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

Answer: (D)

$\frac{\sin \frac{A+\delta_{m}}{2}}{\sin \frac{A}{2}}=\mu$, But $\frac{A+\delta_{m}}{2}=i=45^{\circ}$
So, $\frac{\sin 45^{\circ}}{\sin (A / 2)}=\sqrt{2}$
$ \Rightarrow \frac{1}{2}=\sin \frac{A}{2} \Rightarrow A=60^{\circ}$