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[MCQ’s] Optical Communication [EXTC]

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Module 01

Optical Communication

1. What is the principle of fibre optical communication?
a) Frequency modulation
b) Population inversion
c) Total internal reflection
d) Doppler Effect
Answer: c
Explanation: In optical fibres, the light entering the fibre does not encounter any new surfaces, but repeatedly they hit the same surface. The reason for confining the light beam inside the fibres is the total internal reflection.

2. What is the other name for a maximum external incident angle?
a) Optical angle
b) Total internal reflection angle
c) Refraction angle
d) Wave guide acceptance angle
Answer: d
Explanation: Only this rays which pass within the acceptance angle will be totally reflected. Therefore, light incident on the core within the maximum external incident angle can be coupled into the fibre to propagate. This angle is called a wave guide acceptance angle.

3. A single mode fibre has low intermodal dispersion than multimode.
a) True
b) False
Answer: a
Explanation: In both single and multimode fibres the refractive indices will be in step by step. Since a single mode has less dispersion than multimode, the single mode step index fibre also has low intermodal dispersion compared to multimode step index fibre.

4. How does the refractive index vary in Graded Index fibre?
a) Tangentially
b) Radially
c) Longitudinally
d) Transversely
Answer: b
Explanation: The refractive index of the core is maximum along the fibre axis and it gradually decreases. Here the refractive index varies radially from the axis of the fibre. Hence it is called graded index fibre.

5. Which of the following has more distortion?
a) Single step-index fibre
b) Graded index fibre
c) Multimode step-index fibre
d) Glass fibre
Answer: c
Explanation: When rays travel through longer distances there will be some difference in reflected angles. Hence high angle rays arrive later than low angle rays. Therefore the signal pulses are broadened thereby results in a distorted output.

6. In which of the following there is no distortion?
a) Graded index fibre
b) Multimode step-index fibre
c) Single step-index fibre
d) Glass fibre
Answer: a
Explanation: The light travels with different speeds in different paths because of the variation in their refractive indices. At the outer edge it travels faster than near the centre But almost all the rays reach the exit end at the same time due to the helical path. Thus, there is no dispersion in the pulses and hence the output is not a distorted output.

7. Which of the following loss occurs inside the fibre?
a) Radiative loss
b) Scattering
c) Absorption
d) Attenuation
Answer: b
Explanation: Scattering is a wavelength dependent loss. Since the glass used in the fabrication of fibres, the disordered structure of glass will make some vibrations in the refractive index inside the fibre. This causes Rayleigh scattering.

8. What causes microscopic bend?
a) Uniform pressure
b) Non-uniform volume
c) Uniform volume
d) Non-uniform pressure
Answer: d
Explanation: Micro-bends losses are caused due to non-uniformities inside the fibre. This micro-bends in fibre appears due to non-uniform pressures created during the cabling of fibre.

9. When more than one mode is propagating, how is it dispersed?
a) Dispersion
b) Inter-modal dispersion
c) Material dispersion
d) Waveguide dispersion
Answer: b
Explanation: When more than one mode is propagating through a fibre, then inter modal dispersion will occur. Since many modes are propagating, they will have different wavelengths and will take different time to propagate through the fibre.

10. A fibre optic telephone transmission can handle more than thousands of voice channels.
a) True
b) False
Answer: a
Explanation: Optical fibre has larger bandwidth hence it can handle a large number of channels for communication.

11. Which of the following is known as fibre optic back bone?
a) Telecommunication
b) Cable television
c) Delay lines
d) Bus topology
Answer: d
Explanation: Each computer on the network is connected to the rest of the computers by the optical wiring scheme called bus topology, which is an application known as fibre optic back bone.

12. Calculate the numerical aperture of an optical fibre whose core and cladding are made of materials of refractive index 1.6 and 1.5 respectively.
a) 0.55677
b) 55.77
c) 0.2458
d) 0.647852
Answer: a
Explanation: Numerical aperture = n12n22−−−−−−−−√
Numerical aperture = 0.55677.

13. A step-index fibre has a numerical aperture of 0.26, a core refractive index of 1.5 and a core diameter of 100micrometer. Calculate the acceptance angle.
a) 1.47°
b) 15.07°
c) 2.18°
d) 24.15°
Answer: b
Explanation: sin i = (Numerical aperture)/n
sin i = 15.07°.

