Question

A prism of refractive index \(\sqrt{2}\) has refracting angle of \(60^{\circ}\). At what angle a ray must be incident on it so that it suffers minimum deviation?

• A. \(45^{\circ}\)
• B. \(60^{\circ}\)
• C. \(90^{\circ}\)
• D. \(180^{\circ}\)

Hint:

At minimum deviation, \(i = e\) and \(r = A/2\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
At minimum deviation, \(r = A/2\) and \(i = e\). Snell's law: \(\mu = \frac{\sin i}{\sin r}\).
Step 2: Detailed Explanation:
Given \(A = 60^{\circ}\), \(\mu = \sqrt{2}\). At minimum deviation, \(r = A/2 = 30^{\circ}\).
Snell's law: \(\mu = \frac{\sin i}{\sin r} \Rightarrow \sqrt{2} = \frac{\sin i}{\sin 30^{\circ}} = \frac{\sin i}{0.5}\)
\(\Rightarrow \sin i = 0.5\sqrt{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \Rightarrow i = 45^{\circ}\).
Step 3: Final Answer:
Thus, angle of incidence = \(45^{\circ}\).