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(D) Let the refractive index of the rod be $n = \frac{2}{\sqrt{3}}$.At the point $Q$ on the wall of the rod,the light ray grazes the surface,meaning t

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

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(D) Let the refractive index of the rod be $n = \frac{2}{\sqrt{3}}$.At the point $Q$ on the wall of the rod,the light ray grazes the surface,meaning the angle of refraction is $90^{\circ}$. Let $C$ be the critical angle.Using Snell's law at $Q$: $n \sin C = 1 \cdot \sin 90^{\circ} = 1$.$\sin C = \frac{1}{n} = \frac{1}{2/\sqrt{3}} = \frac{\sqrt{3}}{2}$.Thus,$C = 60^{\circ}$.Now,consider the refraction at the point $P$ on the end face. The angle of incidence is $	heta$ and the angle of refraction is $r = 90^{\circ} - C$.Using Snell's law at $P$: $1 \cdot \sin 	heta = n \cdot \sin(90^{\circ} - C) = n \cos C$.Substituting the values: $\sin 	heta = \frac{2}{\sqrt{3}} \cdot \cos 60^{\circ} = \frac{2}{\sqrt{3}} \cdot \frac{1}{2} = \frac{1}{\sqrt{3}}$.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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