First slide

Prism

Question

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

Moderate

A

30°
B

40°
C

50°
D

60°

Solution

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

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