A small bulb is placed at …
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A small bulb is placed at the bottom of 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. Refractive index of water to be 4/3.
Posted by Dipankar Kundu 6 years, 9 months ago
- 2 answers
Sayeda Iram Ateeq 6 years, 9 months ago
r= h tan ić......
......mu=1/sin ić.......
......Sin ić=1/mu=1/1.33=0.75......
......Now, r=h sin ić/cos ić......
=h sin ić/ (1-sin^2 ić)
= 80×10^-2 × 0.75/(1-sin 0.75^2)....
r=0.91cm....
Area covered by light, A=pi r^2......
A= 3.14× 0.91^2......
= 2.6m
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Dipankar Kundu 6 years, 9 months ago
0Thank You