Interference/diffraction

Consider the superposition of two electromagnetic waves.

\begin{align*} \mathbf E_1 &= A_1 \cos(\omega t - kz + \phi_1) \xhat \\ \mathbf E_2 &= A_2 \cos(\omega t - kz +\phi_2) \xhat \\ \mathbf E &= \mathbf E_1 + \mathbf E_2. \end{align*}

We do not usually measure the electric field of a wave directly, so let’s consider the intensity instead.

\begin{align*} I = |\mathbf S| = \frac{1}{\mu_0}|\mathbf E \times \mathbf B| = \frac{n}{\mu_0 c} |\mathbf E^2| = cn\epsilon_0 |\mathbf E^2| \propto |\mathbf E^2| \end{align*}

where $n = c/v$ is the index of refraction of the material. We calculate $I \propto |\mathbf E^2|$ for our given waves.

\begin{align*} |\mathbf E^2| &= |\mathbf E_1|^2 + |\mathbf E_2|^2 + 2|\mathbf E_1 \mathbf E_2| \\ &= A_1^2 \cos^2(\omega t - kz + \phi_1) + A_2^2 \cos^2(\omega t + kz + \phi_2)\\ &+ 2A_1A_2 \cos(\omega t + kz + \phi_1) \cos (\omega t + kz + \phi_2). \end{align*}

Taking the time average,

\begin{align*} \langle I \rangle \propto \frac{1}{T} \int_0^T |\mathbf E^2| \mathrm dt = \frac{1}{\mu_0} \left(\frac{A_1^2}{2} + \frac{A_2^2}{2} + A_1A_2 \cos(\phi_1 - \phi_2) \right). \end{align*}

The plot below shows this time average intensity in terms of the phase difference $\delta := \phi_1 - \phi_2$ . We see that when $\delta$ is an odd multiple of $\pi$ the amplitude is at a minimum, and when $\delta$ is an even multiple of $\pi$ , a maximum.

Interference at a material boundary

Consider the boundary between two materials, e.g., the boundary between air and water in a soap bubble. The figure below shows three regions, the air outside the bubble to the left, the water in the middle, and the air inside the bubble to the right. An incident light beam is normal to the boundary, shining from the left. We wish to find the intensity of the reflected light.

A component $R_1$ of the incident light reflects off the outer boundary immediately. Another component $R_2$ is transmitted through the first boundary, then reflected off the second, and finally transmitted again through the first. After a few reflections the intensity of the light decreases significantly, so we will ignore all but $R_1$ and $R_2$ .

We use the reflection and transmission coefficients to calculate the amplitude of $R_1$ and $R_2$ relative to the incident beam $E_0$ .

\begin{align*} R_{12} &= \frac{n_1-n_2}{n_1+n_2} \\ T_{12} &= \frac{2n_1}{n_1+n_2}, \quad R_{21} = \frac{n_2-n_1}{n_2+n_1}, \quad T_{21} = \frac{2n_2}{n_2+n_1} \\ R_1 &= R_{12} E_0 \\ R_2 &= T_{12} R_{21} T_{21} E_0. \end{align*}

Since $n_2>n_1$ , the reflection off the first boundary $R_{12} < 0$ , which mean that $R_1$ is negative. This sign flip corresponds to a $180\degree$ phase shift.

$R_2$ has to travel an extra distance $2d$ compared to $R_1$ . We have seen that the phase difference due to extra distance traveled is $\delta = 2dk$ . However, it is important to consider that $k$ depends on the medium.

Let $k_0$ be the wavenumber of the wave in a vacuum. Assuming a non-dispersive wave, $\omega = v_p k$ . The speed of light in material is $v_n = c/n$ , where $n$ is the index of refraction. Then the wavenumber in material $k_n = \omega/v_{p,n} = \omega n/c = n k_0$ .

This means the total phase shift between $R_1$ and $R_2$ is $\delta = 2dkn_2 - \pi$ (or equivalently, $+\pi$ , since a phase shift of $+\pi$ is exactly the same as a phase shift of $-\pi$ ).

If the incident beam is not normal to the surface, the extra distance traveled will be greater than $2d$ . In the case shown below, the extra distance traveled by $R_2$ is $2d/\cos\theta'$ .

Double slit experiment

Consider two small slits a distance $d$ apart in a thin wall allowing light to shine through onto a wall a distance $L$ away. We will consider the electric field at a point $P$ on that wall, an angle $\theta$ from the bottom slit.

We assume that $L \gg d$ , so $\mathbf E_1$ and $\mathbf E_2$ are roughly parallel (the figure above is greatly exagerated). Then we see that the extra distance the light has to travel from $B$ to $P$ is $d\sin\theta$ (shown in $\color{teal}\text{green}$ ). Assume the electric waves take the form $\mathbf E_0 \cos(kr-\omega t)$ , then the phase difference $\delta = kd\sin\theta$ .

As we saw above, we get constructive interference when $\delta$ is an even multiple of $\pi$ , and destructive when it is an odd multiple.

We now extend this experiment to consider the interference from $N$ slits.

As before, each slit is a distance $d$ apart, and we consider a point $P$ far away, such that the line from each slit to $P$ is roughly parallel, at an angle $\theta$ from the horizontal. By the same reasoning as above, the extra distance that light from slit $j$ must travel relative to slit $1$ is $j d \sin \theta$ .