14. 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(radius)


15. 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.

16. 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 (MFD) 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.

17. 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.

18. 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

Solving this equation, we get a=5.05μm. Core-diameter=2a=10.1μm.

19. 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 Bf =ny – nx. Here, ny and nx are refractive indices of two modes.

20. 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: Bf=ny– nx = λ/Lp. Here, λ=wavelength and Lp = beat length. Solving this equation, we will get the answer.

21. 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.

22. 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.

23. 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.

24. 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 20th century where Hondros and debye theoretical and experimental study demonstrated that information can be transferred as a form of light through a dielectric waveguide.

25. 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.

26. Which law gives the relationship between refractive index of the dielectric?
a) Law of reflection
b) Law of refraction (Snell’s Law)
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.

27. 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.

28. 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.

29. 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.

30. 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.

31. 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.

32. 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.

33. 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.

34. 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.

35. Constructive interference occur when total phase change after two successive reflections at upper and lower interfaces is equal to? (Where m is integer)
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.

36. 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.

37. A monochromatic wave propagates along a waveguide in z direction. These points of constant phase travel in constant phase travel at a phase velocity Vp is given by?
a) Vp=ω/β
b) Vp=ω/c
c) Vp=C/N
d) Vp=mass/acceleration
Answer: a
Explanation: Velocity is a function of displacement. Phase velocity Vp is a measure of angular velocity.

38. 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.

39. 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.

40. 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.

41. 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 = (V*V)/2. Here M denotes the number of modes and V denotes normalized frequency.

42. 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(NA)/λ. Substituting and calculating the values, we get option 18.84. Here, V denotes normalized frequency and NA = numerical aperture.

43. 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 2Δ−−−√/V, we get option c. Here, V=2.405, n1 = refractive index of core, a=radius of core.

44. 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.

45. 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.

46. 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.

47. 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.

48. 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.

49. 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.

50. 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.


Module 02

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) Fe2+
b) Fe3+
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 (αr).

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) 8.60μm
b) 9.30μm
c) 9.1μm
d) 10.2μm
Answer: b
Explanation: The radius of curvature of the fiber bend of a multimode fiber is given by

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 8μm, wavelength = 1.5μm cutoff wavelength = 1.214μm. 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-

Where R = radius of curvature,
n1, n2 = refractive indices,
λc = 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.

12. 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.

13. 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.

14. 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.

15. 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.

16+. 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.

17. 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.

18. 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.

19. 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.

20. 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-
PB = 4.4*10-3d2λ2αdBv
Where, PB = threshold optical power
d = diameter of core
λ = wavelength
αdB = attenuation.

21. 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-
PR = 5.9*10-2d2λαdB
Where, PR = optical power for Raman scattering
d = diameter of core
λ = wavelength
αdB = attenuation.

22. 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.

23. 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.

24. 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.

25. 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.

26. 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.

27. 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.

28. 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.
BT < = 12 Γ
Where BT = bit rate
2Γ = duration of pulse.

29. 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.

30. 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 (NRZ), 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.

31. 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
BT = 12 Γ
Where BT = bandwidth.

32. 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.

33. 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.

34. 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.

35. The optical source used in a fiber is an injection laser with a relative spectral width σλ/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 σλ/λ= 0.01 is given. The rms spectral width can be calculated as follows:
σλ/λ = 0.0011
σλ = 0.0011λ
= 0.0011*0.70*10-6
= 0.77 nm.

36. 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.

37. 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.

38. 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 (i.e. critical angle) with the normal.

39. 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
Tmin = Ln1/c
Where L = length of the fiber
n1 = Refractive index of core
c = velocity of light in vacuum.

40. 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.

41. 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.

42. 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.

43. The index profile of a core of multimode graded index fiber is given by?
a) N (r) = n1 [1 – 2Δ(r2/a)2]1/2; r<a
b) N (r) = n1 [3 – 2Δ(r2/a)2]1/2; r<a
c) N (r) = n1 [5 – 2Δ(r2/a)2]1/2; r>a
d) N (r) = n1 [1 – 2Δ(r2/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.

44. 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.

