200+ Optical Communication Solved MCQs


Multimode step index fiber has

A. large core diameter & large numerical aperture
B. large core diameter and small numerical aperture
C. small core diameter and large numerical aperture
D. small core diameter & small numerical aperture
Answer» A. large core diameter & large numerical aperture
Explanation: multimode step-index fiber has large core diameter and large numerical aperture. these parameters provides efficient coupling to inherent light sources such as led’s.

A typically structured glass multimode step index fiber shows as variation of attenuation in range of                        

A. 1.2 to 90 db km-1 at wavelength 0.69μm
B. 3.2 to 30 db km-1 at wavelength 0.59μm
C. 2.6 to 50 db km-1 at wavelength 0.85μm
D. 1.6 to 60 db km-1 at wavelength 0.90μm
Answer» C. 2.6 to 50 db km-1 at wavelength 0.85μm
Explanation: a multimode step index fibers show an attenuation variation in range of 2.6 to 50dbkm-1. the wide variation in attenuation is due to the large differences both within and between the two overall preparation methods i.e. melting and deposition.

Multimode step index fiber has a large core diameter of range is                        

A. 100 to 300 μm
B. 100 to 300 nm
C. 200 to 500 μm
D. 200 to 500 nm
Answer» A. 100 to 300 μm
Explanation: a multimode step index fiber has a core diameter range of 100 to 300μm. this is to facilitate efficient coupling to inherent light sources.

Multimode step index fibers have a bandwidth of                        

A. 2 to 30 mhz km
B. 6 to 50 mhz km
C. 10 to 40 mhz km
D. 8 to 40 mhz km
Answer» B. 6 to 50 mhz km
Explanation: multimode step index fibers have a bandwidth of 6 to 50 mhz km. these fibers with this bandwidth are best suited for short -haul, limited bandwidth and relatively low-cost application.


A. lower purity
B. higher purity than multimode step index fibers.
C. no impurity
D. impurity as same as multimode step index fibers.
Answer» B. higher purity than multimode step index fibers.
Explanation: multimode graded index fibers have higher purity than multimode step index fiber. to reduce fiber losses, these fibers have more impurity.

The performance characteristics of multimode graded index fibers are

A. better than multimode step index fibers
B. same as multimode step index fibers
C. lesser than multimode step index fibers
D. negligible
Answer» A. better than multimode step index fibers
Explanation: multimode graded index fibers use a constant grading factor. performance characteristics of multimode graded index fibers are better than those of multimode step index fibers due to index graded and lower attenuation.

Multimode graded index fibers have overall buffer jackets same as multimode step index fibers but have core diameters

A. larger than multimode step index fibers
B. smaller than multimode step index fibers
C. same as that of multimode step index fibers
D. smaller than single mode step index fibers
Answer» B. smaller than multimode step index fibers
Explanation: multimode graded index fibers have smaller core diameter than multimode step index fibers. a small core diameter helps the fiber gain greater rigidity to resist bending.

Multimode graded index fibers use incoherent source only.

A. true
B. false
Answer» B. false
Explanation: multimode graded index fibers are used for short haul and medium to high bandwidth applications. small haul applications require leds and low accuracy lasers. thus either incoherent or incoherent sources like led’s or injection laser diode are used.

In single mode fibers, which is the most beneficial index profile?

A. step index
B. graded index
C. step and graded index
D. coaxial cable
Answer» B. graded index
Explanation: in single mode fibers, graded index profile is more beneficial as compared to step index. this is because graded index profile provides dispersion-modified-single mode fibers.

Multimode graded index fibers with                        

A. single mode fibers
B. multimode step fibers
C. coaxial cables
D. multimode graded index fibers
Answer» A. single mode fibers
Explanation: single mode fibers are used to produce polarization maintaining fibers which make them expensive. also the alternative to them are multimode fibers which are complex but accurate. so, single-mode fibers are not generally utilized in optical fiber communication.

