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Communications MCQ Set 1

1. __________________ may be improved through the use of frequency-selective feedback so that the cavity loss is different for various longitudinal modes.
a) Frequency selectivity
b) Longitudinal mode selectivity
c) Electrical feedback
d) Dissipated power

View Answer

Answer: b [Reason:] Improved longitudinal mode selectivity can be achieved using structures which gives loss discrimination between the desired and all the unwanted modes. Thus, mode discrimination can be seen. To allow for stable mode operation, the use of frequency-selective feedback approach is undertaken.

2. Device which apply the frequency-selective feedback technique to provide single longitudinal operation are referred to as ________________
a) DSM lasers
b) Nd: YAG lasers
c) Glass fiber lasers
d) QD lasers

View Answer

Answer: a [Reason:] DSM lasers are also known as single frequency lasers. Such devices provide single longitudinal mode operation hence called as dynamic single mode lasers. These lasers reduce fiber intra-modal dispersion within high speed systems.

3. Which of the following does not provide single frequency operation?
a) Short cavity resonator
b) DSM lasers
c) Coupled cavity resonator
d) Fabry-Perot resonator

View Answer

Answer: d [Reason:] DSM lasers, short cavity resonators, coupled cavity resonators employ frequency selective feedback approach and provide single mode operation. However the Fabry-Perot resonator allows several longitudinal modes to exist within the gain spectrum of the device.

4. A method for increasing the longitudinal mode discrimination of an injection laser which is commonly used?
a) Decreasing refractive index
b) Increasing the refractive index
c) Increasing cavity length
d) Shortening of cavity length

View Answer

Answer: d [Reason:] The longitudinal mode discrimination of an injection laser is indirectly proportional to the cavity length. Thus, as the cavity length is shortened, the mode discrimination will get increase. If the cavity length is reduced from 250 to 25 units, the mode spacing is increased from 1 to 10 nm.

5. Conventional cleaved mirror structures are difficult to fabricate with the cavity lengths below
a) 200μmand greater than 150 μm
b) 100 μm and greater than 50 μm
c) 50 μm
d) 150 μm

View Answer

Answer: c [Reason:] cleaved laser mirrors are used in Fabry-Perot resonator which does not give result for shorter cavity lengths. These lengths may vary from 20μmto 50μm. Hence micro-cleaved or etched resonator is used for shorter cavity lens.

6. In the given equation, corrugation period is given by lλb/2Ne. If λb is the Bragg wavelength, then what does ‘l’ stand for?
a) Length of cavity
b) Limitation index
c) Integer order of grating
d) Refractive index

View Answer

Answer: c [Reason:] The period of corrugation is given by Period of corrugation = lλb/2Ne Where, λb= Bragg wavelength L = integer order of grating.

7. The first order grating (l=1) provide the strongest coupling within the device. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] The period of corrugation is given by lλb/2Ne includes order of grating. The second grating provide larger spatial period and thus helps in fabrication. If the order of grating is 1, then the device is coupled at high level.

8. The semiconductor lasers employing the distributed feedback mechanism are classified in _________________ categories
a) One
b) Two
c) Three
d) Four

View Answer

Answer: b [Reason:] Considering the device operation, semiconductor lasers are classified into two broad categories referred to as distributed feedback laser and distributed Bragg reflector laser. In the DFB laser, optical grating is applied over the entire active region, whereas in DBR lasers, the grating is etched only near the cavity ends.

9. DBF-BH lasers exhibit low threshold currents in the range of ________________
a) 40 to 50 mA
b) 21 to 30 mA
c) 2 to 5 mA
d) 10 to 20 mA

View Answer

Answer: d [Reason:] DFB lasers are used to provide single frequency semiconductor optical sources. DFB-BH lasers, developed in laboratories exhibit high modulation speeds, output power but low threshold currents in the range of 10 to 20 mA.

10. Fabry-Perot devices with BH geometries high modulation speeds than DFB-BH lasers. State whether the given statement is true or false
a) True
b) False

View Answer

Answer: b [Reason:] DFB-BH lasers exhibit low threshold currents but high output power and modulation speeds. DFB-BH laser is fabricated by etching or grating. Fabry-perot devices provide modulation speeds of M-bits per seconds whereas, DFB-BH lasers provides modulation speeds of G-bits/ sec.

11. The InGaAsP/InP double channel planar DFB-BH laser with a quarter wavelength shifted first order grating provides a single frequency operation and incorporates a phase shift of ______________
a) π/2 Radians
b) 2π Radians
c) π Radians
d) 3π/2 radians

View Answer

Answer: a [Reason:] The performance of DFB laser is improved by modifying a grating, which in turn introduces an optical phase shift. The phase shift depends on the wavelength used. A quarter wavelength shifted first order grating incorporates the phase shift of π/2 in the corrugation at the center of laser cavity.

