Question

Prism angle and refractive index for a prism are ${60}^{\circ }$ and $1.414$ respectively. Then what will be the Angle of minimum deviation?

• A. ${15}^{\circ }$
• B. ${30}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is B

${30}^{\circ }$

$n=\frac{sin\frac{(A+{\delta }_m)}{2}}{sin{30}^{\circ }}$
where n is the refractive index, ${\delta }_m$ is the angle of minimum deviation, A is the prism angle
$\Rightarrow 1.414=\frac{sin\frac{(60+{\delta }_m)}{2}}{sin{30}^{\circ }}\Rightarrow sin\left(\frac{60+{\delta }_m}{2}\right)=0.707=sin{45}^{\circ }\Rightarrow \frac{60+{\delta }_m}{2}={45}^{\circ }\Rightarrow {\delta }_m={30}^{\circ }$