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 $\frac{4\sqrt2}{5}.$
CBSE 55-1-1 PAPER SET 2019
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When the prism is kept in another medium. we have to take the refractive index of the prism with respect to the given medium:
$^{\text{medium}}\mu_\text{prism}=\frac{\mu_\text{prism}}{\mu_\text{medium}}=\frac{\sin\Big[\Big(\frac{\text{A}+\text{D}_\text{m}}{2}\Big)\Big]}{\sin\Big(\frac{\text{A}}{2}\Big)}$
$\frac{1.6}{\frac{4\sqrt2}{5}}=\frac{\sin\Big[\Big(\frac{60^\circ+\text{D}_\text{m}}{2}\Big)\Big]}{\sin\Big(\frac{60^\circ}{2}\Big)}$
$\sqrt2=\frac{\sin\Big[\Big(\frac{60^\circ+\text{D}_\text{m}}{2}\Big)\Big]}{\frac{1}{2}}$
$\sin^{-1}\Big(\frac{1}{\sqrt2}\Big)=\Big(\frac{60^\circ+\text{D}_\text{m}}{2}\Big)$
$90^\circ=60^\circ+\text{D}_\text{m}$
$\text{D}_\text{m}=30^\circ$
$^{\text{medium}}\mu_\text{prism}=\frac{\mu_\text{prism}}{\mu_\text{medium}}=\frac{\sin\Big[\Big(\frac{\text{A}+\text{D}_\text{m}}{2}\Big)\Big]}{\sin\Big(\frac{\text{A}}{2}\Big)}$
$\frac{1.6}{\frac{4\sqrt2}{5}}=\frac{\sin\Big[\Big(\frac{60^\circ+\text{D}_\text{m}}{2}\Big)\Big]}{\sin\Big(\frac{60^\circ}{2}\Big)}$
$\sqrt2=\frac{\sin\Big[\Big(\frac{60^\circ+\text{D}_\text{m}}{2}\Big)\Big]}{\frac{1}{2}}$
$\sin^{-1}\Big(\frac{1}{\sqrt2}\Big)=\Big(\frac{60^\circ+\text{D}_\text{m}}{2}\Big)$
$90^\circ=60^\circ+\text{D}_\text{m}$
$\text{D}_\text{m}=30^\circ$
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