Question

A small bulb is placed at the bottom of a tank containing water to a depth of $\sqrt{7} m$. The refractive index of water is $\frac{4}{3}$. The area of the surface of water through which light from the bulb can emerge out is $x \pi m ^{2}$. The value of $x$ is ___

Answer 1

$\tan C =\frac{ r }{ h }$
$r = h \tan C$
$\sin C =\frac{1}{\mu}=\frac{3}{4}$
$\tan C =\frac{3}{\sqrt{7}}$
$r =\sqrt{7} \times \frac{3}{\sqrt{7}}=3$
Area of surface $=\pi r ^{2}=9 \pi m ^{2}$