Question

An optical fiber whose core refractive index is 1.5 and refractive index of cladding is 1.47.Determine

1) the critical angle at the core cladding interface

2) the numerical aperture

3) the acceptance angle

Answer 1

Solution

Given data in question

core refractive index of

optical cable

$n_1=$ 1.5

refractive index of cladding

$n_2=$ 1.47

a) critical angle is given by

$\theta_c=\sin^{-1}(\frac{n_2}{n_1})$

Putting all values

$\theta_c=\sin^{-1}(\frac{1.47}{1.5}) =78.52°$

b) Numerical aperture is given

$N. A=\sqrt{n_1^2-n_2^2}$

Putting all values

Then we get

$N. A=\sqrt{(1.5)^2-(1.47)^2}\\=0.298$

c) now acceptance angle is given by

$N. A=\sin\theta_a$

Putting NA value and solve this

$\theta _a=\sin^{-1}(0.298) =17.36°$