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 $h$ and radius of curvature $r$ is $2 \pi r h$
and radius of curvature $r$ is $2 \pi r h$ ]
Correct Option: 3
(3) Given,
Refractive index, $\mu=\frac{4}{3}$
$\frac{4}{3} \sin \theta=1 \sin 90^{\circ}$
$\Rightarrow \sin \theta=\frac{3}{4}$
$\cos \theta=\frac{\sqrt{7}}{4}$
Solid angle, $\Omega=2 \pi(1-\cos \theta)=2 \pi(1-\sqrt{7} / 4)$
Fraction of energy transmitted
$=\frac{2 \pi(1-\cos \theta)}{4 \pi}=\frac{1-\sqrt{7} / 4}{2}=0.17$
Percentage of light emerges out of surface
$=0.17 \times 100=17 \%$
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