Question

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

Answer 1

use 1\sin c=n formula.find c.then tanc=x\4. find x(x= radius of the curcular sheet that needs to b kept) x=2.9996 m\3m

Answer 2

The minimum diameter of disc is 6 meter.

Given:

Refractive index of water is $\frac{5}{3}$

Length of the light source is 4 meter.

To find:

The minimum diameter of the disc on the water surface where it’s cut –off light is coming out from water.

Solution:

By using the formula for,  

The radius of disc is   $r=\frac{h}{\sqrt{n^{2}-1}}$

n means the refractive index of water.

h is the length of the light source,  

Therefore

  $r=\frac{h}{\sqrt{n^{2}-1}}$

$=\frac{4}{\sqrt{\frac{25}{9}-1}}$

$\mathrm{r}=\frac{4}{\sqrt{\frac{16}{9}}}$

$r=\frac{4}{\frac{4}{3}}$

r = 3 m.

We all know that, the diameter is “twice that of the radius”.

Therefore, the diameter becomes, d = 2r = 6 m.