Optical Communications Questions and Answers – Intermodal Dispersion

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This set of Optical Communications Multiple Choice Questions & Answers (MCQs) focuses 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
View Answer

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.
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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
View Answer

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
View Answer

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
View Answer

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
View Answer

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.
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6. The differential attenuation of modes reduces intermodal pulse broadening on a multimode optical link. State whether the given statement is true or false.
a) True
b) False
View Answer

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; r<a
b) N (r) = n1 [3 – 2Δ(r2/a)2]1/2; r<a
c) N (r) = n1 [5 – 2Δ(r2/a)2]1/2; r>a
d) N (r) = n1 [1 – 2Δ(r2/a)2]1/2; r<a
View Answer

Answer: 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. State whether the given statement is true or false.
a) True
b) False
View Answer

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
View Answer

Answer: b
Explanation:
The RMS pulse broadening per km due to intermodal dispersion for multimode step index fiber is given by
s(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
View Answer

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.
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11. The modal noise occurs when uncorrected source frequency is
a) δf>>1/δT
b) δf=1/δT
c) δf<<1/δT
d) Negligible
View Answer

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
View Answer

Answer: 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
View Answer

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. State true or false
a) True
b) False
View Answer

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.

Sanfoundry Global Education & Learning Series – Optical Communications.
To practice all areas of Optical Communications, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn