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Q. A point source of light is placed at a distance $h$ below the surface of a large deep lake. Neglecting the partial reflection of light during refraction, the percentage of light energy that escapes from the water surface is $\left[\right.\mu _{water}=4/3\left]\right.$

NTA AbhyasNTA Abhyas 2020Ray Optics and Optical Instruments

Solution:

Solid angle $\left(\phi\right) = 2 \pi \left(1 - cos \theta \right)$
where $\theta =$ plane angle Solution
$\text{escape energy} = \frac{\text{total energy}}{4 \pi } \times \text{solid angle}$
$\frac{\text{escape energy}}{\text{total energy}} = \frac{1}{4 \pi } \times 2 \pi \left(1 - \text{cosC}\right)$
$= \frac{1}{2} \left(1 - \text{cosC}\right) ....(1)$
from law of refraction
$\mu _{w} sin \, \, i = \mu _{a} \, sin \, \, r$
$\frac{4}{3} \times \text{sin C} = 1 \times \text{sin 90}^{\text{o}}$
$\sin C=\frac{3}{4} \Rightarrow \cos C=\sqrt{1-\sin ^2 C} \quad \Rightarrow \quad \cos C=\frac{\sqrt{7}}{4}$
from (1) and (2)
$\mu _{w} sin \, i = \mu _{a} sin \, r$
$\frac{\text{escaped energy}}{\text{total energy}} = \text{0.17}$
total $\%$ energy enscaped $= 17 \%$