McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Uncategorized topics .
| 101. |
The disc of a siren containing 60holes rotates at a constant speed of 360rpm. The emitted sound is in unison with a tuning fork of frequency. |
| A. | 10hz |
| B. | 360hz |
| C. | 216hz |
| D. | 60hz |
| Answer» B. 360hz | |
| Explanation: frequency of revolution of disc=360rpm=360/60rps=60rps | |
| 102. |
The quantity which does not change, when sound enters from one medium to another |
| A. | wavelength |
| B. | speed |
| C. | frequency |
| D. | velocity |
| Answer» C. frequency | |
| Explanation: frequency remains unchanged when sound travels from one medium to another. | |
| 103. |
If wave y=Acos(ωt+kx) Is moving along x- axis, the shapes of a pulse at t=0 and t=2s. |
| A. | are different |
| B. | are same |
| C. | may not be the same |
| D. | unpredictable |
| Answer» B. are same | |
| Explanation: the shapes of y-x graph remains the same at t=0 and t=2s. | |
| 104. |
y1=4sin(ωt+kx), y2=-4cos(ωt+kx), the phase difference is |
| A. | π/2 |
| B. | 3π/2 |
| C. | π |
| D. | zero |
| Answer» B. 3π/2 | |
| Explanation: y1=4sin(ωt+kx) y2=-4cos(ωt+kx) | |
| 105. |
A wave equation is y=0.1sin[100πt-kx] and wave velocity is 100m/s, its number is equal to |
| A. | 1/m |
| B. | 2/m |
| C. | π/m |
| D. | 2π/m |
| Answer» C. π/m | |
| Explanation: y=0.1sin[100πt-kx] y=asin(ωt-kx) | |
| 106. |
The fundamental frequency of a sonometer wire is n. If the tension is made 3 times and length and diameter are also increased 3 times, the new frequency will be |
| A. | 3n |
| B. | n/3√3 |
| C. | n/3 |
| D. | √3 n |
| Answer» B. n/3√3 | |
| Explanation: n=1/ld×√(t/πρ) n‘=1/(3l×3d)×√(3t/πρ)=√3/9×n=1/(3√3) n. | |
| 107. |
Which of the following Einstein’s coefficient represents spontaneous emission? |
| A. | a12 |
| B. | a21 |
| C. | b12 |
| D. | b21 |
| Answer» B. a21 | |
| Explanation: a21 represents the spontaneous emission of photons. a12 signifies spontaneous absorption.b12 is for stimulated absorption while b21 is for stimulated emission. | |
| 108. |
Which law is used for achieving the relation between the Einstein’s coefficients? |
| A. | heisenberg’s uncertainty principle |
| B. | planck’s radiation law |
| C. | einstein’s equation |
| D. | quantum law |
| Answer» B. planck’s radiation law | |
| Explanation: planck’s radiation law, which | |
| 109. |
The probability of spontaneous emission increases rapidly with the energy difference between the two states. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: from einstein’s relation we know that the ratio of einstein’s coefficients | |
| 110. |
What is the unit for the coefficient of stimulated emission? |
| A. | s-2 |
| B. | m3 s-2 |
| C. | j−1 m3 |
| D. | j−1 m3 s-2 |
| Answer» D. j−1 m3 s-2 | |
| Explanation: for stimulated emission, the | |
| 111. |
Which Einstein’s coefficient should be used in this case? |
| A. | a12 |
| B. | a21 |
| C. | b12 |
| D. | b21 |
| Answer» D. b21 | |
| Explanation: the given figure shows stimulated emission. hence, the einstein coefficient for stimulated emission is b21. if it had been spontaneous emission, then a21 would have been used. | |
| 112. |
A cylindrical cavity resonator can be constructed using a circular waveguide. |
| A. | shorted at both the ends |
| B. | open at both the ends |
| C. | matched at both the ends |
| D. | none of the mentioned |
| Answer» A. shorted at both the ends | |
| Explanation: a cylindrical cavity resonator is formed by shorting both the ends of the cylindrical cavity because open ends may result in radiation losses in the cavity. | |
| 113. |
The dominant mode in the cylindrical cavity resonator is TE101 mode. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: the dominant mode of propagation in a circular waveguide is te111 mode. hence, the dominant mode of resonance in a cylindrical cavity made of a circular waveguide is te111 mode. in a | |
| 114. |
Circular cavities are used for microwave frequency meters. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: circular cavities are used for microwave frequency meters. the cavity is constructed with a movable top wall to allow the mechanical tuning of the resonant frequency. | |
| 115. |
The mode of the circular cavity resonator used in frequency meters is: |
| A. | te011 mode |
| B. | te101 mode |
| C. | te111 mode |
| D. | tm111 mode |
| Answer» A. te011 mode | |
| Explanation: frequency resolution of a frequency meter is determined from its quality factor. q factor of te011 mode is much greater than the quality factor of the dominant mode of propagation. | |
| 116. |