Single mode fibers allow single mode propagation; the cladding diameter must be at least                        

A. twice the core diameter
B. thrice the core diameter
C. five times the core diameter
D. ten times the core diameter
Answer» D. ten times the core diameter
Explanation: the cladding diameter in single mode fiber must be ten times the core diameter. larger ratios contribute to accurate propagation of light. these dimension ratios must be there so as to avoid losses from the vanishing fields.

A fiber which is referred as non- dispersive shifted fiber is?

A. coaxial cables
B. standard single mode fibers
C. o-band
D. c-band and l-band
Answer» C. o-band
Explanation: ssmfs are utilized for operation in o-band only. it shows high dispersion in the range of 16 to 20ps/nm/km in c-band and l-band. so ssmfs are used in o-band.

Fiber mostly suited in single-wavelength transmission in O-band is?

A. low-water-peak non dispersion-shifted fibers
B. standard single mode fibers
C. low minimized fibers
D. non-zero-dispersion-shifted fibers
Answer» B. standard single mode fibers
Explanation: standard single mode fibers with a step index profile are called non dispersion shifted fiber and it is particularly used for single wavelength transmission in o- band and as if has a zero-dispersion wavelength at 1.31μm.

When the dimensions of the guide are reduced, the number of                        also decreases.

A. propagating nodes
B. electrons
C. holes
D. volume of photons
Answer» A. propagating nodes
Explanation: the dimensions of the guide are directly proportional to the number of propagating nodes. as the dimensions are reduced, the number of propagating nodes also decreases.

                       waveguides are plagued by high losses.

A. circular
B. planar
C. depleted
D. metal-clad
Answer» D. metal-clad
Explanation: all suitable waveguide materials are subject to limitations in the confinement. however, metal-clad waveguides are not so limited. hence, they are plagued by high losses.

The planar waveguides may be fabricated from glasses and other isotropic materials such as                        and                              

A. octane and polymers
B. carbon monoxide and diode
C. fluorides and carbonates
D. sulphur dioxide and polymers
Answer» D. sulphur dioxide and polymers
Explanation: these materials are isotropic. however, their properties do not affect the fabrication of planar waveguides. their properties cannot be controlled by external energy sources.

Which of the following devices are less widely used in the field of optical fibre communications?

A. acousto-optic devices
B. regenerators
C. reflectors
D. optical translators
Answer» A. acousto-optic devices
Explanation: acousto-optic devices are less widely used, mainly in the area of field deflection. regenerators, reflectors form a base for the optical fibre communications.

Which of the following materials have refractive index near two?

A. ga as
B. zinc
C. inp
D. alsb
Answer» B. zinc
Explanation: two basic groups are distinguished on the basis of the respective refractive indices near two and near three.

Strip pattern in waveguide structures is obtained through                          

A. lithography
B. cryptography
C. depletion of holes
D. implantation
Answer» A. lithography
Explanation: field strength is an important aspect when it comes to strip patterns in waveguide structures. the electron and laser beam lithography is used to obtain stripe pattern in waveguide structures.

Propagation losses in slab and strip waveguides are smaller than the single mode fibre losses.

A. true
B. false
Answer» B. false
Explanation: the losses are in the range of

A passive Y-junction beam splitter is also used as a switch.

A. true
B. false
Answer» A. true
Explanation: a passive junction beam splitter finds application where equal power division of the incident beam is required. it can be used as a switch if it is fabricated from an electro-optic material.

The linear variation of refractive index with the electric field is known as the

A. linear implantation
B. ionization
C. koppel effect
D. pockels effect
Answer» D. pockels effect
Explanation: the change in refractive index is related by the applied field via the linear and quadratic electro-optic coefficients. the variation of r.i with the electric field is known as pockels effect.

In a transverse electric magnetic wave, which of the following will be true?

A. e is transverse to h
B. e is transverse to wave direction
C. h is transverse to wave direction
D. e and h are transverse to wave direction
Answer» D. e and h are transverse to wave direction
Explanation: in the transverse electric magnetic wave (tem wave), both the electric and magnetic field strengths are transverse to the wave propagation.

