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

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(B) According to Snell's Law at each interface:At the interface between Region $I$ and $II$: $n_0 \sin 	heta = \frac{n_0}{2} \sin r_1$At the interfac

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

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(B) According to Snell's Law at each interface:At the interface between Region $I$ and $II$: $n_0 \sin 	heta = \frac{n_0}{2} \sin r_1$At the interface between Region $II$ and $III$: $\frac{n_0}{2} \sin r_1 = \frac{n_0}{6} \sin r_2$At the interface between Region $III$ and $IV$: $\frac{n_0}{6} \sin r_2 = \frac{n_0}{8} \sin 90^\circ$Since the beam just misses entering Region $IV$,the angle of refraction at the last interface is $90^\circ$.By applying Snell's Law across all interfaces,we can equate the initial and final states: $n_0 \sin 	heta = \frac{n_0}{8} \sin 90^\circ$$\sin 	heta = \frac{1}{8}$$	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 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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