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(D) Let the refractive index of the rod be $\mu = \frac{2}{\sqrt{3}}$.Applying Snell's law at the end face of the rod:$1 \cdot \sin 	heta = \mu \cdot

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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 $\mu = \frac{2}{\sqrt{3}}$.Applying Snell's law at the end face of the rod:$1 \cdot \sin 	heta = \mu \cdot \sin r$$\sin 	heta = \frac{2}{\sqrt{3}} \sin r \quad \dots(1)$For the light ray to graze along the wall,the angle of incidence at the wall must be equal to the critical angle $C$. If $r$ is the angle of refraction at the first surface,the angle of incidence at the wall is $(90^{\circ} - r)$.Thus,$90^{\circ} - r = C$,where $\sin C = \frac{1}{\mu}$.Applying Snell's law at the wall:$\mu \sin(90^{\circ} - r) = 1 \cdot \sin 90^{\circ}$$\mu \cos r = 1$$\cos r = \frac{1}{\mu} = \frac{1}{2/\sqrt{3}} = \frac{\sqrt{3}}{2}$Since $\cos r = \frac{\sqrt{3}}{2}$,we have $r = 30^{\circ}$,so $\sin r = \frac{1}{2}$.Substituting $\sin r = \frac{1}{2}$ into equation $(1)$:$\sin 	heta = \frac{2}{\sqrt{3}} \cdot \frac{1}{2} = \frac{1}{\sqrt{3}}$$	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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