Category: Optical Communications Objective Questions

250+ TOP MCQs on Overall Fiber Dispersion & Dispersion – Modified Single Mode Fibers and Answers

Optical Communications Interview Questions and Answers for freshers on “Overall Fiber Dispersion & Modified Single Mode Fibers”.

1. A multimode step index fiber has source of RMS spectral width of 60nm and dispersion parameter for fiber is 150psnm-1km-1. Estimate rms pulse broadening due to material dispersion.
a) 12.5ns km-1
b) 9.6ns km-1
c) 9.0ns km-1
d) 10.2ns km-1
Answer: c
Explanation: The RMS pulse broadening per km due to material dispersion is given by
σm(1 km) = σλLM
= 60*1* 150pskm-1
= 9.0nskm-1
Where σλ = rms spectral width
L = length of fiber
M = dispersion parameter.

2. A multimode fiber has RMS pulse broadening per km of 12ns/km and 28ns/km due to material dispersion and intermodal dispersion resp. Find the total RMS pulse broadening.
a) 30.46ns/km
b) 31.23ns/km
c) 28.12ns/km
d) 26.10ns/km
Answer: a
Explanation: The overall dispersion in multimode fibers comprises both chromatic and intermodal terms. The total RMS pulse broadening σT is given by
optical-communication-questions-answers-overall-fiber-dispersion-modified single-mode-fibers-q2
Where σm = RMS pulse broadening due to material dispersion
σi = RMS pulse broadening due to intermodal dispersion.

3. Γg = dβ / C*dk. What is β in the given equation?
a) Attenuation constant
b) Propagation constant
c) Boltzmann’s constant
d) Free-space
Answer: b
Explanation: Above given equation is an equation of transit time or a group delay(Γg) for a light pulse. This light pulse is propagating along a unit length of a single mode fiber.

4. Most of the power in an optical fiber is transmitted in fiber cladding.
a) True
b) False
Answer: b
Explanation: Most of the power in optical fiber is transmitted in fiber core. This is because in multimode fibers, majority of modes propagating in the core area are far from cutoff. Hence more power is transmitted.

5. A single mode fiber has a zero dispersion wavelength of 1.21μm and a dispersion slope of 0.08 psnm-2km-1. What is the total first order dispersion at wavelength 1.26μm.
a) -2.8psnm-1 km-1
b) -3.76psnm-1 km-1
c) -1.2psnm-1 km-1
d) 2.4psnm-1 km-1
Answer: b
Explanation: The total first order dispersion for fiber at two wavelength is obtained by
DT(1260 nm) = λS0/4 [1-(λ0/λ)4]
= (1260*0.08*10-12)/4 * (1-[1550/1260]4)
= -3.76psnm-1km-1
Where
λ0 = zero dispersion wavelength
λ = wavelength
S0 = dispersion slope
DT = total first order dispersion.

6. The dispersion due to material, waveguide and profile are -2.8nm-1km-1, 20.1nm-1km-1 and 23.2nm-1km-1respectively. Find the total first order dispersion?
a) 36.2psnm-1 km-1
b) 38.12psnm-1 km-1
c) 40.5psnm-1 km-1
d) 20.9psnm-1 km-1
Answer: c
Explanation: The total dispersion is given by
DT = DM + DW + DP(psnm-1km-1)
Where
DW = waveguide dispersion
DM = Material dispersion
DP = profile dispersion.

7. Dispersion-shifted single mode fibers are created by __________
a) Increasing fiber core diameter and decreasing fractional index difference
b) Decreasing fiber core diameter and decreasing fractional index difference
c) Decreasing fiber core diameter and increasing fractional index difference
d) Increasing fiber core diameter and increasing fractional index difference
Answer: c
Explanation: It is possible to modify the dispersion characteristics of single mode fibers by tailoring of some fiber parameters. These fiber parameters include core diameter and relative index difference.

8. An alternative modification of the dispersion characteristics of single mode fibers involves achievement of low dispersion gap over the low-loss wavelength region between __________
a) 0.2 and 0.9μm
b) 0.1 and 0.2μm
c) 1.3 and 1.6μm
d) 2 and 3μm
Answer: c
Explanation: Dispersion characteristics can be altered by changing fiber parameters and wavelength. The achievement of low dispersion gap over the region 1.3 and 1.6μm modifies the dispersion characteristics of single mode fibers.

