Optical Communications Multiple Choice Questions on “Single-Mode Fibers”.
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. 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
Answer: a
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
Answer: b
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
Answer: b
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 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
Answer: c
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
Answer: c
Explanation: Bf=ny– nx = λ/Lp. Here, λ=wavelength and Lp = beat length. Solving this equation, we will get the answer.
8. How many propagation modes are present in single mode fibers?
a) One
b) Two
c) Three
d) Five
Answer: b
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
Answer: a
Explanation: 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.
10. Plastic fibers are less widely used than glass fibers.
a) True
b) False
Answer: a
Explanation: The majority of the fibers are made up of glass consisting of silica. Plastic fibers are used for short distance transmissions unlike glass fibers which can also be used for long haul applications. Also, plastic fibers have higher attenuation than glass fibers.