Category: Optical Communications Objective Questions

Optical Communications Questions and Answers for Experienced people on “Preparation of Optical Fibers – Liquid Phase Techniques”.

1. What is a fundamental necessity in the fabrication of fibers for light transmission?
a) Same refractive index for both core and cladding
b) Pump source
c) Material composition of fiber
d) Variation of refractive index inside the optical fiber
Answer: d
Explanation: For fabrication of fibers, two different transparent materials to light over a wavelength range of 0.8 to 1.7μm are required. Fiber should exhibit low attenuation, absorption and scattering losses. The variation of refractive indices in a fiber is a necessity for fiber fabrication.

2. Which materials are unsuitable for the fabrication of graded index fiber?
a) Glass-like-materials
b) Mono-crystalline structures
c) Amorphous material
d) Silica based material
Answer: b
Explanation: In case of graded index fiber, it is essential that the refractive index of the material is varied by suitable doping with another compatible material. These two materials should have mutual solubility over a wide range of concentration. This is achieved only in glass-like-materials.

3. How many different categories are available for the methods of preparing optical glasses?
a) 1
b) 2
c) 3
d) 4
Answer: b
Explanation: The methods of preparing optical glasses are divided into two categories. One is the conventional glass refining technique and other is vapor-phase-deposition method.

4. What is the first stage in liquid-phase-technique?
a) Preparation of ultra-pure material powders
b) Melting of materials
c) Decomposition
d) Crystallization
Answer: a
Explanation: In liquid-phase-technique melting, the first stage includes the preparation of ultra-pure material powders. These are usually oxides or carbonates which decomposes during glass melting.

5. Which processes are involved in the purification stage in liquid-phase-technique?
a) Filtration, Co-precipitation, Re-crystallization
b) Decomposition, Filtration, Drying
c) Doping, Drying, Decomposition
d) Filtration, Drying, Doping
Answer: a
Explanation: The compounds such as oxides and carbonates are formed during the glass melting. The purification accounts for a large proportion of material cost. These compounds are commercially available. The purification involves filtration, co-precipitation, re-crystallization and drying.

6. At what temperature range, does the melting of multi components glass systems takes place?
a) 100-300 degree Celsius
b) 600-800 degree Celsius
c) 900-1300 degree Celsius
d) 1500-1800 degree Celsius
Answer: c
Explanation: The glass materials in the powdered form and have relatively low melting point. Thus, the glass materials are melted at relatively low temperatures in the range of 900-1300 degrees Celsius.

7. Fiber drawing using preform was useful for the production of graded index fibers.
a) True
b) False
Answer: b
Explanation: A technique for producing fine optical fiber waveguides is to make a preform using the rod in the tube process. This technique was useful for the production of step-index fibers with large core diameters. In this technique, achievement of low attenuation is not critical as there is a danger of including bubbles at the core-cladding interface.

8. The minute perturbations and impurities in the fiber drawing process using preform technique can result in very high losses of _____________
a) Between 500 and 1000 dB/km
b) Between 100 and 300 dB/km
c) Between 1200 and 1600 dB/km
d) More than 2000 dB/km
Answer: a
Explanation: The minute perturbations and impurities in the fiber include formations of bubbles and involvement of particulate matter. The losses due to such impurities can be between 500 and 1000 dB/km.

9. The liquid-phase melting technique is used for the production of fibers ___________
a) With a core diameter of 50μm
b) With a core diameter less than 100μm
c) With a core diameter more than 200μm
d) With a core diameter of 100μm
Answer: c
Explanation: The multicomponent glass fibers prepared continuously by liquid-phase melting technique have losses in the range of 5 and 20 dB/km at a wavelength of 0.85μm. This method is thus used for preparation of fibers with a large core diameter. Also this technique is used for the continuous production of fibers.

10. Graded index fibers produced by liquid-phase melting techniques are less dispersive than step-index fibers.
a) True
b) False
Answer: a
Explanation: Liquid-phase melting technique does not offer optimum parabolic profile fibers. This parabolic profile yields minimum pulse dispersion. Graded index fibers prepared using liquid-phase melting techniques are less dispersive but do not have the bandwidth-length products of optimum profile fibers.

To practice all areas of Optical Communications for Experienced people, .

Optical Communications Multiple Choice Questions on “Optical Network Concepts”.

