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

In Fig., the source of light (S) is `80 cm` below the surface of water i.e., `SO = 80 cm = 0.8 m`
When `/_ i= C, for SA and SB, /_r = 90^@`
`:.` Area of the surface of water through which light from the bulb can emerge is area of the circle of radius.
`r = (AB)/(2) = OA = OB`
As `mu = (1)/(sin C)`
`sin C = (1)/(mu) = (1)/(1.33) = 0.75`
`C = sin^-1 (0.75) = 48.6^@`
In `Delta OBS`, `tan C = (OB)/(OS) :. OB = OS tan C = 0.8 tan 48.6^@`
`r = 0.8 xx 1.1345 = 0.907 m`
Area of the surface of water through which light emerges `= pi r^2 = 3.14 (0.907)^2 = 2.518 m^2`.
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