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
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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
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