Question

A container contains water up to a height of 20 cm and there is a point source at the centre of the bottom of the container. A rubber ring of radius r floats centrally on the water. The ceiling of the room is 2.0 m in above the water surface.

(a) Find the radius of the shadow of the ring formed on the ceiling if r = 15 cm.

(b) Find the maximum value of r for which the shadow of the ring is formed on the ceiling. Refractive index of water $=\frac{4}{3}$.

Answer 1

(a) Using Snell's Law,

${\mu }_{1}sin{\theta }_1={\mu }_{2}sin{\theta }_2$

$\Rightarrow \frac{4}{3}.\frac{15}{\sqrt{400+225}}=1.\frac{x}{\sqrt{x^2+(200{)}^2}}$

$\Rightarrow \frac{20}{25}=\frac{x}{\sqrt{x^2+(200{)}^2}}$

$\Rightarrow 16.[x^2+(200{)}^2]=25x^2$

$\Rightarrow 9x^2=16\times (200{)}^2$

$\Rightarrow x^2=\frac{16}{9}\times (200{)}^2$

$\Rightarrow x=\frac{4}{3}\times 200\text{cm}$

$\therefore$ Radius of the shadow

$=(\frac{800}{3}+15)\text{cm}$

$=\frac{845}{3}\text{cm}$

= 281.66 cm

= 2.8 m

(b) Using Snell's Law,

${\mu }_{1}sin{\theta }_1={\mu }_{2}sin{\theta }_2$

$\Rightarrow \frac{4}{3}.\frac{r_m}{\sqrt{r_m^2+(20{)}^2}}=1.sin{90}^{\circ }$

$\Rightarrow \frac{4}{3}{r}_m=\sqrt{r_m^2+(20{)}^2}$

$\Rightarrow 16r_{m_2}^2=9[r_m^2+(20{)}^2]$

$\Rightarrow 7r_m^2=9\times (20{)}^2$

$\Rightarrow r_m=\sqrt{\frac{9}{7}\times (20{)}^2}$

$=\frac{3\times 20}{2.64}=22.7\text{cm}$