The cut off frequency of the TEM wave is

A. 0
B. 1 ghz
C. 6 ghz
D. infinity
Answer» A. 0
Explanation: the tem waves have both e and h perpendicular to the guide axis. thus its cut off frequency is zero.

Which component is non zero in a TEM wave?

A. ex
B. hz
C. ez
D. attenuation constant
Answer» A. ex
Explanation: in a tem wave, the wave propagates along the guided axis. thus the components ez and hz are zero. the attenuation is also zero. the non-zero component will be ex.

TEM wave can propagate in rectangular waveguides. State true/false.

A. true
B. false
Answer» B. false
Explanation: the rectangular waveguide does not allow the tem wave. tem mode can exist only in two conductor system and not in hollow waveguide in which the centre conductor does not exist.

The cut off wavelength in the TEM wave will be

A. 0
B. negative
C. infinity
D. 1/6 ghz
Answer» C. infinity
Explanation: the cut off frequency in a tem wave is zero. thus the cut off wavelength will be infinity.

The guided wavelength of a TEM wave in a waveguide having a wavelength of 5 units is

A. 0
B. infinity
C. 5
D. 1/5
Answer» C. 5
Explanation: the guided wavelength is same as the wavelength of the waveguide with a tem wave. thus the guided wavelength is 5 units.

The guided phase constant of a TEM wave in a waveguide with a phase constant of 2.8 units is

A. 2.8
B. 1.4
C. 0
D. infinity
Answer» A. 2.8
Explanation: the guided phase constant is same as the phase constant of the waveguide.

Which type of transmission line accepts the TEM wave?

A. copper cables
B. coaxial cable
C. rectangular waveguides
D. circular waveguides
Answer» B. coaxial cable
Explanation: hollow transmission lines support te and tm waves only. the tem wave is possible only in the coaxial cable transmission line, which is not hollow.

For a TEM wave to propagate in a medium, the medium has to be

A. air
B. insulator
C. dispersive
D. non dispersive
Answer» D. non dispersive
Explanation: the medium in which the tem waves propagate has to be non- dispersive.

Stripline and parallel plate waveguides support the TEM wave. State true/false.

A. true
B. false
Answer» A. true
Explanation: the stripline and parallel plate waveguides are not hollow and the dielectric

Which of the following statements best explain the concept of material absorption?

A. a loss mechanism related to the material composition and fabrication of fiber
B. a transmission loss for optical fibers
C. results in attenuation of transmitted light
D. causes of transfer of optical power
Answer» A. a loss mechanism related to the material composition and fabrication of fiber
Explanation: material absorption is a loss

How many mechanisms are there which causes absorption?

A. one
B. three
C. two
D. four
Answer» B. three
Explanation: absorption is a loss mechanism. it may be intrinsic, extrinsic and also caused by atomic defects.

Absorption losses due to atomic defects mainly include                        

A. radiation
B. missing molecules, oxygen defects in glass
C. impurities in fiber material
D. interaction with other components of core
Answer» B. missing molecules, oxygen defects in glass
Explanation: atomic defects are imperfections in the atomic structure of fiber material. atomic structure includes nucleus, molecules, protons etc. atomic defects thus contribute towards loss of molecules, oxygen, etc.

The effects of intrinsic absorption can be minimized by                        

A. ionization
B. radiation
C. suitable choice of core and cladding components
D. melting
Answer» C. suitable choice of core and cladding components
Explanation: intrinsic absorption is caused by interaction of light with one or more components of the glass i.e. core. thus, if the compositions of core and cladding are chosen suitably, this effect can be minimized.

Which of the following is not a metallic impurity found in glass in extrinsic absorption?

A. fe2+
B. fe3+
C. cu
D. si
Answer» D. si
Explanation: in the optical fibers, prepared by melting techniques, extrinsic absorption can be observed. it is caused from transition metal element impurities. in all these options, si is a constituent of glass and it cannot be considered as an impurity to glass itself.

