Optical Communication Set 3

Electrical Engineering

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This set of Optical Communication Multiple Choice Questions & Answers (MCQs) focuses on Optical Communication Set 3

Q1 | Frequency domain measurement is the preferred method for acquiring the bandwidth of multimode optical fibers.
  • true
  • false
Answer: true
Explaination:
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.
Q2 | Intra-modal dispersion tends to be dominant in multimode fibers.
  • true
  • false
Answer: false
Explaination:
intra-modal dispersion is dominant in case of single mode fibers. in case of multimode fibers, intermodal dispersion comes handy and is dominant.
Q3 | What does ISI stand for in optical fiber communication?
  • invisible size interference
  • infrared size interference
  • inter-symbol interference
  • inter-shape interference
Answer: inter-symbol interference
Explaination:
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.
Q4 | For no overlapping of light pulses down on an optical fiber link, the digital bit rate BT must be                        
  • less than the reciprocal of broadened pulse duration
  • more than the reciprocal of broadened pulse duration
  • same as that of than the reciprocal of broadened pulse duration
  • negligible
Answer: less than the reciprocal of broadened pulse duration
Explaination:
the digital bit rate and pulse duration are always inversely proportional to each other.
Q5 | The maximum bit rate that may be obtained on an optical fiber link is 1/3Γ.
  • true
  • false
Answer: false
Explaination:
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.
Q6 | 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.
  • 4.6 mhz
  • 3.9 mhz
  • 3.3 mhz
  • 4.2 mhz
Answer: 3.3 mhz
Explaination:
the maximum possible bandwidth is equivalent to the maximum possible bitrate. the maximum bit rate assuming no inter-symbol interference is given by
Q7 | Chromatic dispersion is also called as intermodal dispersion.
  • true
  • false
Answer: false
Explaination:
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.
Q8 | Chromatic dispersion is also called as intermodal dispersion.
  • true
  • false
Answer: false
Explaination:
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.
Q9 | 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.
  • 1.2 nm
  • 1.3 nm
  • 0.77 nm
  • 0.98 nm
Answer: 0.77 nm
Explaination:
the relative spectral width σλ/ λ= 0.01 is given. the rms spectral width can be calculated as follows:
Q10 | In waveguide dispersion, refractive index is independent of                              
  • bit rate
  • index difference
  • velocity of medium
  • wavelength
Answer: wavelength
Explaination:
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.
Q11 | 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?
  • 7.31μm
  • 8.71μm
  • 5.26μm
  • 6.50μm
Answer: 6.50μm
Explaination:
normalized frequency v<=2.405 is the value at which the lowest order bessel function j=0. core size(radius)
Q12 | 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?
  • yes
  • no
Answer: yes
Explaination:
v= 2πa.na/λ. calculating this equation, we get the value of v. v is the normalised frequency and should be below
Q13 | What is needed to predict the performance characteristics of single mode fibers?
  • the intermodal delay effect
  • geometric distribution of light in a propagating mode
  • fractional power flow in the cladding of fiber
  • normalized frequency
Answer: geometric distribution of light in a propagating mode
Explaination:
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.
Q14 | Which equation is used to calculate MFD?
  • maxwell’s equations
  • peterman equations
  • allen cahn equations
  • boltzmann’s equations
Answer: peterman equations
Explaination:
mode field diameter is an
Q15 | A single mode fiber has mode field diameter 10.2μm and V=2.20. What is the core diameter of this fiber?
  • 11.1μm
  • 13.2μm
  • 7.6μm
  • 10.1μm
Answer: 10.1μm
Explaination:
for a single mode fiber, mfd=2w0. here, core radius
Q16 | The difference between the modes’ refractive indices is called as                        
  • polarization
  • cutoff
  • fiber birefringence
  • fiber splicing
Answer: fiber birefringence
Explaination:
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.
Q17 | How many propagation modes are present in single mode fibers?
  • one
  • two
  • three
  • five
Answer: two
Explaination:
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.
Q18 | Numerical aperture is constant in case of step index fiber.
  • true
  • false
Answer: true
Explaination:
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.
Q19 | Plastic fibers are less widely used than glass fibers.
  • true
  • false
Answer: true
Explaination:
the majority of the fibers are made up of glass consisting of silica. plastic fibers are used for short distance
Q20 | A perfect semiconductor crystal containing no impurities or lattice defects is called as
  • intrinsic semiconductor
  • extrinsic semiconductor
  • excitation
  • valence electron
Answer: intrinsic semiconductor
Explaination:
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.
Q21 | The energy-level occupation for a semiconductor in thermal equilibrium is described by the                      
  • boltzmann distribution function
  • probability distribution function
  • fermi-dirac distribution function
  • cumulative distribution function
Answer: fermi-dirac distribution function
Explaination:
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
Q22 | The majority of the carriers in a p-type semiconductor are                      
  • holes
  • electrons
  • photons
  • neutrons
Answer: holes
Explaination:
the impurities can be either donor impurities or acceptor impurities.
Q23 |                                    is used when the optical emission results from the application of electric field.
  • radiation
  • efficiency
  • electro-luminescence
  • magnetron oscillator
Answer: electro-luminescence
Explaination:
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.
Q24 | The recombination in indirect band-gap semiconductors is slow.
  • true
  • false
Answer: true
Explaination:
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.
Q25 | 21*10-10cm3s-1.
  • 2ns
  • 1.39ns
  • 1.56ns
  • 2.12ms
Answer: 1.39ns
Explaination:
the radioactive minority carrier lifetime ςrconsidering the p-type region is given by-
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