# Question c8d58

Jul 9, 2016

When two light waves of intensity ${I}_{1} \mathmr{and} {I}_{2}$ respectively interfere with each other the resultant intensity at a point is given by ${I}_{R} = {I}_{1} + {I}_{2} + 2 \sqrt{{I}_{1} {I}_{2}} \cos \phi$
where $\phi$ is the phase difference between the two light waves.

In Young's Double Slit Experiment, the two slits are identical. Therefore, the intensity equation can be shown to be
 I_R=4Icos^2(ϕ/2)
We see that the maximum intensity produced is $4 I$. Putting maximum intensity as ${I}_{\circ}$, we have
 I_R=I_@cos^2(ϕ/2)
=> I_R/I_@=cos^2(ϕ/2)# .....(1)

When two waves meet on the screen, the phase difference between them depends on two factors.

1. the phase difference produced due to the difference in the path taken by the two waves.
2. the initial phase difference when the wave starts at the slit.

For the given point, intensity is $I$ where path difference is $\frac{\lambda}{6}$. We know that for a path length$= \lambda$, phase difference $= 2 \pi$.
$\therefore$ Phase difference $\phi = \frac{2 \pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3}$
Our intensity equation (1) becomes
$\frac{I}{I} _ \circ = {\cos}^{2} \left(\frac{\pi}{6}\right)$
$\implies \frac{I}{I} _ \circ = \frac{3}{4}$

-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.

Suppose two coherent plane waves $1 \mathmr{and} 2$ emerge from the two slits. Let the electric fields of the two is given by
${E}_{1} = {E}_{\circ} \sin \left(\omega t\right)$, and
${E}_{2} = {E}_{\circ} \sin \left(\omega t + \phi\right)$
We have assumed same maximum amplitude for both. And also assumed the point where intensity is to be calculated as the origin. The phase difference between the two is $\phi$.

Assuming both fields pointing in the same direction and using superpostion principle the resultant electric field is
$E = {E}_{1} + {E}_{2} = {E}_{\circ} \left[\sin \omega t + \sin \left(\omega t + \phi\right)\right]$
Using trigonometric identity
$\sin A + \sin B = 2 \sin \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)$
above expression becomes
$E = 2 {E}_{\circ} \sin \left(\omega t + \frac{\phi}{2}\right) \cos \left(\frac{\phi}{2}\right)$

Since the intensity $I$ is proportional to the time average of the square of the total electric field we get

$I \propto \text{<"E^2">"=4E_@^2"<"sin^2(omegat+phi/2)">} {\cos}^{2} \left(\frac{\phi}{2}\right)$
To find the $\text{<"sin^2theta ">}$ term we need to integrate it over one cycle and divide with the cycle interval $2 \pi$. We get $\text{<"sin^2theta">} = \frac{1}{2}$. Inserting $2 {E}_{\circ}^{2} = {I}_{\circ}$ we obtain
$I = {I}_{\circ} {\cos}^{2} \left(\frac{\phi}{2}\right)$