Category: Optical Communications Objective Questions

Optical Communications Multiple Choice Questions on “Optical Bistability, Digital Optics and Optical Computation”.

1. ___________ provides a series of optical processing functions.
a) Wavelength convertors
b) Wavelength amplifiers
c) Detectors
d) Bi-stable optical devices
Answer: d
Explanation: Optical bi-stable devices include optical logic and memory elements, A-D convertors. Their response to light is nonlinear giving the basis of optical communication.

2. ___________ comprise of Fabry-Perot cavity.
a) Wavelength convertors
b) Wavelength amplifiers
c) Bi-stable optical devices
d) Detectors
Answer: c
Explanation: Fabry-Perot cavity consists of a material in which there are variations in refractive index with optical intensity. These variations are nonlinear giving rise to bistability.

3. The optical path length in nonlinear medium is integer number of ______ wavelength.
a) Half
b) Double
c) Three-fourth
d) Single
Answer: a
Explanation: Fabry-Perot cavity exhibits a sharp resonance to optical power passing into and through it. This is achieved when optical path length is integer number of half wavelength in nonlinear medium.

4. As compared to laser, the value of _________ in the cavity controls the optical transmission.
a) Amplification
b) Refractive index
c) Rectification
d) Reflection
Answer: b
Explanation: The refractive index value in the Fabry-Perot cavity controls the optical transmission. This provides high optical output on resonance and low optical output off resonance.

5. ___________ are able to latch between two distinct optical states.
a) Wavelength converters
b) Wavelength amplifiers
c) Detectors
d) Bistable optical devices
Answer: d
Explanation: The transfer characteristic for Bistable optical devices exhibit two state hysteresis resulting from turning in and out of resonance. So they can be latched between two states responding to external signal acting as flip-flop.

6. __________ can act as AND, OR, NOT gate.
a) Wavelength converters
b) Wavelength amplifiers
c) Detectors
d) Bistable optical devices
Answer: d
Explanation: BOD’s exhibit 2-state hysteresis. Thus they are able to latch between two operating states (0 and 1) thereby providing logic functions.

7. _______ proves superior to _______
a) BOD’s, electronic devices
b) Electronic devices, BOD’s
c) BOD’s, convertors
d) Convertors, BOD’s
Answer: a
Explanation: There is also a thing of picosecond switching using only Pico-joules of energy. A BOD comprises of these switching properties. Thus, it proves superior to electronic devices.

8. ________ BOD’s provides optical feedback.
a) Extrinsic
b) Intrinsic
c) Detector
d) Bistable
Answer: b
Explanation: All optical or intrinsic devices which utilize a nonlinear optical medium between a pair of partially reflecting mirrors forming a nonlinear etalon in which feedback is provided optical.

9. ___________ devices employ artificial nonlinearity.
a) Extrinsic
b) Intrinsic
c) Hybrid
d) Bistable
Answer: c
Explanation: Hybrid devices have artificial nonlinearity in an electro-optic medium in the cavity. This produces variations in refractive index through electro-optic effect.

10. Hybrid devices have limited ________ speed.
a) Switching
b) Planar
c) Curvature
d) Electrical
Answer: a
Explanation: Hybrid BOD’s provides flexibility. But at the same time their switching speeds are limited by use of electrical feedback. These devices are interconnected to provide a more complex logic circuit.

11. _______ exhibit optical bistability.
a) Extrinsic lasers
b) Intrinsic lasers
c) Detectors
d) Semiconductor lasers
Answer: d
Explanation: Semiconductor lasers have optical bistability. This is due to nonlinearities in absorption, gain, dispersion, wave guiding and the selection of output polarization.

12. ___________ is fabricated with tandem electrode.
a) Full convertor
b) Semiconductor
c) Detector diode
d) Bistable laser diode
Answer: d
Explanation: Bistable laser diode is fabricated with tandem electrode. The tandem electrode provides two gain sections. Also it has a loss region between them.

13. Optical pulsing can be obtained using _________
a) BODs
b) WDM
c) Detector
d) Semiconductor
Answer: a
Explanation: BODs with a very narrow bi-stable loop can provide optical pulsing. This type of device can be used to shape, clean up and amplify a noisy input pulse.

14. A weak second beam is introduced in _________
a) BOD differential amplifier
b) WDM
c) Detector
d) Semiconductor laser
Answer: a
Explanation: A weak second beam in BOD differential amplifier is introduced into the nonlinear optical cavity. This is used to control the resonance and transmission of the main beam through effects of its own stored energy.

