A newton's ring arrangement is used with a source emitting two wavelengths a1 = 6 x 10 cm and 4.5 x 10 cm and it is found that the nh dark ring due to 1 coincides with (n + 1)th dark ring for 12. If the radius of curvature of the curved surface is 60 cm, find the diameter of nth dark ring
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Answer:
Given: λ₁= 6 x 10 cm
λ₂=4.5 x 10 cm
R=60 cm
(D n )λ₁=(D n+1 )λ₂
Formula:
(D n ²) =4nλ
Solution:
for nth dark ring λ₁,
(D n ²) λ1=4nλ₁ ....(1)
for (n+1)th dark ring λ₂,
(D n+1 ²) λ1=4(n+1)λ₂ ....(2)
from given ,
(D n )λ₁=(D n+1 )λ₂
squaring on both the sides,
(D n² )λ₁=(D n+1 ²)λ₂
4nλ₁=4(n+1)λ₂ ....( from 1 and 2)
nλ₁=(n+1)λ₂
nλ₁=nλ₂+λ₂
n=λ₂/λ₁-λ₂
n= 4.5x10 /6x 10 - 4.5 x10
n=4.5/1.5
n=3
Using equation 1, the diameter of 3th dark ring for λ₁ is,
D₃²= 4x3x60x6x10
D₃²= 43200
D₃=207.84 cm (ACCORDING TO GIVEN DATA)
Explanation:
Use concept of radii of newton's dark rings in reflected light.
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