9. The fibers which relax the spectral requirements for optical sources and allow flexible wavelength division multiplying are known as __________
a) Dispersion-flattened single mode fiber
b) Dispersion-enhanced single mode fiber
c) Dispersion-compressed single mode fiber
d) Dispersion-standardized single mode fiber
Answer: a
Explanation: The dispersion-flattened single mode fibers (DFFS) are obtained by fabricating multilayer index profiles with increased waveguide dispersion. This is tailored to provide overall dispersion say 2psnm-1km-1 over the wavelength range 1.3 to 1.6μm.

10. For suitable power confinement of fundamental mode, the normalized frequency v should be maintained in the range 1.5 to 2.4μm and the fractional index difference must be linearly increased as a square function while the core diameter is linearly reduced to keep v constant. This confinement is achieved by?
a) Increasing level of silica doping in fiber core
b) Increasing level of germanium doping in fiber core
c) Decreasing level of silica germanium in fiber core
d) Decreasing level of silica doping in fiber core
Answer: b
Explanation: The tailoring of fiber parameters provides suitable power confinement. These parameters may be diameter, index-difference, frequency etc. The doping level of germanium contributes to the tailoring of fiber parameters; which in turn provides suitable power confinement.

11. Any amount of stress occurring at the core-cladding interface would be reduced by grading the material composition.
a) True
b) False
Answer: a
Explanation: A problem arises with that of simple step index approach to dispersion shifting is high. The fibers produced exhibit high dopant-dependent losses at operating wavelengths. These losses are caused by induced-stress in the region of core-cladding interface. This can be reduced by grading the material composition of the fiber.

12. The variant of non-zero-dispersion-shifted fiber is called as __________
a) Dispersion flattened fiber
b) Zero-dispersion fiber
c) Positive-dispersion fiber
d) Negative-dispersion fiber
Answer: d
Explanation: The dispersion profile for non-zero dispersion shifted fiber is referred to as bandwidth non-zero-dispersion-shifted fiber. It was introduced to provide wavelength division multiplexed applications to be extended into the s-band. The variant of non-zero-dispersion-shifted fiber can also be referred to as dispersion compensating fiber.

13. Non-zero-dispersion-shifted fiber was introduced in the year 2000.
a) True
b) False
Answer: b
Explanation: Non-zero-dispersion-shifted fiber was introduced in mid-1990s to provide wavelength division multiplexing applications. In the year 2000, the dispersion profile for non-zero-dispersion-shifted fiber was introduced.

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250+ TOP MCQs on Fiber Cutoff Wavelength Measurements and Answers

Optical Communications Multiple Choice Questions on “Fiber Cutoff Wavelength Measurements”.

1. A multimode fiber has many cutoff wavelengths.
a) False
b) True
Answer: b
Explanation: A multimode fiber has many cutoff wavelengths. It is because the number of bound propagating modes is usually large.

2. What does ‘a’ stands for in the given equation?

Mg=(πa/λ)2(n12 - n22)

a) Radius of the core
b) Constant
c) Coefficient of refraction
d) Density
Answer: a
Explanation: The above equation gives the number of guided modes for a parabolic refractive index graded fiber, where a is the core radius and n1, n2 are the core and cladding indices respectively.

3. The _________ wavelength is defined as the wavelength greater than which the ratio of the total power and the fundamental mode power has to be decreased to less than 0.1dB.
a) Magnetic
b) Quasi
c) Cut-off
d) EIA
Answer: c
Explanation: The cut off wavelength is usually measured by increasing the signal wavelength in a fixed length of fiber until the mode is undetectable. It is usually called a effective cut-off wavelength.

4. How many methods are used to determine the effective cutoff wavelength?
a) Two
b) One
c) Four
d) Three
Answer: d
Explanation: Three methods are usually used for the determination of the effective cutoff wavelength. These are bending-reference technique, power step method and alternative test method.

