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(B) Let $	heta$ be the angle of incidence at the air-core interface and $r$ be the angle of refraction inside the core.Applying Snell's Law at the ai

Vedclass · Accepted Answer

$\sin^{-1} \, \sqrt {n^2_1 - n^2_2}$

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(B) Let $	heta$ be the angle of incidence at the air-core interface and $r$ be the angle of refraction inside the core.Applying Snell's Law at the air-core interface:$1 	imes \sin 	heta = n_1 	imes \sin r$$\sin r = \frac{\sin 	heta}{n_1}$  .........$(i)$For total internal reflection at the core-cladding interface,the angle of incidence $i$ must be greater than the critical angle $	heta_c$.From the geometry,$i = 90^\circ - r$.So,$90^\circ - r > 	heta_c \implies \cos r > \sin 	heta_c = \frac{n_2}{n_1}$  .........$(ii)$Using $\sin^2 r + \cos^2 r = 1$,we have $\cos r = \sqrt{1 - \sin^2 r}$.Substituting $(i)$ into $(ii)$:$\sqrt{1 - \left(\frac{\sin 	heta}{n_1}ight)^2} > \frac{n_2}{n_1}$$1 - \frac{\sin^2 	heta}{n_1^2} > \frac{n_2^2}{n_1^2}$$n_1^2 - \sin^2 	heta > n_2^2$$\sin^2 	heta $\sin 	heta Thus,the maximum acceptance angle is $	heta = \sin^{-1} \sqrt{n_1^2 - n_2^2}$.

What should be the maximum acceptance angle at the air-core interface of an optical fibre if $n_1$ and $n_2$ are the refractive indices of the core and the cladding respectively?

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