber dispersion calculation does not take into account the non-regenerative nature of the amplifier repeaters. In this, the pulse spreading and the noise is accumulated.
11. __________ is the ratio of input signal to noise ratio to the output signal to noise ratio of the device.
a) Fiber dispersion
b) Noise figure
c) Transmission rate
d) Population inversion
Answer: b
Explanation: Noise figure judges the performance factor of the devices. It is the in and out the ratio of signal to noise degradation for any device.
12. How many factors govern the noise figure of the device?
a) Four
b) Three
c) Two
d) One
Answer: a
Explanation: Noise figure is governed by factors such as the population inversion, the number of transverse modes in the amplifier cavity, the number of incident photons on the amplifier and the optical bandwidth of the amplified spontaneous emissions.
13. What is the typical range of the noise figure?
a) 1 – 2 dB
b) 3 – 5 dB
c) 7 – 11 dB
d) 12 – 14 dB
Answer: c
Explanation: Typical noise figures range from 7 to 11 dB. The SOAs are generally at the bottom end of the range and the fiber amplifiers towards the top end.
This set of Optical Communications Problems focuses on “Dispersion Management and Soliton Systems”.
1. Calculate second order dispersion coefficient for path length L 2 20km and L 1 160km. Dispersion coefficient for L 2 is 17.
a) -2.125ps/nm km
b) -3.25ps/nm km
c) -3.69ps/nm km
d) -1.28ps/nm km
Answer: a
Explanation: The second order dispersion coefficient for path length is given by-
β 21 = -β 22 L 2 /L 1 . Here, β 22 = Dispersion coefficient forL 2 , L 2 and L 1 are path lengths.
2. Calculate the path length L 2 if L 1 is 160, dispersion coefficient of L 2 is 17, dispersion coefficient of L 1 is -2.25 ps/nmkm.
a) 40 km
b) 20 km
c) 30 km
d) 10 km
Answer: b
Explanation: The path length L 2 is given by-
L 2 = β 21 L 1 /-β 22 . Here, β 22 = Dispersion coefficient forL 2 , β 21 = Dispersion coefficient for L 1 , L 2 and L 1 are path lengths.
3. Calculate path length L 1 if L 2 is 20, dispersion coefficient of L 2 is 17, dispersion coefficient of L 1 is -2.25 ps/nmkm.
a) 180 km
b) 30 km
c) 160 km
d) 44 km
Answer: c
Explanation: The path length L1is given by-
L 1 = β 21 L 2 /-β 22 . Here, β 22 = Dispersion coefficient forL 2 , β 21 = Dispersion coefficient for L 1 , L 2 and L 1 are path lengths.
4. Calculate second order dispersion coefficient for path length L 1 20 km and L 1 160 km. Dispersion coefficient for L 1 is -2.125*10 -12 s/nmkm.
a) 20
b) 19
c) 18
d) 17
Answer: d
Explanation: The second order dispersion coefficient for path length is given by-
β 22 =-β 21 L 2 /L 1 . Here, β21 = Dispersion coefficient forL1, L 2 and L1 are path lengths.
5. Calculate dispersion slope for second path fiber if L 1 is 150, L 2 is 10 and s 1 is 0.075.
a) 1.125
b) 2.125
c) 3.125
d) 1.9
Answer: a
Explanation: Dispersion slope for second path fiber is s 2 = -s 1 (L 1 /L 2 ). Here s1 and s2 are dispersion slopes for L 1 , L 2 . L 2 and L 1 are path lengths.
6. Calculate dispersion slope for first path fiber if L 1 is 160, L 2 is 20 and s 2 is 0.6ps/nm km.
a) 0.1
b) 0.432
c) 0.236
d) 0.075
Answer: d
Explanation: Dispersion slope for first path fiber is s 1 = -s 2 (L 1 /L 2 ). Here s1 and s2 are dispersion slopes for L 1 , L 2 . L 2 And L 1 are path lengths.
7. Calculate L 2 if dispersion slope for first path fiber is 0.075 and L 1 is 160 km and s 2 is -0.6ps/nm km.
a) 20 km
b) 30 km
c) 40 km
d) 50 km
Answer: a
Explanation: L 2 is determined by –
L2 = (-s 1 /s 2 )*L 1 . Here s 1 and s 2 are dispersion slopes for L 1 , L 2 . L 2 and L 1 are path lengths.
8. Calculate L 1 if dispersion slope for first path fiber is 0.075 and L 2 is 20 km and s 2 is -0.6ps/nm km.
a) 170 km
b) 160 km
c) 180 km
d) 175 km
Answer: b
Explanation: L 1 is determined by –
L 2 = (-s 1 /s 2 )* L 2 . Here s 1 and s 2 are dispersion slopes for L 1 , L 2 . L 2 and L 1 are path lengths.
9. Calculate separation of soliton pulses over a bit period length if R 2 pulse width is 6 ps for bit period of 70 ps.
a) 5.9
b) 5.7
c) 5.8
d) 5.4
Answer: c
Explanation: The separation of soliton pulses over a bit period length is calculated by –
q 0 = T 0 /2ς. Here ς = pulse width and T 0 = bit period.
10. Calculate RZ pulse width if bit period is 60ps and separation of soliton pulses is 5.4.
a) 5.5ps
b) 8.1ps
c) 4.3ps
d) 2.3ps
Answer: a
Explanation: RZ pulse width can be calculated by –
ς = T 0 /q 0 . Here ς = pulse width and T 0 = bit period.
11. Calculate bit period if RZ pulse width is 50ps and separation of soliton pulses is 5.6.
a) 570ps
b) 540ps
c) 430ps
d) 560ps
Answer: d
Explanation: Bit period can be calculated by –
T 0 = 2T 2 q 0 . Here T 2 =pulse width and T 0 =bit period.
12. Calculate value of dimensionless parameter if bit period is 45ps and RZ pulse width is 4 ps.
a) 5.625
b) 5.0
c) 4
d) 6.543
Answer: a
Explanation: Dimensionless parameter is given by –
q 0 = T 0 /2ς. Here ς=pulse width and T 0 =bit period.
This set of Optical Communications Questions and Answers for Entrance exams focuses on “Practical Constraints of Coherent Transmission”.
1. Which technology development has helped the field of optical fiber communication?
a) Glass technology
b) Component technology
c) Multiplexing
d) Power
Answer: b
Explanation: Substantial developments in component technology have allowed the initial difficulties in the optical fiber communication to go away. The coherent factor experienced most of the difficulties.
2. __________ dictates the performance characteristics required from components and devices which are to be utilized in coherent optical fiber systems.
a) System considerations
b) Bluetooth technology
c) Multiplexing
d) Practical constraints
Answer: d
Explanation: Practical constraints inhibit the development of coherent optical fiber communications. These constraints are derived from factors associated with the elements of the coherent optical fiber communication.
3. Coherent optical transmission is degraded by the ________ associated with the transmitter and local oscillator lasers.
a) Phase noise
b) White noise
c) Dissipation
d) Power
Answer: a
Explanation: Phase noise is determined by the laser line width. The phase noise associated with both the transmitter and the mid-tier section severely degrades the coherent optical transmission as well as reception.
4. ___________ improves the spectral purity of the device output and noise current.
a) Power dissipation
b) Laser line width reduction
c) Laser line width injection
d) Phase noise
Answer: b
Explanation: Laser line width determines the level of phase noise and long term phase stability. The reduced phase noise is obtained using narrow-line width devices. This improves the spectral purity as well as reduces the noise current.
5. ____________ is the principal cause of line width broadening in the coherent devices.
a) Electromagnetic field
b) Power dissipation
c) Injection laser phase noise
d) Gaussian noise
Answer: c
Explanation: Injection laser phase noise affects the system performance. The system performance considerations include receiver noise, power loss and line width broadening.
6. Which technique was started for narrowing of injection laser line widths?
a) External resonator cavity
b) Long-hauled oscillator
c) Circulator
d) Gyrator
Answer: a
Explanation: Many approaches evolved in time for laser line width problem. The one which sustained and showed effects was the use of external resonator cavity in the lasers.
7. The line width tolerance is wider for heterodyne receivers.
a) False
b) True
Answer: b
Explanation: The laser line width requirements depend on the modulation format, coherent detection mechanism which includes the use of heterodyne and homodyne receivers. The line width tolerance is wider for heterodyne receivers when employing FSK modulation.
8. ___________ is an alternative to reduce phase noise and line width requirements.
a) Homodyne detection
b) Heterodyne detection
c) FSK modulation
d) Phase diversity reception
Answer: d
Explanation: The more sensitive coherent transmission techniques are most affected by phase noise problem. A specially configured reception technique called as phase diversity reception technique is used to overcome phase noise problem.