5. What is the name of the test used to determine the efficient values of the effective cutoff wavelength?
a) Round robin test
b) Mandarin test
c) Hough Werner test
d) Fulton test
Answer: a
Explanation: Round robin test is an effective method to determine the efficient values for the cutoff wavelength. It shows that in some methods, the values through round robin test are the same.

6. The effective cutoff wavelength for a cabled single mode fiber will be generally smaller than that of the un-cabled fiber.
a) True
b) False
Answer: a
Explanation: The effective cutoff wavelength for a cabled single mode fiber is always smaller than that of the un-cabled fiber. This is usually because of the bend effects.

7. How many bend effects are produced in the fiber?
a) One
b) Three
c) Two
d) Four
Answer: c
Explanation: Usually, two bend effects are produced. They are macro-bending and micro-bending. These effects incarcerate certain changes in the fiber efficiency.

8. _______________ method does not require a leaky mode correction factor or equal mode excitation.
a) Bending-reference
b) Power step method
c) Alternative test method
d) Refracted near-field method
Answer: d
Explanation: Refracted near-field method is complementary to the transmitted near-field method. It has the advantage that it does not require a leaky mode correction factor. Moreover, it provides the relative refractive index directly without recourse to external calibration.

9. The _______ method is the most commonly used method for the determination of the fiber refractive index profile.
a) Refracted near-field method
b) Bending-reference
c) Power step method
d) Alternative test method
Answer: a
Explanation: It is the most commonly used technique. Also, it is the EIA reference test method for both multimode and single mode fibers.

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250+ TOP MCQs on The Optical Transmitter Circuit and Answers

Optical Communications Multiple Choice Questions on “The Optical Transmitter Circuit”.

1. _____________ must be operated in stimulated emission region.
a) Injection laser
b) LED’s
c) Detector
d) Receiver
Answer: a
Explanation: Injection laser is a threshold device. In stimulated emission region, continuous optical output power levels are in the range of 1 to 10mW.

2. Coherent radiation is relatively __________
a) Parabolic
b) Elliptic
c) Directional
d) Rectangular
Answer: c
Explanation: Most of the light output is coupled into optical fibre. This is because of the isotropic distribution of narrow-line width, coherent radiation is directional.

3. _____________ are capable of launching powers between 0.5 and several mW.
a) LED’s
b) Injection laser
c) Attenuator
d) Reflector
Answer: b
Explanation: Coupling efficiency up to 30% may be obtained by placing a fiber close to laser mirror. These can approach 90% with suitable lens and optical coupling arrangements. So they can launch 0.5 to several mW of optical power into fiber.

4. LED’s display good linearity.
a) True
b) False
Answer: a
Explanation: LED’s appear to be suited to analog transmission. This is because of its output which is directly proportional to the drive current.

5. Which behaviour may prove as a limitation for injection lasers and LED’s?
a) Isotropic
b) Radioactive
c) Thermal
d) Photosensitive
Answer: c
Explanation: The thermal behaviour of the injection lasers and the LED’s limits their operation within the optical transmitter. The main problem is caused by the variation of injection laser threshold current.

6. Optical output power from an LED is directly proportional to the device junction temperature.
a) False
b) True
Answer: b
Explanation: Output power is dependent on the junction temperature in case of LED’s. Most LED’s exhibit a decrease in the optical output power following an increase in junction temperature.

7. _____________ from the LED is dependent on the effective minority carrier lifetime in the semiconductor material.
a) Spontaneous emission
b) Stimulated emission
c) Absorption
d) Diffusion
Answer: a
Explanation: The speed of the response of the LED is dictated by the respective emission mechanism. Spontaneous emission is related to the carrier lifetime and hence dictating the speed of response.

8. The _________ of the LED is twice that of the effective minority carrier lifetime.
a) Dwell time
b) Reflection scatters
c) Sensitivity
d) Rise time
Answer: d
Explanation: The response of the optical fiber source is specified in terms of the rise time. This rise time is reciprocally related to the device frequency response.

9. The finite spectral width of the optical source causes ___________
a) Depletion
b) Frequency burst
c) Pulse broadening
d) Efficient reflection
Answer: c
Explanation: The finite spectral width causes pulse broadening due to material dispersion on an optical fiber communication link. This results in a limitation on the bandwidth-length product.

