Question

Light of wavelengthis used in a Michelson interferometer. Let$\mathbf{x}$ be the position of the movable mirror, with$\mathbf{x}\mathbf{=}\mathbf{0}$when the arms have equal lengths${\mathbf{d}}_{\mathbf{2}}\mathbf{=}{\mathbf{d}}_{\mathbf{1}}$. Write an expression for the intensity of the observed light as a function of , letting${\mathbf{l}}_{\mathbf{m}}$be the maximum intensity.

Short answer

The expression for the intensity of observed light is $I=I_m{\cos }^2\left(\frac{2\pi x}{\lambda }\right)$.

Step-by-step solution

Given in the question.

Let $l_m$is the maximum intensity. The position of the mirror from the point at which $d_1=d_2$is$x$.

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:

$$\begin{gathered} \mathbf{\Delta \theta }\mathbf{=}\frac{\mathbf{2}\mathbf{\pi d}}{\mathbf{\lambda }} \\ \mathbf{=}\frac{\mathbf{2}\mathbf{\pi xsin\theta }}{\mathbf{\lambda }} \end{gathered}$$

Here,$\mathit{d}\mathbf{=}\mathit{x}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$

According to the question.

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

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

$$\begin{gathered} \Delta \theta =2\left(\frac{2\pi x}{\lambda }\right) \\ =\frac{4\pi x}{\lambda } \end{gathered}$$

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.

$I=I_0{\cos }^2\left(\frac{\Delta \theta }{2}\right)$

Here, $I_0=I_m$is intensity of unpolarised light.

Use the above formula solve as follows:

$$\begin{gathered} I=I_m{\cos }^2\left(\frac{\Delta \theta }{2}\right) \\ I=I_m{\cos }^2\left(\frac{2\pi x}{\lambda }\right) \end{gathered}$$

The intensity $\frac{I}{I_m}$as a function of $\frac{x}{\lambda }$is graphed as below:

From the figure, observe that the intensity of light is maximum at$x=\frac{m}{2}\lambda ,m=0,1,2,...$

From the figure, observe that the intensity of light is minimum at$x=\frac{1}{4}\left(2m+1\right)\lambda ,m=0,1,2,...$

Hence, the expression for the intensity of observed light is $I=I_m{\cos }^2\left(\frac{2\pi x}{\lambda }\right)$.