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 of
the bulb can emerge out? Refractive index of water is
1.33. Consider the bulb to be a point-source.
​

Answer 1

Answer:

Actual depth of the bulb in water (d1) = 80 cm = 0.8 m

Refractive index of water (μ) = 1.33

from the figure

Where,

Angle of incidence = Angle of refraction = 90°

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

R = MP/2 = MO = OP

μ = sin r/ sini

1.33 = sin 90°/sini

i = sin-1 (1/1.33) = 48.75°

From the figure

tan i OP/ON = R/d1

R = tan 48.75° x 0.8 = 0.91 m

Area of the surface of water = πR2 = π (0.91)2 = 2.61 m2

Hence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.61m2