A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. 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.)
Given Data,
Depth of bulb in water = d = 80 cm = 0.8 m
Refractive index of water = μ = 1.33
Construction:
Please consider point “N” at the bottom of the tank where the bulb is placed.
Let the axis line “MOP” be the line on the surface of the water which is 80 cm from the bottom of the tank.
The below is the figure which we would get once we finish our construction.
From the figure we get,
Angle of incidence = Angle of reflection = 90°
Principle: The law of reflection governs the reflection of light-rays off smooth conducting surfaces (here water). The law of reflection states that the incident ray, the reflected ray, and the normal to the surface of the water all lie in the same plane. Furthermore, the angle of reflection is equal to the angle of incidence. Both angles are measured with respect to the normal to the water in the tank.
Since we consider bulb as a point source, the emergent light can be considered as a circle.
The radius of this circle can be obtained by,
R = MP/2 (MP is taken as diameter here)
= MO = MP (from the above figure)
According to Snell’s law, ratio of sine of angle of incidence and sine of angle of refraction is always constant for a given pair of media.
μ = sin r/ sin i
⇒ 
(substituting μ from given data & r from figure)
⇒ i = sin-1 (1/1.33) = 48.75° (from log tables)
From the figure,
tan i = OP/ON = R/d
⇒ R = d × tan(i)
⇒ R = tan 48.75° x 0.8 = 0.91 m
Area of the surface of water (since we considered the point source is bulb and the surface covered is a circle) = 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 2.61 m2