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Optical Communications Mcqs on “Quantum Efficiency , Responsivity and Long – Wavelength Cut-Off”.

1. The fraction of incident photons generated by photodiode of electrons generated collected at detector is known as ___________________
a) Quantum efficiency
b) Absorption coefficient
c) Responsivity
d) Anger recombination
Answer: a
Explanation: Efficiency of a particular device is obtained by ratio of input given to that of output obtained. Thus, similarly, in photodiode, input i.e. incident photon and output generated electrons and their ratio is quantum efficiency.

2. In photo detectors, energy of incident photons must be ________________ band gap energy.
a) Lesser than
b) Greater than
c) Same as
d) Negligible
Answer: b
Explanation: While considering intrinsic absorption process, the energy of incident photon must be greater than band gap energy of material fabricating photo detector.

3. GaAs has band gap energy of 1.93 eV at 300 K. Determine wavelength above which material will cease to operate.
a) 2.431*10^-5
b) 6.424*10^-7
c) 6.023*10³
d) 7.234*10^-7
Answer: b
Explanation: The long wavelength cutoff is given by
λ_c = hc/Eg = 6.6268*10^-34*2.998*10⁸/1.93*1.602*10^-19
= 6.424*10^-7μm.

4. The long cutoff wavelength of GaAs is 0.923 μm. Determine bandgap energy.
a) 1.478*10^-7
b) 4.265*10^-14
c) 2.784*10^-9
d) 2.152*10^-19
Answer: d
Explanation: Long wavelength cutoff of photo detector is given by
λ_c = hc/Eg
Eg = hc/λ_c = 6.6268*10^-34*2.998*10⁸/0.923*10^-6
= 2.152*10^-19eV.

5. Quantum efficiency is a function of photon wavelength.
a) True
b) False
Answer: a
Explanation: Quantum efficiency is less than unity as all of incident photons are not absorbed to create electrons holes pairs. For example quantum efficiency of 60% is equivalent to 60% of electrons collected per 100 photons. Thus efficiency is a function of photon wavelength and must be determined at a particular wavelength.

6. Determine quantum efficiency if incident photons on photodiodes is 4*10¹¹ and electrons collected at terminals is 1.5*10¹¹?
a) 50%
b) 37.5%
c) 25%
d) 30%
Answer: b
Explanation: Quantum efficiency is given by
Quantum Efficiency = No. of electrons collected/No. of incident photons
= 1.5*10¹¹/4*10¹¹
= 0.375 * 100
= 37.5%.

7. A photodiode has quantum efficiency of 45% and incident photons are 3*10¹¹. Determine electrons collected at terminals of device.
a) 2.456*10⁹
b) 1.35*10¹¹
c) 5.245*10^-7
d) 4.21*10^-3
Answer: b
Explanation: Quantum efficiency is given by
Quantum efficiency = No. of electrons collected/No. of incident photons
Electrons collected = Quantum efficiency * number of incident photons
= 45/100 * 3*10¹¹
= 1.35*10¹¹.

8. The quantum efficiency of photodiode is 40% with wavelength of 0.90*10^-6. Determine the responsivity of photodiodes.
a) 0.20
b) 0.52
c) 0.29
d) 0.55
Answer: c
Explanation: Responsivity of photodiodes is given by
R = ηe λ/hc
= 0.4*1.602*10^-19 * 0.90*10^-6/6.626*10^-34 * 3*10⁸
= 0.29 AW^-1.

9. The Responsivity of photodiode is 0.294 AW^-1at wavelength of 0.90 μm. Determine quantum efficiency.
a) 0.405
b) 0.914
c) 0.654
d) 0.249
Answer: a
Explanation: Responsivity of photodiode is
R = ηe λ/hc
η = RXhc/eλ
= 0.294*6.626*10^-34*3*10⁸/ 1.602*10^-19*0.90*10⁸
= 0.405 AW^-1.

10. Determine wavelength of photodiode having quantum efficiency of 40% and Responsivity of 0.304 AW^-1.
a) 0.87 μm
b) 0.91 μm
c) 0.88 μm
d) 0.94 μm
Answer: d
Explanation: The Responsivity of photodiode is
R = ηe λ/hc
λ = Rhc/ηe
= 0.304*6.626*10^-34*3*10⁸/0.4*1.602*10^-19
= 0.94 μm.

