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There is a small source of light at some depth below the surface of water (refractive index $= \frac{4}{3}$ ) in 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)

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21%

34%

17%

50%

answer is C.

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Detailed Solution

Given, Refractive index, $μ = \frac{4}{3}$

$\frac{4}{3} \sin θ = 1 \sin 90^{\circ}$

$\Rightarrow \sin θ = \frac{3}{4} \Rightarrow \cos θ = \frac{\sqrt{7}}{4}$

Solid angle,

$Ω = 2 π (1 - \cos θ) = 2 π (1 - \sqrt{7} / 4)$

Fraction of energy transmitted

$= \frac{2 π (1 - \cos θ)}{4 π} = \frac{1 - \sqrt{7} / 4}{2} = 0 .17$

Percentage of light emerges out of surface $= 0.17 \times 100 = 17 %$

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