Question

A point source of light is placed 4 cm below the surface of a liquid of refractive index 5/3. The minimum diameter of a non transparent disc,which should be placed over the source, on the surface of water to cut off all light coming out of water is :

• A. 2 cm
• B. 6cm
• C. 4m
• D. 3m

Answer 1

The correct option is B 6cm

Since, the light is in the form of a disc, it implies, that the rays after refraction at the interface of the two media, must be going parallel to the surface.

Applying Snell's law at the interface of the two media, we have,

${\mu }_{1}Sini={\mu }_{2}Sinr$

Given, ${\mu }_1={\mu }_{liquid}=\frac{5}{3}$

We know, ${\mu }_2={\mu }_{air}=1$

From the figure,

$i=\theta$ and $r={90}^{\circ }$

$\Rightarrow \frac{5}{3}Sin\theta =1$

$Sin\theta =\frac{3}{5}$

From the figure,

$Sin\theta =\frac{r}{\sqrt{r^2+h^2}}$

Given, $h=4cm$

$\frac{r}{\sqrt{r^2+h^2}}=\frac{3}{5}$

Squaring both sides, we have

$\frac{r^2}{r^2+4^2}=\frac{9}{25}$

$\frac{r^2+16}{r^2}=\frac{25}{9}$

$\frac{16}{r^2}=\frac{25}{9}-1\Rightarrow \frac{16}{r^2}=\frac{16}{9}$

$r^2=\frac{16\times 9}{16}$

$r=3cm$
Diameter, $d=2\times r=2\times 3=6cm$

Hence, the correct answer is OPTION B.