11. Determine wavelength at which photodiode is operating if energy of photons is 1.9*10^-19J?
a) 2.33
b) 1.48
c) 1.04
d) 3.91
Answer: c
Explanation: To determine wavelength,
λ = hc/t
= 6.626*10^-34*3*10⁸/1.9*10^-19
= 1.04 μm.

12. Determine the energy of photons incident on a photodiode if it operates at a wavelength of 1.36 μm.
a) 1.22*10^-34J
b) 1.46*10^-19J
c) 6.45*10^-34J
d) 3.12*10⁹J
Answer: b
Explanation: The wavelength of photodiode is given by
λ = hc/t
E = hc/λ
= 6.626*10^-34*3*10⁸/1.36*10^-6
= 1.46*10^-19J.

13. Determine Responsivity of photodiode having o/p power of 3.55 μm and photo current of 2.9 μm.
a) 0.451
b) 0.367
c) 0.982
d) 0.816
Answer: d
Explanation: The Responsivity of photodiode is
R = Ip/Po
= 2.9*10^-6/3.55*10^-6
= 0.816 A/W.

14. Determine incident optical power on a photodiode if it has photocurrent of 2.1 μA and responsivity of 0.55 A/W.
a) 4.15
b) 1.75
c) 3.81
d) 8.47
Answer: c
Explanation: The Responsivity of photodiode is
R = Ip/Po
Po = Ip/R
= 2.1*10^-6/0.55
= 3.81 μm.

15. If a photodiode requires incident optical power of 0.70 A/W. Determine photocurrent.
a) 1.482
b) 2.457
c) 4.124
d) 3.199
Answer: b
Explanation: The Responsivity of photodiode is given by
R = Ip/Po
Ip = R*Po
= 0.70*3.51*10^-6
= 2.457μm.

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Optical Communications Multiple Choice Questions on “Optical Isolators and Circulators”.

1. An FBG is developed within a fiber core having a refractive index of 1.30. Find the grating period for it to reflect an optical signal with a wavelength of 1.33μm.
a) 0.51 μm
b) 0.58 μm
c) 0.61 μm
d) 0.49 μm
Answer: a
Explanation: The grating period is denoted by Λ. It is given by-
Λ = λ_B/ 2n
Where λ_B = wavelength
n = refractive index.

2. It is a passive device which allows the flow of optical signal power in only one direction and preventing reflections in the backward direction.
a) Fiber slice
b) Optical fiber connector
c) Optical isolator
d) Optical coupler
Answer: c
Explanation: Ideally, an optical isolator transmits the signal power in the desired forward direction. Material imperfections in the isolator medium generate backward reflections. Optical isolators can be implemented by using FBG.

3. Which feature of an optical isolator makes it attractive to use with optical amplifier?
a) Low loss
b) Wavelength blocking
c) Low refractive index
d) Attenuation
Answer: b
Explanation: Optical isolators are made using FBGs. Since FBGs are wavelength dependent, the optical isolators can be designed to allow or block the optical signal at particular wavelength. The wavelength blocking feature makes the optical isolator a very attractive device for use with optical amplifier in order to protect them from backward reflections.

4. Magneto-optic devices can be used to function as isolators.
a) True
b) False
Answer: a
Explanation: Magneto-optic devices use the principle of Faraday rotation. It relates the TM mode characteristics and polarization state of an optical signal with its direction of propagation. The rotation of polarization plane is proportional to the intensity of component of magnetic field in the direction of optical signal. Therefore, it is possible to block and divert an optical signal using magnetic properties which is a function of an isolator.

5. How many implementation methods are available for optical isolators?
a) One
b) Four
c) Two
d) Three
Answer: d
Explanation: Optical isolators can be implemented using three techniques. These are as follows:
-By using FBGs
-By using magnetic oxide materials
-By using semiconductor optical amplifiers (SOAs).

6. A device which is made of isolators and follows a closed loop path is called as a ____________
a) Circulator
b) Gyrator
c) Attenuator
d) Connector
Answer: a
Explanation: Isolator can be connected together to form multiport devices. A circulator is formed from isolators connected together to form a closed circular path. In circulator, the signal continues to travel in closed loop and does not get discarded unlike isolator.

7. The commercially available circulators exhibit insertion losses around ________________
a) 2 dB
b) 0.7 dB
c) 0.2 dB
d) 1 dB
Answer: d
Explanation: A number of isolators can be used to implement a circulator. However, as the number of ports increases, the device complexity increases. Hence, three-or four-port circulators are used for optical interconnection with insertion losses around 1 dB and high isolation in the range of 40-50dB.

