One can get a feel for why there are maxima and minima in case of a **single** slit by considering the following figures:

Here is a discussion of why complex numbers are a powerful way to represent vector addition.

In making the diagram 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 $S_{2}$

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

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

and

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

For a double slit, provided we are sufficiently distant from the slit, $E_{10}=E_{20}=E_{0}$.

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

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

Since

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

we obtain

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

Now we are going to arrive at the same answer by writing 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

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

We will now build on the complex electric field 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).

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 $r-x\sin\theta$

To find the field at a given angle $E_{\theta}$ we consider the contribution from a “slit” of thickness or interval $dx$:

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

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

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

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

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

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

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

Now that we have covered 2-slit interference and single slit diffraction and merged the two concepts together, let us look the demo of how to differentiate between diffraction and interference effects and also how they depend on wavelength.

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

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.