Question

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

• A. $\frac{5}{2}$
• B. $\sqrt{\frac{5}{2}}$
• C. $\sqrt{\frac{3}{2}}$
• D. $\frac{3}{2}$

Answer 1

Answer: (B)

$PQ=QR=2h$
$\therefore \angle i=45^{\circ }$
$\therefore ST=RT=h=KM=MN$
so,$KS=\sqrt{h^2+(2h{)}^2}=h\sqrt{5}$
$\therefore \sin r=\frac{h}{h\sqrt{5}}=\frac{1}{\sqrt{5}}$
$\therefore \mu =\frac{\sin i}{\sin r}=\frac{\sin 45^{\circ }}{1/\sqrt{5}}=\sqrt{\frac{5}{2}}$