A light beam is travelling from Region I to R | vedclass

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(B) According to Snell's law,for a series of parallel interfaces,the product of the refractive index and the sine of the angle with the normal remains

Vedclass · Accepted Answer

$\sin ^{-1}\left(\frac{1}{8}\right)$

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(B) According to Snell's law,for a series of parallel interfaces,the product of the refractive index and the sine of the angle with the normal remains constant at each interface.Let $	heta_1, 	heta_2, 	heta_3, 	heta_4$ be the angles of refraction in regions $I, II, III, IV$ respectively. Here,$	heta_1 = 	heta$.Applying Snell's law at each interface:$n_0 \sin 	heta = n_{II} \sin 	heta_2 = n_{III} \sin 	heta_3 = n_{IV} \sin 	heta_4$For the beam to just miss entering Region $IV$,the angle of refraction in Region $IV$ must be $	heta_4 = 90^\circ$.Thus,we have:$n_0 \sin 	heta = n_{IV} \sin 90^\circ$Substituting the given values:$n_0 \sin 	heta = \frac{n_0}{8} 	imes 1$$\sin 	heta = \frac{1}{8}$$	heta = \sin^{-1}\left(\frac{1}{8}ight)$

$A$ light beam is travelling from Region $I$ to Region $IV$ (Refer Figure). The refractive indices in Regions $I$,$II$,$III$,and $IV$ are $n_0$,$\frac{n_0}{2}$,$\frac{n_0}{6}$,and $\frac{n_0}{8}$,respectively. The angle of incidence $\theta$ for which the beam just misses entering Region $IV$ is:

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