9. ______________ is the progressive spatial separation between the two polarization modes as they propagate along the fiber.
a) Fiber birefringence
b) Fiber dispersion
c) Fiber separation
d) Fiber coupling
Answer: a
Explanation: In a perfectly formed fiber, both modes would travel together. But, in practice, the fiber contains random manufacturing irregularities. This result in a progressive spatial separation called as fiber birefringence.
10. How many compensator devices are required to provide full polarization-state control?
a) Three
b) One
c) Four
d) Two
Answer: d
Explanation: At least two compensator devices are required to provide full polarization-state control. They can be placed in either the incoming signal path or the local oscillator output path.
11. Which technique was found to be providing an infinite range of polarization control?
a) Homodyne detection
b) Fiber squeezers
c) Heterodyne detection
d) Power dissipation
Answer: b
Explanation: Four fiber squeezers provide an infinite range of adjustment or endless polarization control. Stress was applied to the fiber in the local oscillator path using the squeezers which are angled at 45 degrees to each other.
12. What is the main drawback of the squeezer?
a) Damages the fiber
b) Attenuation
c) Dispersion
d) Signal degradation
Answer: a
Explanation: The squeezers are simple to configure. The main drawback of squeezer is that they tend to damage the fiber and could not be engineered into reliable transducers for practical systems.
13. The use of balanced receiver compensates the losses due to coupling optics.
a) True
b) False
Answer: a
Explanation: The losses due to coupling optics and the suppression of the excess noise in the local oscillator signal are eliminated by the use of balanced receiver. It is also called as balanced-mixer receiver.
14. ___________ is the phenomenon which occurs in the single carrier systems due to small refractive index changes induced by the optical power fluctuations.
a) SBS gain
b) Self-phase modulation
c) FSK modulation
d) Birefringence
Answer: b
Explanation: It occurs only in the single-carrier systems. It affects the phase of the transmitted signal.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Modulation Formats”.
1. _____________ is essentially a crude form of Amplitude shift keying.
a) Analog modulation
b) Digital intensity modulation
c) Photodetector
d) Receiver structure
Answer: b
Explanation: Many techniques have been developed to amplitude modulate an optical signal. Digital intensity modulation used in direct detection systems is essentially a crude form of ASK in which the received signal is detected using square law detector.
2. Almost _________ of the transmitter power is wasted in the use of external modulators.
a) Half
b) Quarter
c) One-third
d) Twice
Answer: a
Explanation: All external modulators suffer the drawback that around half of the transmitted power is wasted. To avoid this, non-synchronous detection can be employed.
3. The line width in the range ________ of bit rate is specified for ASK heterodyne detection.
a) 8%
b) 2 to 8%
c) 10 t0 50%
d) 70%
Answer: c
Explanation: The ASK modulation scheme can be used with laser sources exhibiting the line widths comparable with the bit transmission rate. For ASK heterodyne detection, line width range of 10 to 50% is usually specified.
4. ______________ is also referred to as on-off keying .
a) FSK
b) DSK
c) PSK
d) ASK
Answer: d
Explanation: Amplitude shift keying involves the locking and assembling of the amplitude of the wave. It involves the carrier wave along with the amplitude wave or transmitted wave and hence referred to as on-off keying.
5. ________ does not require an external modulator.
a) FSK
b) DSK
c) PSK
d) ASK
Answer: a
Explanation: FSK involves the frequency deviation property of the directly modulated semiconductor laser used in wideband systems. Unlike ASK, it does not require an external modulator, which in turn, avoids the wastage of transmitted power.
6. The frequency deviation at frequencies above 1 MHz is typically ____________
a) 10 to 20 mA -1
b) 100 to 500 mA -1
c) 1000 to 2000 mA -1
d) 30 to 40 mA -1
Answer: b
Explanation: The carrier modulation effect occurs at the frequencies above 1 MHz. At the phase of carrier modulation, the frequency deviation is about 100 to 500 mA -1 .
7. ___________ offers the potential for improving the coherent optical receiver sensitivity by increasing the choice of signalling frequencies.
a) MFSK
b) MDSK
c) MPSK
d) MASK
Answer: a
Explanation: Multilevel FSK includes 4-level or 8-level FSK. It improves the receiver sensitivity by reducing the deviation and increasing the usage of signalling frequencies.
8. Eight level FSK and binary PSK yields an equivalent sensitivity.
a) False
b) True
Answer: b
Explanation: Binary PSK and 8-level FSK provides an equivalent sensitivity. The main drawback of 8-level FSK is that it yields an equivalent sensitivity to binary PSK at the expense of a greater receiver bandwidth requirement.
9. External modulation for ________ modulation format allows the most sensitive coherent detection mechanism.
a) FSK
b) DSK
c) PSK
d) ASK
Answer: c
Explanation: External modulation for PSK is usually straightforward. It is therefore utilized to provide the modulation format which allows the most sensitive coherent detection mechanism.
10. _________ can potentially provide spectral conservation through the use of multilevel signalling.
a) M-ary PSK
b) MFSK
c) ASK
d) DFSK
Answer: a
Explanation: In M-ary schemes, the spectral efficiency is increased by the factor log 2 M.this is purely for M-level schemes which can provide multilevel signalling patterns.
11. The digital transmission on implementation of polarization modulation which involves polarization characteristics of the transmitted optical signal is known as _____________
a) Frequency shift keying
b) Amplitude shift keying
c) Phase shift keying
d) Polarization shift keying
Answer: d
Explanation: Polarization shift keying is abbreviated as PolSK. PolSK requires additional receiver complexity than other modulation formats.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Demodulation Schemes”.
1. _________ heterodyne detection does not require phase matching between the incoming signal and the local oscillator.
a) Synchronous
b) Asynchronous
c) Noisy
d) Spatial
Answer: b
Explanation: For heterodyne detection, a beat note signal produces the IF signal which is obtained using the square law optical detector. Hence, it does not require phase matching.
2. In ___________ detection, the phase of the local oscillator signal is locked to the incoming signal.
a) Homodyne
b) Heterodyne
c) Spatial
d) Noisy
Answer: a
Explanation: Phase diversity and multiport detection is considered to be a form of heterodyne detection. In case of homodyne detection, the incoming signal is bundled with the phase of the local oscillator signal.
3. The introduction of nonlinear element within the ______ is necessary to enable efficient carrier recovery.
a) PSK
b) ASK
c) FSK
d) PLL
Answer: d
Explanation: Carrier recovery is done by slightly reducing the depth of the phase modulation. This is done by the introduction of the nonlinear within the phase locked loop.
4. What is the main attraction of the optical homodyne detection?
a) Receiver sensitivity
b) Transmission power
c) Modulation scheme
d) Line width
Answer: a
Explanation: The main attraction of optical homodyne detection is the potential 3dB improvement in the receiver sensitivity. It also eases the receiver bandwidth requirement considerably.
5. The phase difference between the local oscillator and the source must be ________ for high sensitivity reception.
a) Deep
b) High
c) Direct
d) Near zero
Answer: d
Explanation: The phase difference must be held near zero for high sensitivity reception. As it moves towards 90 degrees, the output signal current becomes zero and the detection process will cease.
6. How many strategies of homodyne demodulation have proved to be successful for coherent optical fiber reception?
a) One
b) Three
c) Four
d) Two
Answer: d
Explanation: Two homodyne demodulation strategies have been successful for optical fiber reception. They are the use of either a pilot carrier or a decision-driven optical phase locked loop.
7. In the __________ receiver, the incoming signal is not precisely shifted to the baseband.
a) Intra-dyne
b) Heterodyne
c) Homodyne
d) Optical
Answer: a
Explanation: In homodyne detection, the incoming signal is precisely shifted to the baseband. In intra-dyne receiver, the incoming signal is shifted to a frequency lower than the transmission rate.
8. The use of PLL can be avoided in the intra-dyne detection.
a) False
b) True
Answer: b
Explanation: In intra-dyne receiver, the incoming signal is lower than the transmission rate. In place of PLL, a baseband filter in the form of electrical filtering can be used.
9. Phase diversity reception is also referred to as ______________
a) Homodyne detection
b) Noisy detection
c) Homodyne detection
d) Multiport detection
Answer: d
Explanation: In phase diversity reception, phase is not locked. It also has bandwidth advantage over homodyne detection as it converts the incoming signal directly to baseband.
10. The received signal through polarization diversity reception is linearly polarized.
a) False
b) True
Answer: a
Explanation: The received optical signal through polarization diversity reception is elliptically polarized. It is also uncontrollable unlike coherent receivers.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Receiver Sensitivities”.
1. ASK with heterodyne detection can achieve same SNR limit as ASK with asynchronous detection.
a) True
b) False
Answer: a
Explanation: The coherent optical systems’ receiver sensitivities depend on the SNR limit and the 3dB improvement in the receiver bandwidth. The ASK with heterodyne and asynchronous detection allows for the same receiver SNR limit.