We can write the resulting electric field at $P$ as a superposition of the field from each slit

\begin{align*} E_\text{total} &= E_0 \left[e^{i(\mathbf k \cdot \mathbf r-\omega t)} + e^{i(\mathbf k \cdot \mathbf r - \omega t + \delta)} + \cdots + e^{i(\mathbf k \cdot \mathbf r - \omega t + (N-1)\delta)}\right] \\ &= E_0 e^{i(\mathbf k \cdot \mathbf r - \omega t)} \sum_{j=0}^{N-1} \left(e^{i\delta}\right)^j = E_0 e^{i(\mathbf k \cdot \mathbf r - \omega t)} \left(\frac{1-e^{i\delta N}}{1-e^{i\delta}} \right) \\ \text{where } \delta &= \frac{2\pi d\sin\theta}{\lambda}. \end{align*}

We factor the fraction and simplify to find

\begin{align*} \frac{1-e^{i\delta N}}{1-e^{i\delta}} &= \frac{e^{i\delta N/2}}{e^{i\delta/2}} \cdot \frac{e^{-i\delta N/2} - e^{i\delta N/2}}{e^{-i\delta/2} - e^{i\delta/2}} \\ &= e^{i\delta(N-1)/2} \left(\frac{\sin(\delta N/2)}{\sin(\delta/2)}\right) \\ E_\text{total} &= E_0 e^{i(\mathbf k \cdot \mathbf r - \omega t)} e^{i\delta(N-1)/2} \left(\frac{\sin(\delta N/2)}{\sin(\delta/2)}\right). \end{align*}

The resulting intensity at $P$ is then

\begin{align*} \langle I\rangle &\propto |E|^2 = I_0 \left(\frac{\sin(N\delta/2)}{\sin(\delta/2)}\right)^2. \end{align*}

The intensity as a function of $\delta$ (with $N=4$ ) is plotted below.

Note a few things about this plot:

We have “principle maxima” at multiples of $2\pi$ , and for $N>2$ we also have smaller maxima and minima in between.
We have $N-1$ minima between adjacent principle maxima.
As $N$ increases, the width of the principle maxima decreases.
We can find the width of the principle maxima by finding the first minima.

Infinite slits (diffraction)

We have considered slits of infinitesimal width, which isn’t very physical. We would like to represent slits with some finite width. One way to do this is to consider an infinite set of these infinitesimal slits very close together. This is similar to what we did when moving from discrete coupled oscillators to continuous oscillators.

We start with the same sum as before

E_{\text{total}} = E_0e^{i(\mathbf k \cdot \mathbf r - \omega t)} \sum_{j=0}^{N-1} \left(e^{i\delta} \right)^j.

Instead of a discrete index $j$ we will use a continuous variable $y$ to represent the position of each infinitesimal slit.

\begin{align*} E_\text{total} &= E_0 e^{i(\mathbf k \cdot \mathbf r - \omega t)} \sum_y^\infty e^{i\delta(y)} \\ \text{where } \delta(y) &:= \frac{2\pi y\sin\theta}{\lambda}. \end{align*}

We now define a function $S(y)$ to describe the shape of our slit. When $S(y)=0$ none of the incident electric field passes through, and when $S(y)=1$ all of the field passes through. We use $S$ to write the electric field as an integral

\begin{align*} E_\text{total} &= E_0 e^{i(\mathbf k \cdot \mathbf r - \omega t)} \int_{-\infty} ^\infty S(y) e^{-i \mathcal K y} \d y \\ \text{where } \mathcal K &= -\frac{2\pi \sin\theta}{\lambda}. \end{align*}

We see that $E_\text{total}$ is the Fourier transform of $S$ .

As usual, we are more interested in intensity than electric field. We can write intensity as

I = I_0 \left[ \int_{-\infty}^\infty S(y) e^{-i\mathcal Ky} \d y\right]^2.

Let’s evaluate this intensity for a few different slit functions and see what it gives us.

When we have a single wide slit, we get a $\mathrm{sinc}$ pattern.

\begin{align*} S(y) &= \begin{cases} 1 & |y| < a/2 \\ 0 & \text{otherwise} \end{cases} \\ I &= I_0 \, \mathrm{sinc}^2 (a\pi\sin\theta/\lambda). \end{align*}

When we have two infinitesimal slits, we get an oscillating intensity. (The plot is not able to show the delta functions well).

\begin{align*} S(y) &= \delta(y-d/2) + \delta(y+d/2) \\ I &= I_0 \cos\left(\frac{\pi d \sin\theta}{\lambda}\right). \end{align*}

When we have two thin slits, the result is a product of the previous two results. The width of the slits determines the large-scale pattern, while the distance between them determines the small-scale oscillations.

\begin{align*} S(y) &= \begin{cases} 1 & |y \pm d/2| < a/2 \\ 0 & \text{otherwise} \end{cases} \\ I &= I_0\, \mathrm{sinc}^2\left(\frac{a\pi\sin\theta}{\lambda}\right)\cos^2\left(\frac{\pi d \sin\theta}{\lambda}\right). \end{align*}

Notice a few things about the solutions:

$I$ and $S$ are Fourier pairs, so $y$ and $\mathcal K$ are reciprocal variables. This means that a greater $y$ corresponds to a smaller $\mathcal K$ , and vice-versa. For example, a narrower slit will lead to a greater oscillation period in the intensity plot. Two slits closer together will lead to a wider $\mathrm{sinc}$ pattern.
Intuitively, if two slits are far apart then it takes only a small change in $\theta$ to cause a significant difference in the distance light has to travel from one slit compared to the other. If they are close together, then $\theta$ needs to change significantly to cause the same phase difference. This shows why the separation of the slits is inversely related to the period of the intensity oscillations.

Diffraction in two dimensions

A one dimensional slit as we have considered above is still not very physical. Let’s consider how we can extend our calculations to consider a 2D slit function $S(x,y)$ .