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Q.
A prism is made of a glass having refractive index $\sqrt{2}$. If the angle of minimum deviation is equal to angle of the prism, then the angle of prism is.
TS EAMCET 2020
Solution:
Let $\delta_{m}$ be the angle of minimum deviation.
As we know that,
$\mu=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \frac{A}{2}}$
Given, $\mu=\sqrt{2}$ and $A=\delta_{m}$
$\Rightarrow \sqrt{2}=\frac{\sin \left(\frac{A+A}{2}\right)}{\sin \frac{A}{2}}=\frac{\sin A}{\sin \frac{A}{2}}$
$\Rightarrow \sqrt{2}=\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}}$
$[\because \sin 2 A=2 \sin A \cos A]$
$\sqrt{2}=2 \cos \frac{A}{2}$
$\Rightarrow \cos \frac{A}{2}=\frac{1}{\sqrt{2}}$
$\Rightarrow \frac{A}{2}=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)$
$\Rightarrow \frac{A}{2}=45^{\circ}$
$\Rightarrow A=90^{\circ}$