Question

A prism having refractive index \(\sqrt{2}\) and refracting angle \(30^\circ\) has one of the refracting surfaces silvered. The beam of light incident on the other refracting surface will retrace its path, if angle of incidence is \([\sin(\pi/6)=0.5]\)

• A. \( \sin^{-1}\!\left(\dfrac{3}{4}\right) \)
• B. \( \sin^{-1}\!\left(\dfrac{1}{2}\right) \)
• C. \( \sin^{-1}\!\left(\dfrac{\sqrt{3}}{2}\right) \)
• D. \( \sin^{-1}\!\left(\dfrac{1}{\sqrt{2}}\right) \)

Hint:

For retracing of light, incidence on the reflecting surface must be normal.

Answer 1

The Correct Option is D

Step 1: Condition for retracing the path.
For retracing, the ray must strike the silvered surface normally after refraction at the first surface.
Step 2: Refraction at first surface.
Let the angle of refraction inside the prism be \(r\). For normal incidence on the silvered face, \[ r = A = 30^\circ. \]
Step 3: Applying Snell’s law.
\[ \sin i = \mu \sin r = \sqrt{2} \times \sin 30^\circ. \] \[ \sin i = \sqrt{2} \times 0.5 = \frac{1}{\sqrt{2}}. \]
Step 4: Conclusion.
\[ i = \sin^{-1}\!\left(\frac{1}{\sqrt{2}}\right). \]