A silica optical fiber with core diameter large enough to be considered by Ray theory has a core refractive index of 1.5 and cladding refractive index of 1.47.

Determine:

$i. The \ critical \ angle. \\ ii.The \ NA. \\ iii.The \ acceptance \ angle$

Mumbai University > Electronics and Telecommunication > Sem7 > Optical Communication and Networks

Marks: 10M

Year: Dec 2014

Given: $n_1=1.5 \ \ n_2=1.47$

To Find:

1) Critical angle ($Ө_c$)

2) Numerical aperture (N.A)

3) Acceptance angle ($Ө_a$)

Solution:

• $θ_{c }=sin^{-1} \frac{n_2}{n_1}=sin^{-1}\left(\frac{1.47}{1.5}\right)$

• $θ_{c }$=78.52°

$$\boxed{Critical \ \ angle ( θ_{c }) = {78.52}^° } $$

• N.A = $\sqrt{n_1^2-n_2^2 }\\ =\sqrt{1.5^2-1.47^2 }$

  N.A= 0.2984

$$\boxed{Numerical \ \ aperture \ \ (N.A) =0.2984}$$

• N.A = sin $Ө_a$

• 0.2984=sin $Ө_a$

• $Ө_a$ = 17.36°

$$\boxed{Acceptance \ angle\ (Өa) = 17.36°}$$