Question

An object is placed at a depth of 0.5m inside a liquid. A circular patch of light of radius 0.5675m is formed on the surface. Find the critical angle and refractive index for the liquid

Answer 1

Explanation:

Light of the source emerges out of water only when the incident angle at the water air boundary equals critical angle. (θ

c

)

Let depth of lake =H

and radius of Illuminated circle =R

Given : πR

2

=3H

2

⇒H

2

=

3

π

R

2

...(1)

From figure,

sinθ

c

=

R

2

+H

2

R

also we know, sinθ

c

=(

μ

1

) [μ: Refractive index of water]

So we get, from above:-

⇒μ=

R

R

2

+H

2

=

R

R

1+

3

π

[from (1)]

⇒μ=

1+

3

π

hope you understand it bro do hardwrok all the best