Question

When an equilateral prism of refractive index \(\sqrt{3}\) produces minimum deviation the angle of incidence at the first face must be

• A. 30\(^\circ\)
• B. 42\(^\circ\)
• C. 60\(^\circ\)
• D. 75\(^\circ\)

Hint:

At minimum deviation in a prism: \(i = e\) and \(r_1 = r_2 = A/2\). Use Snell's law: \(\sin i = n \sin(A/2)\).

Answer 1

The Correct Option is C

Step 1: Basic Principle
At minimum deviation, angle of refraction inside prism \(r = A/2\).
Step 2: Solution Procedure:
For equilateral prism: \(A = 60^\circ\), \(r = 30^\circ\)
Snell's law at first face: \(\sin i = n \sin r = \sqrt{3} \times \sin 30^\circ = \sqrt{3} \times 0.5 = \dfrac{\sqrt{3}}{2}\)
\(i = 60^\circ\)
Step 3: Required Answer:
Angle of incidence \(= \mathbf{60^\circ}\).