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.)

Explanation

Given data :
Actual depth of the bulb in water (`d_1`) = 80 cm v = 0.80 m
Refractive index of water (`mu`) = 1.33
The light rays starting from bulb can pass through the surface if angle of incidence at surface is less than or equal to critical angle (C) for water - air interface.
h - depth of bulb from the surface,
The light will emerge only through a circle of radius r given by
r = h tan C,
Where h = 80 cm = 0.80 m
But sin C =` (1)/(a^(h) omega) = 3/4`
`tan C = (3)/(sqrt7)`
` r = 0.80 xx ((3)/(sqrt7))`
` therefore ` Area of circular surface of water,
` A = pir^2 = 3.14 xx (0.8 xx (3)/(sqrt7))^2`
` = 3.14 xx (0.64 xx 9/7)`
` A = 2.6 m^2` .