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(B) Let $\mu_1 = \frac{4}{\sqrt{3}}$ be the refractive index of the rod and $\mu_2 = 2$ be the refractive index of the surrounding medium.Applying Sne

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$\sin^{-1}\left(\frac{2}{\sqrt{3}}\right)$

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(B) Let $\mu_1 = \frac{4}{\sqrt{3}}$ be the refractive index of the rod and $\mu_2 = 2$ be the refractive index of the surrounding medium.Applying Snell's law at the interface of the rod and the surrounding medium for the grazing ray:$\mu_1 \sin(90^{\circ} - r) = \mu_2 \sin(90^{\circ})$$\mu_1 \cos r = \mu_2$$\frac{4}{\sqrt{3}} \cos r = 2$$\cos r = \frac{2 \sqrt{3}}{4} = \frac{\sqrt{3}}{2}$Thus,$r = 30^{\circ}$.Now,applying Snell's law at the incident face of the rod (assuming air as the external medium with $\mu = 1$):$1 \cdot \sin 	heta = \mu_1 \sin r$$\sin 	heta = \frac{4}{\sqrt{3}} \sin 30^{\circ}$$\sin 	heta = \frac{4}{\sqrt{3}} \cdot \frac{1}{2} = \frac{2}{\sqrt{3}}$Therefore,$	heta = \sin^{-1}\left(\frac{2}{\sqrt{3}}ight)$.

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