1. Each stage of information transfer is required to follow the fundamentals of ____________
a) Optical interconnection
b) Optical hibernation
c) Optical networking
d) Optical regeneration
Answer: c
Explanation: Optical networking uses optical fiber as a transmission medium. It provides a connection between users to enable them to communicate with each other by transporting information from a source to a destination.

2. ____________ is a multi-functional element of optical network.
a) Hop
b) Optical node
c) Wavelength
d) Optical attenuation
Answer: b
Explanation: An optical node is a multi-functional element which acts as a transceiver unit capable of receiving, transmitting and processing the optical signal. The optical nodes are interconnected with optical fiber links.

3. A signal carried on a dedicated wavelength from source to destination node is known as a ___________
a) Light path
b) Light wave
c) Light node
d) Light source
Answer: a
Explanation: A light path is a dedicated path from a source to a destination. The data can be sent over the light paths as soon as connections are set up. A controlling mechanism is present to control the data flow.

4. The fundamentals of optical networking are divided into _______ areas.
a) Two
b) One
c) Four
d) Three
Answer: d
Explanation: The fundamentals divided into three areas contain mainly optical network terminology. The other two areas include functions and types of optical network node and switching elements and the wavelength division multiplexed optical networks.

5. The optical networking fundamentals are _____________ of the transmission techniques.
a) Dependent
b) Independent
c) Similar
d) Dissimilar
Answer: b
Explanation: The optical networking fundamentals include transfer of data. Irrespective of the difference in the transmission techniques, the fiber networking fundamentals remain the same.

6. The network structure formed due to the interconnectivity patterns is known as a ____________
a) Network
b) Struck
c) Topology
d) D-pattern
Answer: c
Explanation: A topology is a combination of patterns interconnected to each other. It provides connection patterns to users at different places. It embarks on the principle of multi-usability.

7. In the __________ topology, the data generally circulates bi-directionally.
a) Mesh
b) Bus
c) Star
d) Ring
Answer: b
Explanation: In a bus topology, data is input via four port couplers. The couplers couples and stations the data bi-directionally and are removed from the same ports.

8. The ring and star topologies are combined in a ________ configuration.
a) Mesh
b) Fringe
c) Data
d) Singular
Answer: a
Explanation: The mesh configuration is a combination of ring and star topologies. It is referred to as full-mesh when each network node is interconnected with all nodes in the network.

9. The full-mesh configuration is complex.
a) False
b) True
Answer: b
Explanation: The full-mesh topology is a combination of two or more topologies. It is often preferred for the provision of either a logical or virtual topology due to its high flexibility and interconnectivity features.

10. How many networking modes are available to establish a transmission path?
a) Three
b) One
c) Two
d) Four
Answer: c
Explanation: There are two networking modes often referred to the networking. These are connection-oriented and connectionless networking modes. These include an end-to-end and bidirectional communication environment between source and destination.

11. Packet switching is also called as ___________
a) Frame switching
b) Cell switching
c) Trans-switching
d) Buffer switching
Answer: b
Explanation: In packet or cell switching, messages are sent in small packets called cells. Cells from different sources are statistically multiplexed and are sent to the destinations.

12. ___________ mode is temporary, selective and continuous.
a) Cell switching
b) Buffer switching
c) Cache
d) Circuit switching
Answer: d
Explanation: An end-to-end connection is required for a circuit switching to take place. The transmissions are continuous and are in real time. Once the transmission is complete, the connection is ended.

13. A _______________ is a series of logical connections between the source and destination nodes.
a) Cell circuit
b) Attenuation circuit
c) Virtual circuit
d) Switched network
Answer: c
Explanation: A virtual circuit consists of different routes which provide connections between sending and receiving devices. These routes can change at any time and the incoming return route does not have to mirror the outgoing route.

14. ____________ refers to the process whereby a node finds one or more paths to possible destinations in a network.
a) Routing
b) Framing
c) Lightning
d) Cloning
Answer: a
Explanation: Routing refers to the path finding process in a network. In this, the control and data functions are performed to identify the route and to handle the data during the journey from source to destination.

15. How many stages are possessed by the control plane?
a) Two
b) Three
c) Four
d) Five
Answer: b
Explanation: The routing process called as control plane has three stages. These are neighbor discovery, topology discovery and path selection. These stages enable the network in routing mechanisms efficiently.

