Question

When light rays are incident on a prism at an angle of ${45}^{\circ }$, the minimum deviation is obtained. If refractive index of the material of prism is $\sqrt{2}$ then the angle of prism will be ​

• A. ${30}^{\circ }$
• B. ${40}^{\circ }$
• C. ${50}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is D ${60}^{\circ }$

$\mu =\frac{sini}{sin\frac{A}{2}}\Rightarrow \sqrt{2}=\frac{sin45}{sin\frac{A}{2}}\Rightarrow sin\frac{A}{2}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac{1}{2}\Rightarrow \frac{A}{2}={30}^{\circ }\Rightarrow A={60}^{\circ }$