Category: Optical Communications Objective Questions

Optical Communications Multiple Choice Questions on “Fiber Numerical Aperture and Fiber Diameter Measurements”.

1. The ____________ affects the light gathering capacity and the normalized frequency of the fiber.
a) Numerical aperture
b) Amplitude modulation
c) Responsivity
d) Quantum efficiency
Answer: a
Explanation: Numerical aperture is an important optical parameter as it dictates the important characteristics of the optical fiber. This in turn dictates the number of propagating modes within the fiber.

2. The numerical aperture for a step index fiber is sine angle of the ____________
a) Efficient angle
b) Aperture
c) Acceptance angle
d) Attenuation
Answer: c
Explanation: The numerical aperture of a step index fiber is given by –
NA = sinθ_a, where θ_a is the acceptance angle and NA is the numerical aperture.

3. The calculations of the numerical aperture from a refractive index data are less accurate for the graded index fibers than for step index fibers.
a) False
b) True
Answer: b
Explanation: The refractive indices of the core and cladding are fluctuating, thus causing the data to be less efficient. For graded index fibers, it is usually less accurate than for the step index fibers.

4. Far field pattern measurements with regard to multimode fibers are dependent on the _____________ of the fiber.
a) Amplitude
b) Frequency
c) Diameter
d) Length
Answer: d
Explanation: The accuracy of the measurement technique is dependent upon the visual assessment of the far-field pattern from the fiber. In case of multimode fibers, far field pattern measurements are dependent on the length of the fiber.

5. The screen is positioned 10 cm from the fiber end face. When illuminated from a wide angled visible source the measured output pattern size is 6.2 cm. Calculate the approximate numerical aperture of the fiber.
a) 0.21
b) 0.30
c) 0.9
d) 1.21
Answer: b
Explanation: The numerical aperture can be obtained from a trigonometric relationship given by-
NA = A/(A²+4D²)^1/2, where A is constant(38.44) and D is the distance of the fiber end face from the screen in mm.

6. During the fiber drawing process, the fiber outer diameter is maintained constant to within ________
a) 2%
b) 1%
c) 5%
d) 10%
Answer: b
Explanation: During the fiber manufacturing stage, all processes needs to be performed efficiently. Especially, in the drawing process, the outer diameter should be compiled to within 1 % to avoid miscommunication through fibers.

7. What is the minimum value of accuracy in diameter is needed to avoid radiation losses in the fiber?
a) 0.1%
b) 0.2%
c) 0.3%
d) 0.03%
Answer: c
Explanation: Any diameter variations can cause excessive radiation losses and make accurate fiber-fiber connection difficult. Hence, on-line diameter measurement systems are used which provides accuracy of 0.3%.

8. Which of the following is a non-contacting optical method of on-line diameter measurement?
a) Brussels’s method
b) Velocity differentiator method
c) Photo detector method
d) Image projection method
Answer: d
Explanation: On-line diameter measurement technique uses fiber image projection method. It is also known as non-contacting optical method and shadow method.

9. The shadow method is used for measurement of the outer diameter of an optical fiber. The apparatus employs a rotating mirror with an angular velocity of 4 rad/s which is located at 10 cm from the photo detector. Compute the shadow velocity.
a) 0.1 μm μs^-1
b) 0.4 μm μs^-1
c) 0.87 μm μs^-1
d) 1 μm μs^-1
Answer: b
Explanation: The shadow velocity is obtained by the below equation:
ds/dt = l. dϕ/dt, where l is the distance of the apparatus from the photodetector and dϕ/dt is the angular velocity.

10. The shadow velocity is given by 0.4 μm μs^-1 and shadow pulse of width 300 μs is registered at an instant by the photodetector. Determine the outer diameter of the optical fiber in μm.
a) 100 μm
b) 120 μm
c) 140 μm
d) 90 μm
Answer: b
Explanation: The fiber outer diameter is given by-
d₀ = W_e.Ds/dt, where W_e = pulse width and ds/dt = shadow velocity.

11. The techniques used to determine the refractive index profile can also be used to determine the core diameter.
a) True
b) False
Answer: a
Explanation: Some of the techniques used to determine the refractive index profile are interferometry, near field scanning and refracted ray technique. The core diameter for step index fibers is defined by the step change in the refractive index profile. Hence, they can be used to measure the core diameter.

.

Optical Communications Multiple Choice Questions on “The Optical Receiver Circuit”.

1. ____________ limits receiver sensitivity.
a) Noise
b) Depletion layer
c) Avalanche
d) Current
Answer: a
Explanation: Receiver noise affects receiver sensitivity. It can dictate the overall system design. The noise can be temperature, environmental factor or due to components.