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Optical Communications Multiple Choice Questions on “Digital System Planning Considerations”.

1. Sampling rate for each speech channel on 32-channel PCM is 8 KHz each encoded into 8 bits. Determine number of bits in a frame.
a) 64
b) 128
c) 32
d) 256
Answer: d
Explanation: Number of bits in a frame can be calculated as follows:
Bits in a frame = No. of channels * Sampling rate for each channel.

2. Sampling rate for each speech channel on 32-channel PCM is 8 KHz each encoded into 8 bits. Determine the transmission rate for system with 256 bits in a frame.
a) 2.96 Mbits/s
b) 2.048 Mbits/s
c) 3.92 Mbits/s
d) 4 Mbits/s
Answer: b
Explanation: Transmission rate can be determined by-
Transmission rate = Sampling rate * No. of bits in a frame.

3. Sampling rate for each speech channel on 32-channel PCM is 8 KHz each encoded into 8 bits. Determine the bit duration with transmission rate of 2.048 M bits/s.
a) 388 ns
b) 490 ns
c) 488 ns
d) 540 ns
Answer: c
Explanation: Bit duration is the reciprocal of the transmission rate. Thus, it is given by-
Bit duration = 1/transmission rate.

4. The bit duration is 488 ns. Sampling rate for each channel on 32-channel PCM is 8 KHz encoded into 8 bits. Determine the time slot duration.
a) 3.2 μs
b) 3.1 μs
c) 7 μs
d) 3.9 μs
Answer: d
Explanation: Time slot duration is given by –
Time slot duration = Encoded bits * bit duration.

5. Sampling rate for each speech channel on 32-channel PCM is 8 KHz each encoded into 8 bits. Determine duration of frame with time slot duration of 3.9μs.
a) 125 μs
b) 130 μs
c) 132 μs
d) 133 μs
Answer: a
Explanation: Duration of a frame is determined by –
Duration of a frame = 32 * time slot duration.

6. Sampling rate for each speech channel on 32-channel PCM is 8 KHz each encoded into 8 bits. Determine the duration of multi-frame if duration of a frame is 125μs.
a) 2ms
b) 3ms
c) 4ms
d) 10ms
Answer: a
Explanation: Multi-frame duration can be determined by –
Multi-frame duration = 16 * Duration of a single frame.

7. Determine excess avalanche noise factor F(M) if APD has multiplication factor of 100, carrier ionization rate of 0.02.
a) 3.99
b) 3.95
c) 4.3
d) 4
Answer: b
Explanation: Excess avalanche noise factor is computed by –
F (M) = k*M + (2-1/M) (1-k), where k is ionization rate and M is the multiplication factor.

8. Compute average number of photons incident at receiver in APD if quantum efficiency is 80%, F (M) = 4, SNR = 144.
a) 866
b) 865
c) 864
d) 867
Answer: c
Explanation: Average number of photons arez_m=[2β_ςF(M)]*[S/N*η]
Here, η = quantum efficiency, S/N = signal to noise ratio.

9. Determine incident optical power if z_m=864, wavelength = 1μm.
a) -85 dBm
b) -80 dBm
c) -69.7 dBm
d) -60.7 dBm
Answer: d
Explanation: Incident optical power is P0=z_mh_cB_T/2λ. Here z_m=average number of photons, h_c=Planck’s constant.

10. Determine wavelength of incident optical power if z_m=864, incident optical power is -60.7 dB, B_T=1 * 10⁷.
a) 1 μs
b) 2 μs
c) 3 μs
d) 4 μs
Answer: a
Explanation: Wavelength is determined by λ=z_mh_cB_T/2P₀. Here z_m=average number of photons, h_c=Planck’s constant, P₀=incident optical power.

11. Determine total channel loss if connector loss at source and detector is 3.5 and 2.5 dB and attenuation of 5 dB/km.
a) 34 dB
b) 35 dB
c) 36 dB
d) 38 dB
Answer: a
Explanation: The total channel loss is C_L=(α_fc+α_j)L + α_cr. Here α_cr=loss at detector and source combined, α_fc = attenuation in dB/km.

12. Determine length of the fiber if attenuation is 5dB/km, splice loss is 2 dB/km, connector loss at source and detector is 3.5 and 2.5.
a) 5 km
b) 4 km
c) 3 km
d) 8 km
Answer: b
Explanation: Length of the fiber is L = C_L/(α_fc+α_j) – α_cr. Here α_cr = loss at detector and source combined, α_fc = attenuation in dB/km.

