# Lecture 35 - Diffraction

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## Single slit diffraction

In our last lecture we 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 (IMHO), 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 $r-x\sin\theta$

## Intensity for single slit diffraction

To find the field at a given angle $E_{\theta}$ we need to take all our contributions from the slit

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

and integrate

$E_{\theta}=\int_{-\frac{D}{2}}^{\frac{D}{2}}E_{0}e^{ik(r-x\sin\theta)}\,dx=E_{0}e^{ikr}\int_{-\frac{D}{2}}^{\frac{D}{2}}e^{-ikx\sin\theta}\,dx$

$=E_{0}e^{ikr}\frac{1}{-ik\sin\theta}[e^{-ikx\sin\theta}]_{-D/2}^{D/2}=E_{0}e^{ikr}\frac{1}{-ik\sin\theta}[e^{-ik\frac{D}{2}\sin\theta}-e^{ik\frac{D}{2}\sin\theta}]$

$=E_{0}e^{ikr}\frac{1}{-ik\sin\theta}[-2i\sin(\frac{Dk}{2}\sin\theta)]=E_{0}e^{ikr}\frac{2}{k\sin\theta}\sin(\frac{Dk}{2}\sin\theta)$

The intensity is found from $I_{\theta}=|E_{\theta}|^{2}$

$I_{\theta}=E_{0}^{2}\frac{4}{k^{2}\sin^{2}\theta}\sin^{2}(\frac{Dk}{2}\sin\theta)=E_{0}^2D^{2} \mathrm{sinc^{2}}(\frac{Dk}{2}\sin\theta)$

We have just defined the intensity in terms of the sinc function. We can now normalize the intensity with respect to the central intensity using the fact that $\mathrm{sinc}(0)=1$

$I_{0}=E_{0}^{2}{D}$

so

$I_{\theta}=I_{0}\mathrm{sinc^{2}}(\frac{Dk}{2}\sin\theta)$

## Intensity pattern for a single slit

Rather than writing our equation in terms of k

$I_{\theta}=I_{0}\mathrm{sinc^{2}}(\frac{Dk}{2}\sin\theta)$

it is convenient to write it in terms of the wavelength, using $k=\frac{2\pi}{\lambda}$

$I_{\theta}=I_{0}\mathrm{sinc^{2}}(\frac{D\pi}{\lambda}\sin\theta)$

This function has a central maxima and then minima at

$D\sin\theta=m\lambda$       $m=\pm 1,\pm 2,\pm 3,..$

## Diffraction in the double slit

Now we have an expression for the intensity from a slit of width $D$

$I_{\theta}=I_{0}\mathrm{sinc^{2}}(\frac{D\pi}{\lambda}\sin\theta)$

we can consider slits as sources instead of the point sources we considered earlier which gave

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

The combination of these gives

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

## Diffraction from a circular aperture

Diffraction also occurs when light is shone through a circular aperture resulting in an Airy pattern.

## Diffraction limit

The diffraction effect that occurs for a small aperture places a limit on the resolution that can be achieved based on the size of the lens that is used.

phy142/lectures/35.txt · Last modified: 2014/05/02 18:21 by mdawber 