91Ó°ÊÓ

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

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:

$$\begin{gathered} \mathbf{Δθ}\mathbf{=}\frac{\mathbf{2}\mathbf{Ï€»å}}{\mathbf{λ}} \\ \mathbf{=}\frac{\mathbf{2}\mathbf{Ï€³\ae ²õ¾\pm ²Ôθ}}{\mathbf{λ}} \end{gathered}$$

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

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:

$$\begin{gathered} \mathrm{Δ}\mathrm{θ}=2\left(\frac{2\mathrm{π}x}{\mathrm{λ}}\right) \\ =\frac{4\mathrm{π}x}{\mathrm{λ}} \end{gathered}$$

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.

$I=I_0{\cos }^2\left(\frac{\mathrm{Δ}\mathrm{θ}}{2}\right)$

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

Use the above formula solve as follows:

$$\begin{gathered} I=I_m{\cos }^2\left(\frac{\mathrm{Δ}\mathrm{θ}}{2}\right) \\ I=I_m{\cos }^2\left(\frac{2\mathrm{π}x}{\mathrm{λ}}\right) \end{gathered}$$

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

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

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

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