2. In a ______________ receiver, the analyses of signal and noise phenomena are more complicated than in the IM-DD case.
a) Homodyne
b) Heterodyne
c) Circular
d) Heterogeneous
Answer: b
Explanation: The optical output appearance shows the analysis of signal and noise phenomenon. In a heterodyne receiver, the analyses of signal and noise phenomena are more complicated than in the IM-DD case because the optical detector output appears as an IF signal and not as a baseband signal.
3. FSK modulation is attributed to the use of __________ frequencies unlike ASK modulation.
a) One
b) Three
c) Two
d) Four
Answer: c
Explanation: FSK heterodyne detection has receiver sensitivity of 3dB more than tha ASK. Hence, it is attributed to the use of two frequencies unlike ASK where only one frequency is used.
4. BER of FSK modulation scheme is ________ as/to the ASK modulation scheme.
a) Twice
b) Thrice
c) Unequal
d) Same
Answer: d
Explanation: FSK modulation uses two frequencies. The similar BER can be obtained with the two modulation schemes when the average power is transmitted.
5. ___________ can also be used in place of multilayered filters in the dual filter direct detection FSK receiver.
a) Bragg gratings
b) Ceramic gratings
c) Aluminum arsenide
d) Bragg diodes
Answer: a
Explanation: A dual optical filter uses two multilayered dielectric filters. Bragg gratings are an excellent alternative to the dual dielectric filters as they perform the same function as dual dielectric multilayer filters.
6. FSK synchronous detection is _________ more sensitive than asynchronous heterodyne detection.
a) 0.24 dB
b) 0.45 dB
c) 0.9 dB
d) 0.12 dB
Answer: b
Explanation: During FSK detection, the probability of error is increased with respect to direct detection as a result of amplified noise from the orthogonal polarization. Thus, the synchronous detection is more sensitive than the asynchronous heterodyne detection.
7. The asynchronous heterodyne detection is _________ more sensitive than the dual filter direct detection FSK receiver.
a) 0.9 dB
b) 0.23 dB
c) 0.43 dB
d) 0.40 dB
Answer: d
Explanation: The orthogonal polarization leads to the amplified noise structures and behavior in the FSK receiver. This leads to the loss in the sensitivity and accuracy of the FSK receiver.
8. The use of __________ was undertaken to separate the polarization in an analog to digital conversion.
a) FSK
b) DSP
c) ASK
d) DP-FSK
Answer: b
Explanation: DSP is abbreviated as digital signal processing. Digital signal processing involves the cutting, trimming and manipulating of signals in the process of analog to digital conversion. The signals are converted in the form of numerous pulses.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Multicarrier Systems”.
1. A major attribute of coherent optical transmission was its ability to provide _______________ for future multicarrier systems and networks.
a) Attenuation
b) Dispersion
c) Frequency selectivity
d) Noisy carriers
Answer: c
Explanation: A coherent optical transmission involves the wavelength, frequency and the distance as its main factors. It provides frequency and wavelength selectivity with narrow channel spacing’s for future multicarrier systems and networks.
2. The channel width is narrow for the coherent systems than the WDM systems.
a) True
b) False
Answer: a
Explanation: The coherent optical systems involve wavelength selectivity with narrow channel spacing. The conventional WDM systems use a far more relaxed channel spacing than the coherent systems.
3. The technique within the coherent multicarrier systems used to broadcast the optical signals over the network is the use of passive _____________
a) Star coupler
b) Optical resonator
c) Optical regenerator
d) Local oscillator
Answer: a
Explanation: In multicarrier systems, the channels are separated via various techniques. The use of passive star coupler ensures that the optical signals are broadcasted over the network within the coherent system.
4. Estimate the minimum transmitter power if number of photons per bit are 150, wavelength 1.3 micrometer with an optical bandwidth of 20 THz.
a) 0.2 mW
b) 0.5mW
c) 1 mW
d) 2.3mW
Answer: b
Explanation: The minimum transmitter power is given by –
P tx = N p hf B
Here, P tx = transmitter power, f = frequency, h = Planck’s constant and B = bandwidth.
5. The performance degradation due to nonlinear phase noise is referred to as _____________
a) Munich effect
b) Linear bipolar effect
c) Gordon Mollenauer effect
d) Delta effect
Answer: c
Explanation: The phase noise is a variant due to the RZ signal. The nonlinear phase noise causes severe performance degradation in terms of bandwidth and frequency.
6. ______________ facilitates the doubling of the feasible spectral efficiency through the transmission of independent information in each of the two orthogonal polarizations.
a) WDM
b) Gordon Mollenauer effect
c) EDFA control
d) POLMUX
Answer: d
Explanation: POLMUX is abbreviated as Polarization multiplexing. It provides a different approach to the multilevel modulation. It requires polarization control at the receiver side.
7. ____________ is a transparent multiplexing technique.
a) POLMUX
b) PDM
c) WDM
d) Munich
Answer: a
Explanation: It is a transparent technique as it is not dominated by polarization mode dispersion or polarization-dependant loss. It provides many advantages to the multilevel modulation format in comparison to the non-POLMUX signals at the same data rate.
8. Which of the following is not a drawback of POLMUX?
a) Polarization-sensitive detection at the receiver
b) Receiver complexity
c) Cross-polarization nonlinearities
d) Multilevel modulation
Answer: d
Explanation: The drawbacks of the POLMUX include receiver complexity, polarization sensitivity at the receiver side and the cross polarization nonlinearities. It is advantageous to the multilevel modulation scheme.
9. A multicarrier modulation format in which there has been growing interest to compensate for impairments in optical fiber transmission systems is _______________
a) OFDM
b) EDM
c) WDM
d) ADM
Answer: a
Explanation: OFDM is abbreviated as orthogonal frequency division multiplexing. It combats both fiber chromatic dispersion and polarization mode dispersion.
10. It is suggested that the technique with high white noise is an attractive option for use in long haul systems.
a) False
b) True
Answer: a
Explanation: Long haul systems require a technique which boosts the distance covered by the signals with less use of carrier signal. The technique which exhibits high spectral density is an attractive option for the long haul systems.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Fiber Attenuation Measurements”.
1. A technique used for determining the total fiber attenuation per unit length is ________ method.
a) Frank
b) Cut-off
c) cut-back
d) Erlangen
Answer: c
Explanation: Fiber attenuation techniques have been developed in order to determine the total fiber attenuation. This fiber attenuation is from both scattering and absorption losses. Cut-back method is used as per unit length medium.
2. The system designer finds greatest interest in the ______________
a) Overall fiber attenuation
b) Fiber dispersion
c) Latitude of the fiber
d) Durability
Answer: a
Explanation: Fiber attenuation results from various aspects such as signal degradation and physical factors such as the environment. The system designer has to look upon the overall fiber attenuation while the relative magnitude is important for the development team.
3. How many parameters are usually worked upon by the measurement techniques in attenuation?
a) Three
b) Two
c) One
d) Five
Answer: b
Explanation: The measurement techniques used to obtain fiber attenuation usually gives two parameters. One is spectral loss characteristic and the other is the spot measurement. The spot measurement is usually referred to as the loss at a single wavelength.
4. What type of light source is usually present in the cut-back method?
a) Tungsten or xenon
b) LED
c) Laser
d) Photo-sensor
Answer: a
Explanation: A cut-back method usually calculates the spectral loss which eventually is used as a parameter to obtain total fiber attenuation. It consists of a white light source which is a tungsten halogen or xenon lamp.
5. The device used to remove any scattered optical power from the core is __________
a) Mode setup terminator
b) Nodal spectrum
c) Mode stripper
d) Attenuator
Answer: c
Explanation: A mode stripper is usually incorporated at the output end of the fiber. It removes the optical power that is scattered from the core to cladding down the fiber length.
6. What is the hierarchy of the process at the receiving end of the cut-back technique?
a) Photodiode – photo-detector – lock-in amplifier
b) Photodiode – lock-in amplifier
c) Photodiode – photo-detector – Attenuator
d) Photo-detector – lock-in amplifier – receiver
Answer: a
Explanation: The optical power at the receiving end is detected using the p-i-n or avalanche photodiode. The photo-detector is then index matched to the fiber output end. It is then fed to the lock-in amplifier.
7. What is the unit of measurement of the optical attenuation per unit length?
a) dB-km
b) dB/km
c) km/dB
d) V
Answer: b
Explanation: The optical fiber attenuation per unit length is measured by unit dB/km. dB refers to the electrical parameter used to calculate the attenuation in the form of losses and spot measurements.
8. Determine the attenuation per kilometer for a fiber whose length is 2 km, output voltage is 2.1 V at a wavelength of 0.85μm. The output voltage increases to 10.7 V when the fiber is cut-back to leave 2 metres. Determine the attenuation per km for the fiber at wavelength 0.85μm.
a) 2.8dB/km
b) 3.1dB/km
c) 3.5dB/km
d) 8dB/km
Answer: c
Explanation: The attenuation per kilometer can be obtained by –
α dB = 10 log 10 (V 2 /V 1 )/(L 1 -L 2 ) dB/km where L 1 -L 2 = 1.998 and V 2 , V 1 are output voltages.
