Question

The light ray is incident at angle of $60^{\circ}$ on a prism of angle $45^{\circ}$. When the light rays falls on the other surface at $90^{\circ}$, the refractive index of the material of prism $\mu$ and the angle of deviation $\delta$ are given by

• A. $\mu = \sqrt{2},\delta = 30^{\circ}$
• B. $\mu = 1.5,\delta = 15^{\circ}$
• C. $\mu = \frac{\sqrt{3}}{2},\delta = 30^{\circ}$
• D. $\mu = \sqrt{\frac{3}{2}},\delta = 15^{\circ}$

Hint:

When a ray emerges normally, $e = 0$ and $r_2 = 0$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
When a ray emerges normally, $e = 0$ and $r_2 = 0$.
Step 2: Detailed Explanation:
At the second surface, the ray falls at $90^{\circ}$, so angle of incidence at second surface $r_2 = 0$ (since it emerges normally). Then $r_1 = A - r_2 = 45^{\circ}$. Using Snell's law at first surface: $\mu = \frac{\sin i}{\sin r_1} = \frac{\sin 60^{\circ}}{\sin 45^{\circ}} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \sqrt{\frac{3}{2}}$. Deviation $\delta = i + e - A = 60^{\circ} + 0 - 45^{\circ} = 15^{\circ}$.
Step 3: Final Answer:
$\mu = \sqrt{\frac{3}{2}}$ and $\delta = 15^{\circ}$.