Optical fibers suffer radiation losses at bends or curves on their paths.

A. true
B. false
Answer» A. true
Explanation: optical fibers suffer radiation losses due to the energy in the bend or curves exceeding the velocity of light in the cladding. hence, guiding mechanism is inhibited, which in turn causes light energy to be radiated from the fiber.

In the given equation, state what αr suggests?

A. radius of curvature
B. refractive index difference
C. radiation attenuation coefficients
D. constant of proportionality
Answer» C. radiation attenuation coefficients
Explanation: above equation represents the fiber loss. this loss is seen at bends and curves as the fibers suffer radiation losses at curves. these radiation losses are represented by a radiation attenuation coefficient (αr).

15, n2 = 1.11 and an operating wavelength of 0.7μm. Find the radius of curvature?

A. 8.60μm
B. 9.30μm
C. 9.1μm
D. 10.2μm
Answer» B. 9.30μm
Explanation: the radius of curvature of the fiber bend of a multimode fiber is given by

A single mode fiber has refractive indices n1=1.50, n2 = 2.23, core diameter of 8μm, wavelength = 1.5μm cutoff wavelength = 1.214μm. Find the radius of curvature?

A. 12 mm
B. 20 mm
C. 34 mm
D. 36 mm
Answer» C. 34 mm
Explanation: the radius of curvature of the fiber bend of a single mode fiber is given by-

How the potential macro bending losses can be reduced in case of multimode fiber?

A. by designing fibers with large relative refractive index differences
B. by maintaining direction of propagation
C. by reducing the bend
D. by operating at larger wavelengths
Answer» A. by designing fibers with large relative refractive index differences
Explanation: in the case of multimode fibers, radius of curvature is directly proportional to

Sharp bends or micro bends causes significant losses in fiber.

A. true
B. false
Answer» A. true
Explanation: sharp bends usually have a radius of curvature almost near to the critical radius. the fibers with the radius near to the critical radius cause significant losses and hence they are avoided.

                       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. dispersion
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.

How many types of mechanisms are present which produce dispersion in optical fibers?

A. three
B. two
C. one
D. four
Answer» A. three
Explanation: there are three major mechanisms which produce dispersion in optical fibers. these are: material dispersion, waveguide dispersion and intermodal dispersion.

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. intra-modal dispersion
Explanation: in single mode case, the dominant dispersion mechanism is chromatic. chromatic dispersion is called as intra-modal dispersion.

Devices such as                        are used to simulate the steady-state mode distribution.

A. gyrators
B. circulators
C. mode scramblers
D. attenuators
Answer» C. mode scramblers
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.

How many domains support the measurements of fiber dispersion?

A. one
B. three
C. four
D. two
Answer» D. two
Explanation: fiber dispersion measurements can be made in two domains. these are time domain and frequency domain.

The time domain dispersion measurement setup involves                            as the photo detector.

A. avalanche photodiode
B. oscilloscope
C. circulator
D. gyrator
Answer» A. avalanche photodiode
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.

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. 41.9 mhz km
Explanation: the optical bandwidth for the fiber is given by –

Frequency domain measurement is the preferred method for acquiring the bandwidth of multimode optical fibers.

A. true
B. false
Answer» A. true
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.

Intra-modal dispersion tends to be dominant in multimode fibers.

A. true
B. false
Answer» B. false
Explanation: intra-modal dispersion is dominant in case of single mode fibers. in case of multimode fibers, intermodal dispersion comes handy and is dominant.

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. inter-symbol interference
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.

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. less than the reciprocal of broadened pulse duration
Explanation: the digital bit rate and pulse duration are always inversely proportional to each other.

The maximum bit rate that may be obtained on an optical fiber link is 1/3Γ.

A. true
B. false
Answer» B. false
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.

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. 3.3 mhz
Explanation: the maximum possible bandwidth is equivalent to the maximum possible bitrate. the maximum bit rate assuming no inter-symbol interference is given by

Chromatic dispersion is also called as intermodal dispersion.