13. Determine total RMS pulse broadening over 8 km if RMS pulse broadening is 0.6ns/km.
a) 3.6 ns
b) 4 ns
c) 4.8 ns
d) 3 ns
Answer: c
Explanation: Total RMS pulse broadening is given by –
σ_T = σ*L Where σ = rms pulse broadening and L = length of the fiber.

14. Determine RMS pulse broadening over 8 km if total RMS pulse broadening is 5.8ns/km.
a) 0.2ns/km
b) 0.1ns/km
c) 0.4ns/km
d) 0.72ns/km
Answer: d
Explanation: RMS pulse broadening is given by –
σ = σ_T/L where σ = rms pulse broadening and L = length of the fiber.

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Optical Communications online test on “Phototransistors and Metal – Semiconductor – Metal Photodetectors”.

1. The _____________ is photosensitive to act as light gathering element.
a) Base-emitter junction
b) Base-collector junction
c) Collector-emitter junction
d) Base-collector junction and Base-emitter junction
Answer: a
Explanation: Base-collector junction is photosensitive in n-p-n phototransistor and act as light gathering element. This light absorbed affects the base current and gives multiplication of primary photocurrent in device.

2. A large secondary current _________________ in n-p-n InGaAs phototransistor is achieved.
a) Between base and collector
b) Between emitter and collector
c) Between base and emitter
d) Plasma
Answer: b
Explanation: The photo-generated holes are swept to the base. This increases the forward bias device. This generates secondary current between emitter and collector.

3. _______ emitter-base and collector-base junction capacitances is achieved by use of hetero-structure along with _________ base resistance.
a) Low, high
b) High, low
c) Low, low
d) High, negligible
Answer: c
Explanation: In hetero-structure, there is low doping level in emitter and collector which is coupled with heavy doping base. This is due low emitter-base and collector-base junction capacitance and low base resistance. This allows large current gain.

4. A ________ is created by hetero-junction at collector-base junction.
a) Potential barrier
b) Depletion region
c) Parasitic capacitance
d) Inductance
Answer: a
Explanation: Potential barrier is created at emitter-base junction by hetero-junction. This eliminates hole junction from base. This is achieved when junction is forward-biased and provides good emitter-base efficiency.

5. Phototransistors based on hetero-junction using _________ material are known as waveguide phototransistors.
a) InGaP
b) InGaAs
c) InGaAsP/ InAlAs
d) ErGaAs
Answer: c
Explanation: Phototransistor using InGaAsP/ InAlAs are known as waveguide phototransistors. They function as waveguide phototransistors. They function as high performance photo-detectors at 1.3 micro-meter wavelength. They utilize a passive waveguide layer under active transistor region.

6. A phototransistor has collector current of 18 mA, incident optical power of 128 μW with a wavelength of 1.24 μm. Determine an optical gain.
a) 1.407 *10²
b) 19.407 *10²
c) 2.407 *10²
d) 3.407 *10²
Answer: a
Explanation: The optical gain is given by-
G₀=hcI_c/λeP₀, where h=Planck’s constant, I_c=collector current, λ=wavelength, P₀=incident optical power.

7. For a phototransistor having gain of 116.5, wavelength of 1.28 μm, optical power 123μW. Determine collector current.
a) 0.123 mA
b) 0.0149 mA
c) 1.23 mA
d) 0.54 mA
Answer: b
Explanation: The collector current is given by-
I_c= G₀λeP₀/ hc, where h=Planck’s constant, I_c=collector current, λ=wavelength, P₀=incident optical power.

8. The detection mechanism in the ____________ photo-detector includes inter sub-band transitions.
a) Dwell
b) Set
c) Avalanche
d) Futile
Answer: a
Explanation: The inter sub-band transitions are also known as type-2 transitions. It comprises of mini-bands within a single energy band, The detection mechanism in DWELL photo-detector includes inter sub-band transitions.

9. Which of the following is the difference between the n-p-n and conventional bipolar transistor?
a) Electric property
b) Magnetic property
c) Unconnected base
d) Emitter base efficiency
Answer: c
Explanation: The n-p-n bipolar transistor differs in the following ways: base is unconnected, base-collector junction is photosensitive as a light gathering element.

