The Total solution for NCERT class 6-12
Asmall 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 bulbcan emerge out? Refractive index of water is 1.33. (Consider the bulb to be apoint source.)
Actual depth of the bulb in water,┬аd1┬а= 80 cm = 0.8 m
The given situation isshown in the following figure:
Where,
i┬а= Angle of incidence
r┬а= Angle of refraction = 90┬░
Sincethe bulb is a point source, the emergent light can be considered as a circle ofradius,┬а
Using SnellтАЩ law, we canwrite the relation for the refractive index of water as:
Using the given figure, wehave the relation:
тИ┤R┬а= tan 48.75┬░ ├Ч 0.8 = 0.91m
тИ┤Area of the surface ofwater = ╧АR2┬а= ╧А (0.91)2┬а= 2.61 m2
Hence,the area of the surface of water through which the light from the bulb canemerge is approximately 2.61 m2.