This is the Multiples Choice Questions Part 8 of the Series in Fiber Optics Communications as one of the Communications Engineering topics. In Preparation for the ECE Board Exam make sure to expose yourself and familiarize yourself with each and every question compiled here taken from various sources including but not limited to past Board Examination Questions in Electronic Systems and Technologies, Communications Books, Journals, and other Communications References.
MCQ Topic Outline included in ECE Board Exam Syllabi
- MCQ in Principles of Light Transmission
- MCQ in Types of Light Sources, Laser, LED
- MCQ in Light Detectors
- MCQ in Modulation and Waveform
- MCQ in System Design
- MCQ in General application
- MCQ in System Bandwidth
- MCQ in Splicing Techniques
Continue Practice Exam Test Questions Part 8 of the Series
⇐ MCQ in Fiber Optics Communications Part 7 | ECE Board Exam
Choose the letter of the best answer in each questions.
351. Frequency domain measurement is the preferred method for acquiring the bandwidth of multimode optical fibers.
A. true
B. false
Answer: Option 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.
352. Intra-modal dispersion tends to be dominant in multimode fibers.
A. true
B. false
Answer: Option 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.
353. 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: Option 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.
354. 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: Option A
Explanation: the digital bit rate and pulse duration are always inversely proportional to each other.
355. The maximum bit rate that may be obtained on an optical fiber link is 1/3Γ.
A. true
B. false
Answer: Option 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.
356. A multimode graded index fiber exhibits a total pulse broadening of 0.15 μs over 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: Option 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
BT = 12 Γ
Where BT = bandwidth.
357. Chromatic dispersion is also called as intermodal dispersion.
A. true
B. false
Answer: Option 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.
358. Chromatic dispersion is also called as intermodal dispersion.
A. true
B. false
Answer: Option 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.
359. 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 0.70 μm. Estimate the RMS spectral width.
A. 1.2 nm
B. 1.3 nm
C. 0.77 nm
D. 0.98 nm
Answer: Option C
Explanation: the relative spectral width σλ/ λ= 0.01 is given. the rms spectral width can be calculated as follows:
σλ/λ = 0.0011
σλ = 0.0011λ
= 0.0011*0.70*10-6
= 0.77 nm.
360. In waveguide dispersion, refractive index is independent of
A. bit rate
B. index difference
C. velocity of medium
D. wavelength
Answer: Option 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.
361. 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 1310 nm?
A. 7.31 μm
B. 8.71 μm
C. 5.26 μm
D. 6.50 μm
Answer: Option D
Explanation:
362. 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: Option A
Explanation: V= 2πa.na/λ. calculating this equation, we get the value of v. v is the normalized 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.
363. 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: Option B
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.
364. Which equation is used to calculate MFD?
A. maxwell’s equations
B. peterman equations
C. Allen Cahn equations
D. Boltzmann’s equations
Answer: Option B
Explanation: 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.
365. A single mode fiber has mode field diameter 10.2 μm and V = 2.20. What is the core diameter of this fiber?
A. 11.1 μm
B. 13.2 μm
C. 7.6 μm
D. 10.1 μm
Answer: Option D
Explanation:
366. The difference between the modes’ refractive indices is called as
A. polarization
B. cutoff
C. fiber birefringence
D. fiber splicing
Answer: Option C
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.
367. How many propagation modes are present in single mode fibers?
A. one
B. two
C. three
D. five
Answer: Option B
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.
368. Numerical aperture is constant in case of step index fiber.
A. true
B. false
Answer: Option A
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.
369. Plastic fibers are less widely used than glass fibers.
A. true
B. false
Answer: Option A
Explanation: The majority of the fibers are made up of glass consisting of 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.
370. 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: Option A
Explanation: an intrinsic semiconductor is usually un-doped. it is a pure semiconductor. the number of charge carriers is determined by the semiconductor material properties and not by the impurities.
