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

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(B) According to Snell's law,for a light beam passing through multiple parallel interfaces,the product of the refractive index and the sine of the ang

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

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(B) According to Snell's law,for a light beam passing through multiple parallel interfaces,the product of the refractive index and the sine of the angle with the normal remains constant at each interface.Let $n_1, n_2, n_3, n_4$ be the refractive indices and $	heta, 	heta_1, 	heta_2, 	heta_3$ be the angles of refraction in regions $I, II, III, IV$ respectively.Applying Snell's law:$n_0 \sin 	heta = n_1 \sin 	heta_1 = n_2 \sin 	heta_2 = n_3 \sin 	heta_3$Given:$n_1 = n_0, n_2 = \frac{n_0}{2}, n_3 = \frac{n_0}{6}, n_4 = \frac{n_0}{8}$Therefore:$n_0 \sin 	heta = \frac{n_0}{8} \sin 	heta_3$For the beam to just miss entering Region $IV$,the angle of refraction in Region $IV$ must be $	heta_3 = 90^\circ$ (or $\frac{\pi}{2}$ radians).Substituting $	heta_3 = 90^\circ$:$n_0 \sin 	heta = \frac{n_0}{8} \sin(90^\circ)$$n_0 \sin 	heta = \frac{n_0}{8} (1)$$\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 Region $I, II, III$ 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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