Question 200: Ray Optics

Question

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]:

Accepted Answer

$${4 \over 3}$$ sin $$	heta $$ = 1sin90
o
sin $$	heta $$ = $${3 \over 4}$$
cos $$	heta $$ = $${{\sqrt 7 } \over 4}$$
Surface area in solid angle d$$\Omega $$ = 2$$\pi $$R
2
(1 - cos $$	heta $$)
= 2$$\pi $$R
2
(1 - $${{\sqrt 7 } \over 4}$$)
Percentage of light = $${{2\pi {R^2}\left( {1 - {{\sqrt 7 } \over 4}} ight)} \over {4\pi {R^2}}}$$ $$ 	imes $$ 100%
= $${{{4 - \sqrt 7 } \over 8}}$$ $$ 	imes $$ 100%
= 17%

Answer

17%

Answer

34%

Answer

50%

Answer

21%

Ray Optics

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