10. The n-p-n hetero-junction phototransistor is grown using ______________
a) Liquid-phase tranquilizers
b) Liquid-phase epistaxis
c) Solid substrate
d) Hetero poleax
Answer: b
Explanation: The technique LPE consists of a thin layer of n-type collector based on a p-type base layer. Liquid phase epistaxis is used in hetero-junction technology.

11. The _____________ at emitter-base junction gives good emitter base injection efficiency.
a) Homo-junction
b) Depletion layer
c) Holes
d) Hetero-junction
Answer: d
Explanation: The hetero-junction at the emitter-base junction effectively eliminates hole injection from the base when the junction is forward biased. This gives good emitter-base injection efficiency.

12. Waveguide phototransistors utilize a ___________ waveguide layer under the _________ transistor region.
a) Active, passive
b) Passive, active
c) Homo, hetero
d) Hetero, homo
Answer: b
Explanation: Waveguide phototransistors are based on hetero-junction structure. They function as high-performance photo-detectors and thus utilize a passive waveguide layer under the active transistor region.

13. What is the main benefit of the waveguide structure over conventional hetero-junction phototransistor?
a) High depletion region
b) Depletion width
c) Increased photocurrent, responsivity
d) Low gain
Answer: c
Explanation: Waveguide structure offers increased photocurrent. Photocurrent is directly proportional to the responsivity; thus in turn increases responsivity.

14. Waveguide structure provides high quantum efficiency.
a) True
b) False
Answer: b
Explanation: Responsivity and quantum efficiency follow a different path. They are indirectly proportional to each other. Thus, in waveguide structure, as the responsivity increases, quantum efficiency remains low.

15. Metal-semiconductor-metal (MSM) photo-detectors are photoconductive detectors.
a) True
b) False
Answer: a
Explanation: MSM photo-detectors are the simplest of photo-detectors. It provides the simplest form of photo-detection within optical fiber communications and are photoconductive.

tests, .

Optical Communications Multiple Choice Questions on “Some Injection Laser Structures”.

1. In multimode injection lasers, the construction of current flow to the strip is obtained in structure by __________
a) Covering the strip with ceramic
b) Intrinsic doping
c) Implantation outside strip region with protons
d)Implantation outside strip region with electrons
Answer: c
Explanation: The current flow is realized by implanting the region outside strip with protons. This implantation makes the laser highly resistive and gives superior thermal properties due to absence of silicon dioxide layer.

2. What is the strip width of injection laser?
a) 12 μm
b) 11.5 μm
c) Less than 10 μm
d) 15 μm
Answer: c
Explanation: A strip width less than or equal to 10 μm is usually preferred in injection lasers. This width range provides the lasers highly efficient coupling into multimode fibers as comapred to single mode fibers.

3. Some refractive index variation is introduced into lateral structure of laser.
a) True
b) False
Answer: a
Explanation: Gain guided lasers possess several undesirable characteristics, nonlinearities in light output versus current characteristics, high threshold current, low differential quantum efficiency, movement of optical a;ong junction plane. This problems can be reduced by introducing refractive index variations into lateral structure of lasers so that optical mode is determined along the junction plane.

4. Buried hetero-junction (BH) device is a type of _____________ laser where the active volume is buried in a material of wider band-gap and lower refractive index.
a) Gas lasers.
b) Gain guided lasers.
c) Weak index guiding lasers.
d) Strong index guiding lasers.
Answer: d
Explanation: In strong index guiding lasers, a uniformly thick, planar active waveguide is achieved by lateral variations in confinement layer thickness or refractive index. In Buried hetero-junction (BH) devices, strong index guiding along junction plane introduces transverse mode control in injection lasers.

5. In Buried hetero-junction (BH) lasers, the optical field is confined within __________
a) Transverse direction
b) Lateral direction
c) Outside the strip
d) Both transverse and lateral direction
Answer: d
Explanation: Optical field is strongly confined in both transverse and lateral direction. This provides strong index guiding of optical mode along with good carrier confinement.

6. A double-channel planar buried hetero-structure (DCP BH) has a planar active region, the confinement material is?
a) Alga AS
b) InGaAsP
c) GaAs
d) SiO₂
Answer: b
Explanation: The planar active region made up of InGaAsP can be seen in double-channel planar buried hetero-structure (DCP BH). This material confinement provides a very high power operation with CW output power up to 40 mW in longer wavelength region.

