Question

A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)

Answer 1

The light rays from the small bulb S which are incident at an angle i > i_c are totally internally reflected and cannot emerge out of water surface. The light from the bulb ‘S’ comes out through a circles path of radius r is given by

`tan"i"_"c" = "OA"/"OS" = "r"/"h" ("or") "r" = "h" tan "i"_"c"`

`sin "i"_"c" = 1/mu = 1/1.33 = 3/4`

`cos "i"_"c" = sqrt(1 - (3/4)^2) = sqrt7/4`

`tan"i"_"c" = 3/4 xx 4/sqrt7 = 3/sqrt7`

Area of the patch = πr²

= π h² tan² i_c

= 3.14 × 0.80² × `9/7`

= 2.58 m²

≈ 2.6 m²