A light beam is traveling from Region I to Re | vedclass

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

Vedclass · Accepted Answer

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

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(B) According to Snell's Law,for a light ray 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 $	heta_1, 	heta_2, 	heta_3, 	heta_4$ be the angles of refraction in regions $I, II, III, IV$ respectively.By Snell's Law:$n_0 \sin 	heta = n_1 \sin 	heta_1 = n_2 \sin 	heta_2 = n_3 \sin 	heta_3 = n_4 \sin 	heta_4$Here,the beam just misses entering Region $IV$,which means the angle of refraction in Region $IV$ is $90^{\circ}$ (grazing emergence).Therefore,we equate the refractive index and angle of Region $I$ to Region $IV$:$n_0 \sin 	heta = n_4 \sin 90^{\circ}$Given $n_4 = \frac{n_0}{8}$:$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 traveling 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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