9. ___________ are used to allow measurements at a selection of different wavelengths.
a) Diaphragms
b) Spot attenuators
c) Belts
d) Interference filters
Answer: d
Explanation: The interference filters are located on a wheel at a length. These allow measurement at different wavelengths simultaneously and are accurate to a known level.
10. Cut-back technique is destructive.
a) True
b) False
Answer: a
Explanation: The cut-back method has some drawback. It is suitable for laboratory measurements but is far from ideal for attenuation measurements in the field. Hence it is termed as destructive.
11. Which technology is used by the backscatter measurement method?
a) Refraction
b) Francis flat recovery
c) Optical time domain reflectometry
d) Optical frequency
Answer: c
Explanation: Backscatter method for attenuation measurement is the most popular non-destructive method. It uses optical time domain reflectometry and provides best results in laboratory as well as field measurements.
12. ________________ measurements checks the impurity level in the manufacturing process.
a) Material reflectometry
b) Material absorption loss
c) Material attenuation loss
d) Calorimetric loss
Answer: b
Explanation: The material absorption loss measurements check the absorption losses. It checks the impurity level in the fiber at the manufacturing stage. Thus, it is efficient than the attenuation measurement methods.
13. _____________ may be achieved by replacing the optical fibers with thin resistance wires.
a) Diffraction
b) Segmentation
c) Calorimetric calibration
d) Electrical calibration
Answer: d
Explanation: Electrical measurement measurements are also efficient. The calibration is done by many methods. Electrical calibration involves the use of resistance wires in place of optical fibers.
14. A scattering cell consists of ______ square solar cells called as Tynes cell.
a) Five
b) Four
c) Six
d) Three
Answer: c
Explanation: Fiber scattering measurements use scattering cell to detect the light which is scattered. This cell consists of six square solar cells which are called as Tynes cell. It contains index-matching fluid.
15. ___________ removes the light propagating in the cladding.
a) Cladding mode strippers
b) Core strippers
c) Mode enhancers
d) Attenuators
Answer: a
Explanation: The inaccuracies in the measurements resulting from the scattered light are removed by cladding mode strippers. These strippers are placed before and after the scattering cell.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Fiber Dispersion Measurements”.
1. ___________ measurements give an indication of the distortion to the optical signals as they propagate down optical fibers.
a) Attenuation
b) Dispersion
c) Encapsulation
d) Frequency
Answer: b
Explanation: Dispersion measurements provide the exact parameters to truly determine the quality and degradation to the optical signals. It gives an indication of the distortion to the optical signals as they propagate down the optical fibers.
2. The measurement of dispersion allows the _________ of the fiber to be determined.
a) Capacity
b) Frequency
c) Bandwidth
d) Power
Answer: c
Explanation: Dispersion measurements give an indication of distortion, which in turn determines the information carrying capacity of the fiber. This information carrying capacity of the fiber is purely dependent on the bandwidth of the fiber.
3. How many types of mechanisms are present which produce dispersion in optical fibers?
a) Three
b) Two
c) One
d) Four
Answer: a
Explanation: There are three major mechanisms which produce dispersion in optical fibers. These are: Material dispersion, waveguide dispersion and intermodal dispersion.
4. Intermodal dispersion is nonexistent in ________ fibers.
a) Multimode
b) Single mode
c) Step index- multimode
d) Al-GU
Answer: b
Explanation: Intra-modal as the name suggests need multimode fibers to propagate. In single mode fibers, only one mode is there to propagate. Hence, Intermodal dispersion is nonexistent in single mode fibers.
5. In the single mode fibers, the dominant dispersion mechanism is ____________
a) Intermodal dispersion
b) Frequency distribution
c) Material dispersion
d) Intra-modal dispersion
Answer: d
Explanation: In single mode case, the dominant dispersion mechanism is chromatic. Chromatic dispersion is called as intra-modal dispersion.
6. Devices such as ___________ are used to simulate the steady-state mode distribution.
a) Gyrators
b) Circulators
c) Mode scramblers
d) Attenuators
Answer: c
Explanation: The dispersion measurements on the fiber are performed only when the equilibrium mode distribution is set up within the fiber. Hence, filters or scramblers are used to simulate the steady state mode distribution.
7. How many domains support the measurements of fiber dispersion?
a) One
b) Three
c) Four
d) Two
Answer: d
Explanation: Fiber dispersion measurements can be made in two domains. These are time domain and frequency domain.
8. The time domain dispersion measurement setup involves _____________ as the photo detector.
a) Avalanche photodiode
b) Oscilloscope
c) Circulator
d) Gyrator
Answer: a
Explanation: The time domain fiber dispersion measurement involves the pulses to be received by the photo detector in order to determine the distortion in the optical signals. These pulses are received by avalanche photodiode.
9. In pulse dispersion measurements, the 3dB pulse broadening for the fiber is 10.5 ns/km and the length of the fiber is 1.2 km. Calculate the optical bandwidth for the fiber.
a) 32 MHz km
b) 45 MHz km
c) 41.9 MHz km
d) 10 MHz km
Answer: c
Explanation: The optical bandwidth for the fiber is given by –
B opt = 0.44/3dB pulse broadening
Where, 0.44 = constant.
10. Frequency domain measurement is the preferred method for acquiring the bandwidth of multimode optical fibers.
a) True
b) False
Answer: a
Explanation: Bandwidth is usually the difference in the frequency. Frequency domain measurement is usually the best method in order to find the bandwidth of the multimode optical fibers.
11. Intra-modal dispersion tends to be dominant in multimode fibers.
a) True
b) False
Answer: b
Explanation: Intra-modal dispersion is dominant in case of single mode fibers. In case of multimode fibers, intermodal dispersion comes handy and is dominant.
This set of Optical Communications Questions and Answers for Campus interviews focuses on “Fiber Refractive Index Profile Measurements”.
1. The detailed knowledge of the refractive index profile predicts the __________ of the fiber.
a) Nodal response
b) Variation in frequency
c) Impulse response
d) Amplitude
Answer: c
Explanation: Refractive index profile plays an important role in characterizing the properties of optical fibers. These properties include numerical aperture, intermodal response and number of modes propagating in the fiber. This makes the impulse response of the fiber.
2. ______________ of the fiber is strongly dependent on the refractive index profile.
a) Amplitude
b) Tuning frequency
c) Diameter
d) Information carrying capacity
Answer: d
Explanation: Information carrying capacity is strongly dependent on the refractive index profile same as that of the impulse response of the fiber. These characteristics predict the dispersion in the fiber.
3. ______________ is required in case of graded index fibers.
a) High amplitude
b) High frequency
c) High impulse response
d) Optimum profile
Answer: d
Explanation: In case of graded index fibers, it is essential that the fiber manufacturer is able to produce particular profiles with higher accuracy. This profile is known as optimum profile as it minimizes the dispersion rate.
4. _______________ have been widely used to determine the refractive index profiles of optical fibers.
a) Interference microscopes
b) Gyro meters
c) Mode-diameter device
d) Tunable microscopes
Answer: a
Explanation: Interference microscopes are used to determine the refractive index profiles of the fibers. These microscopes use an index matching technique to determine the index profiles.
5. Which of the following is not an interference microscope?
a) Gerry Buzynski
b) Michelson
c) Mach-Zehnder
d) L.G. Cohen
Answer: a
Explanation: Mach-Zehnder and Michelson are the widely used interference microscopes. L.G.Cohen reused a modification of the Mach-Zehnder.
6. What stands for x in the given equation?
δn = qλ/x
a) Displacement
b) Thickness
c) Constant
d) Coefficient of refraction
Answer: b
Explanation: The given equation is an equation for the difference in the refractive index which is denoted by δn. q is the fringe shift, λ is the wavelength and x is the thickness of the fiber slab.
7. The ________ method gives an accurate measurement of the refractive index profile.
a) Slab
b) Biometric
c) GSLB
d) Tuning
Answer: a
Explanation: The slab method gives an accurate measurement of the refractive index profile. Also, there are some drawbacks to it. The computation of individual points is somewhat tedious unless an automated technique is used.
8. Which of the following is the main drawback of the slab technique?
a) Efficiency
b) Amplitude
c) Time
d) Accuracy
Answer: c
Explanation: The main drawback of the slab method is the time required to prepare the fiber slab. There are some interferometry methods which require no sample preparation.
9. ________________ method is used to measure the nonlinear refractive index of the silica fiber.
a) Grating
b) Non-linear
c) Silica-refraction
d) IGA
Answer: d
Explanation: IGA is abbreviated as induced-grating autocorrelation. This method requires no slab preparation and involves electro-optic effect where measuring the electric field autocorrelation function determines the refractive index of the optical fiber.
