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 the refractive index of water to be \frac{4}{3}.

CBSE Class 12 Physics 2013 Paper

Question : 20

Total: 29

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

Solution:

Using Snell's law,
‌µ=‌

sin‌i

sin‌r

‌‌ Or, ‌‌‌‌

=‌

sin‌90∘

sin‌r

‌∴‌‌i=sin‌−1‌

=48.59∘
‌‌ Now, ‌‌‌tan‌i=‌

=‌

0.8

‌∴‌‌R=0.8×tan‌48.59∘
‌=0.8×1.134=0.9m
Area of surface of water through which light will emerge
‌=πR2=π×(0.9)2
‌=2.54m2

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