Question

A disc is placed on the surface of a pond filled with liquid of refractive index $\frac{5}{3}$. A source of light is placed 4 m below the surface of the liquid. The minimum area of the disc so that light does not come out of the liquid into air is about

• A. $28.26m^2$
• B. $26.28m^2$
• C. $28.26c{m}^2$
• D. $26.28c{m}^2$

Answer 1

The correct option is A $28.26m^2$

Let radius of the disc be r.
Distance of source of light from the surface of liquid = AO = h = 4 m
Refractive index, $\mu =\frac{5}{3}$
Light does not come out of the liquid into the air if:
$\text{sin c}=\frac{1}{\mu }$ [c = critical angle]
$\Rightarrow \frac{r}{AB}=\frac{1}{\mu }$ $[In\Delta AOB,\sin c=\frac{OB}{AB}=\frac{r}{AB}]$
$\Rightarrow \frac{r}{\sqrt{r^2+h^2}}=\frac{1}{\left(\frac{5}{3}\right)}$ $[AB=\sqrt{O{B}^2+A{O}^2}=\sqrt{r^2+h^2}]$
$\Rightarrow \frac{r}{\sqrt{r^2+4^2}}=\frac{3}{5}$
$\Rightarrow r=3m$
$\therefore \text{Area of the disc}=\pi r^2$
$=3.14\times 3\times 3$
$=28.26m^2$
Hence, the correct answer is option (a).