10. Near field scanning method provides a rapid method for acquiring the refractive index profile.
a) True
b) False
Answer: a
Explanation: The near field scanning method provides no time lag for slab preparation, thus it is not time bound. Hence, it is accurate and rapid mode of measuring refractive index profile.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Fiber Cutoff Wavelength Measurements”.
1. A multimode fiber has many cutoff wavelengths.
a) False
b) True
Answer: b
Explanation: A multimode fiber has many cutoff wavelengths. It is because the number of bound propagating modes is usually large.
2. What does ‘a’ stands for in the given equation?
Mg=(πa/λ)2(n12 - n22)
a) Radius of the core
b) Constant
c) Coefficient of refraction
d) Density
Answer: a
Explanation: The above equation gives the number of guided modes for a parabolic refractive index graded fiber, where a is the core radius and n1, n2 are the core and cladding indices respectively.
3. The _________ wavelength is defined as the wavelength greater than which the ratio of the total power and the fundamental mode power has to be decreased to less than 0.1dB.
a) Magnetic
b) Quasi
c) Cut-off
d) EIA
Answer: c
Explanation: The cut off wavelength is usually measured by increasing the signal wavelength in a fixed length of fiber until the mode is undetectable. It is usually called a effective cut-off wavelength.
4. How many methods are used to determine the effective cutoff wavelength?
a) Two
b) One
c) Four
d) Three
Answer: d
Explanation: Three methods are usually used for the determination of the effective cutoff wavelength. These are bending-reference technique, power step method and alternative test method.
5. What is the name of the test used to determine the efficient values of the effective cutoff wavelength?
a) Round robin test
b) Mandarin test
c) Hough Werner test
d) Fulton test
Answer: a
Explanation: Round robin test is an effective method to determine the efficient values for the cutoff wavelength. It shows that in some methods, the values through round robin test are the same.
6. The effective cutoff wavelength for a cabled single mode fiber will be generally smaller than that of the un-cabled fiber.
a) True
b) False
Answer: a
Explanation: The effective cutoff wavelength for a cabled single mode fiber is always smaller than that of the un-cabled fiber. This is usually because of the bend effects.
7. How many bend effects are produced in the fiber?
a) One
b) Three
c) Two
d) Four
Answer: c
Explanation: Usually, two bend effects are produced. They are macro-bending and micro-bending. These effects incarcerate certain changes in the fiber efficiency.
8. _______________ method does not require a leaky mode correction factor or equal mode excitation.
a) Bending-reference
b) Power step method
c) Alternative test method
d) Refracted near-field method
Answer: d
Explanation: Refracted near-field method is complementary to the transmitted near-field method. It has the advantage that it does not require a leaky mode correction factor. Moreover, it provides the relative refractive index directly without recourse to external calibration.
9. The _______ method is the most commonly used method for the determination of the fiber refractive index profile.
a) Refracted near-field method
b) Bending-reference
c) Power step method
d) Alternative test method
Answer: a
Explanation: It is the most commonly used technique. Also, it is the EIA reference test method for both multimode and single mode fibers.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Fiber Numerical Aperture and Fiber Diameter Measurements”.
1. The ____________ affects the light gathering capacity and the normalized frequency of the fiber.
a) Numerical aperture
b) Amplitude modulation
c) Responsivity
d) Quantum efficiency
Answer: a
Explanation: Numerical aperture is an important optical parameter as it dictates the important characteristics of the optical fiber. This in turn dictates the number of propagating modes within the fiber.
2. The numerical aperture for a step index fiber is sine angle of the ____________
a) Efficient angle
b) Aperture
c) Acceptance angle
d) Attenuation
Answer: c
Explanation: The numerical aperture of a step index fiber is given by –
NA = sinθ a , where θ a is the acceptance angle and NA is the numerical aperture.
3. The calculations of the numerical aperture from a refractive index data are less accurate for the graded index fibers than for step index fibers.
a) False
b) True
Answer: b
Explanation: The refractive indices of the core and cladding are fluctuating, thus causing the data to be less efficient. For graded index fibers, it is usually less accurate than for the step index fibers.
4. Far field pattern measurements with regard to multimode fibers are dependent on the _____________ of the fiber.
a) Amplitude
b) Frequency
c) Diameter
d) Length
Answer: d
Explanation: The accuracy of the measurement technique is dependent upon the visual assessment of the far-field pattern from the fiber. In case of multimode fibers, far field pattern measurements are dependent on the length of the fiber.
5. The screen is positioned 10 cm from the fiber end face. When illuminated from a wide angled visible source the measured output pattern size is 6.2 cm. Calculate the approximate numerical aperture of the fiber.
a) 0.21
b) 0.30
c) 0.9
d) 1.21
Answer: b
Explanation: The numerical aperture can be obtained from a trigonometric relationship given by-
NA = A/(A 2 +4D 2 ) 1/2 , where A is constant and D is the distance of the fiber end face from the screen in mm.
6. During the fiber drawing process, the fiber outer diameter is maintained constant to within ________
a) 2%
b) 1%
c) 5%
d) 10%
Answer: b
Explanation: During the fiber manufacturing stage, all processes needs to be performed efficiently. Especially, in the drawing process, the outer diameter should be compiled to within 1 % to avoid miscommunication through fibers.
7. What is the minimum value of accuracy in diameter is needed to avoid radiation losses in the fiber?
a) 0.1%
b) 0.2%
c) 0.3%
d) 0.03%
Answer: c
Explanation: Any diameter variations can cause excessive radiation losses and make accurate fiber-fiber connection difficult. Hence, on-line diameter measurement systems are used which provides accuracy of 0.3%.
8. Which of the following is a non-contacting optical method of on-line diameter measurement?
a) Brussels’s method
b) Velocity differentiator method
c) Photo detector method
d) Image projection method
Answer: d
Explanation: On-line diameter measurement technique uses fiber image projection method. It is also known as non-contacting optical method and shadow method.
9. The shadow method is used for measurement of the outer diameter of an optical fiber. The apparatus employs a rotating mirror with an angular velocity of 4 rad/s which is located at 10 cm from the photo detector. Compute the shadow velocity.
a) 0.1 μm μs -1
b) 0.4 μm μs -1
c) 0.87 μm μs -1
d) 1 μm μs -1
Answer: b
Explanation: The shadow velocity is obtained by the below equation:
ds/dt = l. dϕ/dt, where l is the distance of the apparatus from the photodetector and dϕ/dt is the angular velocity.
10. The shadow velocity is given by 0.4 μm μs -1 and shadow pulse of width 300 μs is registered at an instant by the photodetector. Determine the outer diameter of the optical fiber in μm.
a) 100 μm
b) 120 μm
c) 140 μm
d) 90 μm
Answer: b
Explanation: The fiber outer diameter is given by-
d 0 = W e .Ds/dt, where W e = pulse width and ds/dt = shadow velocity.
11. The techniques used to determine the refractive index profile can also be used to determine the core diameter.
a) True
b) False
Answer: a
Explanation: Some of the techniques used to determine the refractive index profile are interferometry, near field scanning and refracted ray technique. The core diameter for step index fibers is defined by the step change in the refractive index profile. Hence, they can be used to measure the core diameter.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Field Measurements”.
1. ____________ affects both the fiber attenuation and dispersion.
a) Refractive index
b) Micro-bending
c) Connectors
d) Splices
Answer: b
Explanation: Effects such as micro-bending with a resultant mode coupling affect both the fiber attenuation and dispersion. It does not provide the overall characteristics of the transmission link.
2. Which of the following is not included in the optical fiber link measurement test?
a) Attenuation measurement
b) Dispersion measurement
c) Splice loss measurement
d) Receiver sensitivity
Answer: d
Explanation: It is necessary to perform some tests on the optical fiber link to enhance productivity. Apart from receiver sensitivity, other measurement methods are required to test the fiber link.
3. In case of field measurements, the equipment must have ___________ power consumption keeping in mind the battery operation.
a) Low
b) High
c) Negligible
d) Maximum
Answer: a
Explanation: The design criteria allows you to distinguish in parameters required for adaptation of battery operation and equipment handling. The power consumption must be low for an equipment to handle.
4. Which of the following are not considered as environmental conditions required for field measurements?
a) Temperature
b) Humidity
c) Mechanical load
d) Power
Answer: d
Explanation: The equipment must be reliable and provide accurate measurements under extreme environmental conditions such as humidity, temperature and mechanical load. Power is an internal factor.
5. Complicated and involved fiber connection arrangements should be _________ in case of field measurements.
a) Provided
b) Avoided
c) Maximized
d) Minimized
Answer: b
Explanation: The equipment must be connected to the fiber in a simple manner. It should be connected without the need for fine or critical adjustment.
6. Which of the following cannot be used in equipment for field measurements?
a) Fiber
b) Connector
c) External triggering
d) Environmental factor
Answer: c
Explanation: The equipment cannot usually make use of external triggering and regulating circuits between the transmitter and receiver. This is because of their wide spacing on the majority of the optical links.
