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Question: What should be the maximum acceptance angle at the air core interface of an optical fibre if n<sub>1...

What should be the maximum acceptance angle at the air core interface of an optical fibre if n1 and n2 are the refractive indices of the core and the cladding, respectively

A

sin⁡−1(n2/n1)\sin^{- 1}(n_{2}/n_{1})

B

sin⁡−1n12−n22\sin^{- 1}\sqrt{n_{1}^{2} - n_{2}^{2}}

C

[tan⁡−1n2n1]\left\lbrack \tan^{- 1}\frac{n_{2}}{n_{1}} \right\rbrack

D

[tan⁡−1n1n2]\left\lbrack \tan^{- 1}\frac{n_{1}}{n_{2}} \right\rbrack

Answer

sin⁡−1n12−n22\sin^{- 1}\sqrt{n_{1}^{2} - n_{2}^{2}}

Explanation

Solution

Core of acceptance angle θ=sin⁡−1n12−n22\theta = \sin^{- 1}\sqrt{n_{1}^{2} - n_{2}^{2}}