The propagation constant of TEmn mode of propagation for a cylindrical cavity resonator is: |
| A. | 2) |
| B. | √ pnm/a |
| C. | √ (k2+(pnm/a)2) |
| D. | none of the mentioned |
| Answer» A. 2) | |
| Explanation: the propagation constant for a circular cavity depends on the radius of the cavity, and the wave number. if the mode of propagation is known and the dimension of the cavity is known then the propagation constant can be found out. | |
| 117. |
A circular cavity resonator is filled with a dielectric of 2.08 and is operating at 5GHz of frequency. Then the wave number is: |
| A. | 181 |
| B. | 151 |
| C. | 161 |
| D. | 216 |
| Answer» B. 151 | |
| Explanation: wave number for a circular cavity resonator is given by the expression 2πf011√∈r/c. substituting the given values in the above expression; the wave number of the cavity resonator is 151. | |
| 118. |
Given that the wave number of a circular cavity resonator is 151 (TE011 mode), and the length of the cavity is twice the radius of the cavity, the radius of the circular cavity operating at 5GHz frequency is: |
| A. | 2.1 cm |
| B. | 1.7 cm |
| C. | 2.84 cm |
| D. | insufficient data |
| Answer» D. insufficient data | |
| Explanation: for a circular cavity resonator, wave number is given by √( (p01/a)2 +(π/d)2). p01 for the given mode of resonance is 3.832. | |
| 119. |
A circular cavity resonator has a wave number of 151, radius of 2.74 cm, and surface resistance of 0.0184Ω. If the cavity is filled with a dielectric of 2.01, then unloaded Q due to conductor loss is: |
| A. | 25490 |
| B. | 21460 |
| C. | 29390 |
| D. | none of the mentioned |
| Answer» C. 29390 | |
| Explanation: unloaded q of a circular resonator due to conductor loss is given by ka/2rs. is the intrinsic impedance of the medium. substituting the given values in the equation for loaded q, value is 29390. | |
| 120. |
If unloaded Q due to conductor loss and unloaded Q due to dielectric loss is 29390 and 2500 respectively, then the total unloaded Q of the circular cavity is: |
| A. | 2500 |
| B. | 29390 |
| C. | 2300 |
| D. | 31890 |
| Answer» C. 2300 | |
| Explanation: the total unloaded q of a circular cavity resonator is given by the expression (q -1+ q -1)-1. substituting the | |
| 121. |
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 particular energy level e, is given by the fermi-dirac distribution. it is given by- | |
| 122. |
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 |
| Answer» B. doping the material with impurities | |
| Explanation: 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. | |
| 123. |
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. | |
| 124. |
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. | |
| 125. |
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. | |
| 126. |
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 |
| Answer» B. 1.39ns | |
| Explanation: the radioactive minority carrier lifetime ςrconsidering the p-type region is given by- | |
| 127. |
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 | |
| 128. |
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 doping with acceptor impurities causes a lowering of the fermi-level between the filled and empty states into the valence band. | |
| 129. |
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- | |
| 130. |
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- | |
| 131. |
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 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. | |
| 132. |
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. | |
| 133. |
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). | |
| 134. |
What is the principle of fibre optical communication? |
| A. | frequency modulation |
| B. | population inversion |
| C. | total internal reflection |
| D. | doppler effect |
| Answer» C. total internal reflection | |
| Explanation: in optical fibres, the light entering the fibre does not encounter any new surfaces, but repeatedly they hit the same surface. the reason for confining the light beam inside the fibres is the total internal reflection. | |
| 135. |
What is the other name for a maximum external incident angle? |
| A. | optical angle |
| B. | total internal reflection angle |
| C. | refraction angle |
| D. | wave guide acceptance angle |
| Answer» D. wave guide acceptance angle | |
| Explanation: only this rays which pass within the acceptance angle will be totally reflected. therefore, light incident on the core within the maximum external incident angle | |
| 136. |
