A small bulb (assumed to be a point source) is placed at the bottom of a tank containing water to a depth of `80 cm`. Find out the area of the surface of water through which light from the bulb can emerge. Take the value of refractive index of water to be `4//3`.

Explanation

Actual depth of the bulb in water, d1=80cm=0.8m
Refractive index of water, `mu=1.33`
The given situation is shown in the following figure.

Where, i=Angle of incidence
r = Angle of refraction = `90^(@)`.
Since,the bulb is a point source, the emergent light can be considered as a circle of
`R=AC/2= AO=OB`
radius, Using Snell's law, we can write the relation for the refractive index of water as:
`mu=(sinr)/(sini)`
`1.33 =(sin90^(@))/(sini)`
`therefore i=sin^(-1)(1/1.33) = 48.75^(@)`
using the given figure, we have the relation,
`tani=(OC)/(OB) = R/(d_(1))`
`R=tan 48.75^(@) xx 0.8 = 0.91m`
Area of the surface of water = nR2=n(0.91)2=2.61m2
Hence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.62 m2.