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When a glass prism of refracting angle $60^{\circ}$ is immersed in a liquid its angle of minimum deviation is $30^{\circ}$. The critical angle of glass with respect to the liquid medium is
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$45^{\circ}$
$A=60^{\circ}, \delta=30^{\circ}$
Refractive index of prism with respect to liquid,
$\begin{aligned} \mu & =\frac{\sin \frac{A+\delta_n}{2}}{\sin \frac{A}{2}}=\frac{\sin \frac{\left(60^{\circ}+30^{\circ}\right)}{2}}{\sin \frac{60^{\circ}}{2}} \\ & =\frac{\sin 45^{\circ}}{\sin 30^{\circ}}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}=\sqrt{2}\end{aligned}$
$\therefore$ Critical angle $=\sin ^{-1}\left(\frac{1}{\mu}\right)=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$
$=\sin ^{-1}\left(\sin 45^{\circ}\right)=45^{\circ}$
Refractive index of prism with respect to liquid,
$\begin{aligned} \mu & =\frac{\sin \frac{A+\delta_n}{2}}{\sin \frac{A}{2}}=\frac{\sin \frac{\left(60^{\circ}+30^{\circ}\right)}{2}}{\sin \frac{60^{\circ}}{2}} \\ & =\frac{\sin 45^{\circ}}{\sin 30^{\circ}}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}=\sqrt{2}\end{aligned}$
$\therefore$ Critical angle $=\sin ^{-1}\left(\frac{1}{\mu}\right)=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$
$=\sin ^{-1}\left(\sin 45^{\circ}\right)=45^{\circ}$
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