12. The narrow line-width obtained under the CW operation for quarter wavelength shifted DFB laser is ________________
a) 2 MHz
b) 10 MHz
c) 3 MHz
d) 1 MHz

View Answer

Answer: c [Reason:] A quarter wavelength shifted DFB laser provides a large gain difference between the central mode and side modes. It provides improved dynamic single mode stability. Narrow line-width of around 3 MHz can be obtained under CW operation.

13. Line-width narrowing is achieved in DFB lasers by a strategy referred as _______________
a) Noise partition
b) Grating
c) Tuning
d) Bragg wavelength detuning

View Answer

Answer: d [Reason:] Line-width narrowing is achieved in DFB lasers by detuning the lasing wavelength towards the shorter wavelength side of gain peak. It increases the differential gain between the central mode and nearest side mode. This strategy is called as Bragg wavelength detuning.

14. _________________ is a technique used to render the non-conducting material around the active cavity by producing permanent defects in the implanted area
a) Dispersion
b) Ion de-plantation
c) Ion implantation
d) Attenuation

View Answer

Answer: c [Reason:] Ion implantation approach concentrates the injection current in active region. Current confinement is realized by ion implantation. Ions are implanted into a selective area of a semiconducting material to make it non-conducting.

Communications MCQ Set 2

1. An optical fiber has core-index of 1.480 and a cladding index of 1.478. What should be the core size for single mode operation at 1310nm?
a) 7.31μm
b) 8.71μm
c) 5.26μm
d) 6.50μm

View Answer

Answer: d [Reason:] Normalized frequency V<=2.405 is the value at which the lowest order Bessel function J=0. Core size(radius) optical-communication-questions-answers-single-mode-fibers-q1.

2. An optical fiber has a core radius 2μmand a numerical aperture of 0.1. Will this fiber operate at single mode at 600 nm?
a) Yes
b) No

View Answer

Answer: a [Reason:] V= 2πa.NA/λ. Calculating this equation, we get the value of V. V is the normalised frequency and should be below 2.405 in order to operate the fiber at single mode. Here, V=2.094, is less than 2.405. Thus, this optical fiber exhibit single mode operation.

3. What is needed to predict the performance characteristicsics 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

View Answer

Answer: b [Reason:] 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.

4. Which equation is used to calculate MFD?
a) Maxwell’s equations
b) Peterman equations
c) Allen Cahn equations
d) Boltzmann’s equations

View Answer

Answer: b [Reason:] Mode field diameter is an important parameter for single mode fibers because it is used to predict fiber properties such as splice loss, bending loss. The standard technique is to first measure the far-field intensity distribution and then calculating mode field diameter using Peterman equations.

5. A single mode fiber has mode field diameter 10.2μmand V=2.20. What is the core diameter of this fiber?
a) 11.1μm
b) 13.2μm
c) 7.6μm
d) 10.1μm

View Answer

Answer: d [Reason:] For a single mode fiber, MFD=2w0. Here, core radius optical-communication-questions-answers-single-mode-fibers-q5 Solving this equation, we get a=5.05μm. Core-diameter =2a=10.1μm.

6. The difference between the modes’ refractive indices is called as
a) Polarization
b) Cutoff
c) Fiber birefringence
d) Fiber splicing

View Answer

Answer: c [Reason:] 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.

7. A single mode fiber has a beat length of 4cm at 1200nm. What is birefringence?
a) 2*10-5
b) 1.2*10-5
c) 3*10-5
d) 2

View Answer

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

8. How many propagation modes are present in single mode fibers?
a) One
b) Two
c) Three
d) Five

View Answer

Answer: b [Reason:] For a given optical fiber, the number of modes depends on the dimensions of the cable and the variations of the indices of refraction of both core and cladding across the cross section. Thus, for a single mode fiber, there are two independent, degenerate propagation modes with their polarization planes orthogonal.

9. Numerical aperture is constant in case of step index fiber. State whether the statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] Numerical aperture is a measure of acceptance angle of a fiber. It also gives the light gathering capacity of the fiber. For a single mode fiber, core is of constant refractive index. There is no variation with respect to core. Thus, Numerical aperture is constant for single mode fibers.

10. Plastic fibers are less widely used than glass fibers. State whether the statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] The majority of the fibers are made up of glass consisting 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.

Communications MCQ Set 3

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.

View Answer

Answer: c [Reason:] 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. The strip width of injection laser is
a) 12 μm
b) 11.5 μm
c) Less than 10 μm
d) 15 μm

View Answer

Answer: c [Reason:] 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. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] 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.

View Answer

Answer: d [Reason:] 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.

View Answer

Answer: d [Reason:] 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

View Answer

Answer: b [Reason:] 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.

View Answer

Answer: a [Reason:] 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.

View Answer

Answer: b [Reason:] 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.

View Answer

Answer: c [Reason:] 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. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] 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.

View Answer

Answer: a [Reason:] 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.

