Question

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.

• A. $45^{\circ }$
• B. $90^{\circ }$
• C. $60^{\circ }$
• D. $30^{\circ }$

Answer 1

Answer: (B)

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 $A={\delta }_m$
$\Rightarrow \sqrt{2}=\frac{\sin \left(\frac{A+A}{2}\right)}{\sin \frac{A}{2}}=\frac{\sin A}{\sin \frac{A}{2}}$
$\Rightarrow \sqrt{2}=\frac{2\sin \frac{A}{2}\cos \frac{A}{2}}{\sin \frac{A}{2}}$
$[\because \sin 2A=2\sin A\cos A]$
$\sqrt{2}=2\cos \frac{A}{2}$
$\Rightarrow \cos \frac{A}{2}=\frac{1}{\sqrt{2}}$
$\Rightarrow \frac{A}{2}={\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)$
$\Rightarrow \frac{A}{2}=45^{\circ }$
$\Rightarrow A=90^{\circ }$