A cylindrical transparent rod of refractive i | vedclass

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Question

(B) Let $\mu = \frac{2}{\sqrt{3}}$ be the refractive index of the rod.For total internal reflection at the side wall,the angle of incidence at the wal

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $\mu = \frac{2}{\sqrt{3}}$ be the refractive index of the rod.For total internal reflection at the side wall,the angle of incidence at the wall must be greater than or equal to the critical angle $i_c$.At the critical angle $i_c$,$\sin i_c = \frac{1}{\mu} = \frac{\sqrt{3}}{2}$,so $i_c = 60^{\circ}$.The angle of refraction $r$ at the first surface is $r = 90^{\circ} - i_c = 90^{\circ} - 60^{\circ} = 30^{\circ}$.Applying Snell's law at the first surface: $1 \cdot \sin 	heta = \mu \cdot \sin r$.$\sin 	heta = \left(\frac{2}{\sqrt{3}}ight) \cdot \sin 30^{\circ} = \left(\frac{2}{\sqrt{3}}ight) \cdot \left(\frac{1}{2}ight) = \frac{1}{\sqrt{3}}$.Therefore,$	heta = \sin^{-1}\left(\frac{1}{\sqrt{3}}ight)$.

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