A single mode fibre has low intermodal dispersion than multimode. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: in both single and multimode fibres the refractive indices will be in step by step. since a single mode has less dispersion than multimode, the single mode step index fibre also has low intermodal dispersion compared to multimode step index fibre. | |
| 137. |
How does the refractive index vary in Graded Index fibre? |
| A. | tangentially |
| B. | radially |
| C. | longitudinally |
| D. | transversely |
| Answer» B. radially | |
| Explanation: the refractive index of the core is maximum along the fibre axis and it gradually decreases. here the refractive index varies radially from the axis of the fibre. | |
| 138. |
Which of the following has more distortion? |
| A. | single step-index fibre |
| B. | graded index fibre |
| C. | multimode step-index fibre |
| D. | glass fibre |
| Answer» C. multimode step-index fibre | |
| Explanation: when rays travel through longer distances there will be some difference in reflected angles. hence high angle rays arrive later than low angle rays. therefore the signal pulses are broadened thereby results in a distorted output. | |
| 139. |
In which of the following there is no distortion? |
| A. | graded index fibre |
| B. | multimode step-index fibre |
| C. | single step-index fibre |
| D. | glass fibre |
| Answer» A. graded index fibre | |
| Explanation: the light travels with different speeds in different paths because of the variation in their refractive indices. at the outer edge it travels faster than near the centre but almost all the rays reach the exit end at the same time due to the helical path. thus, there is no dispersion in the pulses and hence the output is not a distorted output. | |
| 140. |
Which of the following loss occurs inside the fibre? |
| A. | radiative loss |
| B. | scattering |
| C. | absorption |
| D. | attenuation |
| Answer» B. scattering | |
| Explanation: scattering is a wavelength dependent loss. since the glass used in the fabrication of fibres, the disordered structure of glass will make some vibrations in the refractive index inside the fibre. this causes rayleigh scattering. | |
| 141. |
What causes microscopic bend? |
| A. | uniform pressure |
| B. | non-uniform volume |
| C. | uniform volume |
| D. | non-uniform pressure |
| Answer» D. non-uniform pressure | |
| Explanation: micro-bends losses are caused due to non-uniformities inside the fibre. this micro-bends in fibre appears due to non- uniform pressures created during the cabling of fibre. | |
| 142. |
When more than one mode is propagating, how is it dispersed? |
| A. | dispersion |
| B. | inter-modal dispersion |
| C. | material dispersion |
| D. | waveguide dispersion |
| Answer» B. inter-modal dispersion | |
| Explanation: when more than one mode is propagating through a fibre, then inter modal dispersion will occur. since many modes are propagating, they will have different wavelengths and will take different time to propagate through the fibre. | |
| 143. |
A fibre optic telephone transmission can handle more than thousands of voice channels. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: optical fibre has larger bandwidth hence it can handle a large number of channels for communication. | |
| 144. |
Which of the following is known as fibre optic back bone? |
| A. | telecommunication |
| B. | cable television |
| C. | delay lines |
| D. | bus topology |
| Answer» D. bus topology | |
| Explanation: each computer on the network is connected to the rest of the computers by the optical wiring scheme called bus topology, which is an application known as fibre optic back bone. | |
| 145. |
Calculate the numerical aperture of an optical fibre whose core and cladding are made of materials of refractive index 1.6 and |
| A. | 0.55677 |
| B. | 55.77 |
| C. | 0.2458 |
| D. | 0.647852 |
| Answer» A. 0.55677 | |
| Explanation: numerical aperture = | |
| 146. |
A step-index fibre has a numerical aperture of 0.26, a core refractive index of 1.5 and a core diameter of 100micrometer. Calculate the acceptance angle. |
| A. | 1.47° |
| B. | 15.07° |
| C. | 2.18° |
| D. | 24.15° |
| Answer» B. 15.07° | |
| Explanation: sin i = (numerical aperture)/n sin i = 15.07°. | |
| 147. |
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. | |
| 148. |
9 LOSSES ASSOCIATED WITH OPTICAL FIBERS |
| 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 | |
| 149. |
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. | |
| 150. |
Multimode step index fibers have a |
| 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. | |
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