371. 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: Option C
Explanation: For a semiconductor in thermal equilibrium, the probability P(E) that an electron gains sufficient thermal energy at an absolute temperature so as to occupy a particular energy level E, is given by the Fermi-Dirac distribution. It is given by-
P(E) = 1/(1+exp(E-EF/KT))
Where K = Boltzmann constant, T = absolute temperature, EF = Fermi energy level.
372. The majority of the carriers in a p-type semiconductor are
A. holes
B. electrons
C. photons
D. neutrons
Answer: Option A
Explanation: the impurities can be either donor impurities or acceptor impurities.
373. _________ is used when the optical emission results from the application of electric field.
A. radiation
B. efficiency
C. electro-luminescence
D. magnetron oscillator
Answer: Option C
Explanation: electro-luminescence is encouraged by selecting an appropriate semiconductor material. direct band-gap semiconductors are used for this purpose. in band-to-band recombination, the energy is released with the creation of photon. this emission of light is known as electroluminescence.
374. The recombination in indirect band-gap semiconductors is slow.
A. true
B. false
Answer: Option A
Explanation: in an indirect band-gap semiconductor, the maximum and minimum energies occur at different values of crystal momentum. however, three-particle recombination process is far less probable than the two-particle process exhibited by direct band-gap semiconductors. hence, the recombination in an indirect band-gap semiconductor is relatively slow.
375. A _________ is a series of logical connections between the source and destination nodes.
A. cell circuit
B. attenuation circuit
C. virtual circuit
D. switched network
Answer: Option C
Explanation: a virtual circuit consists of different routes which provide connections between sending and receiving devices. these routes can change at any time and the incoming return route does not have to mirror the outgoing route.
376. Which impurity is added to gallium phosphide to make it an efficient light emitter?
A. silicon
B. hydrogen
C. nitrogen
D. phosphorus
Answer: Option C
Explanation: an indirect band-gap semiconductor may be made into an electro- luminescent material by the addition of impurity centers which will convert it into a direct band-gap material. the introduction of nitrogen as an impurity into gallium phosphide makes it an effective emitter of light. such conversion is only achieved in materials where the direct and indirect band- gaps have a small energy difference.
377. 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: Option D
Explanation: Population inversion at p-n junction is obtained by heavy doping of both p-type and n-type material. Heavy p-type doping with acceptor impurities causes a lowering of the Fermi-level between the filled and empty states into the valence band. Similarly n-type doping causes Fermi-level to enter the conduction band of the material.
378. A GaAs injection laser has a threshold current density of 2.5*103 Acm-2 and length and width of the cavity is 240 μm and 110 μm respectively. Find the threshold current for the device.
A. 663 ma
B. 660 ma
C. 664 ma
D. 712 ma
Answer: Option B
Explanation: 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.
379. 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: Option C
Explanation: The reflectivity for normal incidence of the plane wave on the GaAs-air interface is given by-
r = ((n-1)/(n+1))2 where r=reflectivity and n=refractive index.
380. A homo-junction is an interface between two adjoining single-crystal semiconductors with different band-gap energies.
A. true
B. false
Answer: Option B
Explanation: The photo-emissive properties of a single p-n junction fabricated from a single-crystal semiconductor material are called as homo-junction. A hetero-junction is an interface between two single-crystal semiconductors with different band-gap energies. The devices which are fabricated with hetero-junctions are said to have hetero-structure.
381. How many types of hetero-junctions are available?
A. two
B. one
C. three
D. four
Answer: Option A
Explanation: hetero-junctions are classified into an isotype and an-isotype. the isotype hetero-junctions are also called as n-n or p-p junction. the an-isotype hetero-junctions are called as p-n junction with large band-gap energies.
382. 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: Option D
Explanation: for dh device fabrication, materials such as GaAs, alga as are used. the band-gap in this material may be tailored to span the entire wavelength band by changing the alga composition. thus, GaAs / alga as dh system is used for fabrication of lasers and LEDs for shorter wavelength region (0.8μm-0.9μm).