7. Problems resulting from parasitic capacitances can be overcome __________
a) Through regrowth of semi-insulating material
b) By using oxide material
c) By using a planar InGaAsP active region
d) By using a AlGaAs active region
Answer: a
Explanation: The use of reverse-biased current confinement layers introduces parasitic capacitances which reduces high speed modulation of BH lasers. This problem can be reduced by regrowth of semi-insulating material or deposition of dielectric material. This causes increase in modulation speeds of 20 GHz.

8. Quantum well lasers are also known as __________
a) BH lasers
b) DH lasers
c) Chemical lasers
d) Gain-guided lasers
Answer: b
Explanation: DH lasers are known as Quantum well lasers. The carrier motion normal to active layer is restricted in these devices. This results in quantization of kinetic energy into discrete energy levels for carriers moving in that direction. This phenomenon is similar to quantum mechanical problem of one dimensional potential well which is seen in DH lasers.

9. Quantum well lasers are providing high inherent advantage over __________
a) Chemical lasers
b) Gas lasers
c) Conventional DH devices
d) BH device
Answer: c
Explanation: Quantum well lasers exhibit high incoherent advantage over conventional DH lasers. In Quantum well laser structures, the thin active layer results in drastic changes in electronic and optical properties over conventional DH laser structures. This changes are due to quantized nature of discrete energy levels with step-like density and also allow high gain and low carrier density.

10. Strip geometry of a device or laser is important.
a) True
b) False
Answer: a
Explanation: Near fluid intensity distribution corresponding to single optical output power level in plane of junction can be seen in GaAs or AlGaAs lasers. This distribution is in lateral direction and is determined by the nature of lateral waveguide. The single intensity maximum shows the fundamental lateral mode is dominant.

11. Better confinement of optical mode is obtained in __________
a) Multi Quantum well lasers
b) Single Quantum well lasers
c) Gain guided lasers
d) BH lasers
Answer: a
Explanation: As compared to all lasers including single quantum well lasers, multi-Quantum well lasers are having better confinement of optical mode. This results in a lower threshold current density for these devices.

12. Multi-quantum devices have superior characteristics over __________
a) BH lasers
b) DH lasers
c) Gain guided lasers
d) Single-quantum-well devices
Answer: b
Explanation: Lower threshold currents, narrower bandwidths, high modulation speeds, lower frequency chirps and less temperature dependence are parameters determining characteristics of a particular laser. All the above parameters make multi-quantum devices superior over DH lasers.

13. Dot-in-well device is also known as __________
a) DH lasers
b) BH lasers
c) QD lasers
d) Gain guided lasers
Answer: c
Explanation: Quantum well lasers are devices in which device contains a single discrete atomic structure or Quantum-dot. These are elements that contain electron tiny droplets which forms a quantum well structure.

14. A BH can have anything from a single electron to several electrons.
a) True
b) False
Answer: b
Explanation: Quantum-dot lasers are fabricated using semiconductor crystalline materials. They have a particular dimension ranging from nm to few microns. The size, shape of these structures and number of electrons they contain are precisely controlled.

15. QD lasers have a very low threshold current densities of range __________
a) 0.5 to 5 A cm^-2
b) 2 to 10 A cm^-2
c) 10 to 30 A cm^-2
d) 6 to 20 A cm^-2
Answer: d
Explanation: Low-threshold current density between 6 to 20 A cm^-2 is obtained with InAs/InGaAs QD lasers which emit at a wavelength of 1.3 μm and 1.5 μm Such low values of threshold current densities make these lasers possible to create stacked or cascaded QD structures. These structures provide high optical gain for short-cavity transmitters and vertical cavity surface-emitting lasers.

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Optical Communications Multiple Choice Questions on “Non-Linear Effects”.

1. The nonlinear effects in optical fibers are large.
a) True
b) False
Answer: b
Explanation: The nonlinear effect arises from the interactions between light waves and the material transmitting them and thus affects the optical signals. The nonlinear effects are usually small in optical fibers. They have power levels of up to few milliWatts.

2. How many categories of nonlinear effects are seen in optical fibers?
a) One
b) Two
c) Three
d) Four
Answer: b
Explanation: The nonlinear effects are separated on the basis of their characteristics. There are two such categories; one is scattering effect and the other is Kerr effect.

