Question

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? The refractive index of water is 1.33. (Consider the bulb to be a point source.)

Answer 1

The actual depth of the bulb in water,d1=80cm=0.8m
Refractive index of water,μ=1.33
The given situation is shown in the following figure: Where i=Angle of incidence r=Angle of refraction=90º
Since the bulb is a point source, the emergent light can be considered as a circle of radius, R=\(\frac{AC}{2}\)AO=OB
Using Snell law, we can write the relation for the refractive index of water as:μ=\(\frac{sin\,r}{sin\,i}\)=1.33=\(\frac{sin90^{\circ}}{sin\,i}\)
∴i=sin^-1(\(\frac{1}{3.33}\))=48.75º
Using the given figure, we have the relation:tani=\(\frac{OC}{OB}\)=\(\frac{R}{d_1}\)
R=tan48.75^º×0.8=0.91m
Area of the structure of water\(\pi R^2\)=\(\pi\)(0.91)²=2.61m²
Hence, the area of the structure of water through which the light from the bulb can emerge is approximately 2.61m².