7. Which sensors are used for alteration of spectral range in equipment?
a) Wide-area photodiodes
b) Circulators
c) Gyrators
d) Photogenic sensors
Answer: a
Explanation: Wide area photodiodes such as silicon, germanium diodes are used for alteration of spectral range. It is generally preferred to have a measurement range from -100 dBm.
8. The handheld optical power meter has a measurement accuracy of ________
a) 0.01 dB
b) 0.25 dB
c) 0.8 dB
d) 1 dB
Answer: b
Explanation: The optical power meter detects the fiber type and switches to optical power measurements. It provides an accuracy of about + 25 dB.
9. _____________ may be used for measurement of the absolute optical attenuation on a fiber link.
a) Silicon photodiodes
b) InGaAsP photodiodes
c) Optical power meters
d) Gyrators
Answer: c
Explanation: Optical power meter employs cut-back technique. It is used for the measurement of the optical attenuation.
10. A large-area photodiode is utilized in the receiver to eliminate any effects from differing fiber and faces.
a) True
b) False
Answer: a
Explanation: The modulating voltage maintains the equilibrium between the transmitter and the receiver side. A large area photodiode is required to eliminate differing fiber and faces to maintain the equilibrium.
11. Optical time domain reflectometry is also called a backscatter measurement method.
a) False
b) True
Answer: b
Explanation: OTDR technique is used in both on-field and laboratory applications. It is also called a backscatter measurement method as it provides the measurement of attenuation on an optical link down its entire length.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Optical Network Concepts”.
1. Each stage of information transfer is required to follow the fundamentals of ____________
a) Optical interconnection
b) Optical hibernation
c) Optical networking
d) Optical regeneration
Answer: c
Explanation: Optical networking uses optical fiber as a transmission medium. It provides a connection between users to enable them to communicate with each other by transporting information from a source to a destination.
2. ____________ is a multi-functional element of optical network.
a) Hop
b) Optical node
c) Wavelength
d) Optical attenuation
Answer: b
Explanation: An optical node is a multi-functional element which acts as a transceiver unit capable of receiving, transmitting and processing the optical signal. The optical nodes are interconnected with optical fiber links.
3. A signal carried on a dedicated wavelength from source to destination node is known as a ___________
a) Light path
b) Light wave
c) Light node
d) Light source
Answer: a
Explanation: A light path is a dedicated path from a source to a destination. The data can be sent over the light paths as soon as connections are set up. A controlling mechanism is present to control the data flow.
4. The fundamentals of optical networking are divided into _______ areas.
a) Two
b) One
c) Four
d) Three
Answer: d
Explanation: The fundamentals divided into three areas contain mainly optical network terminology. The other two areas include functions and types of optical network node and switching elements and the wavelength division multiplexed optical networks.
5. The optical networking fundamentals are _____________ of the transmission techniques.
a) Dependent
b) Independent
c) Similar
d) Dissimilar
Answer: b
Explanation: The optical networking fundamentals include transfer of data. Irrespective of the difference in the transmission techniques, the fiber networking fundamentals remain the same.
6. The network structure formed due to the interconnectivity patterns is known as a ____________
a) Network
b) Struck
c) Topology
d) D-pattern
Answer: c
Explanation: A topology is a combination of patterns interconnected to each other. It provides connection patterns to users at different places. It embarks on the principle of multi-usability.
7. In the __________ topology, the data generally circulates bi-directionally.
a) Mesh
b) Bus
c) Star
d) Ring
Answer: b
Explanation: In a bus topology, data is input via four port couplers. The couplers couples and stations the data bi-directionally and are removed from the same ports.
8. The ring and star topologies are combined in a ________ configuration.
a) Mesh
b) Fringe
c) Data
d) Singular
Answer: a
Explanation: The mesh configuration is a combination of ring and star topologies. It is referred to as full-mesh when each network node is interconnected with all nodes in the network.
9. The full-mesh configuration is complex.
a) False
b) True
Answer: b
Explanation: The full-mesh topology is a combination of two or more topologies. It is often preferred for the provision of either a logical or virtual topology due to its high flexibility and interconnectivity features.
10. How many networking modes are available to establish a transmission path?
a) Three
b) One
c) Two
d) Four
Answer: c
Explanation: There are two networking modes often referred to the networking. These are connection-oriented and connectionless networking modes. These include an end-to-end and bidirectional communication environment between source and destination.
11. Packet switching is also called as ___________
a) Frame switching
b) Cell switching
c) Trans-switching
d) Buffer switching
Answer: b
Explanation: In packet or cell switching, messages are sent in small packets called cells. Cells from different sources are statistically multiplexed and are sent to the destinations.
12. ___________ mode is temporary, selective and continuous.
a) Cell switching
b) Buffer switching
c) Cache
d) Circuit switching
Answer: d
Explanation: An end-to-end connection is required for a circuit switching to take place. The transmissions are continuous and are in real time. Once the transmission is complete, the connection is ended.
13. A _______________ is a series of logical connections between the source and destination nodes.
a) Cell circuit
b) Attenuation circuit
c) Virtual circuit
d) Switched network
Answer: c
Explanation: A virtual circuit consists of different routes which provide connections between sending and receiving devices. These routes can change at any time and the incoming return route does not have to mirror the outgoing route.
14. ____________ refers to the process whereby a node finds one or more paths to possible destinations in a network.
a) Routing
b) Framing
c) Lightning
d) Cloning
Answer: a
Explanation: Routing refers to the path finding process in a network. In this, the control and data functions are performed to identify the route and to handle the data during the journey from source to destination.
15. How many stages are possessed by the control plane?
a) Two
b) Three
c) Four
d) Five
Answer: b
Explanation: The routing process called as control plane has three stages. These are neighbor discovery, topology discovery and path selection. These stages enable the network in routing mechanisms efficiently.
This set of Optical Communications Questions and Answers for Aptitude test focuses on “Optical Network Transmission Modes, Layers and Protocols”.
1. Electrical devices in optical network are basically used for _____________
a) Signal degradation
b) Node transfer
c) Signal control
d) Amplification
Answer: c
Explanation: The optical infrastructure in networks constitutes a transparent network in which electronic devices are present. They are basically used for signal control. The other use includes providing interconnection to other networks.
2. Signals are defined as ________________ if their significant instants occur at nominally the same rate, any variation being constrained within specific limits.
a) Plesiochronous
b) Opt-electric
c) Physiometric
d) Opt-immune
Answer: a
Explanation: With any multiplexing strategies, come some setbacks. This includes the differentiation in the frequencies occurring throughout a network. This is called as plesiochronous transmission.
3. The bit stuffing in the plesiochronous digital hierarchy is complex and uneconomic.
a) False
b) True
Answer: b
Explanation: The bit stuffing is a complex process as it does not provide individual channel extraction. For individual channel extraction, the whole de-multiplexing procedure is to be performed again. This is both uneconomic and complex.
4. A ____________ digital hierarchy was required to enable the international communications network to evolve in the optical fiber era.
a) Asynchronous
b) Dedicated
c) Seismic
d) Synchronous
Answer: d
Explanation: The plesiochronous digital hierarchy was uneconomic and complex in execution. To reduce the complexity and efficient bit stuffing purpose, a synchronous digital hierarchy was required. It transformed the international communications into an optical fiber era.
5. The standardization towards a synchronous optical network termed SONET commenced in US in _______
a) 1985
b) 1887
c) 2001
d) 1986
Answer: a
Explanation: Synchronous optical network mechanism was efficient and its standardization process mainly started in 1985. Some modifications in the plesiochronous hierarchy were retained and some new features were added in the optical era to achieve efficient bit stuffing.
6. ______________ is a packetized multiplexing and switching technique which combines the benefits of circuit and packet switching.
a) Synchronous mode
b) Asynchronous transfer mode
c) Circuit packet
d) Homogeneous mode
Answer: b
Explanation: ATM transfers the information in fixed size units called cells. Each cell contains the information identifying the source of the transmission. It generally contains less data than packets.
7. The ___________ sits at the top of hierarchy of the OSI layer model.
a) Session layer
b) Transport layer
c) Application layer
d) Data link layer
Answer: c
Explanation: Application layer is the seventh layer and sits at the top of the hierarchy. It provides a means for a user to access information on or utilize the network by receiving a service.
8. The ____________ controls the dialogs between intelligent devices.
a) Physical layer
b) Transport layer
c) Application layer
d) Session layer
Answer: d
Explanation: The session layer is fifth in the OSI layer model. It controls the sessions between the devices and manages the connections between the remote and local application.
9. The network layer looks after the flow and error control mechanism.
a) True
b) False
Answer: b
Explanation: The network layer is the third level in the OSI layer model. It provides procedural and functional method for transferring data sequences from source to destination.
10. The physical layer is located at the bottom of the OSI model.
a) True
b) False
Answer:a
Explanation: The physical layer defines all the electrical, optical and media specifications for devices. Hence, it is located at the bottom of the OSI model. It establishes and terminates the connection between the media devices.
