## Optical Communications Questions and Answers Part-3

1. 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?
a) 7.31μm
b) 8.71μm
c) 5.26μm
d) 6.50μm

Explanation: Normalized frequency V<=2.405 is the value at which the lowest order Bessel function J=0. Core size(radius)
$=a=V\lambda/2\pi\sqrt{n_1^2-n_2^2}=6.50 \mu m$

2. 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?
a) Yes
b) No

Explanation: V= 2πa.NA/λ. Calculating this equation, we get the value of V. V is the normalised frequency and should be below 2.405 in order to operate the fiber at single mode. Here, V=2.094, is less than 2.405. Thus, this optical fiber exhibit single mode operation.

3. What is needed to predict the performance characteristics of single mode fibers?
a) The intermodal delay effect
b) Geometric distribution of light in a propagating mode
c) Fractional power flow in the cladding of fiber
d) Normalized frequency

Explanation: 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.

4. Which equation is used to calculate MFD?
a) Maxwell’s equations
b) Peterman equations
c) Allen Cahn equations
d) Boltzmann’s equations

Explanation: Mode field diameter is an important parameter for single mode fibers because it is used to predict fiber properties such as splice loss, bending loss. The standard technique is to first measure the far-field intensity distribution and then calculating mode field diameter using Peterman equations.

5. A single mode fiber has mode field diameter 10.2μm and V=2.20. What is the core diameter of this fiber?
a) 11.1μm
b) 13.2μm
c) 7.6μm
d) 10.1μm

Explanation: For a single mode fiber, MFD=2w0. Here, core radius
$a=w_{0}\sqrt{0.65+1.619V^{-3/2}+2.879V^{-6}}$
Solving this equation, we get a=5.05μm. Core-diameter=2a=10.1μm.

6. The difference between the modes’ refractive indices is called as ___________
a) Polarization
b) Cutoff
c) Fiber birefringence
d) Fiber splicing

Explanation: 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.

7. A single mode fiber has a beat length of 4cm at 1200nm. What is birefringence?
a) 2*10-5
b) 1.2*10-5
c) 3*10-5
d) 2

Explanation: Bf=ny– nx = λ/Lp. Here, λ=wavelength and Lp = beat length. Solve this equation to get the answer.

8. How many propagation modes are present in single mode fibers?
a) One
b) Two
c) Three
d) Five

Explanation: 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.

9. Numerical aperture is constant in case of step index fiber.
a) True
b) False