45. 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
The RMS pulse broadening per km due to intermodal dispersion for multimode step index fiber is given by
(σs(1 km)/L = 80.6/4 = 20.15
Where L = length of fiber
σs = pulse broadening.

46. 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.

47. 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.

48. 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.

49. 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.

50. 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.

51. A multimode step index fiber has source of RMS spectral width of 60nm and dispersion parameter for fiber is 150psnm-1km-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
σm(1 km) = σλLM
= 60*1* 150pskm-1
= 9.0nskm-1
Where σλ = rms spectral width
L = length of fiber
M = dispersion parameter.

52. 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 σT is given by

Where σm = RMS pulse broadening due to material dispersion
σi = RMS pulse broadening due to intermodal dispersion.

53. Γg = 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(Γg) for a light pulse. This light pulse is propagating along a unit length of a single mode fiber.

54. 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.

55. A single mode fiber has a zero dispersion wavelength of 1.21μm and a dispersion slope of 0.08 psnm-2km-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
DT(1260 nm) = λS0/4 [1-(λ0/λ)4]
= (1260*0.08*10-12)/4 * (1-[1550/1260]4)
= -3.76psnm-1km-1
λ0 = zero dispersion wavelength
λ = wavelength
S0 = dispersion slope
DT = total first order dispersion.

56. The dispersion due to material, waveguide and profile are -2.8nm-1km-1, 20.1nm-1km-1 and 23.2nm-1km-1respectively. 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
DT = DM + DW + DP(psnm-1km-1)
DW = waveguide dispersion
DM = Material dispersion
DP = profile dispersion.

57. 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.

58. 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.

59. 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 (DFFS) are obtained by fabricating multilayer index profiles with increased waveguide dispersion. This is tailored to provide overall dispersion say 2psnm-1km-1 over the wavelength range 1.3 to 1.6μm.

60. 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.

61. 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.

62. 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.

63. 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.

64. 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.

65. 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.

66. 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.

67. 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-
BF = λ/LB = 0.8×10-6/0.08
= 1×10-5
λ = peak wavelength
LB = beat length.

68. 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:
βx – βy = 2Π/LB = 2×3.14/0.06
= 104.66
βx & βy are propagation constants for slow & fast modes resp.
LB = beat length.

69. 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/BF Where λ is operating wavelength, BF = Birefringence, T = period of perturbation.

70. When two components are equally excited at the fiber input, then for polarization maintaining fibers δΓg 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 δΓg is related to polarization mode dispersion (PMD) of fiber. This linear relationship to fiber length however applies only to short fiber-lengths in which birefringence are uniform.

71. 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.

72.Which type of scattering occurs due to interaction of light in a medium with time
dependent optical density variations thereby resulting into the change of energy
(frequency) & path?
a. Stimulated Brilliouin Scattering (SBS)
b. Stimulated Raman Scattering (SRS)
c. Mie Scattering
d. Rayleigh Scattering
ANSWER: Stimulated Brilliouin Scattering (SBS)

Module 03

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 (holes) 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 τi.

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 = τrτnrrnr = 70× 100/70 + 100 ns = 41.17 ns
τr = radiative recombination lifetime of minority carriers
τnr = 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
ηint = τ/τr = 40/80 ×100 = 80%
τ = total carrier recombination lifetime
τr = 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
Pint = ηint hci/eλ = 0.645×6.626×10-34×3×108×40×10-3/ 1.602×10-19 × 0.82 × 10-6
= 0.039 W
ηint = 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 I0is 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
Pe = PintFn2/4nx2 = Pint (0.680×1/4×(2.5)2) = 0.0272 Pint.
Hence power emitted is only 2.72 % of optional power emitted internally.
Fn2 = 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
ηcp = (0.018Pint/2Pint)*100
Pint = power emitted
ηcp = 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 = -10log10 ηc
ηc = 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 * 10-4Pint)
a) 34 dB
b) 32.59 dB
c) 42 dB
d) 33.1 dB
Answer: b
Explanation: The loss in decibels relative to Pint is given by-
Loss = -10log10Pc/Pint
Pc = 5.5 * 10-4Pint.

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
ηc = (NA)2 = (0.3)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.

16. 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.

17. 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.

