Q.
An observer can see through a pin-hole the top end of a thin rod of height A, 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
IIT JEEIIT JEE 2002
Solution:
$PQ = QR = 2h$
$\therefore \angle i=45^{\circ}$
$\therefore ST = RT = h = KM = MN$
so,$KS=\sqrt{h^2+(2h)^2}=h\sqrt{5}$
$\therefore \sin r=\frac{h}{h\sqrt{5}}=\frac{1}{\sqrt{5}}$
$\therefore \mu=\frac{\sin i}{\sin r}=\frac{\sin 45^{\circ}}{1/\sqrt{5}}=\sqrt{\frac{5}{2}}$
