Question

A torch light kept at a depth h below the surface of an ocean (n = 4/3) facing upward is turned ON. Determine the illuminated surface area at the free surface of ocean.

• A. $\frac{9\pi h^2}{16}$
• B. $\frac{9\pi h^2}{7}$
• C. $\frac{3\pi h^2}{4}$
• D. $\frac{\pi h^2}{3}$

Answer 1

The correct option is B $\frac{9\pi h^2}{7}$

Given,
Depth of torch light = h
Refractive index of ocean (n) = $\frac{4}{3}$

The boundary of the illuminated area at the free surface of water is determined by the ray which strikes the interface just at critical angle.
All the rays from the torch light which strikes the interface at an angle greater than crictical angle will reflect back in the sea and will not illuminate the free surface of water.
So, at the critical angle

$\mu =\frac{1}{sin{\theta }_c}$

Putting the known values,

$\frac{4}{3}=\frac{1}{sin{\theta }_c}$

$sin{\theta }_c=\frac{3}{4}$

Also,
$sin{\theta }_c=\frac{perpendicular\left(p\right)}{hypotenuse\left(h\right)}$

So,
Perpendicular (p) = 3 unit
hypotenuse (h) = 4 unit
Applying pythagoras theoram,
Base (b) = $\sqrt{Hypotenus{e}^2-Perpendicula{r}^2}=\sqrt{h^2-p^2}$

Which gives,
base (b) = $\sqrt{7}$ unit

Now,
$tan{\theta }_c=\frac{p}{b}$
So,
$tan{\theta }_c=\frac{3}{\surd 7}$

Now, let the radius of the circle which appears bright at the free surface be r.
Then,
From the above triangle
$tan{\theta }_c=\frac{r}{h}$

$r=htan{\theta }_c=\frac{3h}{\surd 7}$

Area (A) of circle,
$A=\pi r^2=\pi (\frac{3h}{\surd 7}{)}^2=$ $\frac{9\pi h^2}{7}$