A. true
B. false
Answer» B. false
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.

The optical source used in a fiber is an injection laser with a relative spectral width σλ/λ of 0.0011 at a wavelength of 0.70μm. Estimate the RMS spectral width.

A. 1.2 nm
B. 1.3 nm
C. 0.77 nm
D. 0.98 nm
Answer» C. 0.77 nm
Explanation: the relative spectral width σλ/ λ= 0.01 is given. the rms spectral width can be calculated as follows:

In waveguide dispersion, refractive index is independent of                              

A. bit rate
B. index difference
C. velocity of medium
D. wavelength
Answer» D. wavelength
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.

An optical fiber has core-index of 1.480 and a cladding index of 1.478. What should be the core size for single mode operation at 1310nm?

A. 7.31μm
B. 8.71μm
C. 5.26μm
D. 6.50μm
Answer» D. 6.50μm
Explanation: normalized frequency v<=2.405 is the value at which the lowest order bessel function j=0. core size(radius)

An optical fiber has a core radius 2μm and a numerical aperture of 0.1. Will this fiber operate at single mode at 600 nm?

A. yes
B. no
Answer» A. yes
Explanation: v= 2πa.na/λ. calculating this equation, we get the value of v. v is the normalised frequency and should be below

What is needed to predict the performance characteristics of single mode fibers?

A. the intermodal delay effect
B. geometric distribution of light in a propagating mode
C. fractional power flow in the cladding of fiber
D. normalized frequency
Answer» B. geometric distribution of light in a propagating mode
Explanation: a mode field diameter (mfd) is a fundamental parameter of single mode fibers. it tells us about the geometric distribution of light. mfd is analogous to core diameter in multimode fibers, except in single mode fibers not all the light that propagates is carried in the core.

Which equation is used to calculate MFD?

A. maxwell’s equations
B. peterman equations
C. allen cahn equations
D. boltzmann’s equations
Answer» B. peterman equations
Explanation: mode field diameter is an

A single mode fiber has mode field diameter 10.2μm and V=2.20. What is the core diameter of this fiber?

A. 11.1μm
B. 13.2μm
C. 7.6μm
D. 10.1μm
Answer» D. 10.1μm
Explanation: for a single mode fiber, mfd=2w0. here, core radius

The difference between the modes’ refractive indices is called as                        

A. polarization
B. cutoff
C. fiber birefringence
D. fiber splicing
Answer» C. fiber birefringence
Explanation: there are two propagation modes in single mode fibers. these two modes are similar but their polarization planes are orthogonal. in actual fibers, there are imperfections such as variations in refractive index profiles. these modes propagate with different phase velocities and their difference is given by bf =ny – nx. here, ny and nx are refractive indices of two modes.

How many propagation modes are present in single mode fibers?

A. one
B. two
C. three
D. five
Answer» B. two
Explanation: for a given optical fiber, the number of modes depends on the dimensions of the cable and the variations of the indices of refraction of both core and cladding across the cross section. thus, for a single mode fiber, there are two independent, degenerate propagation modes with their polarization planes orthogonal.

Numerical aperture is constant in case of step index fiber.

A. true
B. false
Answer» A. true
Explanation: numerical aperture is a measure of acceptance angle of a fiber. it also gives the light gathering capacity of the fiber. for a single mode fiber, core is of constant refractive index. there is no variation with respect to core. thus, numerical aperture is constant for single mode fibers.

Plastic fibers are less widely used than glass fibers.

A. true
B. false
Answer» A. true
Explanation: the majority of the fibers are made up of glass consisting of silica. plastic fibers are used for short distance

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. intrinsic semiconductor
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.

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. fermi-dirac distribution function
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

The majority of the carriers in a p-type semiconductor are                      

A. holes
B. electrons
C. photons
D. neutrons
Answer» A. holes
Explanation: the impurities can be either donor impurities or acceptor impurities.

