Q. If eye is kept at a depth $h$ inside water of refractive index and viewed outside, then the diameter of the circle through which the outer objects become visible, will be
A) $\frac{h}{\sqrt{\mu^{2} - 1}}$
B) $\frac{h}{\sqrt{\mu^{2} + 1}}$
C) $\frac{2h}{\sqrt{\mu^{2} - 1}}$
D) $\frac{h}{\sqrt{\mu^{2}}}$
Explanation
(c) Let $r$ be the radius of circle through which other objects become visidble. The rays of light must be incident at critical angle C
$\sin{C = \frac{1}{\mu} = \frac{r}{\sqrt{r^{2} + h^{2}}}}$
$\mu^{2}r^{2} = r^{2} + h^{2}$
$\left( \mu^{2} - 1 \right)r^{2} = h^{2}$
$r = \frac{h}{\sqrt{\mu^{2} - 1}}$
Diameter $2r = \frac{2h}{\sqrt{\mu^{2} - 1}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.