Question

There is a small source of light at some depth below the surface of water $(\text{refractive index}=\frac{4}{3})$, in a tank of large cross-sectional surface area. Neglecting any reflection from the bottom and absorption by water, the percentage of light that emerges out of the 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. $50\%$

Answer 1

The correct option is C $17\%$


Given,
Refractive index, $\mu =\frac{4}{3}$
Consider the cone of light formed on the surface of water, as shown in the figure. $C$ is the critical angle.

Now, $\sin C=\frac{1}{\mu }=\frac{3}{4}$

$\Rightarrow \cos C=\frac{\sqrt{7}}{4}[\because \cos C=\sqrt{1-{\sin }^{2}C}]$

The solid angle subtended by the cone of light, at the source is,

$\Omega =2\pi (1-\cos C)=2\pi (1-\frac{\sqrt{7}}{4})$

$\therefore$ Fractional energy transmitted is,

$=\frac{\text{solid angle subtended by the cone of light}}{\text{total solid angle}}$

$=\frac{2\pi (1-\cos \theta )}{4\pi }=\frac{(1-\frac{\sqrt{7}}{4})}{2}=0.17$

So, percentage of light emerging out of the water surface is,
$=0.17\times 100=17\%$

Hence, $\left(C\right)$ is the correct answer.