Determine:

i. The critical angle.

ii. The NA.

iii. The acceptance angle

Mumbai University > Electronics and telecommunication > Sem 7 > optical communication and networks

Marks: 10

Years: MAY 2016

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

To Find:

1) Critical angle $(Өc)$

2) Numerical aperture (N.A)

3) Acceptance angle $(Өa)$

Solution:

$\Rightarrow θ_c =\sin^{-1} \dfrac {n_2}{n_1}= \sin^{-1}( \dfrac {1.47}{1.5}) \\ θ_c =78.52°$

Critical angle $( θ_c ) = 78.52°$

$\Rightarrow N.A = \sqrt{n_1^2 -n_2^2 } \\ =\sqrt{1.5^2-1.47^2 } \\ N.A= 0.2984$

Numerical aperture $(N.A) =0.2984$

$\Rightarrow N.A = \sin Өa \\ 0.2984=\sin Өa \\ Өa = 17.36° $

Acceptance angle $(Өa) = 17.36°$