Question

A small bulb is placed at the bottom of a tank containing water to a depth of √7 m. The refractive index of water is 4/3. The area of the surface of water through which light from the bulb can emerge out is x π m2. The value of x is ________.

Answer 1

The correct answer is 9π

Fig.

$r = h \frac{sin i_c} {\sqrt1-sin² i_c}$
So A = πr²
$= \frac{πh²sin² i_c}{1-sin² i_c}$
$= \frac{π7× \frac{9}{16}}{1 - \frac{9}{16}}$
$= \frac{π × 7× 9 }{7}$
= 9π