8. A combination of a FBG and optical isolators can be used to produce non-blocking optical wavelength division add/draw multiplexers.
a) True
b) False
Answer: b
Explanation: Optical wavelength divisions add/draw multiplexers can be produced by a combination of a FBG and a circulator. Non-blocking NXM optical wavelengths divisions add/draw multiplexer is produced where N and M denotes the number of wavelength channels and add/drop channels.

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Optical Communications Multiple Choice Questions on “Dispersion – Chromatic Dispersion “.

1. What is dispersion in optical fiber communication?
a) Compression of light pulses
b) Broadening of transmitted light pulses along the channel
c) Overlapping of light pulses on compression
d) Absorption of light pulses
Answer: b
Explanation: Dispersion of transmitted optical signal causes distortion of analog as well as digital transmission. When the optical signal travels along the channel, the dispersion mechanism causes broadening of light pulses and thus in turn overlaps with their neighboring pulses.

2. What does ISI stand for in optical fiber communication?
a) Invisible size interference
b) Infrared size interference
c) Inter-symbol interference
d) Inter-shape interference
Answer: c
Explanation: Dispersion causes the light pulses to broaden and overlap with other light pulses. This overlapping creates an interference which is termed as inter-symbol interference.

3. For no overlapping of light pulses down on an optical fiber link, the digital bit rate BT must be ___________
a) Less than the reciprocal of broadened pulse duration
b) More than the reciprocal of broadened pulse duration
c) Same as that of than the reciprocal of broadened pulse duration
d) Negligible
Answer: a
Explanation: The digital bit rate and pulse duration are always inversely proportional to each other.
B_T < = (frac{1}{2}) Γ
Where B_T = bit rate
2Γ = duration of pulse.

4. The maximum bit rate that may be obtained on an optical fiber link is 1/3Γ.
a) True
b) False
Answer: b
Explanation: The digital bit rate is function of signal attenuation on a link and signal to noise ratio. For the restriction of interference, the bit rate should be always equal to or less than 1/2Γ.

5. 3dB optical bandwidth is always ___________ the 3dB electrical bandwidth.
a) Smaller than
b) Larger than
c) Negligible than
d) Equal to
Answer: b
Explanation: Optical bandwidth is half of the maximum data rate. For non-return:0 (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.

6. A multimode graded index fiber exhibits a total pulse broadening of 0.15μsover a distance of 16 km. Estimate the maximum possible bandwidth, assuming no intersymbol interference.
a) 4.6 MHz
b) 3.9 MHz
c) 3.3 MHz
d) 4.2 MHz
Answer: c
Explanation: The maximum possible bandwidth is equivalent to the maximum possible bitrate. The maximum bit rate assuming no inter-symbol interference is given by
B_T = (frac{1}{2}) Γ
Where B_T = bandwidth.

7. What is pulse dispersion per unit length if for a graded index fiber, 0.1μs pulse broadening is seen over a distance of 13 km?
a) 6.12ns/km
b) 7.69ns/km
c) 10.29ns/km
d) 8.23ns/km
Answer: b
Explanation: The dispersion mechanism causes broadening of light pulses. The pulse dispersion per unit length is obtained by dividing total dispersion of total length of fiber.
Dispersion = 0.1*10^-6/13 = 7.69 ns/km.

8. Chromatic dispersion is also called as intermodal dispersion.
a) True
b) False
Answer: b
Explanation: Intermodal delay is a result of each mode having a different group velocity at a single frequency. The intermodal delay helps us to know about the information carrying capacity of the fiber.

9. Chromatic dispersion is also called as intermodal dispersion.
a) True
b) False
Answer: b
Explanation: Intermodal delay, the name only suggests, includes many modes. On the other hand chromatic dispersion is pulse spreading that takes place within a single mode. Chromatic dispersion is also called as intermodal dispersion.

10. The optical source used in a fiber is an injection laser with a relative spectral width σ_λ/λ of 0.0011 at a wavelength of 0.70μm. Estimate the RMS spectral width.
a) 1.2 nm
b) 1.3 nm
c) 0.77 nm
d) 0.98 nm
Answer: c
Explanation: The relative spectral width σ_λ/λ= 0.01 is given. The rms spectral width can be calculated as follows:
σ_λ/λ = 0.0011
σ_λ = 0.0011λ
= 0.0011*0.70*10^-6
= 0.77 nm.

11. In waveguide dispersion, refractive index is independent of ______________
a) Bit rate
b) Index difference
c) Velocity of medium
d) Wavelength
Answer: d
Explanation: In material dispersion, refractive index is a function of optical wavelength. It varies as a function of wavelength. In wavelength dispersion, group delay is expressed in terms of normalized propagation constant instead of wavelength.

