Q. An observer can see through a pin-hole the top end of a thin rod of height $h$, placed as shown in the figure. The beaker height is $3h$ and its radius $h$. When the beaker is filled with a liquid up to a height $2h$, he can see the lower end of the rod. Then the refractive index of the liquid is
A) 5/2
B) $\sqrt{(5/2)}$
C) $\sqrt{(3/2)}$
D) 3/2
Explanation
(b) The line of sight of the observer remains constant making an angle of $45{^\circ}$ with the normal
$\sin\theta = \frac{h}{\sqrt{h^{2} + (2h)^{2}}} = \frac{1}{\sqrt{5}}$
$\mu = \frac{\sin{45{^\circ}}}{\sin\theta}$
$= \frac{1/\sqrt{2}}{1/\sqrt{5}} = \sqrt{\left( \frac{5}{2} \right)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.