11. In order to access for end-to-end networking of optical channels to transparently convey information, the _____________ is employed in the OTN structure.
a) Presentation layer
b) Session layer
c) OPU
d) OCh layer
Answer: d
Explanation: The OCh stands for optical channel. It provides end-to-end access in networking of optical channels. It includes the multiplexing section to support multi-user networking.
12. An advanced type of reconfigurable OTN is referred to as an _______________
a) Automatic OTN
b) Auto-generated photon
c) Automatically switched optical network
d) Optical reimbursement
Answer: c
Explanation: Automatically switched optical network is capable of switching the optical channels automatically when requested. It is specified in the ITU-T Recommendation G.8080 and it is basically a transport layer.
13. The __________ is a network layer that contains both addressing and control information to enable packets to be routed within a network.
a) TCP
b) Internet protocol
c) UDP
d) SONET/SDH protocol
Answer: b
Explanation: Internet protocol forms a part of the network layer. It controls the logical architecture within the network and addresses the issues accordingly. It routes the packets within a network.
14. The mapping of IP frames in SDH/SONET is accomplished in ___________ stages.
a) Four
b) Two
c) Three
d) One
Answer: c
Explanation: Mapping requires three stages. In the first stage, point-to-point protocol is used. The second and the third stage includes synchronous mapping of data onto the SDH/SONET frame.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Wavelength Routing Networks”.
1. Which of the following is used to provide wavelength signal service among the nodes?
a) Regularization
b) Optical enhancing
c) Hopping
d) Pulse breakdown
Answer: c
Explanation: The optical layer is dependent on wavelength. The entire physical interconnected network provides wavelength signal service among the nodes using hopping technique.
2. How many types of hopping are present?
a) Two
b) One
c) Three
d) Four
Answer: a
Explanation: There are two types of hopping. They are single hop and multihop. These techniques provide wavelength dependent service for interconnected physical network among the nodes.
3. How many switching layers are possessed by MG-OXC?
a) Two
b) Three
c) One
d) Six
Answer: b
Explanation: An MG-OXC has three switching layers. They are wavelength cross-connect , waveband cross-connects , and fiber cross-connects . These layers help to terminate the wavebands and individual wavelength channels.
4. _____________ supports a great number of wavelength channels and reduces the number of switches within the optical network.
a) Waveband switching
b) Optical remuneration
c) Optical genesis
d) Wavelength multiplexing
Answer: a
Explanation: Waveband switching reduces the number of ports within the optical network. It reduces the complexity of numerous wavelength-driven channels and makes it efficient.
5. Individual wavelength channels and wavebands are terminated through ________________ layers.
a) WXC and PXC
b) WXC and FXC
c) BXC and FXC
d) WXC and BXC
Answer: d
Explanation: The individual wavelength channels are terminated and the terminated waveband is then de-multiplexed. The de-multiplexing is in the form of individual channels which are sent to WXC layer as inputs.
6. The routing and wavelength assignment problem addresses the core issue of _____________
a) Traffic patterns in a network
b) Wavelength adjustment
c) Wavelength continuity constraint
d) Design problem
Answer: c
Explanation: The routing and wavelength assignment problem includes selecting a suitable path and allocating an available wavelength. These problems fall into two categories of sequential or combinational selections.
7. How many techniques of implementation are there for routing wavelength assignment ?
a) Two
b) Six
c) Three
d) Four
Answer: a
Explanation: The implementation of RWA can be static and dynamic. This depends on the traffic patterns in the network. Static RWA techniques are semi-permanent and dynamic RWA techniques are random in nature.
8. ____________ deals with establishing the light path in frequently varying traffic patterns.
a) Wavelength routing
b) Wavelength multiplexing
c) Static RWA
d) Dynamic RWA
Answer: d
Explanation: In dynamic RWA, the traffic patterns are not known. Thus, the connection requests are initiated in random fashion. Its random nature depends on the network state at the time of request.
9. Static RWA problem is also known as _____________
a) Routing problem
b) Virtual topology problem
c) Static wavelength problem
d) Light path problem
Answer: b
Explanation: Static RWA problem refers to the connection problems which remain connected for a smaller duration of time. Thus, network resources are assigned to each connection. It is also called as virtual topology design problem.
10. The ___________ provides information about the physical path and wavelength assignment for all active light paths.
a) Network state
b) RWA
c) LAN topology
d) Secluded communication protocol
Answer: a
Explanation: The physical path i.e. route is associated with the routing problem. Each connection is provided with network resources to reduce complexity in functioning. The network state is basically required to provide information related to routing and assignment problems.
11. ________________ plays an important role in determining the blocking probability of a network.
a) CGA algorithm
b) Semi-pristine environment
c) RWA algorithm
d) Pass key protocol
Answer: c
Explanation: RWA algorithm’s efficiency is calculated on the basis of no blocking or lowest blocking probability. It also provides the information about the availability of the path between the source and destination.
12. Wavelength assignment in RWA is independent on the network topology.
a) True
b) False
Answer: b
Explanation: RWA algorithm deals with the wavelength assignment, physical path and blocking probability. Network topology plays a crucial role in the wavelength assignment. The network state and topology enables the RWA algorithm to function smoothly.
13. Static RWA technique is semi-permanent.
a) True
b) False
Answer: a
Explanation: The connections employs in static RWA are semi-permanent but remain active for a relatively longer period of time. The traffic patterns are known in advance and thus the optimization can be done by assigning network resources to each connection.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Optical Switching Networks”.
1. Optical switching can be classified into ________ categories.
a) Two
b) Three
c) Four
d) One
Answer: a
Explanation: Optical switching is classified into two categories same as that of electronic switching.
The two categories are circuit switching and packet switching.
2. ___________________ are the array of switches which forms circuit switching fabrics.
a) Packet arrays
b) Optical cross connects
c) Circuit arrays
d) Optical networks
Answer: b
Explanation: Optical cross-connects incorporate switching connections or light paths. These larger arrays can switch signals from one port to another.
3. ___________ is an example of a static circuit-switched network.
a) OXC
b) Circuit regenerator
c) Packet resolver
d) SDH/SONET
Answer: d
Explanation: The circuit is said to be static when the network resources remain dedicated to the circuit connection. This should be followed during the entire transfer and the complete message follows the same path.
4. What is the main disadvantage of OCS?
a) Regenerating mechanism
b) Optical session
c) Time permit
d) Disability to handle burst traffic
Answer: d
Explanation: In traffic conditions, data is sent in the form of bursts of different lengths. Thus, the resources cannot be readily assigned. The OCS cannot efficiently handle burst traffic.
5. Optical electro-conversions takes place in _________________ networks.
a) Sessional
b) Optical packet-switched
c) Optical circuit-switched
d) Circular
Answer: c
Explanation: In an optical packet-switched network, data is transported in the optical domain. This is done without intermediate optic-electrical conversions. Optical electro-conversions takes place in circuit-switched networks.
6. How many functions are performed by an optical packet switch?
a) Four
b) Three
c) Two
d) One
Answer: a
Explanation: An optical packet switch performs four basic functions. These include routing, forwarding, switching and buffering.
7. ____________ provides data storage for packets to resolve contention problems.
a) Switching
b) Routing
c) Buffering
d) Reversing
Answer: c
Explanation: Switching involves directing the packets. Routing provides network connectivity while forwarding and reversing involves defining a packet. Buffering usually provides data storage for packets.
8. What is usually required by a packet to ensure that the data is not overwritten?
a) Header
b) Footer
c) Guard band
d) Payload
Answer: c
Explanation: A packet consists of a header and the payload. The label points to an entry in the lookup table. A guard band is usually included to ensure the data is not overwritten.
9. Routing technique is faster than the labeling technique. State whether the given statement is true or false.
a) False
b) True
Answer: a
Explanation: Labeling suggests where the packet should be directed. Routing routes the data in the given direction. Thus, labeling technique is efficient and faster than the routing technique.
10. ______________ provides efficient designation, routing, forwarding, switching of traffic through an optical packet-switched network.
a) Label correlation
b) Multiprotocol label switching
c) Optical correlation
d) Routing
Answer: b
Explanation: Multiprotocol label switching was first proposed by CISCO systems. Earlier, it was called as tag switching. MPLS uses labels to forward, switch, designate the traffic.
11. MPLS is independent of layer 2 and 3 in the OSI model. State whether the given statement is true or false.
a) True
b) False
Answer: a
Explanation: MPLS is flexible in the current protocol landscape. It supports Ethernet, frame relay as a data link layer but is independent of layer 2 and 3 in the OSI model.
12. Which of the following service is provided by Multiprotocol label switching ?
a) Data forwarding
b) Routing
c) VPN’s
d) Switching
Answer: c
Explanation: One of the important services provided by MPLS is IP virtual private networks. All others are provided by packet switched networks. These VPN’s provide a secure, dedicated wide area network in order to connect the offices all over the world.
