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

When the angle of incidence is equal to the critical angle, the ray of light grazes the surface. Hence, the light will appear to emerge out from a circle whose edges make angle i c with the vertical.

 

The area of surface of water through which light will come out, a = πr²

µ = 1/sini_c ∴ sini_c = 1/µ = 1/1.33 = 0.752

∴ i_c = 48.76°. tani_c = AB/AO ⇒ AB = AO tani_c

r = 80 x tan 48.76° = 91.2cm.

∴ Area, a = πr² = π × (0.912)² = 2.61m²