18. The frequency of the absorbed or emitted radiation is related to difference in energy E between the higher energy state E2 and the lower energy state E1. 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 (f) 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 = 6.626×10-34Js.

19. The radiation emission process (emission of a proton at frequency) 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.

20. 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.

21. An incandescent lamp is operating at a temperature of 1000K at an operating frequency of 5.2×1014 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 (hf/KT)-1.

22. 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.

23. __________________ 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.

24. 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) 1×102
b) 3×106
c) 2.9×105
d) 2.2×105
Answer: d
Explanation: The number of longitudinal modes is given by-
q = 2nL/λ
q = Number of longitudinal modes
n = Refractive index
L = Length of the crystal
λ = Peak emission wavelength.

25. 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
c = velocity of light
n = Refractive index
L = Length of the crystal.

26. 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.

27. 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 –
gth = α + 1/L ln (1/r)
α = active cavity loss
L = Length of the cavity
r = reflectivity.

28. 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.

29. Considering the values given below, calculate the mode separation in terms of free space wavelength for a laser. (Frequency separation = 2GHz, Wavelength = 0.5 μm)
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 = λ2/c δf
δf = frequency separation
λ = wavelength
c = velocity of light.

30. 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.

31. 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.

32. 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.

33. 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.

34. 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.

35. 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.

36. 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.

37. 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*10-9cm2.

38. Estimate optical power coupled into fiber of DH SLED having emission area of 1.96*10-5, radiance of 40 W/rcm2, 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*102
d) 9.551*10-5
Answer: d
Explanation: The optical power coupler in the step index fiber of SLED is given by
Pc = π(1-r) A RD(NA) 2
= 3.14 (1-0.03)*1.96*10-5*40*(0.2) 2
= 9.551*10-5W.

39. 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.

40. 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*102
c) 4.25*10-3
d) 2.14*10-3
Answer: c
Explanation: The overall power conversion efficiency is determined by
η pc = Pc/P = 170*10-6/20*10-3*2
= 4.25*10-3.

41. 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 = Pc/P
Pc = η pc* P
= 0.8 * 0.0375
= 0.03.

42. 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.

43. 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.

44. 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.

45. 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.

46. 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.

47. 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.

48. 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.

49. 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.

50. 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.

51. 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.

52. The optical 3 dB point occurs when currents ratio is equal to _____________
a) 83
b) 22
c) 12
d) 34
Answer: c
Explanation: In optical regime, the bandwidth is defined by frequency at which output current has dropped to ½ output input current system.

53. 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.

54. 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(30 MHz) = Pdc/(1+(wΓi)2)1/2
= 320*10-6/(1+(2π*30*10-6*5*10-9)2)1/2
= 2.29*10-4W.

55. 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(20 MHz) = Pdc/(1+(WTi)2)1/2
246.2*10-6 = Pdc/(1+(2π*20*10-6*5*10-9)2)1/2
Pdc = 3.48*10-4.

56. 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.

57. 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.

58. 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
βr = β0exp (-Ea/KT)
= 1.89*107exp (-1*1.602*10-19/1.38*10-23*290)
= 7.79*10-11 h-1.

59. 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.

60. 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.

61. 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(E) 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(E) = 1/(1+exp(E-EF/KT))
Where K = Boltzmann constant, T = absolute temperature, EF = Fermi energy level.

62. 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.

63. 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.

64. _________________ 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.

65. 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.

66. 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.

67. 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 1018cm-3. The recombination coefficient for gallium arsenide is 7.21*10-10cm3s-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-
ςr = [BrN]-1 where Br = Recombination coefficient in cm3s-1 and N = carrier concentration in n-region.

68. 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.

69. 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.

70. A GaAs injection laser has a threshold current density of 2.5*103Acm-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 Ith. It is given by-
Ith = Jth * area of the optical cavity
Where Jth = threshold current density
Area of the cavity = length and width.

71. 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 = ((n-1)/(n+1))2 where r=reflectivity and n=refractive index.

72. 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.

73. 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.

74. 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 (0.8μm-0.9μm).

75. 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.

76. _______________ 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.

77. 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.

78. 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.

79. 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.

80. 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.

81. How many techniques are used to tune monolithic integrated devices (lasers)?
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.