10. The coherent emission from an injection laser has a line width of ________
a) 2 nm
b) 3nm
c) 8 nm
d) 1nm
Answer: d
Explanation: An optical source such as an injection laser is a narrow line width device as compared to the LED. It has a narrow line width of 1 nm or less.

11. Extinction ratio is denoted by symbol __________
a) ε
b) σ
c) β
d) ρ
Answer: a
Explanation: Extinction ratio is defined as the ratio of the optical energy emitted in the 0 bit period to that emitted during the 1 bit period. It is denoted by ε.

12. The use of low impedance driving circuit may increase _____________
a) Noise
b) Width
c) Intensity
d) Switching speed
Answer: d
Explanation: Pulse shaping is usually required to increase the switching speed. However, increased switching speed may be obtained from an LED without a speed-up element by use of a low-impedance driving circuit.

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250+ TOP MCQs on Semiconductor Photodiodes Without Internal Gain and Answers

Optical Communications Multiple Choice Questions & Answers on “Semiconductor Photodiodes Without Internal Gain”.

1. The width of depletion region is dependent on ___________ of semiconductor.
a) Doping concentrations for applied reverse bias
b) Doping concentrations for applied forward bias
c) Properties of material
d) Amount of current provided
Answer: a
Explanation: The depletion region is formed by immobile positively and immobile negatively charged donor and acceptor atoms in n- and p-type respectively. When carriers are swept towards majority side under electric field, lower the doping, wider the depletion region.

2. Electron-hole pairs are generated in ___________
a) Depletion region
b) Diffusion region
c) Depletion region
d) P-type region
Answer: c
Explanation: Photons are absorbed in both depletion and diffusion regions. The position and width of absorption region depends on incident photons energy. The absorption region may extend throughout device in weakly absorption of photons. Thus carriers are generated in both regions.

3. The diffusion process is _____________ as compared with drift.
a) Very fast
b) Very slow
c) Negligible
d) Better
Answer: b
Explanation: None.

4. Determine drift time for carrier across depletion region for photodiode having intrinsic region width of 30μm and electron drift velocity of 105 ms-1.
a) 1×10-10 Seconds
b) 2×10-10 Seconds
c) 3×10-10 Seconds
d) 4×10-10 Seconds
Answer: c
Explanation: The drift time is given by
tdrift = w/vd = 30×10-6/1×10-10 = 3×10-10 seconds.

5. Determine intrinsic region width for a photodiode having drift time of 4×10-10 s and electron velocity of 2×10-10ms-1.
a) 3×10-5M
b) 8×10-5M
c) 5×10-5M
d) 7×10-5M
Answer: b
Explanation: The drift time is given by
tdrift = w/vd
4×10-10 = w/2×105
= 4×10-10×2×105
= 8×10-5m.

6. Determine velocity of electron if drift time is 2×10-10s and intrinsic region width of 25×10-6μm.
a) 12.5×104
b) 11.5×104
c) 14.5×104
d) 13.5×104
Answer: a
Explanation: The drift time is given by
tdrift = w/vd
vd = 25×10-6/2×10-10 = 12.5×104ms-1.

7. Compute junction capacitance for a p-i-n photodiode if it has area of 0.69×10-6m2, permittivity of 10.5×10-13Fcm-1 and width of 30μm.
a) 3.043×10-5
b) 2.415×10-7
c) 4.641×10-4
d) 3.708×10-5
Answer: b
Explanation: The junction capacitance is given by,
Cj = εsA/w = 10.5×10-13×0.69×10-6/30×10-13
= 2.415×10-7F.

8. Determine the area where permittivity of material is 15.5×10-15Fcm-1 and width of 25×10-6 and junction capacitance is 5pF.
a) 8.0645×10-5
b) 5.456×10-6
c) 3.0405×10-2
d) 8.0645×10-3
Answer: d
Explanation: The junction capacitance is given by,
Cj = εsA/ w = 5×10-12×25×10-6/15.5×10-15
= 8.0645×10-3m2.

