Question

A light source is placed at the bottom of a water tank 1m deep. It is found that a circle of illuminated surface is formed at the top of the tank. The radius of the circle is

Answer 1

Given:

A light source is placed at the bottom of a water tank 1m deep. It is found that a circle of illuminated surface is formed at the top of the tank.

To find:

Radius of the circle.

Calculation:

In critical condition , applying Snell's Law:

$\therefore \: \mu \times \sin( \theta) = 1 \times \sin(90 \degree)$

$= > \: \sin( \theta) = \dfrac{1}{ \mu}$

$= > \dfrac{r}{ \sqrt{ {r}^{2} + {h}^{2} } } = \dfrac{1}{ \mu}$

Squaring both sides:

$= > \dfrac{ {r}^{2} }{ {r}^{2} + {h}^{2} } = \dfrac{1}{ { \mu}^{2} }$

$= > \: { \mu}^{2} {r}^{2} = {r}^{2} + {h}^{2}$

$= > \: ({ \mu}^{2} - 1) {r}^{2} = {h}^{2}$

$= > \: {r}^{2} = \dfrac{ {h}^{2} }{ { \mu}^{2} - 1}$

$= > \: r= \sqrt{\dfrac{ {h}^{2} }{ { \mu}^{2} - 1} }$

Putting h = 1 metre ;

$= > \: r= \dfrac{ 1 }{ \sqrt{ { \mu}^{2} - 1}}$

So , final answer is :

$\boxed{ \red{ \sf{ \: r= \dfrac{ 1 }{ \sqrt{ { \mu}^{2} - 1}} }}}$