# Lecture 35 - From interference to diffraction

## Two-Slit Interference

The condition for constructive interference (bright fringes) is

$d\sin\theta=m\lambda$       (m=0,1,2,..)

For small angles the angles $\theta_{1}$ and $\theta_{2}$ can be approximated as

$\sin\theta_1\approx\theta_{1}=\frac{\lambda}{d}$ and $\sin\theta_2\approx\theta_{2}=\frac{2\lambda}{d}$

The positions of the fringes are

$x_{1}=l\tan\theta_{1}\approx l\theta_{1}\approx \frac{l\lambda}{d}$ and $x_{2}=l\tan\theta_{2}\approx l\theta_{2}\approx\frac{2l\lambda}{d}$

## A Pioneering Application of Interference

The Michelson Interferometer is sometimes called the most famous failed experiment.

## Another look at Huygen's principle

We have used Huygen's principle for two point sources to explain the positions of the minima and maxima in a two slit interference pattern.

To explain diffraction based on Huygen's principle we will need to make some additions to it, which were originally made by Fresnel.

It should be noted that both Huygen's and Fresnel's work preceded Maxwell, Huygen by about 200 years, Fresnel by about 40, so while we can take Maxwell to be the great unifier of electromagnetism and optics the idea of a wave picture of light already existed before his equations.

While Fresnel proposed his theory on an empirical basis, it was later shown by Kirchoff that the theory could be derived from Maxwell's equations.

## How to deal with phase

The key modifications to Huygen's principle that Fresnel made were to be able to deal adequately with the phase of the waves coming from different source points that make up the intensity at a certain point.

These phase modifications are essential to be able to look at diffraction effects.

In making the diagram for for two slit interference we simply added together two spherically propagating wave from two point sources taking into account their phase difference.

We can consider these waves as electric field waves

$E_{1}=E_{10}\sin\omega t$

and

$E_{2}=E_{20}\sin(\omega t + \delta)$

The electric field at a given point we will define as $E_{\theta}$ is the sum of the electric field from the waves propagating from the two sources $S_{1}$ and $S_{2}$

$E_{\theta}=E_{1}+E_{2}$

The figure is simply a plot of the total field, but it would actually be more useful for us to have a way to calculate the average intensity of the light at a given point.

## Phasor method for calculating intensity

We saw when we looked at AC circuits that there are a couple of approaches for adding time varying quantities that may have a phase difference together, a graphical phasor approach, which helps get an intuitive picture or an approach involving complex math which is ultimately much more powerful.

$E_{1}=E_{10}\sin\omega t$

and

$E_{2}=E_{20}\sin(\omega t + \delta)$

For a double slit, provided we are sufficiently distant from the slit, $E_{10}=E_{20}=E_{0}$. This assumption distinguishes Fraunhofer diffraction from near-field diffraction which is much more complicated.

It can now be seen that

$E_{\theta}=E_{\theta 0}\sin({\omega t +\frac{\delta}{2}})$

and

$E_{\theta_{0}}=2E_{0}\cos\frac{\delta}{2}$

The average intensity is given by the square of the magnitude of the electric field at a point

$I_{\theta}=E_{\theta 0}^{2}=4E_{0}^2\cos^{2}\frac{\delta}{2}$

A useful reference point for intensity in terms of angle is the central intensity when the waves from both sources are always in phase

$I_{0}=(E_{10}+E_{20})^2=4E_{0}^2$

And we can combine the previous equations to obtain

$I_{\theta}=I_{0}\cos^{2}\frac{\delta}{2}$

## Intensity for double slit interference

As we now have the intensity in terms of the phase difference $\delta$

$I_{\theta}=I_{0}\cos^{2}\frac{\delta}{2}$

we should return to our diagram to find the intensity in terms of $d$ and $\theta$

We can see that

$\frac{\delta\lambda}{2\pi}=d \sin \theta$

giving

$\delta=\frac{2\pi}{\lambda}d\sin\theta$

and the intensity is

$I_{\theta}=I_{0}\cos^{2}(\frac{\pi d \sin \theta}{\lambda})$

## Complex approach

A different approach to the same problem is to write our electric fields as complex functions

$E_{1}=E_{0}e^{i\omega t}$

$E_{2}=E_{0}e^{i(\omega t + \delta)}$

$E_{\theta}=E_{1}+E_{2}=E_{0}e^{i\omega t}(1+e^{i\delta})$

and the intensity is

$I_{\theta0}=E_{\theta}^{*}E_{\theta}=E_{0}^2(1+e^{-i\delta})(1+e^{i\delta})=E_{0}^2(2+2\cos\delta)=4E_{0}^2\cos^{2}{\frac{\delta}{2}}$

## The Poisson Spot

Fresnel's theory came at a time when the nature of light was hotly debated.

Fresnel presented it a competition organize in 1818 by the French Academy of Sciences.

One of the judges was Siméon-Denis Poisson who was a great believer in the corpuscular theory of light and challenged the theory claiming it would predict the ridiculous observation of a bright spot at the center of a beam blocked by a circular object.

But it turned out to be the case!

This is due to diffraction, which we discuss in detail at the next lecture .

## The Arago Spot

Unfortunately for Poisson, Dominique-François-Jean Arago (who was the chief judge) decided to try the experiment and found what is known alternatively as the Poisson or Arago Spot.

Here is the basic phenomenon.

Here is a better picture than in the video we just saw.

## Single slit diffraction

We have looked at how to deal with the phase of light waves coming from two point sources in order to calculate the intensity at a given angle.

We will now build on this approach to be able to find the intensity distribution when light falls on a single slit.

We will only consider the case where the screen where the pattern is seen is very far away from the slit (normally referred to as Fraunhofer diffraction).

For this problem the complex approach is going to be much easier than the phasor approach, so we'll do it this way.

The phasor approach can be found in your textbook.

As the field for a travelling wave can be expressed as

$E=E_{0}e^{i(kz-\omega t)}$

we can say that the field contribution from a point on the slit can be approximated by

$E(x)=E_{0}e^{ik(r-x\sin\theta)}$

here we have assumed that the screen is far enough away that we don't need to worry about the $\frac{1}{r}$ dependence of the field. We are also assuming that the rays are virtually parallel so that the path distance from each point is $r-x\sin\theta$