Question

A small bulb is placed at the bottom of a tank
containing water to a depth of 80 cm. What is the
area of the surface of water through which light
from the bulb can emerge out :- INCERT)
( =1.33)
(1) 2.54 m
(2) 3.71 m
(3) 1.11 m2​

Answer 1

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

r=AB2

=OA=OB

μ=1sinc

=sinc=1μ

=11.33

=0.75

C=sin−1(0.75)

=48.6∘

Consider ΔOBS

tanC=OBOS

OB=OStanc

=0.8tan48.6

r=0.8×1.1345

=0.907  

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

=3.14×(0/907)2

=2.518m2