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

Answer: 9

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