JEE MAIN - Physics (2020 - 9th January Evening Slot - No. 17)
There is a small source of light at some depth
below the surface of water (refractive
index = $${4 \over 3}$$) in a tank of large cross sectional
surface area. Neglecting any reflection from the
bottom and absorption by water, percentage of
light that emerges out of surface is (nearly) :
[Use the fact that surface area of a spherical cap of height h and radius of curvature r is 2$$\pi $$rh]:
[Use the fact that surface area of a spherical cap of height h and radius of curvature r is 2$$\pi $$rh]:
17%
34%
50%
21%
Explanation
_9th_January_Evening_Slot_en_17_1.png)
$${4 \over 3}$$ sin $$\theta $$ = 1sin90o
sin $$\theta $$ = $${3 \over 4}$$
cos $$\theta $$ = $${{\sqrt 7 } \over 4}$$
Surface area in solid angle d$$\Omega $$ = 2$$\pi $$R2(1 - cos $$\theta $$)
= 2$$\pi $$R2(1 - $${{\sqrt 7 } \over 4}$$)
Percentage of light = $${{2\pi {R^2}\left( {1 - {{\sqrt 7 } \over 4}} \right)} \over {4\pi {R^2}}}$$ $$ \times $$ 100%
= $${{{4 - \sqrt 7 } \over 8}}$$ $$ \times $$ 100%
= 17%
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