Question

A point source is placed at the bottom of a tank containing a transparent liquid (refractive index \( n \)) to a depth H. The area of the surface of the liquid through which light from the source can emerge out is:

• A. \( \frac{\pi H^2}{(n-1)} \)
• B. \( \frac{\pi H^2}{(n^2-1)} \)
• C. \( \frac{\pi H^2}{\sqrt{n^2-1}} \)
• D. \( \frac{\pi H^2}{(n^2+1)} \)

Hint:

The phenomenon of total internal reflection is widely used in optical fibers, where light is confined within the core due to repeated reflections at the boundary.

Answer 1

The Correct Option is B

Total internal reflection causes light from the point source to exit the liquid surface within a circle of radius \( r \). The critical angle \( \theta_c \) for this phenomenon is defined by \( \sin \theta_c = \frac{1}{n} \). The radius \( r \) of this surface circle is calculated as \( r = H \tan \theta_c \). Substituting \( \tan \theta_c = \frac{\sin \theta_c}{\sqrt{1 - \sin^2 \theta_c}} \), the radius becomes \( r = H \cdot \frac{1}{\sqrt{n^2 - 1}} \). Consequently, the area \( A \) of the circle is \( A = \pi r^2 = \pi \left( \frac{H}{\sqrt{n^2 - 1}} \right)^2 = \frac{\pi H^2}{n^2 - 1} \). Therefore, option (C) is the correct answer.