Question

A ray of light incidents at an angle of \(60^\circ\) on the first face of a prism. The angle of the prism is \(30^\circ\) and its second face is silvered. If the light ray inside the prism retraces its path after reflection from the second face, then the refractive index of the material of the prism is

• A. \( \dfrac{2}{\sqrt{3}} \)
• B. \( \dfrac{3}{2} \)
• C. \( \sqrt{2} \)
• D. \( \sqrt{3} \)

Hint:

If a prism face is silvered and the ray retraces its path, the emergent ray retraces the incident ray direction. Use Snell's law and the geometry of the prism accordingly.

Answer 1

The Correct Option is D

Step 1: Given: \[ \text{Angle of incidence } i = 60^\circ,
\text{Prism angle } A = 30^\circ \] Step 2: Since the ray retraces its path, the emergent angle inside the prism is equal to the angle of incidence on the second face: \[ \text{Total deviation } = 0^\circ \Rightarrow \text{ray is normal to second face} \] Step 3: Use Snell’s law at the first surface: \[ \mu = \frac{\sin i}{\sin r} \] Use geometry in prism: if angle of prism \( A = 30^\circ \), and the ray goes normally to the second face (internal angle of refraction = 90° to face), then internal angle at first face: \[ r = \frac{A}{2} = 15^\circ \] Step 4: \[ \mu = \frac{\sin 60^\circ}{\sin 15^\circ} = \frac{\sqrt{3}/2}{\sin 15^\circ} \] But for retracing, \( r = 30^\circ \) leads to: \[ \mu = \frac{\sin 60^\circ}{\sin 30^\circ} = \frac{\sqrt{3}/2}{1/2} = \boxed{\sqrt{3}} \]