Question

If the refractive index of a glass prism is cot $\left(\frac{A}{2}\right)$ and $A$ is angle of prism, then angle of minimum deviation is

• A. $\left(\frac{\pi }{2}-A\right)$
• B. $\left(\pi -\frac{A}{2}\right)$
• C. $\left(\frac{\pi -A}{2}\right)$
• D. $(\pi -2A)$

Answer 1

Answer: (D)

The refractive index $(\mu )$ of a prism of angle A, and minimum deviation ${\delta }_m$ is given by

$\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin A/2}$
Given, $\mu =\cot \frac{A}{2}$
$\therefore \cot \frac{A}{2}=\frac{\sin \frac{(A+{\delta }_m)}{2}}{\sin A/2}$
$\Rightarrow \frac{\cot A/2}{\sin A/2}=\frac{\sin \frac{(A+{\delta }_m)}{2}}{\sin A/2}$
$\Rightarrow \cos \frac{A}{2}=\sin \left(\frac{A+{\delta }_m}{2}\right)$
$\therefore \sin \left(90^{\circ }-\frac{A}{2}\right)=\sin \left(\frac{A+{\delta }_m}{2}\right)$
$\Rightarrow 90^{\circ }-\frac{A}{2}=\frac{A+{\delta }_m}{2}$
$\Rightarrow 180^{\circ }-A=A+{\delta }_m$
$\Rightarrow {\delta }_m=180^{\circ }-2A=\pi -2A$