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 is1.33.(Consider the bulb to be appoint source.)

Answer

Definite depth of the bulb in waterd1 = 80cm = 0.8m

Refractive index of water μ = 1.33

The given circumstances are shown in the following figure:

Where

i=Angle of incidence

r=Angle of refraction=90°

Given that the bulb is a point source the emergent light can be considered as a circle of radius,

Using Snell law we are able to write the relation for the refractive index of water as:

Using the given equitation we have the relation:

→R=tan 48.75°× 0.8=0.91m
→Area of the surface of water = π R² = π(0.91)² = 2.61m² thus the area of the surface of water through which the light from the bulb can emerge is roughly 2.61 m².