Question

A prism has a refractive index $\sqrt{\frac{3}{2}}$ and refracting angle 90${}^o$. Find the minimum deviation produced by prism.

• A. 40${}^o$
• B. 45${}^o$
• C. 30${}^o$
• D. 49${}^o$

Answer 1

Answer: (C)

30${}^o$

Let $\mu =\frac{\sqrt{3}}{\sqrt{2}}$ be the refractive index of prism.

Refracting angle of prism is$A={90}^{\circ }$

At minimum deviation ,angle of incidence =angle of emergent

Therefore, $1\times \sin I=\mu \times \sin r^1$

$1\times \sin e=\mu \times \sin r^2$

Since, I=e

Therefore, $r^1=r^2$

$A+{90}^{\circ }-r^1+{90}^{\circ }-r^2={180}^{\circ }$

Therefore, $r^1={45}^{\circ }$

$1\times \sin I=\mu \times \sin {45}^{\circ }$

Therefore, I=${60}^{\circ }$

minimum deviation=$2\times i-90$

minimum deviation=${30}^{\circ }$

Hence, Option c is answer