Question

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 $2\pi rh$]

• A. $34\%$
• B. $17\%$
• C. $50\%$
• D. $21\%$

Answer 1

Answer: (B)

$\frac{4}{3}sin\theta =1sin90^{\circ }$
$sin\theta =\frac{3}{4}$
Area of sphere in which light spread $=4\pi R^2$
$\Omega =2\pi \left(1-cos\theta \right)$
$\Omega =2\pi \left(1-\frac{\sqrt{7}}{4}\right)$
$P\to 4\pi$ steradians
$P^{\prime }\to \frac{P}{4\pi }\left(1-cos\theta \right)$
Ratio $=\frac{P^{\prime }}{P}=\frac{2\pi \left(1-cos\theta \right)}{4\pi }=\frac{\left(1-cos\theta \right)}{2}=\frac{1-\frac{\sqrt{7}}{4}}{2}$
$=\frac{0.33}{2}=0.17$