Question 62 of 295

There is a small source of light at some depth below the surface of water (refractive index $=\dfrac{4}{3}$ ) 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 $2\pi rh$ ]

A$21$B$34$C$17$D$50$

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Answer: (C) $17$

$sin\beta =\dfrac{3}{4},\cos ⁡\beta =\dfrac{\sqrt{7}}{4}$
Solid angle $d\Omega =2\pi {R}^{2}\left(1-\cos ⁡\beta \right)$
Percentage of light $=\dfrac{2\pi {R}^{2}\left(1-\cos ⁡\beta \right)}{4\pi {R}^{2}}\times 100$
$=\dfrac{1-cos\beta }{2}\times 100=\left(\dfrac{4-\sqrt{7}}{8}\right)\times 100\approx 17$