                                   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. electro-luminescence
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.

The recombination in indirect band-gap semiconductors is slow.

A. true
B. false
Answer» A. true
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.


A. 2ns
B. 1.39ns
C. 1.56ns
D. 2.12ms
Answer» B. 1.39ns
Explanation: the radioactive minority carrier lifetime ςrconsidering the p-type region is given by-

Which impurity is added to gallium phosphide to make it an efficient light emitter?

A. silicon
B. hydrogen
C. nitrogen
D. phosphorus
Answer» C. nitrogen
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.

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. heavy doping of both p-type and n-type material
Explanation: population inversion at p-n junction is obtained by heavy doping of both p-type and n-type material. heavy p-type

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. 660 ma
Explanation: the threshold current is denoted by ith. it is given by-

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. 0.32
Explanation: the reflectivity for normal incidence of the plane wave on the gaas-air interface is given by-

A ho*mo-junction is an interface between two adjoining single-crystal semiconductors with different band-gap energies.

A. true
B. false
Answer» B. false
Explanation: the photo-emissive properties

How many types of hetero-junctions are available?

A. two
B. one
C. three
D. four
Answer» A. two
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.

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. gaas/alga as dh
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).

The amount of radiance in planer type of LED structures is                          

A. low
B. high
C. zero
D. negligible
Answer» A. low
Explanation: planer leds are fabricated using liquid or vapor phase epitaxial processes. here p-type is diffused into n-type substrate which creates junction. forward current flow through junction provides lambertian spontaneous emission. thus, device emits light from all surfaces. however a limited amount of light escapes the structure due to total internal reflection thus providing low radiance.

As compared to planar LED structure, Dome LEDs have                              External power efficiency                        effective emission area and                            radiance.

A. greater, lesser, reduced
B. higher, greater, reduced
C. higher, lesser, increased
D. greater, greater, increased
Answer» B. higher, greater, reduced
Explanation: in dome leds, the diameter of dome is selected so as to maximum the internal emission reaching surface within critical angle of gaas. thus, dome leds have high external power efficiency. the geometry of dome leds is such that dome is much larger than active recombination area, so it has greater emission era and reduced of radiance.

The techniques by Burros and Dawson in reference to ho*mo structure device is to use an etched well in GaAs structure.

A. true
B. false
Answer» A. true
Explanation: burros and dawson provided a technique to restrict emission to small active region within device thus providing high radiance. etched well in a gaas substrate is used to prevent heavy absorption of emitted region and physically accommodating the fiber. these structures provide low thermal impedance allowing high current densities of high radiance.

In surface emitter LEDs, more advantage can be obtained by using                          

A. bh structures
B. qc structures
C. dh structures
D. gain-guided structure
Answer» C. dh structures
Explanation: dh structures provide high efficiency from electrical and optical confinement. along with efficiency, they provide less absorption of emitted radiation.

Internal absorption in DH surface emitter Burros type LEDs is                          

A. cannot be determined
B. negligible
C. high
D. very low
Answer» D. very low
Explanation: the larger band gap confining layers and the reflection coefficient at the back crystal space is high in dh surface emitter burros type leds. this provides good forward radiance. thus these structure leds have very less internal absorption.

DH surface emitter generally give

A. more coupled optical power
B. less coupled optical power
C. low current densities
D. low radiance emission into-fiber
Answer» A. more coupled optical power
Explanation: the optical power coupled into a fiber depends on distance, alignment between emission area and fiber, sled emission pattern and medium between emitting area and fiber. all these parameters if considered, reduces refractive index mismatch and increases external power efficiency thus providing more coupled optical power.

In a multimode fiber, much of light coupled in the fiber from an LED is

A. increased
B. reduced
C. lost
D. unaffected
Answer» C. lost
Explanation: optical power from an incoherent source is initially coupled into large angle rays falling within acceptance angle of fiber but have more energy than meridional rays. energy from these rays goes into the cladding and thus may be lost.

