Question

What is the refractive index of a prism whose angle $A={60}^{\circ }$ and angle of minimum deviation $d_m={30}^{\circ }$?

• A. $\sqrt{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $1$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is A $\sqrt{2}$

Refractive index of prism
$\mu =\frac{\sin \frac{A+{\delta }_m}{2}}{\sin \frac{A}{2}}$
$=\frac{\sin \frac{{60}^{\circ }+{30}^{\circ }}{2}}{\sin \frac{{60}^{\circ }}{2}}$
$=\frac{\sin {45}^{\circ }}{\sin {30}^{\circ }}=\frac{1/\sqrt{2}}{1/2}$
$=\frac{1}{\sqrt{2}}\times \frac{2}{1}=\sqrt{2}$