Answer: d
Explanation: Burst header cell consists of information regarding switching and destination address. It works with the use of transmission units called as data bursts.
This set of Optical Communications Assessment Questions and Answers focuses on “Optical Network Deployment”.
1. A _________________ is a network connecting several regional or national networks together.
a) Long-haul network
b) Domain network
c) Short-haul network
d) Erbium network
Answer: a
Explanation: Long-haul networks are also called as core or backbone networks. These networks connect regional or national networks together on a large scale. This can be extended to extended long haul networks.
2. What is the range of transmission of extended long haul network?
a) 200-400 km
b) 600-1000 km
c) 1000-2000 km
d) 2000-4000 km
Answer: c
Explanation: Extended long haul networks comprise of DWDM links. The transmission ranges may vary depending upon the complexity of the network. The extended long haul network’s transmission range varies from 1000 to 2000 km.
3. What is the range of transmission of ultra-long haul network?
a) 200-400 km
b) 600-1000 km
c) 1000-2000 km
d) 2000-4000 km
Answer: d
Explanation: Ultra haul networks comprise of DWDM links which provides them maximum range. The transmission ranges may vary depending upon the complexity of the network. The ultra-long haul network’s transmission range varies from 2000 to 4000 km.
4. Which feature plays an important role in making the longer haul networks feasible?
a) Channeling
b) Forward error control
c) Backward error control
d) Interconnection
Answer: b
Explanation: The longer haul networks can be made feasible by improvements in the DWDM systems and using forward error control mechanism. Such networks operate at channel rates in G-bits.
5. Which of the following is not an element of a submerged cable system?
a) Repeater
b) Branching unit
c) Gain equalizer
d) Attenuator
Answer: d
Explanation: The submerged cable system consists of a dry and wet plant. The elements associated with it include a repeater, branching unit, gain equalizer and a line amplifier. Attenuator is not present in cable system.
6. ___________ provides interconnection between the United States and European countries.
a) TAT
b) WTE
c) PFE
d) POP
Answer: a
Explanation: TAT is abbreviated as transatlantic optical fiber cables. TAT-14 is the newest version which is used as a medium of interconnection between the countries.
7. TAT-14 employs a DWDM bidirectional ring configuration.
a) False
b) True
Answer: b
Explanation: TAT-14 was first used in the year 2000. Its transmission capacity is more that the previous TAT versions. It’s DWDM configuration enables it to connect the various countries of Europe with the United States.
8. A single fiber in TAT-14 can carry _________ wavelength channels.
a) One
b) Twelve
c) Sixteen
d) Ten
Answer: c
Explanation: TAT-14 ‘ s single fiber carries a total of 16 wavelength channels. Each channel can allow a transmission rate of 10 Gigabits. It possesses a high operational capacity.
9. Optical MAN’S are usually structured in _______ topologies.
a) Ring
b) Bus
c) Mesh
d) Star
Answer: a
Explanation: MAN’s are characterized by changing traffic patterns requiring the networks to be fast. Also, MAN must be cost effective in terms of both operation and maintenance. Hence, they are structured in ring topologies.
10. The ________ network is an element of public telecommunication network that connects access nodes to individual users or MAN’s.
a) Ring
b) Access
c) Mesh
d) Nodal
Answer: b
Explanation: Access network is usually a last link in the network. It provides the strategies to connect to end-point users on both sides of connection.
11. _____________ is a technique that combines two or more network resources for redundancy or higher throughput.
a) Signal bonding
b) Attenuation
c) Re-signaling
d) Channel bonding
Answer: d
Explanation: Channel bonding combines two interfaces. It increases the overall bandwidth by the number of channels bonded. The data-rates are similar in this technique.
12. The upstream traffic in EPON is managed by employing a TDM approach.
a) True
b) False
Answer: a
Explanation: EPON upstream traffic is divided into time slots. The time slots are dedicated to each ONU in order not to interfere with the data.
This set of Optical Communications Multiple Choice Questions & Answers focuses on “Optical Ethernet”.
1. What is the exception in the similarities between the optical Ethernet and the Ethernet LAN?
a) Physical layer
b) Data-link layer
c) Refractive index
d) Attenuation mechanism
Answer: a
Explanation: Optical Ethernet is similar to the conventional Ethernet LAN with the exception of the physical layer. Physical layer includes the flow of data in the form of binary digits. This transmission takes place on the bit level.
2. Which technology is used by optical Ethernet?
a) GP-technology
b) HJ-technology
c) IP-technology
d) GB-technology
Answer: c
Explanation: Optical Ethernet is the fourth generation of the Ethernet family. The earlier generations include X.25, Frame Relay and ATM. Unlike these technologies, optical Ethernet uses IP-based technology.
3. When was the Gigabit Ethernet network developed?
a) 1977
b) 1988
c) 1990
d) 2002
Answer: b
Explanation: The Gigabit Ethernet network was developed in 1988. It was developed by merging two technologies namely 802.3 Ethernet and ANSI X3T11 fiber channel.
4. Optical Ethernet can operate at the transmission rates as low as ______________
a) 10 M bits per second
b) 40 M bits per second
c) 100 M bits per second
d) 1000 M bits per second
Answer: a
Explanation: Usually, high transmission rates define optical Ethernet. The ITU-T Recommendation specifies the physical layer for optical Ethernet. It can operate at transmission rates as low as 10 M bits per second.
5. How many types of optical Ethernet connections are developed?
a) Two
b) One
c) Four
d) Three
Answer: d
Explanation: There are three different types of optical Ethernet connections. They are point-to-point, point-to-multipoint, and multipoint-to-multipoint. Multipoint refers to more number of connections on either side.
6. Which type of connection can be used as an Ethernet switch?
a) Point-to-point
b) Multipoint-to-multipoint
c) Multipoint-to-point
d) Point-to-multipoint
Answer: b
Explanation: The multipoint-to-multipoint configuration refers to the bus, tree or mesh topology. Such a mesh can be made to work as a switching hub with non-blocking switching features. It facilitates switching between different optical Ethernet users.
7. How many aspects are included in the standard Ethernet protocol?
a) One
b) Two
c) Four
d) Three
Answer: c
Explanation: Optical Ethernet follows standard Ethernet protocol. This protocol includes four different aspects: Frame, MAC, signaling components and the physical medium.
8. Which of the following is not included in the Ethernet frame format?
a) MAC
b) Preamble
c) Destination address
d) Source address
Answer: a
Explanation: The Ethernet frame format includes preamble, destination and source addresses, length, data and the check sequence. MAC is the protocol which is used for sharing the network nodes.
9. The _______________ provides point-to-point access to a bidirectional single-mode optical fiber.
a) Optical regenerator
b) Optical session
c) Optical distribution node
d) Optical buffer
Answer: c
Explanation: The ODN is abbreviated as Optical distribution node. It can access an optical fiber on a point-to-point basis. Thus, a single mode bidirectional optical fiber can be accessed by an Optical distribution node.
10. _______________ is the de-multiplexing technique used to split SONET bandwidth into logical groups.
a) SDH
b) Virtual concatenation
c) STS-1
d) Optical breakdown
Answer: b
Explanation: Virtual concatenation is basically a splitting technique. It can split the SONET bandwidth into groups. These groups may be transported or routed independently.
11. Ethernet switches support multiprotocol label switching.
a) True
b) False
Answer: a
Explanation: Ethernet switches support multiprotocol label switching. This feature is desired mainly in MAN. The use of such switches in LAN exceeds the network capabilities.
12. Length field in MAC frame ensures that the frame signals stay on the network in order to detect the frame within the correct time limit.
a) True
b) False
Answer: b
Explanation: MAC frame includes Length field to identify the type or length of the network protocols. The data field is used to ensure that the frame signals stay on the network long enough to detect the frame within the desired limit.
13. The ___________ protocol is not used when the Ethernet connections are configured for a full duplex operation.
a) TCP/IP
b) MAC
c) CSMA/CD
d) DTH
Answer: c
Explanation: In Ethernet connections, the full-duplex operation situation may lead to an increased frame dropping rate. The dropped frame cannot be detected without collision. Thus, CSMA protocol is not used in full duplex mode.
14. Optical Ethernet provides switching capabilities in layers ________
a) 1 and 2
b) 2 and 3
c) 3 and 4
d) 1 and 4
Answer: b
Explanation: Layer 2 and 3 are data link and network layers. IP routing is usually considered to be ⅔ switched network. Thus, unlike conventional LAN, optical Ethernet provides switching capabilities between layers 2 and 3.
15. The ______________ approach can provide interconnection among multiple site locations within 40 km range.
a) 3 Gbe
b) 5 Gbe
c) 1 Gbe
d) 10 Gbe
Answer: d
Explanation: The 10Gbe approach uses a 40 km window range. It can be used in LAN’s and MAN. It provides switch-to-switch network within data centers.