0375 V. Determine optical power launched into fiber.

A. 0.03
B. 0.05
C. 0.3
D. 0.01
Answer» A. 0.03
Explanation: optical power launched can be computed by

Mesa structured SLEDs are used

A. to reduce radiance
B. to increase radiance
C. to reduce current spreading
D. to increase current spreading
Answer» C. to reduce current spreading
Explanation: the planar structures of burros-type led allow lateral current spreading specially for contact diameters less than 25 μm.this results in reduced current density and effective emission area greater than contact area. this technique to reduce current spreading in very small devices is mesa structured sleds.

The InGaAsP is emitting LEDs are realized in terms of restricted are

A. length strip geometry
B. radiance
C. current spreading
D. coupled optical power
Answer» A. length strip geometry
Explanation: the short striped structure of these leds around 100 μmimproves the external efficiency of leds by reducing internal absorption of carriers. these are also called truncated strip e-leds.

The active layer of E-LED is heavily doped with                          

A. zn
B. eu
C. cu
D. sn
Answer» A. zn
Explanation: zn doping reduces the minority carrier lifetime. thus this improves the device modulation bandwidth hence active layer is doped in zn in e-leds.

A device which converts electrical energy in the form of a current into optical energy is called as                        

A. optical source
B. optical coupler
C. optical isolator
D. circulator
Answer» A. optical source
Explanation: an optical source is an active component in an optical fiber communication system. it converts electrical energy into optical energy and allows the light output to be efficiently coupled into the optical fiber.

How many types of sources of optical light are available?

A. one
B. two
C. three
D. four
Answer» C. three
Explanation: three main types of optical light sources are available. these are wideband sources, monochromatic incoherent

Which process gives the laser its special properties as an optical source?

A. dispersion
B. stimulated absorption
C. spontaneous emission
D. stimulated emission
Answer» D. stimulated emission
Explanation: in stimulated emission, the photon produced is of the same energy to the one which cause it. hence, the light associated with stimulated photon is in phase and has same polarization. therefore, in contrast to spontaneous emission, coherent radiation is obtained. the coherent radiation phenomenon in laser provides amplification thereby making laser a better optical source than led.

An incandescent lamp is operating at a temperature of 1000K at an operating frequency of 5.2×1014 Hz. Calculate the ratio of stimulated emission rate to spontaneous emission rate.

A. 3×10-13
B. 1.47×10-11
C. 2×10-12
D. 1.5×10-13
Answer» B. 1.47×10-11
Explanation: the ratio of the stimulated emission rate to the spontaneous emission rate is given by-

The lower energy level contains more atoms than upper level under the conditions of                                  

A. isothermal packaging
B. population inversion
C. thermal equilibrium
D. pumping
Answer» C. thermal equilibrium
Explanation: under the conditions of thermal equilibrium, the lower energy level contains more atoms than the upper level. to achieve optical amplification, it is required to create a non-equilibrium distribution such that the population of upper energy level is more than

                                     in the laser occurs when photon colliding with an excited atom causes the stimulated emission of a second photon.

A. light amplification
B. attenuation
C. dispersion
D. population inversion
Answer» A. light amplification
Explanation: laser emits coherent radiation of one or more discrete wavelength. lasers produce coherent light through a process called stimulated emission. light amplification is obtained through stimulated emission. continuation of this process creates avalanche multiplication.

43 μm. Determine the number of longitudinal modes.

A. 1×102
B. 3×106
C. 2.9×105
D. 2.2×105
Answer» D. 2.2×105
Explanation: the number of longitudinal modes is given by-

A semiconductor laser crystal of length 5 cm, refractive index 1.8 is used as an optical source. Determine the frequency separation of the modes.

A. 2.8 ghz
B. 1.2 ghz
C. 1.6 ghz
D. 2 ghz
Answer» C. 1.6 ghz
Explanation: the modes of laser are separated by a frequency internal δf and this separation is given by-
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