UNIT 1
INTRODUCTION TO OPTICAL FIBERS
1. A light ray is incident from medium-1 to medium-2. If the refractive indices of medium-1 and
medium 2 are 1.5 and 1.36 respectively then determine the angle of refraction for an angle of
incidence of 30 o.
2. Calculate the NA, acceptance angle and critical angle of the fiber having
n1(Core refractive index)=1.50 and refractive index of cladding n2 = 1.45.
5.Calculate the number of modes of an optical fiber having diameter of 50µm, n1=1.48,
n2=1.46 and λ=0.82µm.
Solution: d=50µm
n1=1.4830
n2=1.46
λ=0.82µm
6.A fiber has normalized frequency V=26.6 and the operating wavelength
is1300nm. If the radius of the fiber core is 25µm. Compute the numerical
aperture.
Solution: V=26.6 λ=1300nm=1300X10-9m
a=25µm=25X10-6m
7. A multimode step index fiber with a core diameter of 80µm and a relative
index difference of 1.5% is operating at a wavelength of 0.85µm. If the core
refractive index is 1.48, estimate the normalized frequency for the fiber and
number of guided modes.
Solution:
Given:
MMstepindexfiber, 2a=80µm
Core radians a=40µm
Relative index difference=1.5% = 0.015
Wavelength, λ=0.85µm
Core refractive index, n1 =1.48
Normalized frequency,V=?
Number of modes,M=?
Optical Communication-Unit 1- Problems

Optical Communication-Unit 1- Problems

  • 1.
    UNIT 1 INTRODUCTION TOOPTICAL FIBERS
  • 3.
    1. A lightray is incident from medium-1 to medium-2. If the refractive indices of medium-1 and medium 2 are 1.5 and 1.36 respectively then determine the angle of refraction for an angle of incidence of 30 o.
  • 4.
    2. Calculate theNA, acceptance angle and critical angle of the fiber having n1(Core refractive index)=1.50 and refractive index of cladding n2 = 1.45.
  • 7.
    5.Calculate the numberof modes of an optical fiber having diameter of 50µm, n1=1.48, n2=1.46 and λ=0.82µm. Solution: d=50µm n1=1.4830 n2=1.46 λ=0.82µm
  • 8.
    6.A fiber hasnormalized frequency V=26.6 and the operating wavelength is1300nm. If the radius of the fiber core is 25µm. Compute the numerical aperture. Solution: V=26.6 λ=1300nm=1300X10-9m a=25µm=25X10-6m
  • 9.
    7. A multimodestep index fiber with a core diameter of 80µm and a relative index difference of 1.5% is operating at a wavelength of 0.85µm. If the core refractive index is 1.48, estimate the normalized frequency for the fiber and number of guided modes. Solution: Given: MMstepindexfiber, 2a=80µm Core radians a=40µm Relative index difference=1.5% = 0.015 Wavelength, λ=0.85µm Core refractive index, n1 =1.48 Normalized frequency,V=? Number of modes,M=?