82. _________________ laser can be produced when a coupler section is introduced between the amplifier and phase sections of a structure.
c) Y 4-shifted
d) DSM
Answer: b
Explanation: DBR lasers are capable of wavelength tuning. Grating assisted co-directional coupler with sampled reflector (GCSR). 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.

83. 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.

84. 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.

85. 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.

86. 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.

87. 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.

88. How many sections are included in a sampling grating distributed Bragg-reflector laser (SG-DBR)?
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.

89. 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.

Module 04

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(photodiode).
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.

9. _____________ 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.

10. 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.

11. ______________ 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.

12. 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.

13. ___________ 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.

14. The probability of zero pairs being generated when a light pulse is present is given by which of the following equation?
a) P(0/1) = exp(-Zm)
b) P(x) = exp (Zm)
c) P(y) = x (0) + x(1)
d) P(z) = P(-Zm)
Answer: a
Explanation: The probability of zero pairs being generated when a light pulse is present is given by equation –
P (0/1) = exp(-Zm)
Where, P (0/1) represents the system error probability p(e) and Zm is variance of the probability distribution.

15. The minimum pulse energy needed to maintain a given bit-error-rate (BER) 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.

16. 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(e) = exp(-Zm)
Where, Zm = No. of photons
Here P(e) = 10-9, therefore Zm is calculated from above relation.

17. For a given optical fiber communication system, P(e) = 10-9, Zm = 20.7, f = 2.9×1014, 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 –
Emin = Zmhf/η
Where, Zm = Number of photons
h = Planck’s constant
f = frequency
η = Quantum efficiency.

18. 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 = 2.99×1014Hz).
a) 46
b) 40
c) 50
d) 52
Answer: c
Explanation: The SNR is given by –
S/N = ηP0/2hfB
Where, η = 1 (for ideal detector)
P0 = incident power
h = Planck’s constant
B = Bandwidth.

19. 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 P0. It is given by –
P0 = 10log10(P0(watts))
Where P0(watts) = incident power in Watts/milliWatt.

20. 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, Zm 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 Zm. Hence, τ is the time and P0 is the incident power, ηis the quantum efficiency and f is the frequency.

21. 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.

22. 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.

23. 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
Ip = ηP0eλ/hc
Where η = Quantum efficiency
P0 = Incident optical power
e = electron charge
λ = Wavelength
h = Planck’s constant
c = Velocity of light.

24. 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 × 10-16A2
b) 1.23 × 10-17A2
c) 1.65 × 10-16A2
d) 1.61 × 10-17A2
Answer: d
Explanation: The thermal noise in the load resistor is given by –
it2 = 4KTB/RL
Where T = Temperature
B = Bandwidth
RL = Load resistance.

25. ________________ 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.

26. ______________ 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.

27. 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-
RL = 1/2πCdB
B = Post detection bandwidth
Cd = Input capacitance
RL = Load resistance.

28. 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.

29. ____________ 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 (M). Its value depends upon the detector material, shape of electric field profile and holes and electrons inclusion. It is a function of multiplication factor.

30. 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 (K) 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.

31. __________ 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.

32. _________________ 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.

33. 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.

34. ___________ 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.

35. 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.

36. _______ 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 104V/m. This causes all the drifting of carriers at saturated limited velocities.

37. 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.

38. 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.

39. 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.

40. 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.

41. 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.

42. 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.

43. 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 byIP=P0R. Here IP = photocurrent, P0=Power, R = responsivity.

44. 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 IP = P0R. Here IP = photocurrent, P0 = Power, R = responsivity.
P0 = IP/R gives the optical power.

45. 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 IP = P0R. Here IP = photocurrent, P0 = Power, R = responsivity.
R = IP/P0 gives the responsivity.

46. 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/IP where I = output current, IP = photocurrent.

47. 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/IP where I = output current, IP = photocurrent. I = M*IP gives the output current inμA.

48. 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/IP where I = output current, IP = photocurrent. IP = I/M Gives the output current inμA.

Module 05

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 (melting)
b) Radio frequency induction
c) Optical attenuation
d) Vapor Phase Deposition (VPD)
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) SiCl4
b) GeCl4
c) O2
d) B2O3
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. B2O3 is used as a dopant.

