McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Electrical Engineering .
| 51. |
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. | |
| 52. |
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. | |
| 53. |
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. | |
| 54. |
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. | |
| 55. |
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. | |
| 56. |
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 | |
| 57. |
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. | |
| 58. |
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: | |
| 59. |
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. | |
| 60. |
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) | |
| 61. |
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 | |
| 62. |
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. | |
| 63. |
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 | |
| 64. |
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 | |
| 65. |
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. | |
| 66. |
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. | |
| 67. |
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. | |
| 68. |
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 | |
| 69. |
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. | |
| 70. |
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 | |
| 71. |
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. | |
| 72. |
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. | |
| 73. |
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. | |
| 74. |
21*10-10cm3s-1. |
| 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- | |
| 75. |
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. | |
| 76. |
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 | |
| 77. |
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- | |
| 78. |
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- | |
| 79. |
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 | |
| 80. |
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. | |
| 81. |
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). | |
| 82. |
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. | |
| 83. |
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. | |
| 84. |
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. | |
| 85. |
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. | |
| 86. |
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. | |
| 87. |
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. | |
| 88. |
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. | |
| 89. |
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 | |
| 90. |
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. | |
| 91. |
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. | |
| 92. |
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. | |
| 93. |
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. | |
| 94. |
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 | |
| 95. |
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. | |
| 96. |
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- | |
| 97. |
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 | |
| 98. |
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. | |
| 99. |
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- | |
| 100. |
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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