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Optical Communications Multiple Choice Questions on “Fiber Dispersion Measurements”.

1. ___________ measurements give an indication of the distortion to the optical signals as they propagate down optical fibers.
a) Attenuation
b) Dispersion
c) Encapsulation
d) Frequency
Answer: b
Explanation: Dispersion measurements provide the exact parameters to truly determine the quality and degradation to the optical signals. It gives an indication of the distortion to the optical signals as they propagate down the optical fibers.

2. The measurement of dispersion allows the _________ of the fiber to be determined.
a) Capacity
b) Frequency
c) Bandwidth
d) Power
Answer: c
Explanation: Dispersion measurements give an indication of distortion, which in turn determines the information carrying capacity of the fiber. This information carrying capacity of the fiber is purely dependent on the bandwidth of the fiber.

3. How many types of mechanisms are present which produce dispersion in optical fibers?
a) Three
b) Two
c) One
d) Four
Answer: a
Explanation: There are three major mechanisms which produce dispersion in optical fibers. These are: Material dispersion, waveguide dispersion and intermodal dispersion.

4. Intermodal dispersion is nonexistent in ________ fibers.
a) Multimode
b) Single mode
c) Step index- multimode
d) Al-GU
Answer: b
Explanation: Intra-modal as the name suggests need multimode fibers to propagate. In single mode fibers, only one mode is there to propagate. Hence, Intermodal dispersion is nonexistent in single mode fibers.

5. In the single mode fibers, the dominant dispersion mechanism is ____________
a) Intermodal dispersion
b) Frequency distribution
c) Material dispersion
d) Intra-modal dispersion
Answer: d
Explanation: In single mode case, the dominant dispersion mechanism is chromatic. Chromatic dispersion is called as intra-modal dispersion.

6. Devices such as ___________ are used to simulate the steady-state mode distribution.
a) Gyrators
b) Circulators
c) Mode scramblers
d) Attenuators
Answer: c
Explanation: The dispersion measurements on the fiber are performed only when the equilibrium mode distribution is set up within the fiber. Hence, filters or scramblers are used to simulate the steady state mode distribution.

7. How many domains support the measurements of fiber dispersion?
a) One
b) Three
c) Four
d) Two
Answer: d
Explanation: Fiber dispersion measurements can be made in two domains. These are time domain and frequency domain.

8. The time domain dispersion measurement setup involves _____________ as the photo detector.
a) Avalanche photodiode
b) Oscilloscope
c) Circulator
d) Gyrator
Answer: a
Explanation: The time domain fiber dispersion measurement involves the pulses to be received by the photo detector in order to determine the distortion in the optical signals. These pulses are received by avalanche photodiode.

9. In pulse dispersion measurements, the 3dB pulse broadening for the fiber is 10.5 ns/km and the length of the fiber is 1.2 km. Calculate the optical bandwidth for the fiber.
a) 32 MHz km
b) 45 MHz km
c) 41.9 MHz km
d) 10 MHz km
Answer: c
Explanation: The optical bandwidth for the fiber is given by –
B_opt = 0.44/3dB pulse broadening
Where, 0.44 = constant.

10. Frequency domain measurement is the preferred method for acquiring the bandwidth of multimode optical fibers.
a) True
b) False
Answer: a
Explanation: Bandwidth is usually the difference in the frequency. Frequency domain measurement is usually the best method in order to find the bandwidth of the multimode optical fibers.

11. Intra-modal dispersion tends to be dominant in multimode fibers.
a) True
b) False
Answer: b
Explanation: Intra-modal dispersion is dominant in case of single mode fibers. In case of multimode fibers, intermodal dispersion comes handy and is dominant.

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Optical Communications Question Bank on “Optoelectronic Integration and Photonic Integrated Circuits”.

1. Monolithic integration for optical sources are confined to the use of __________ semiconductors.
a) Ⅲ-Ⅴ
b) Ⅱ-Ⅲ
c) Ⅰ-Ⅱ
d) Ⅶ-Ⅷ
Answer: a
Explanation: Ⅲ-Ⅴsemiconductor compounds are much useful. They possess both optical and electronic properties. These properties can be exploited to produce high performance devices.