383. The amount of radiance in planer type of LED structures is
A. low
B. high
C. zero
D. negligible
Answer: Option A
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.
384. 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: Option B
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.
385. The techniques by Burros and Dawson in reference to homo structure device is to use an etched well in GaAs structure.
A. true
B. false
Answer: Option A
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.
386. 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: Option C
Explanation: dh structures provide high efficiency from electrical and optical confinement. along with efficiency, they provide less absorption of emitted radiation.
387. Internal absorption in DH surface emitter Burros type LEDs is
A. cannot be determined
B. negligible
C. high
D. very low
Answer: Option D
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.
388. 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: Option A
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.
389. In a multimode fiber, much of light coupled in the fiber from an LED is
A. increased
B. reduced
C. lost
D. unaffected
Answer: Option C
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.
390. The overall power conversion efficiency of electrical lens coupled LED is 0.8% and power applied 0.0375 V. Determine optical power launched into fiber.
A. 0.03
B. 0.05
C. 0.3
D. 0.01
Answer: Option A
Explanation: Optical power launched can be computed by
η pc = Pc/P
Pc = η pc* P
= 0.8 * 0.0375
= 0.03.
391. Mesa structured SLEDs are used
A. to reduce radiance
B. to increase radiance
C. to reduce current spreading
D. to increase current spreading
Answer: Option C
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.
392. 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: Option A
Explanation: the short striped structure of these LEDs around 100 μm improves the external efficiency of LEDs by reducing internal absorption of carriers. these are also called truncated strip e-leds.
393. The active layer of E-LED is heavily doped with
A. zn
B. eu
C. cu
D. sn
Answer: Option A
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.
394. 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: Option A
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.
395. How many types of sources of optical light are available?
A. one
B. two
C. three
D. four
Answer: Option C
Explanation: Three main types of optical light sources are available. These are wideband sources, monochromatic incoherent sources. Ideally the optical source should be linear.
396. Which process gives the laser its special properties as an optical source?
A. dispersion
B. stimulated absorption
C. spontaneous emission
D. stimulated emission
Answer: Option D
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.
397. An incandescent lamp is operating at a temperature of 1000 K 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: Option B
Explanation: The ratio of the stimulated emission rate to the spontaneous emission rate is given by-
Stimulated emission rate/ Spontaneous emission rate = 1/exp (hf/KT)-1.
398. 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: Option C
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 the lower energy level. This process of excitation of atoms into the upper level is achieved by using an external energy source and is called as pumping.
399. __________ 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: Option A
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.
400. A ruby laser has a crystal of length 3 cm with a refractive index of 1.60, wavelength 0.43 μm. Determine the number of longitudinal modes.
A. 1×102
B. 3×106
C. 2.9×105
D. 2.2×105
Answer: Option D
Explanation: The number of longitudinal modes is given by-
q = 2nL/λ
Where
q = Number of longitudinal modes
n = Refractive index
L = Length of the crystal
λ = Peak emission wavelength.
- MCQ in Fiber Optics Communications Part 10 | ECE Board Exam
- MCQ in Fiber Optics Communications Part 9 | ECE Board Exam
- MCQ in Fiber Optics Communications Part 8 | ECE Board Exam
- MCQ in Fiber Optics Communications Part 7 | ECE Board Exam
- MCQ in Fiber Optics Communications Part 6 | ECE Board Exam
- MCQ in Fiber Optics Communications Part 5 | ECE Board Exam
- MCQ in Fiber Optics Communications Part 4 | ECE Board Exam
- MCQ in Fiber Optics Communications Part 3 | ECE Board Exam
- MCQ in Fiber Optics Communications Part 2 | ECE Board Exam
- MCQ in Fiber Optics Communications Part 1 | ECE Board Exam
Complete List of MCQ in Communications Engineering per topic
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