Question

A newtons rings setup is used with a source emitting two wavelength λ1=6000A0 and λ2=4500 A0. It is found that nth dark ring due to 6000A0 coincides with (n+2)th dark ring due to 4500 A0. If the radius of curvature of the lens is 90 cm, find the diameter of nth dark ring of 6000A0.

Answer 1

Given:- λ1=6000Aº

λ2=4500 Aº

R=90 cm

(𝐷_𝑛)_λ1=(𝐷_𝑛+2)_λ2

Formula:- 𝐷_𝑛²=4𝑛𝑅λ

For nth dark ring λ1
(𝑫_𝒏^𝟐)_𝛌𝟏=𝟒𝒏𝑹𝛌𝟏------------(1)

And for (n+1)th dark ring λ2

(𝑫_𝒏+𝟐^𝟐)_𝛌𝟐=𝟒(𝒏+𝟐)𝑹𝛌𝟐---------------(2)

4𝑛𝑅λ1=4(𝑛+2)𝑅λ2

𝑛λ1=(𝑛+2)λ2

𝑛=`(2λ2)/(λ1−λ2)`

𝑛=`(2×4500)/(6000−4500)`

𝒏=𝟔

Using Equation 1, the diameter of 6th dark ring for λ1 is

𝐷₆²=4×6×90×6×10⁻⁵

`D_6=sqrt(4×6×90×6×10^(−5))`

𝑫_𝟔=𝟎.𝟑𝟔 𝒄𝒎