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(C) Let $\mu_1 = \frac{2}{\sqrt{3}}$ be the refractive index of the rod and $\mu_2 = 1$ be the refractive index of the surrounding medium (air).Applyi

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

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(C) Let $\mu_1 = \frac{2}{\sqrt{3}}$ be the refractive index of the rod and $\mu_2 = 1$ be the refractive index of the surrounding medium (air).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})$$\mu_1 \cos r = 1$$\frac{2}{\sqrt{3}} \cos r = 1$$\cos r = \frac{\sqrt{3}}{2}$Thus, $r = 30^{\circ}$.Now, applying Snell's law at the incident face of the rod:$1 \cdot \sin 	heta = \mu_1 \sin r$$\sin 	heta = \frac{2}{\sqrt{3}} \sin 30^{\circ}$$\sin 	heta = \frac{2}{\sqrt{3}} \cdot \frac{1}{2} = \frac{1}{\sqrt{3}}$Therefore, $	heta = \sin^{-1}\left(\frac{1}{\sqrt{3}}ight)$.

$A$ transparent solid cylindrical rod has a refractive index of $\frac{2}{\sqrt{3}}$. It is surrounded by air. $A$ light ray is incident at the mid-point of one end of the rod as shown in the figure. The incident angle $\theta$ for which the light ray grazes along the wall of the rod is

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