9. Compute intrinsic region width of p-i-n photodiode having junction capacitance of 4pF and material permittivity of 16.5×10-13Fcm-1 and area of 0.55×10-6m2.
a) 7.45×10-6
b) 2.26×10-7
c) 4.64×10-7
d) 5.65×10-6
Answer: b
Explanation: The junction capacitance is given by,
Cj = εsA/ W
w = εsA/Cj
= 16.5×10-13 × 0.55×10-6/4×10-12
= 2.26×10-7.

10. Determine permittivity of p-i-n photodiode with junction capacitance of 5pF, area of 0.62×10-6m2 and intrinsic region width of 28 μm.
a) 7.55×10-12
b) 2.25×10-10
c) 5×10-9
d) 8.5×10-12
Answer: b
Explanation: The junction capacitance is given by,
Cj = εsA/ W
εs = Cj w/A = 5×10-12×28×10-6/0.62×10-6
= 2.25×10-10Fcm-1.

11. Determine response time of p-i-n photodiode if it has 3 dB bandwidth of 1.98×108Hz.
a) 5.05×10-6sec
b) 5.05×10-7Sec
c) 5.05×10-7sec
d) 5.05×10-8Sec
Answer: c
Explanation: The maximum response time is
Maximum response time = 1/Bm = 1/1.98×108 = 5.05×10-9sec.

12. Compute maximum 3 dB bandwidth of p-i-n photodiode if it has a max response time of 5.8 ns.
a) 0.12 GHz
b) 0.14 GHz
c) 0.17 GHz
d) 0.13 GHz
Answer: c
Explanation: The maximum response time is
Maximum response time = 1/Bm
= 1/5.8×10-9 = 0.17 GHz.

13. Determine maximum response time for a p-i-n photodiode having width of 28×10-6m and carrier velocity of 4×104ms-1.
a) 105.67 MHz
b) 180.43 MHz
c) 227.47 MHz
d) 250.65 MHz
Answer: c
Explanation: Maximum 3 dB bandwidth of photodiode is given by
Bm = Vd/2ΠW = 4×10-4/2×3.14×28×10-6 = 227.47 MHz.

14. Determine carrier velocity of a p-i-n photodiode where 3dB bandwidth is1.9×108Hz and depletion region width of 24μm.
a) 93.43×10-5
b) 29.55×10-3
c) 41.56×10-3
d) 65.3×10-4
Answer: b
Explanation: Maximum 3 dB bandwidth of photodiode is given by
Bm = Vd/2ΠW
Vd = Bm × 2Π × W
= 1.98×108×2Π×24×10-6
= 29.55×10-3.

15. Compute depletion region width of a p-i-n photodiode with 3dB bandwidth of 1.91×108and carrier velocity of 2×104ms-s.
a) 1.66×10-5
b) 3.2×10-3
c) 2×10-5
d) 2.34×104
Answer: a
Explanation: Maximum 3 dB bandwidth of photodiode is given by
Bm = Vd/2ΠW
W = Vd/Bm2Π
= 2×10-5/1.91×108×2Π
= 1.66×10-5m.

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250+ TOP MCQs on Optical Sources : Laser Basics and Answers

Optical Communications Multiple Choice Questions on “Optical Sources : Laser Basics”.

1. 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
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.

2. How many types of sources of optical light are available?
a) One
b) Two
c) Three
d) Four
Answer: 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.

3. The frequency of the absorbed or emitted radiation is related to difference in energy E between the higher energy state E2 and the lower energy state E1. State what h stands for in the given equation?

E = E2 - E1 = hf

a) Gravitation constant
b) Planck’s constant
c) Permittivity
d) Attenuation constant
Answer: b
Explanation: In the given equation, difference in the energy E is directly proportional to the absorbed frequency (f) where h is used as a constant and is called as Planck’s constant. The value of h is measured in Joules/sec & is given by-
h = 6.626×10-34Js.

4. The radiation emission process (emission of a proton at frequency) can occur in __________ ways.
a) Two
b) Three
c) Four
d) One
Answer: a
Explanation: The emission process can occur in two ways. First is by spontaneous emission in which the atom returns to the lower energy state in a random manner. Second is by stimulated emission where the energy of a photon is equal to the energy difference and it interacts with the atom in the upper state causing it to return to the lower state along with the creation of a new photon.

