Q. A point source of light is kept at a depth of $h$ in water of refractive index 4/3. The radius of the circle at the surface of water through which light emits is
A) $\frac{3}{\sqrt{7}}h$
B) $\frac{\sqrt{7}}{3}h$
C) $\frac{\sqrt{3}}{7}h$
D) $\frac{7}{\sqrt{3}}h$
Explanation
(a) $\frac{\sin{90{^\circ}}}{\sin C} = \mu$
$\sin C = \frac{1}{\mu} \Rightarrow \frac{R}{\sqrt{R^{2} + h^{2}}} = \frac{3}{4}$
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Squaring, $16R^{2} = 9R^{2} + 9h^{2}$
$7R^{2} = 9h^{2} \Rightarrow R = \frac{3}{\sqrt{7}}h$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.