Question

The angle of a prism is ${60}^{\circ }$ and its refractive index is $\sqrt{2}$. The angle of minimum deviation suffered by a ray of light in passing through it is:

• A. About 20°
• B. 30°
• C. 60°
• D. 45°

Answer 1

The correct option is B 30°

$n=\frac{sin\left(\frac{A+{\delta }_m}{2}\right)}{sin\left(\frac{A}{2}\right)}where{\delta }_{m}istheangleofminimumdeviation$
$\Rightarrow \sqrt{2}=\frac{sin\left(\frac{60+{\delta }_m}{2}\right)}{sin\left(\frac{60}{2}\right)}$
$\Rightarrow \sqrt{2}\times sin30=sin\left(\frac{60+{\delta }_m}{2}\right)\Rightarrow {45}^{\circ }$
$=sin\left(\frac{60+{\delta }_m}{2}\right)\Rightarrow {\delta }_m={30}^{\circ }$