4. P2O5 is used as a _____________
a) Dopant
b) Starting material
c) Cladding glass
d) Core glass
Answer: a
Explanation: P2O5 is a non silica material. Dopants are formed from non silica materials so that refractive index modification is achieved. Other dopants include Ti O2, Ge O2, 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 (CVD). 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) 2Cl2
d) 4Cl2
Answer: b
Explanation: SiCl4 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 SiO2 (solid compound) along with 4HCl(gas) 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 (OVPO)
b) Vapor Axial Deposition (VAD)
c) Modified Chemical Vapor Deposition (MCVD)
d) Plasma-activated Chemical Vapor Deposition (PCVD)
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 (MCVD) process is also called as an inside vapor phase oxidation (IVPD) 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 (IVPO) as it takes place inside the silica tube at the temperatures between 1400 and 1600 degrees Celsius.

12. 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 = PF/Ps
Where PF = power coupled into the fiber
Ps = power emitted by the source.

13. The ratio r = (n1 – n)/(n1 – n) indicates ____________
a) Fresnel reflection
b) Reflection coefficient
c) Refraction coefficient
d) Angular power distribution coefficient
Answer: b
Explanation: The ratio, r = (n1-n)/(n1-n) is known as Reflection coefficient. It relates the amplitude of the reflected ray to the amplitude of the incident wave.

14. 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 = ((n1-n)/(n1-n))2
Multiplying r by 100, we get the value of r in terms of percentage.

15. 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 = -10log10(1-r)
Where L = loss due to Fresnel reflection
R = magnitude of Fresnel reflection.

16. Two joined step index fibers are perfectly aligned. What is the coupling loss of numerical aperture are NAR= 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
LF(NA) = -10 log (NAR/NAE)2
Where LF = coupling loss
NAR = Numerical aperture of receiving fiber
NAE = Numerical aperture of emitting fiber.

17. Two joined graded index fibers that are perfectly aligned have refractive indices αR = 1.93 for receiving fiber αE = 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
LF(α) = -10log10αRE+2)/αER+2)
Where LF(α) = Coupling loss
αR = refractive index of receiving fiber
αE = refractive index of emitting fiber.

18. 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.

19. 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.

20. 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
Loss10t = -10log10t η10t
Where, Loss10t = insertion loss due to lateral misalignment
η10t = Coupling efficiency.

21. A graded index fiber has a parabolic refractive index profile (α=2) 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 (assuming there is uniform illumination of all guided modes) is given by
Lt = 0.85(y/a)
Where y = lateral misalignment
a = core radius.

22. 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-Lt
Where η = coupling efficiency
Lt = misalignment loss.

23. 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
TT = TL + Ta
Where, TT = total insertion loss
TL = lateral offset loss
Ta = Angular misalignment loss.

24. 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.

25. 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.

26. 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 (welding) and mechanical splicing.

27. 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.

28. 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.

29. 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.

30. 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.

31. 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.

32. 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.

33. _____________ 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.

34. 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.

35. _____________ 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.

36. 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.

37. 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.

38. 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.

39. 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.

40. 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.

41. 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.

42. 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.

43. 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.

44. 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.

45. 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.

46. 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.

47. 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.

48. 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.

49. 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.

50. 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.

51. 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 (WDM) couplers. These couplers distribute light among the branch fibers with no scattering loss.

52. 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.

53. ________ 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.

54. 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.

55. 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 (FBT) 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.

56. 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.

57. 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 = 10log10 P1/(P3+P4)
WhereP1, P3, P4 = output power at ports 1,3 and 4 resp.

58. 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 = [P3/(P3+P4)]*100%
Where P3 and P4 are output powers at ports 3 and 4 respectively.

59. 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.

60. 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 (Star coupler) = 10log10N (dB).

61. 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.

62. 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 2M=16. Hence M=4.
Thus, excess loss is given by-
Excess loss = (M×loss in each 3-port coupler) + (Number of splices×Loss in each stage)
Where number of splices = 3 (as the value of M is equal to 4).

63. 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 = λB/ 2n
Where λB = wavelength
n = refractive index.

64. 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.

65. 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.

66. 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.

67. 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 (SOAs).

68. 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.

