Question

A point source of light is placed at a distance $2\text{m}$ below the surface of a large and deep lake. The fraction of light that escapes directly from water surface will be $({\mu }_{water}=4/3$)

$[$ Absorbtion of light within the water and reflection at surface except where it is total, have been neglected $]$

• A. $0.5$
• B. $0.3$
• C. $0.2$
• D. $0.4$

Answer 1

The correct option is C $0.2$

Due to total internal reflection some of the light rays incident at the interface will return back into water if,
angle of incidence $\left(i\right)>$ critical angle $\left({\theta }_c\right)$.

The light rays that reach beyond the base of the cone whose vertical angle is $\left(2{\theta }_c\right)$ will suffer total internal reflection.

Hence, Only the light incident on the base of the cone refracts and escapes.


For fraction of light escaping ,
We know,
$\text{Escape energy}=\frac{\text{Total energy}}{\text{Solid angle of sphere}}\times \text{Solid angle of cone}$

$\Rightarrow \text{Fraction of light escaping (f)}=\frac{\text{Escape energy}}{\text{Total energy}}=\frac{\text{Solid angle of cone}}{\text{Solid angle of sphere}}$

$\therefore$$f=\frac{\text{solid angle of cone}}{\text{solid angle of sphere}}$

$\Rightarrow f=\frac{2\pi (1-\cos {\theta }_c)}{4\pi }$

$\Rightarrow f=\frac{1-\sqrt{1-si{n}^2{\theta }_c}}{2}$

$[\sin {\theta }_c=\frac{{\mu }_{air}}{{\mu }_{water}}=\frac{1}{\left(\frac{4}{3}\right)}]$

$\Rightarrow f=\frac{1-\sqrt{1-\frac{9}{16}}}{2}$

$\Rightarrow f=\frac{4-\sqrt{7}}{8}\approx 0.2$

Why this question? This question gives you an intuition about why a point source of light seems to be dim in water. Not all the light comes out of water, thus intensity decreases. Hence, it looks dim.