5. 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
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.

6. 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
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.

7. 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
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.

8. __________________ 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
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.

9. 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: 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.

10. 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
Explanation: The modes of laser are separated by a frequency internal δf and this separation is given by-
δf = c/2nL
Where
c = velocity of light
n = Refractive index
L = Length of the crystal.

11. Doppler broadening is a homogeneous broadening mechanism.
a) True
b) False
Answer: b
Explanation: Doppler broadening is a inhomogeneous broadening mechanism. In this broadening, the individual groups of atoms have different apparent resonance frequencies. Atomic collisions usually provide homogeneous broadening as each atom in collection has same resonant frequency and spectral spread.

12. An injection laser has active cavity losses of 25 cm-1 and the reflectivity of each laser facet is 30%. Determine the laser gain coefficient for the cavity it has a length of 500μm.
a) 46 cm-1
b) 51 cm-1
c) 50 cm-1
d) 49.07 cm-1
Answer: d
Explanation: The laser gain coefficient is equivalent to the threshold gain per unit length and is given by –
gth = α + 1/L ln (1/r)
Where
α = active cavity loss
L = Length of the cavity
r = reflectivity.

13. Longitudinal modes contribute only a single spot of light to the laser output.
a) True
b) False
Answer: a
Explanation: Laser emission includes the longitudinal modes and transverse modes. Transverse modes give rise to a pattern of spots at the output. Longitudinal modes give only a spot of light to the output.

14. Considering the values given below, calculate the mode separation in terms of free space wavelength for a laser. (Frequency separation = 2GHz, Wavelength = 0.5 μm)
a) 1.4×10-11
b) 1.6×10-12
c) 1×10-12
d) 6×10-11
Answer: b
Explanation: The mode separation in terms of free space wavelength is given by-
δλ = λ2/c δf
Where
δf = frequency separation
λ = wavelength
c = velocity of light.

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250+ TOP MCQs on Intermodal Dispersion and Answers

Optical Communications Multiple Choice Questions on “Intermodal Dispersion”.

1. Intermodal dispersion occurring in a large amount in multimode step index fiber results in ____________
a) Propagation of the fiber
b) Propagating through the fiber
c) Pulse broadening at output
d) Attenuation of waves
Answer: c
Explanation: Pulse broadening due to intermodal dispersion is caused due to difference in propagation delay between different modes in the multimode fiber. As different modes travel with different group velocities, the pulse width at output depends on transmission time of all modes. This creates difference in overall dispersion which results in pulse broadening.

2. After Total Internal Reflection the Meridional ray __________
a) Makes an angle equal to acceptance angle with the axial ray
b) Makes an angle equal to critical angle with the axial ray
c) Travels parallel equal to critical angle with the axial ray
d) Makes an angle equal to critical angle with the axial ray
Answer: d
Explanation: The Meridional ray travels along the axis of the fiber. When the ray is incident, makes an angle equal to acceptance angle and thus it propagates through the fiber. As the propagating ray gets refracted from the boundary, it makes an angle (i.e. critical angle) with the normal.

3. Consider a single mode fiber having core refractive index n1= 1.5. The fiber length is 12m. Find the time taken by the axial ray to travel along the fiber.
a) 1.00μsec
b) 0.06μsec
c) 0.90μsec
d) 0.30μsec
Answer: b
Explanation: The time taken by the axial ray to travel along the fiber gives the minimum delay time
Tmin = Ln1/c
Where L = length of the fiber
n1 = Refractive index of core
c = velocity of light in vacuum.

4. A 4 km optical link consists of multimode step index fiber with core refractive index of 1.3 and a relative refractive index difference of 1%. Find the delay difference between the slowest and fastest modes at the fiber output.
a) 0.173 μsec
b) 0.152 μsec
c) 0.96 μsec
d) 0.121 μsec
Answer: a
Explanation: The delay difference is given by
δTs = Ln1/c
Where δTs = delay difference
n1 = core refractive index
Δ = Relative refractive index difference
c = velocity of light in vacuum.

