Question

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

• A. $30^\circ$
• B. $75^\circ$
• C. $45^\circ$
• D. $60^\circ$

Hint:

At minimum deviation, the angle of prism is twice the angle of refraction inside the prism.

Answer 1

The Correct Option is D

Step 1: Condition for minimum deviation.
At minimum deviation, angle of incidence equals angle of emergence, and refraction angles are equal: \[ i = e = 45^\circ \] Step 2: Apply Snell’s law at first surface.
\[ \mu = \dfrac{\sin i}{\sin r} \] \[ \sqrt{2} = \dfrac{\sin 45^\circ}{\sin r} \] Step 3: Calculate angle of refraction.
\[ \sin r = \dfrac{1/\sqrt{2}}{\sqrt{2}} = \dfrac{1}{2} \] \[ r = 30^\circ \] Step 4: Find angle of prism.
\[ A = 2r = 60^\circ \]