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)`
`= pih^(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`.