5. A multimode step-index fiber has a core refractive index of 1.5 and relative refractive index difference of 1%. The length of the optical link is 6 km. Estimate the RMS pulse broadening due to intermodal dispersion on the link.
a) 92.6 ns
b) 86.7 ns
c) 69.3 ns
d) 68.32 ns
Answer: b
Explanation: The RMS pulse broadening due to intermodal dispersion is obtained by the equation is given below:
σs = Ln1Δ/2√3c
Where σs = RMS pulse broadening
L = length of optical link
C = velocity of light in vacuum
n1 = core refractive index.

6. The differential attenuation of modes reduces intermodal pulse broadening on a multimode optical link.
a) True
b) False
Answer: a
Explanation: Intermodal dispersion may be reduced by propagation mechanisms. The differential attenuation of various modes is due to the greater field penetration of the higher order modes into the cladding of waveguide. These slower modes exhibit larger losses at any core-cladding irregularities.

7. The index profile of a core of multimode graded index fiber is given by?
a) N (r) = n1 [1 – 2Δ(r2/a)2]1/2; rb) N (r) = n1 [3 – 2Δ(r2/a)2]1/2; rc) N (r) = n1 [5 – 2Δ(r2/a)2]1/2; r>a
d) N (r) = n1 [1 – 2Δ(r2/a)2]1/2; rAnswer: d
Explanation: In multimode graded index fibers, many rays can propagate simultaneously. The Meridional rays follow sinusoidal trajectories of different path length which results from index grading.

8. Intermodal dispersion in multimode fibers is minimized with the use of step-index fibers.
a) True
b) False
Answer: b
Explanation: As multimode graded index fibers show substantial bandwidth improvement over multimode step index fibers. So, inter-modal dispersion in multimode fiber is minimized with the use of multimode graded index fibers.

9. Estimate RMS pulse broadening per km due to intermodal dispersion for multimode step index fiber where length of fiber is 4 km and pulse broadening per km is 80.6 ns.
a) 18.23ns/km
b) 20.15ns/km
c) 26.93ns/km
d) 10.23ns/km
Answer: b
Explanation:
The RMS pulse broadening per km due to intermodal dispersion for multimode step index fiber is given by
(σs(1 km)/L = 80.6/4 = 20.15
Where L = length of fiber
σs = pulse broadening.

10. Practical pulse broadening value for graded index fiber lies in the range of __________
a) 0.9 to 1.2 ns/km
b) 0.2 to 1 ns/km
c) 0.23 to 5 ns/km
d) 0.45 to 8 ns/km
Answer: b
Explanation: As all optical fiber sources have a finite spectral width, the profile shape must be altered to compensate for this dispersion mechanism. The minimum overall dispersion for graded index fiber is also limited by other intermodal dispersion mechanism. Thus pulse broadening values lie within range of 0.2 to 1 ns/km.

11. The modal noise occurs when uncorrected source frequency is?
a) δf>>1/δT
b) δf=1/δT
c) δf<<1/δT
d) Negligible
Answer: a
Explanation: Modal noise is dependent on change in frequency. Frequency is inversely proportional to time. The patterns are formed by interference of modes from a coherent source when coherence time of source is greater than intermodal dispersion time δT within fiber.

12. Disturbance along the fiber such as vibrations, discontinuities, connectors, splices, source/detectors coupling result in __________
a) Modal noise
b) Inter-symbol interference
c) Infrared interference
d) Pulse broadening
Answer: a
Explanation: Disturbance along the fiber cause fluctuations in specific pattern. These speckle patterns have characteristics time longer than resolution time of detector and is known as modal noise.

13. The modal noise can be reduced by __________
a) Decreasing width of signal longitudinal mode
b) Increasing coherence time
c) Decreasing number of longitudinal modes
d) Using fiber with large numerical aperture
Answer: d
Explanation: Disturbances along fiber cause fluctuations in speckle patterns. Fibers with large numerical apertures support the transmission of large number of modes giving greater number of speckle, thereby reducing modal noise.

14. Digital transmission is more likely to be affected by modal noise.
a) True
b) False
Answer: b
Explanation: Analog transmission is more affected by modal noise due to higher optical power levels which is required at receiver when quantum noise effects are considered. So it is important to look into design considerations.

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