Question

The angle of acceptance of an optical fibre is 30 degree when kept in air. find the angle of acceptance when it is in a medium of refractive index 1.33

Answer 1

Acceptance angle= $\sin^{-1}(\frac{NA}{n_o})=\sin^{-1}(\frac{n_1^2-n_2^2}{n_o})$

in air $n_o=1$

$30^o=\sin^{-1}(NA)$

NA=0.5

acceptance angle when refractive index of medium is 1.33

new acceptance angle = $\sin^{-1}(\frac{0.5}{1.33})=\sin^{-1}0.37$ =$21.7^o$