3. Which of the following is not related to Kerr effects?
a) Self-phase modulation
b) Cross-phase modulation
c) Four-wave mixing
d) Stimulated Raman Scattering
Answer: d
Explanation: Stimulated Raman Scattering is related to scattering. The other effects include modulation and mixing which are parts of Kerr effect.

4. Linear scattering effects are _______ in nature.
a) Elastic
b) Non-Elastic
c) Mechanical
d) Electrical
Answer: a
Explanation: Linear scattering effects are elastic because the scattered wave frequency is equal to incident wave frequency. Nonlinear scattering effects are purely inelastic.

5. Which thing is more dominant in making a fiber function as a bidirectional optical amplifier?
a) Core material
b) Pump source
c) Cladding material
d) Diameter of fiber
Answer: b
Explanation: Brillouin gain is always greater than Raman gain. It exists for light propagation in opposite direction to the pump source. Also Brillouin frequency shifts and gain bandwidth are much smaller than Raman. Raman amplification occurs for light propagating in either direction. Thus, pump source is more important in making a fiber function as bidirectional optical amplifier.

6. _________ semiconductor laser sources generally have broader bandwidths.
a) Injection
b) Pulsed
c) Solid-state
d) Silicon hybrid
Answer: b
Explanation: Pulsed semiconductor lasers have broader bandwidths. Therefore, these sources prove to be inefficient pump sources. They prove inefficient especially for narrow gain spectrum.

7. Nonlinear effects which are defined by the intensity – dependent refractive index of the fiber are called as ________
a) Scattering effects
b) Kerr effects
c) Raman effects
d) Tomlinson effects
Answer: b
Explanation: Kerr effects are nonlinear effects. Nonlinear effects are divided into scattering and Kerr effects. Scattering effects include scattering of phonon whereas Kerr effects include intensity refractive index parameters.

8. Self-phase modulation causes modifications to the pulse spectrum.
a) True
b) False
Answer: a
Explanation: Kerr effect results in different transmission phase for the peak of the pulse compared with leading and trailing edges. Self-phase modulation can broaden the frequency spectrum of the pulse as the time varying phase creates a time varying frequency.

9. Self-phase modulation can be used for _____________
a) Enhancing the core diameter
b) Wavelength shifting
c) Decreasing the attenuation
d) Reducing the losses in the fiber
Answer: b
Explanation: Self phase modulation is related to phase change. It imposes a positive frequency sweep on the pulse which in turn enables wavelength or frequency shifting.

10. The beating between light at different frequencies or wavelengths in multichannel fiber transmission causes ________
a) Attenuation
b) Amplitude modulation of channels
c) Phase modulation of channels
d) Loss in transmission
Answer: c
Explanation: Phase modulation is related to frequency and wavelength shifting. In multichannel fiber transmission, phase modulation causes generation of modulation sidebands at new frequencies. This phenomenon is called as four-wave mixing.

11. What is different in case of cross-phase modulation from self-phase modulation?
a) Overlapping but same pulses
b) Overlapping but distinguishable pulses
c) Non-overlapping and same pulses
d) Non-overlapping but distinguishable pulses
Answer: b
Explanation: In cross phase modulation, variation in intensity of one pulse width modulates the refractive index of the fiber which causes phase modulation of the overlapping phases. In self-phase modulation, this phase modulation broadens the pulse spectrum.

12. When three wave components co-propagate at angular frequency w1, w2, w3, then a new wave is generated at frequency w4, which is given by?
a) w4 = w1 – w2 – w3
b) w4 = w1 + w2 + w3
c) w4 = w1 + w2 – w3
d) w4 = w1 – w2 + w3
Answer: c
Explanation: This type of frequency mixing is called as four-wave mixing. This frequency combination is problematic for multichannel optical communication as they become phase matched if the channel wavelengths are near to zero dispersion wavelengths.

13. _____________ results from a case of nonlinear dispersion compensation in which the nonlinear dispersion compensation in which the nonlinear chirp caused by self-phase modulation balances, postpones, the temporal broadening induced by group velocity delay.
a) Four wave mixing
b) Phase modulation
c) Soliton propagation
d) Raman scattering
Answer: c
Explanation: Soliton propagation is a nonlinear dispersion phenomenon. It limits the propagation distance that can be achieved when acting independently. It balances broadening of light pulse.

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