69. 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.

70. 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.

71.Which among the following do/does not support/s the soot formation process?
d. All of the above

Module 06

1. Link budget consists of calculation of
a) Useful signal power
b) Interfering noise power
c) Useful signal & Interfering noise power
d) None of the mentioned
Answer: c
Explanation: The link analysis and its output, the link budget consists of calculations and tabulations of useful signal power and interfering noise power at the receiver.

2. Link budget can help in predicting
a) Equipment weight and size
b) Technical risk
c) Prime power requirements
d) All of the mentioned
Answer: d
Explanation: Link budget can help to predict equipment weight, size, prime power requirements, technical risk and cost. Link budget is one of the system manager’s useful document.

3. Which is the primary cost for degradation of error performance?
a) Loss in signal to noise ratio
b) Signal distortion
c) Signal distortion & Loss in signal to noise ratio
d) None of the mentioned
Answer: c
Explanation: There are two primary causes for the degradation of error performance. They are loss in signal to noise ratio and the second is signal distortion caused by intersymbol interference.

4. Which factor adds phase noise to the signal?
a) Jitter
b) Phase fluctuations
c) Jitter & Phase fluctuations
d) None of the mentioned
Answer: c
Explanation: When a local oscillator is used in signal mixing, phase fluctuations and jitter adds phase noise to the signal.

5. Antennas are used
a) As transducer
b) To focus
c) As transducer & To focus
d) None of the mentioned
Answer: c
Explanation: Antennas are used as transducer that converts electronic signals to electromagnetic fields and vice versa. They are also used to focus the electromagnetic energy in the desired direction.

6. Mechanism contributing to a reduction in efficiency is called as
a) Amplitude tapering
b) Blockage
c) Edge diffraction
d) All of the mentioned
Answer: d
Explanation: Mechanism contributing to a reduction in efficiency is called as amplitude tapering, spillover, edge diffraction, blockage, scattering, re-radiation and dissipative loss.

7. Space loss occurs due to a decrease in
a) Electric field strength
b) Efficiency
c) Phase
d) Signal power
Answer: a
Explanation: Due to the decrease in electric field strength there will be a decrease in signal strength as a function of distance. This is called as space loss.

8. Antenna’s efficiency is given by the ratio of
a) Effective aperture to physical aperture
b) Physical aperture to effective aperture
c) Signal power to noise power
d) Losses
Answer: a
Explanation: The larger the antenna aperture the larger is the resulting signal power density in the desired direction. The ratio of effective aperture to physical aperture is the antenna’s efficiency.

9. Effective radiated power of an isotropic radiator can be given as a product of
a) Radiated power and received power
b) Effective area and physical area
c) Transmitted power and transmitting gain
d) Receiving power and receiving gain
Answer: c
Explanation: An effective radiated power with respect to an isotropic radiator EIRP can be defined as the product of transmitted power and the gain of the transmitting antenna.

10. According to reciprocity theorem, _____ and _____ are identical.
a) Transmitting power and receiving power
b) Transmitting gain and receiving gain
c) Effective area and physical area
d) None of the mentioned
Answer: b
Explanation: The reciprocity theorem states that for a given antenna and carrier wavelength the transmitting and receiving gain are identical.

11. 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.

12. 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.

13. _________ 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.

14. 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.

15. __________ 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.

16. 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.

17. 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.

18. 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.

19. 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.

20. Electronic circuits have a practical limitation on speed of operation at a frequency of around _________
a) 1010Hz
b) 1012Hz
c) 1014Hz
d) 1011Hz
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 1010Hz.

21. The use of light as an electromagnetic wave of high frequency provides high speed operation around ____________ times the conceivable employing electronic circuits.
a) 108Hz
b) 105Hz
c) 106Hz
d) 104Hz
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 1014to 1015Hz.

22. How many layers are possessed by waveguide structures of silica-on-silicon(SOS)?
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.

23. 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.

24. Which method determines the dispersion limitation of an optical link?
a. Link power budget
b. Rise time budget
c. Both a and b
d. None of the above

25. A link budget is accounting of all __________
a) Gain and losses from the transmitter
b) Power transmitted by transmitter
c) Power received by receiver
d) Power transmitted and received
Answer: a

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