2. A ____________ performs the linear conversion of the received optical signal into an electric current.
a) Receiver
b) Converter
c) Detector
d) Reflector
Answer: c
Explanation: An optical signal is always fed to a detector. A detector is an optoelectronic converter which linearly converts the received optical signal into an electric current.

3. __________ are provided to reduce distortion and to provide a suitable signal shape for the filter.
a) Detector
b) Equalizer
c) Filters
d) Amplifier
Answer: b
Explanation: Optical detectors are linear devices. They do not introduce distortion themselves but other components may exhibit nonlinear behaviour. To compensate for distortion, an equalizer is provided in the receiver circuit.

4. A _________ maximizes the received signal-to-noise ratio in the receiver circuitry.
a) Filter
b) Equalizer
c) Detector
d) Reflector
Answer: a
Explanation: A filter reduces the noise bandwidth as well as inbounds noise levels. A filter maximizes the received signal-to-noise ratio while preserving the essential features of the signal. It also reduces ISI.

5. ________ can be operated in three connections.
a) Reflectors
b) Diodes
c) LED’s
d) FET’s
Answer: d
Explanation: FET’s or bipolar transistors are operated in three useful connections. These are the common emitter, the common base or gate, and the emitter or source follower.

6. How many structures of pre-amplifiers exist?
a) Two
b) Three
c) Four
d) One
Answer: b
Explanation: The basic structures of pre-amplifiers are observed in three forms. These are low-impedance, high-impedance and trans-impedance front end preamplifier structures.

7. What is the main factor contributing to the choice of the operational amplifier?
a) Gain
b) Impedance
c) Conductance
d) Gain-Bandwidth product
Answer: d
Explanation: A TTL interface stage is always used with the operational amplifier. A device that requires higher accuracy often tends to depend on gain-bandwidth product.
The choice of amplifier for receiver accuracy is dependant on gain-bandwidth product.

8. The multiplication factor for the APD varies with the device temperature.
a) True
b) False
Answer: a
Explanation: Optimum multiplication factor is required for smooth voltage variance. The multiplication factor for APD varies with the device temperature thus making provision of fine control for bias voltage.

9. How many categories of dynamic gain equalizers are available?
a) One
b) Two
c) Three
d) Four
Answer: b
Explanation: Dynamic gain equalizers are categorized into two types. These are single-channel and multichannel equalizers, thus providing operation using single or multiple wavelengths.

10. How many simultaneous channels can be provided in a band DGE(Dynamic gain equalizer)?
a) Six
b) Two
c) Eight
d) Ten
Answer: c
Explanation: Generally, eight channels are provided simultaneously in a band DGE. These are for the attenuation purpose of channels along with gain equalization.

.

Optical Communications Multiple Choice Questions on “Semiconductor Photodiodes With Internal Gain”.

1. ___________ has more sophisticated structure than p-i-n photodiode.
a) Avalanche photodiode
b) p-n junction diode
c) Zener diode
d) Varactor diode
Answer: a
Explanation: Avalanche photodiode is second major type of detector in optical communications. This diode is more sophisticated so as to create a much higher electric field region.

2. The phenomenon leading to avalanche breakdown in reverse-biased diodes is known as _______
a) Auger recombination
b) Mode hopping
c) Impact ionization
d) Extract ionization
Answer: c
Explanation: In depletion region, almost all photons are absorbed and carrier pairs are generated. So there comes a high field region where carriers acquire energy to excite new carrier pairs. This is impact ionization.

3. _______ is fully depleted by employing electric fields.
a) Avalanche photodiode
b) P-I-N diode
c) Varactor diode
d) P-n diode
Answer: a
Explanation: APD is fully depleted by electric fields more than 10⁴V/m. This causes all the drifting of carriers at saturated limited velocities.

4. At low gain, the transit time and RC effects ________
a) Are negligible
b) Are very less
c) Dominate
d) Reduce gradually
Answer: c
Explanation: Low gain causes the dominance of transit time and RC effects. This gives a definitive response time and thus device obtains constant bandwidth.

5. At high gain, avalanche buildup time ________
a) Is negligible
b) Very less
c) Increases gradually
d) Dominates
Answer: d
Explanation: High gain causes avalanche buildup time to dominate. Thus the bandwidth of device decreases as increase in gain.

6. Often __________ pulse shape is obtained from APD.
a) Negligible
b) Distorted
c) Asymmetric
d) Symmetric
Answer: c
Explanation: Asymmetric pulse shape is acquired from APD. This is due to relatively fast rise time as electrons are collected and fall time dictated by transit time of holes.

7. Fall times of 1 ns or more are common.
a) False
b) True
Answer: b
Explanation: The use of suitable materials and structures give rise times between 150 and 200 ps. Thus fall times of 1 ns or more are common which in turn limits the overall response of device.

