An equilateral glass prism has a refractive index $1.6$ in air. calculate the angle of minimum deviation of the prism, when kept in a medium of refractive index $4\sqrt{2}/5$.

$n=\frac{\sin \frac{\delta m +A}{2}}{\sin \frac{A}{2}}$

Where, $'\delta m'$ is the angle of minimum deviation,

$'n'$  is the refractive index of the prism with respect to the medium

$'A'$ is the angle of the prism

Hence,

The refractive index of the prism with respect to the medium will be - $n=\frac{1.6}{\frac{4\sqrt{2}}{5}}=\frac{1.6}{4\sqrt{2}}\times \frac{5}{1}=\frac{8}{4\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$

Now, substituting the above values, in the formula used:-

We have,

$\sqrt{2}=\frac{\sin (\frac{\delta m}{2}+\frac{60^{\circ}}{2})}{\sin \frac{60^{\circ}}{2}}=\frac{\sin \left ( \frac{\delta m}{2}+30^{\circ} \right )}{\sin 30^{\circ}}$ $\frac{\sqrt{2}}{2}=\sin \left ( \frac{\delta m}{2}+30^{\circ} \right )$            $\left [ \because \sin 30^{\circ} =\frac{1}{2}\right ]$

$\frac{\delta m}{2}+30^{\circ}=sin^{-1}(\frac{1}{\sqrt{2}})$

$\\\Rightarrow \frac{\delta m}{2}+30^{\circ}=45^{\circ}$

$\delta m=30^{\circ}$

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