A small bulb (assumed to be a point source) is placed at the bottom of a tank containing water to a depth of `80 cm`. Find out the area of the surface of water through which light from thr bulb can emerge. Take the value of refractive index of water to be `4//3`.

Answer

`25837.7 sq. cm`

Explanation

Here, `h = 80 cm, pi r^2 = ?`
`mu = 4//3 = (1)/(sin C)`
`tan C = ( r)/(h), r = h tan C`
`pi r^2 = pi h^2 tan^2 C = pi h^2 (sin^2 C)/(1 - sin^2 C)`
= `pi h^2 ((1//mu^2)/(1 - 1//mu^2)) = pi h^2 xx (1)/(mu^2 - 1)`
=`(3.14(80)^2)/((4//3)^2 - 1) = 25837.7 sq.cm`.
.