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

Answer 1

Solution :

Area of surface of water through which light from the bulb can emerge is the area of the circle of radius

r=AB2r=AB2

=OA=OB=OA=OB

μ=1sincμ=1sin⁡c

=sinc=1μ=sin⁡c=1μ

=11.33=11.33

=0.75=0.75

C=sin−1(0.75)C=sin−1⁡(0.75)

=48.6∘=48.6∘

Consider ΔOBSΔOBS

tanC=OBOStan⁡C=OBOS

OB=OStancOB=OStan⁡c

=0.8tan48.6=0.8tan⁡48.6

r=0.8×1.1345r=0.8×1.1345

=0.907=0.907

Area of the surface of water through which light emerge =πr2=πr2

=3.14×(0/907)2=3.14×(0/907)2

=2.518m2