Light of wavelengthis used in a Michelson interferometer. Letx be the position of the movable mirror, withx=0when the arms have equal lengthsd2=d1. Write an expression for the intensity of the observed light as a function of , lettinglmbe the maximum intensity.

Short Answer

Expert verified

The expression for the intensity of observed light is I=Imcos22πxλ.

Step by step solution

01

Given in the question.

Let lmis the maximum intensity. The position of the mirror from the point at which d1=d2isx.

02

Definition of wave interference between two plane waves.

Assume that the two waves are in phase at the point P, then the relative phase changes along the x-axis. The phase difference at the point Q is given as:

Δθ=2πdλ=2πxsinθλ

Here,d=xsinθ

03

According to the question.

Let the beam splitting mechanism for x=0in two waves interfere constructively.

Eliminate the factor . The phase difference between two light paths is written as follows:

Δθ=22πxλ=4πxλ

04

The intensity of observed light. 

By Maul’s law, the intensity of plane-polarized light that passes through an analyzer varies as the square of the cosine of the angle between the plane of the two plane that is the polarizer and the transmission axes of the analyzer.

I=I0cos2Δθ2

Here, I0=Imis intensity of unpolarised light.

Use the above formula solve as follows:

I=Imcos2Δθ2I=Imcos22πxλ

The intensity IImas a function of xλis graphed as below:

From the figure, observe that the intensity of light is maximum atx=m2λ,m=0,1,2,...

From the figure, observe that the intensity of light is minimum atx=142m+1λ,m=0,1,2,...

Hence, the expression for the intensity of observed light is I=Imcos22πxλ.

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