This set of Optical Communications Multiple Choice Questions & Answers (MCQs) focuses on “Cylindrical Fiber”.
1. A multimode step index fiber has a normalized frequency of 72. Estimate the number of guided modes.
Explanation: A step-index fiber has a constant refractive index core. The number of guided modes in a step-index fiber are given by M = (V*V)/2. Here M denotes the number of modes and V denotes normalized frequency.
2. A graded-index fiber has a core with parabolic refractive index profile of diameter of 30μm, NA=0.2, λ=1μm. Estimate the normalised frequency.
Explanation: Normalized frequency for a graded index fiber is given by V= 2Πa(NA)/λ. Substituting and calculating the values, we get option b. Here, V denotes normalized frequency and NA= numerical aperture.
3. A step-index fiber has core refractive index 1.46 and radius 4.5μm. Find the cutoff wavelength to exhibit single mode operation. Use relative index difference as 0.25%.
Explanation: The cutoff wavelength is the wavelength beyond which no single mode operation takes place. On solving λc= 2Πan12∆/ V, we get option c. Here, V=2.405, n1= refractive index of core, a=radius of core.
4. A single-mode step-index fiber or multimode step-index fiber allows propagation of only one transverse electromagnetic wave.
Explanation: Single mode step index fiber is also called as mono-mode step index fiber. As the name suggests, only one mode is transmitted and hence it has the distinct advantage of low intermodal dispersion.
5. One of the given statements is true for intermodal dispersion. Choose the right one.
a) Low in single mode and considerable in multimode fiber
b) Low in both single mode and multimode fiber
c) High in both single mode and multimode fiber
d) High in single mode and low in multimode fiber
Explanation: Single mode propagates only one wave or only one mode is transmitted. Therefore, intermodal dispersion is low in single mode. In multimode fibers, higher dispersion may occur due to varying group velocities of propagating modes.
6. For lower bandwidth applications,
a) Single mode fiber is advantageous
b) Photonic crystal fibers are advantageous
c) Coaxial cables are advantageous
d) Multimode fiber is advantageous
Explanation: In multimode fibers, intermodal dispersion occurs. The group velocities often differ which gradually restricts maximum bandwidth attainability in multimode fibers.
7. Most of the optical power is carried out in core region than in cladding. State true or false:
Explanation: In an ideal multimode fiber, there is no mode coupling. The optical power launched into a particular mode remains in that mode itself. The majority of these modes are mostly confined to fiber core only.
8. Meridional rays in graded index fibers follow
a) Straight path along the axis
b) Curved path along the axis
c) Path where rays changes angles at core-cladding interface
d) Helical path
Explanation: Meridional rays pass through axis of the core. Due to the varying refractive index at the core, the path of rays is in curved form.
9. What is the unit of normalized frequency?
d) It is a dimensionless quantity
Explanation: Normalized frequency of optical fiber is the frequency which exists at cut-off condition. There is no propagation and attenuation above cut-off. It is directly proportional to numerical aperture which is a dimensionless quantity; hence itself is a dimensionless quantity.
10. Skew rays follow a
a) Hyperbolic path along the axis
b) Parabolic path along the axis
c) Helical path
d) Path where rays changes angles at core-cladding interface
Explanation: The ray which does not pass through the fiber axis is termed as skew ray. Unlike Meridional rays, skew rays are more in number which makes them follow a round path called as helical path.
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.