McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Electrical Engineering .
| 1. |
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. | |
| 2. |
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. | |
| 3. |
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. | |
| 4. |
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. | |
| 5. |
2 BASIC OPTICAL LAWS AND DEFINITIONS, OPTICAL MODES AND CONFIGURATIONS |
| 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. | |
| 6. |
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. | |
| 7. |
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. | |
| 8. |
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. | |
| 9. |
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. | |
| 10. |
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. | |
| 11. |
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. | |
| 12. |
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. | |
| 13. |
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. | |
| 14. |
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. | |
| 15. |
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. | |
| 16. |
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. | |
| 17. |
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. | |
| 18. |
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. | |
| 19. |
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. | |
| 20. |
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 | |
| 21. |
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. | |
| 22. |
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. | |
| 23. |
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. | |
| 24. |
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. | |
| 25. |
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. | |
| 26. |
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. | |
| 27. |
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. | |
| 28. |
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. | |
| 29. |
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. | |
| 30. |
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. | |
| 31. |
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. | |
| 32. |
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 | |
| 33. |
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 | |
| 34. |
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. | |
| 35. |
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. | |
| 36. |
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. | |
| 37. |
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. | |
| 38. |
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. | |
| 39. |
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). | |
| 40. |
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 | |
| 41. |
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- | |
| 42. |
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 | |
| 43. |
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. | |
| 44. |
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. | |
| 45. |
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. | |
| 46. |
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. | |
| 47. |
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. | |
| 48. |
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. | |
| 49. |
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. | |
| 50. |
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 – | |
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