8. Determine Responsivity of a silicon RAPD with 80% efficiency, 0.7μm wavelength.
a) 0.459
b) 0.7
c) 0.312
d) 0.42
Answer: a
Explanation: The Responsivity of a RAPD is given by-
R = ηeλ/hc A/w where, η=efficiency, λ = wavelength, h = Planck’s constant.

9. Compute wavelength of RAPD with 70% efficiency and Responsivity of 0.689 A/w.
a) 6μm
b) 7.21μm
c) 0.112μm
d) 3μm
Answer: c
Explanation: The wavelength can be found from the Responsivity formula given by-
R = ηeλ/hc. The unit of wavelength isμm.

10. Compute photocurrent of RAPD having optical power of 0.7 μw and responsivity of 0.689 A/W.
a) 0.23 μA
b) 0.489 μA
c) 0.123 μA
d) 9 μA
Answer: b
Explanation: The photocurrent is given byI_P=P₀R. Here I_P = photocurrent, P₀=Power, R = responsivity.

11. Determine optical power of RAPD with photocurrent of 0.396 μAand responsivity of 0.49 A/w.
a) 0.91 μW
b) 0.32 μW
c) 0.312 μW
d) 0.80 μW
Answer: d
Explanation: The photocurrent is given by I_P = P₀R. Here I_P = photocurrent, P₀ = Power, R = responsivity.
P₀ = IP/R gives the optical power.

12. Determine the Responsivity of optical power of 0.4μW and photocurrent of 0.294 μA.
a) 0.735
b) 0.54
c) 0.56
d) 0.21
Answer: a
Explanation: The photocurrent is given by I_P = P₀R. Here I_P = photocurrent, P₀ = Power, R = responsivity.
R = I_P/P0 gives the responsivity.

13. Compute multiplication factor of RAPD with output current of 10 μAand photocurrent of 0.369μA.
a) 25.32
b) 27.100
c) 43
d) 22.2
Answer: b
Explanation: The multiplication factor of photodiode is given by-
M = I/I_P where I = output current, I_P = photocurrent.

14. Determine the output current of RAPD having multiplication factor of 39 and photocurrent of 0.469μA.
a) 17.21
b) 10.32
c) 12.21
d) 18.29
Answer: d
Explanation: The multiplication factor of photodiode is given by-
M = I/I_P where I = output current, I_P = photocurrent. I = M*I_P gives the output current inμA.

15. Compute the photocurrent of RAPD having multiplication factor of 36.7 and output current of 7μA.
a) 0.01 μA
b) 0.07 μA
c) 0.54 μA
d) 0.9 μA
Answer: a
Explanation: The multiplication factor of photodiode is given by-
M = I/I_P where I = output current, I_P = photocurrent. I_P = I/M Gives the output current inμA.

.

Optical Communications test on “Optical Emission From Semiconductors”.

1. A perfect semiconductor crystal containing no impurities or lattice defects is called as __________
a) Intrinsic semiconductor
b) Extrinsic semiconductor
c) Excitation
d) Valence electron
Answer: a
Explanation: An intrinsic semiconductor is usually un-doped. It is a pure semiconductor. The number of charge carriers is determined by the semiconductor material properties and not by the impurities.

2. The energy-level occupation for a semiconductor in thermal equilibrium is described by the __________
a) Boltzmann distribution function
b) Probability distribution function
c) Fermi-Dirac distribution function
d) Cumulative distribution function
Answer: c
Explanation: For a semiconductor in thermal equilibrium, the probability P(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-E_F/KT))
Where K = Boltzmann constant, T = absolute temperature, E_F = Fermi energy level.

3. What is done to create an extrinsic semiconductor?
a) Refractive index is decreased
b) Doping the material with impurities
c) Increase the band-gap of the material
d) Stimulated emission
Answer: b
Explanation: An intrinsic semiconductor is a pure semiconductor. An extrinsic semiconductor is obtained by doping the material with impurity atoms. These impurity atoms create either free electrons or holes. Thus, extrinsic semiconductor is a doped semiconductor.

4. The majority of the carriers in a p-type semiconductor are __________
a) Holes
b) Electrons
c) Photons
d) Neutrons
Answer: a
Explanation: The impurities can be either donor impurities or acceptor impurities. When acceptor impurities are added, the excited electrons are raised from the valence band to the acceptor impurity levels leaving positive charge carriers in the valence band. Thus, p-type semiconductor is formed in which majority of the carriers are positive i.e. holes.