2. Circuits fabricated from GaAs or AlGaAs operate in wavelength region of __________
a) 0.1 and 0.2 μm
b) 0.8 and 0.9 μm
c) 0.4 and 0.6 μm
d) 0.6 and 0.7 μm
Answer: b
Explanation: Circuits fabricated from GaAs use injection laser which is fabricated on GaAs with a MESFET. This is used to bias and modulate the laser.

3. The OEICs realization __________ as compared to the other developments in IO.
a) Scripted
b) Decreased
c) Lagged behind
d) Increased
Answer: c
Explanation: IO devices use dielectric materials such as lithium niobate. This lagging behind is caused by inherent difficulties in fabrication of OEICs even if Ⅲ-Ⅴ semiconductors are used.

4. Compositional and structural differences between photonic and electronic devices __________
a) Provide high efficiency
b) Provide low efficiency
c) Highly used
d) Create problems
Answer: d
Explanation: Compositional and structural differences cause epitaxial crystal growth, planarization for lithography, electrical interconnections. They also cause thermal and chemical stability of materials, electric matching and heat dissipation.

5. To avoid large chip __________ devices are used.
a) InGaAsP
b) InGa
c) GaAs
d) InGaAs
Answer: a
Explanation: To avoid large chip, InGaAsP devices are used with directly modulated semiconductor lasers. This gives good dynamic characteristics at 40 Gbit/s at 1.55 μmwavelength.

6. Devices operating at transmission rates greater than 40 Gb/s are _________
a) GaAs and InP
b) GaAs
c) InGa
d) InGaAs
Answer: a
Explanation: Optoelectronic integrated circuits are based on heterojunction bipolar transistor and electron mobility transistor use GaAs and InP. These are capable of operating at transmission rates higher than 40 Gb/s.

7. HEMT based __________ have a spot-size convertor with a photodiode.
a) p-n junction diode
b) p-i-n photoreceiver
c) IGBT
d) BJT
Answer: b
Explanation: P-I-N photoreceiver comprises of spot-size convertor with a photodiode. Spot-size convertor increases fiber alignment tolerances by one order of magnitude. This enables use of cleaved instead of lensed fiber.

8. P-I-N photoreceiver based on HEMT is integrated with _________ guiding layers.
a) GaAs and InP
b) GaAs
c) InGa
d) InGaAsP
Answer: d
Explanation: P-I-N photoreceiver is integrated with InGaAsP guiding layers. In this HEMT based technology, InGaAsP provides more confinement.

9. An optical power splitter integrated with optical waveguide amplifier is more useful.
a) True
b) False
Answer: a
Explanation: The aim of optical waveguide amplifier is to reduce the number of amplifiers in system. Alongwith, it also reaches maximum number of nodes.

10. The use of intelligent optical switches is necessary.
a) False
b) True
Answer: b
Explanation: Most applications of OEICs in optical networks require large switching capacity to support a large number of WDM channels. This also provides control of both optical signal wavelength and signal power.

11. The wafer scale replication technology uses ____________
a) SOL gel
b) GaAs
c) InGa
d) InGaAsP
Answer: a
Explanation: Replication technology employs hot embossing, molding and ultraviolet lithography. Ultraviolet curable SOL gel enables refractive and diffractive micro-optical elements to be replicated directly on glass substrates.

12. ___________ is useful for production of both planar micro-optical elements and stacked optical microsystems.
a) Wavelength amplifier
b) Wavelength convertor
c) Replication technology
d) Optical switching matrix
Answer: c
Explanation: SOL gel materials used in replication technology allows combination of replication with lithography. This leaves selected areas material-free for sawing and bunding.

13. Optical interconnection between optoelectronic device is achieved in _________
a) Wavelength amplifier
b) Wavelength convertor
c) Replication technology
d) Chip-to-chip interconnection
Answer: d
Explanation: The chip-to-chip interconnection of optical components have a vertical cavity surface-emitting laser. These are assembled in micro-trenches in which embedded electrodes are connected through passive junction of poliver waveguide on alignment pits.

14. Multilevel interconnections are incorporated in _______
a) PIC
b) AWG based coupler
c) Convertors
d) OEIC technologies
Answer: a
Explanation: PIC reduces the overall size of optical functions. This causes the interconnection of several modules growing on same substrate.

15. When there is M number of WDM channels present at N input ports, then the output port 1 produces a _________
a) CW signal
b) WDM signal
c) Amplified signal
d) Distorted signal
Answer: b
Explanation: The reconstituted spectrum of WDM signal at any output port consists of a different set of wavelength channels with at least one wavelength channel from each input port producing a WDM signal having wavelength signal from each of input ports.

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