9.5 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.)

Hint: Recall total internal reflection.

Step 1: Find the angle of incidence.
The actual depth of the bulb in water, dl= 80 cm = 0.8 m
Refractive Index of water, = 1.33

Where,
i = Angle of incidence
r = Angle of refraction = 90°
Using Snell’s law,

μ=sinrsini1.33=sin90sinii=sin-111.33=48.75°

Step 2: Find the radius of the circle.
Since the bulb is a point source, the emergent light can be considered as a circle of radius, R=AC2=OA
Using the given figure,
The actual depth of the bulb in water, dl = 80 cm = 0.8 m

tani=OAOB=Rdl
 R=tan48.75×0.8=0.91m

Step 3: Find the area of the circle.
Area of the circle = \(\pi r^{2}=\pi (0.91)^{2}=2.60\) m2
Hence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.60 m2.