There is a small source of light at some depth below the surface of water (refractive index \(=\frac{4}{3}\) ) 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 \(\mathrm{h}\) and radius of curvature \(\mathrm{r} \text { is } 2 \pi \mathrm{rh}]\)

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Answer & Solution

Correct Answer

(A) \(17\)

Detailed explanation

\(\frac{4}{3} \sin \theta=1 \sin 90^{\circ}\)

\(\sin \theta=\frac{3}{4}\)

Area of sphere in which light spread \(=4 \pi \mathrm{R}^{2}\) \(\Omega=2 \pi(1-\cos \theta)\)

\(\Omega=2 \pi\left(1-\frac{\sqrt{7}}{4}\right)\)

\(\mathrm{P} \rightarrow 4 \pi\) steradians

\(\mathrm{P}^{\prime} \rightarrow \frac{\mathrm{P}}{4 \pi}(1-\cos \theta)\)

ratio \(=\frac{P'}{P}=\frac{2 \pi(1-\cos \theta)}{4 \pi}=\frac{(1-\cos \theta)}{2}=\frac{1-\frac{\sqrt{7}}{4}}{2}\)

\(=\frac{0.33}{2}=0.17\)

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