View Answer

Answer: b [Reason:] 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.

View Answer

Answer: c [Reason:] 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. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: b [Reason:] 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

View Answer

Answer: d [Reason:] 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 μmand 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.

Communications MCQ Set 4

1. __________ is the unique property of the glass fiber.
a) Transmission
b) Opaque property
c) Ductile
d) Malleable

View Answer

Answer: a [Reason:] Glass fibres have a unique property as a transmission medium which enables their use in the communication. The major transmission characteristics are dispersion and attenuation.

2. __________ limits the maximum distance between the optical fiber transmitter and receiver.
a) Attenuation
b) Transmission
c) Equipment
d) Fiber length

View Answer

Answer: a [Reason:] Attenuation along with dispersion and the conductor size are some of the factors that limit the maximum distance between the optical transmitter and the receiver. The associated constraints within the equipment also affect the distance.

3. The ___________ incorporates a line receiver in order to convert the optical signal into the electrical regime.
a) Attenuator
b) Transmitter
c) Repeater
d) Designator

View Answer

Answer: c [Reason:] Repeaters are a mediator between transmitter and receiver. The weak signal is strengthened back by the repeaters on its path to the receiver.

4. A regenerative repeater is called as ____________
a) Repetitive repeater
b) Regenerator
c) Attenuator
d) Gyrator

View Answer

Answer: b [Reason:] When digital transmission techniques are used, the repeater also regenerates the original digital signal in the electrical signal before it is retransmitted as an optical signal via a line transmitter.

5. The wavelength range of __________ will be fruitful for the operating wavelength of the system referring to the system performance.
a) 0.8 – 0.9 μm
b) 1.1 – 2 μm
c) 5.2 – 5.7 μm
d) 3.1 – 3.2 μm

View Answer

Answer: a [Reason:] It is useful if the operating wavelength of the system is established to range of 0.8-0.9μm. This will be dictated by the overall requirements for the system performance, cost etc.

6. How many encoding schemes are used in optical fiber communication system design requirements?
a) Three
b) One
c) Two
d) Four

View Answer

Answer: c [Reason:] Encoding schemes are used for digital transmission of data. These are bi-phase and delay modulation codes. They are also called as Manchester and Miller codes respectively.

7. In ________ the optical channel bandwidth is divided into non-overlapping frequency bands.
a) Time division multiplexing
b) Frequency division multiplexing
c) Code division multiplexing
d) De-multiplexing

View Answer

Answer: b [Reason:] In FDM, the non-overlapping frequency bands are divided to the individual frequencies. These individual signals can be extracted from the combined FDM signal by electrical filtering at the receiver terminal.

8. A multiplexing technique which does not involve the application of several message signals onto a single fiber is called as _________
a) Time division multiplexing
b) Frequency division multiplexing
c) Code division multiplexing
d) Space division multiplexing

View Answer

Answer: d [Reason:] In SDM, each signal channel is carried on a separate fiber within a fiber bundle or multi-fiber cable form. The cross coupling between channels is negligible.

9. Which of the following is not an optical fiber component?
a) Fiber
b) Connector
c) Circulator
d) Detector

View Answer

Answer: c [Reason:] Circulator is a device used in electromagnetic theory. All others are optical components.

10. ________technique involves increase in the number of components required.
a) Time division multiplexing
b) Space division multiplexing
c) Code division multiplexing
d) Frequency division multiplexing

View Answer

Answer: SDM involves good optical isolation due to the negligible cross coupling between channels. It uses separate fiber and thus requires more number of components.

11. Time division multiplexing is inverse to that of frequency division multiplexing. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: a [Reason:] TDM involves distribution of channels in time slots whereas FDM involves bands that are run on different frequencies. Both these techniques improve accuracy and reduce complexity.

Communications MCQ Set 5

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

View Answer

Answer: a [Reason:] 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

View Answer

Answer: c [Reason:] 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.

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

View Answer

Answer: b [Reason:] 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

View Answer

Answer: a [Reason:] 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

View Answer

Answer: c [Reason:] 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

View Answer

Answer: d [Reason:] 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. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: True [Reason:] 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

View Answer

Answer: b [Reason:] The radioactive minority carrier lifetime ςrconsidering the p-type region is given by- ςr= [BrN]-1 where Br=Recombination coefficient in cm3s-1and 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

View Answer

Answer: c [Reason:] 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

View Answer

Answer: d [Reason:] 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-2and length and width of the cavity is 240μmand 110μm respectively. Find the threshold current for the device.
a) 663 mA
b) 660 mA
c) 664 mA
d) 712 mA

View Answer

Answer: b [Reason:] 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.

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

View Answer

Answer: c [Reason:] 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. State whether the given statement is true or false.
a) True
b) False

View Answer

Answer: b [Reason:] 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

View Answer

Answer: a [Reason:] 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

View Answer

Answer: d [Reason:] 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).