5. _________________ is used when the optical emission results from the application of electric field.
a) Radiation
b) Efficiency
c) Electro-luminescence
d) Magnetron oscillator
Answer: c
Explanation: Electro-luminescence is encouraged by selecting an appropriate semiconductor material. Direct band-gap semiconductors are used for this purpose. In band-to-band recombination, the energy is released with the creation of photon. This emission of light is known as electroluminescence.

6. In the given equation, what does p stands for?

p = 2πhk

a) Permittivity
b) Probability
c) Holes
d) Crystal momentum
Answer: d
Explanation: The given equation is a relation of crystal momentum and wave vector. In the given equation, h is the Planck’s constant, k is the wave vector and p is the crystal momentum.

7. The recombination in indirect band-gap semiconductors is slow.
a) True
b) False
Answer: a
Explanation: In an indirect band-gap semiconductor, the maximum and minimum energies occur at different values of crystal momentum. However, three-particle recombination process is far less probable than the two-particle process exhibited by direct band-gap semiconductors. Hence, the recombination in an indirect band-gap semiconductor is relatively slow.

8. Calculate the radioactive minority carrier lifetime in gallium arsenide when the minority carriers are electrons injected into a p-type semiconductor region which has a hole concentration of 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-
ς_r = [B_rN]^-1 where Br = Recombination coefficient in cm³s^-1 and N = carrier concentration in n-region.

9. Which impurity is added to gallium phosphide to make it an efficient light emitter?
a) Silicon
b) Hydrogen
c) Nitrogen
d) Phosphorus
Answer: c
Explanation: An indirect band-gap semiconductor may be made into an electro-luminescent material by the addition of impurity centers which will convert it into a direct band-gap material. The introduction of nitrogen as an impurity into gallium phosphide makes it an effective emitter of light. Such conversion is only achieved in materials where the direct and indirect band-gaps have a small energy difference.

10. Population inversion is obtained at a p-n junction by __________
a) Heavy doping of p-type material
b) Heavy doping of n-type material
c) Light doping of p-type material
d) Heavy doping of both p-type and n-type material
Answer: d
Explanation: Population inversion at p-n junction is obtained by heavy doping of both p-type and n-type material. Heavy p-type doping with acceptor impurities causes a lowering of the Fermi-level between the filled and empty states into the valence band. Similarly n-type doping causes Fermi-level to enter the conduction band of the material.

11. A GaAs injection laser has a threshold current density of 2.5*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 I_th. It is given by-
I_th = J_th * area of the optical cavity
Where J_th = threshold current density
Area of the cavity = length and width.

12. A GaAs injection laser with an optical cavity has refractive index of 3.6. Calculate the reflectivity for normal incidence of the plane wave on the GaAs-air interface.
a) 0.61
b) 0.12
c) 0.32
d) 0.48
Answer: c
Explanation: The reflectivity for normal incidence of the plane wave on the GaAs-air interface is given by-
r = ((n-1)/(n+1))² where r=reflectivity and n=refractive index.

13. A homo-junction is an interface between two adjoining single-crystal semiconductors with different band-gap energies.
a) True
b) False
Answer: b
Explanation: The photo-emissive properties of a single p-n junction fabricated from a single-crystal semiconductor material are called as homo-junction. A hetero-junction is an interface between two single-crystal semiconductors with different band-gap energies. The devices which are fabricated with hetero-junctions are said to have hetero-structure.

14. How many types of hetero-junctions are available?
a) Two
b) One
c) Three
d) Four
Answer: a
Explanation: Hetero-junctions are classified into an isotype and an-isotype. The isotype hetero-junctions are also called as n-n or p-p junction. The an-isotype hetero-junctions are called as p-n junction with large band-gap energies.

15. The ______________ system is best developed and is used for fabricating both lasers and LEDs for the shorter wavelength region.
a) InP
b) GaSb
c) GaAs/GaSb
d) GaAs/Alga AS DH
Answer: d
Explanation: For DH device fabrication, materials such as GaAs, Alga AS are used. The band-gap in this material may be tailored to span the entire wavelength band by changing the AlGa composition. Thus, GaAs/ Alga As DH system is used for fabrication of lasers and LEDs for shorter wavelength region (0.8μm-0.9μm).

To practice all areas of Optical Communications for tests, .

Optical Communications Interview Questions and Answers for freshers on “Overall Fiber Dispersion & Modified Single Mode Fibers”.

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

2. A multimode fiber has RMS pulse broadening per km of 12ns/km and 28ns/km due to material dispersion and intermodal dispersion resp. Find the total RMS pulse broadening.
a) 30.46ns/km
b) 31.23ns/km
c) 28.12ns/km
d) 26.10ns/km
Answer: a
Explanation: The overall dispersion in multimode fibers comprises both chromatic and intermodal terms. The total RMS pulse broadening σ_T is given by

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