# Interference from two wide slits

#### Prerequisites

We have analyzed what happens when waves from two point sources combine. At different points in space the result may be a strengthening of what each source would do by itself (constructive interference) or a weakening of what each source would do by itself (destructive interference), depending on whether the two waves are in phase with each other or out of phase with each other (Two-slit interference). We've also considered what happens if a single slit is wide enough that the interference from different parts of the slit have to be taken into account (Diffraction). This is what the set-ups and patterns look like in the two cases:

Now our question is: What happens if we have two slits that are each wide enough so that the width of the slit has to be taken into account?

## Thinking about it conceptually

If you think of what is happening physically in term of wavelets from the sources at the slits, it is fairly straightforward to see what's happening. If we are at a particular position, y, on the screen up (if positive) or down (if negative) from the center line, we receive oscillating signals from each of the slits. These signals are the result of all the wavelets coming out of that slit adding together. Now we have to add those two results together -- with an additional phase shift as a result of the extra path difference between the two slits. This will produce the interference pattern on the left (the two-slit pattern) multiplied by the pattern on the right (the one-slit pattern). We are getting interference of the waves from the two slits, but the waves from each of the slits have a non-uniform strength as y changes due to the one-slit pattern.

In particular, at that position on the screen where the one slit pattern cancels (the "squash point") then the total interference pattern goes to zero there as well. You are interference waves from two slits, but waves of 0 intensity -- so the total result is zero. The regular fringes produced by the two-slit pattern are "squashed" by the one slit pattern.

## Doing the math

These are a lot of ideas to keep in one's head at once. One of the things math is good for in physics is in showing how the conceptual ideas work, a step at a time and then letting you scan over it to hold the entire picture in your head at once. Let's give it a try.

- From our analysis of the interference produced by two very narrow slits, we got the result as a function of $y$ (the position along the screen) was

$$I = A_0^2 \cos^2\bigg({\frac{2\pi ay}{\lambda L}\bigg)}$$

where $a$ is the separation of the slits, $\lambda$ is the wavelength, and $L$ is the distance to the screen, and $A_0$ is the amplitude of the oscillating light. The graph of this is shown in figure A in the set of graphs below. It's an oscillation that peaks at $y=0$ (the midpoint of the screen, where the distance from each slit is the same) and oscillates with a constant amplitude.

- From our analysis of the variation in the intensity of the light produced by a single slit, we found that the amplitude received at the screen from each slit is not constant, but rather is given by

$$A^2 = A_0^2 \bigg(\frac{\sin{(\pi d y / \lambda L)}}{\pi d y / \lambda L}\bigg)^2$$

where $d$ is the width of the slit. This graph is shown in graph B below. It has a broad central peak. After the first 0 on either side of the central peak it continues to oscillate, but the peaks get smaller and smaller, as a result of the $1/y^2$ factor in the equation.

- Now we need to put these together. Since the result we found in part 2 is providing the actual (non-constant) amplitude, we should replace the $A_0$ in the equation in part 1 by the $A$ we found in part 2. This gives a total amplitude

$$I = A_0^2 \bigg(\frac{\sin{(\pi d y / \lambda L)}}{\pi d y / \lambda L}\bigg)^2 \cos^2\bigg({\frac{2\pi ay}{\lambda L}\bigg)} $$

This result is shown in graph C below. The uniformly oscillating pattern we saw in A is modified by multiplying by the function shown in B. For the parameters shown below, the 4th peaks from the center that would be expected from A are in the same place as the 0 of function B, so they don't appear in the product. They are "squashed" by the fact that although they are at a place where the two waves add constructively, each wave itself is actually 0 due to the interference from within each slit.

All three of the graphs are shown overlapping in figure D. It's a little hard to read, but it shows how they coordinate.

The equations above are pretty messy. We can make better sense of them if we focus on the parameters we are interested in: the separation of the slits, $a$, and the width of the slits, $d$.

Let's use our repackaging tool to make these equations look more like math. Since we're really only interested in the functional dependence — seeing how changing $a$ and $d$ changes our graph — let's temporarily replace the constants $\pi /\lambda L$ by $1$.

The result looks like this:

$$I = \bigg[A_0^2 \bigg(\frac{\sin{(d y)}}{d y}\bigg)^2\bigg]\bigg[ \cos^2({2 ay)}\bigg] $$

So we can see that if we decrease $a$, the $\cos^2$ term will wiggle more slowly, moving the fringes farther out. If we decrease $d$, the $(\sin{y} / y)^2$ term will oscillate more slowly, moving the squash point farther out.

Both the separation between the slits and the width of the slits get "turned inside out" when they get to the screen. Making the slits narrower widens the squash pattern and making the slits farther apart pushes the interference fringes closer together. You can explore this using a sim in the workout link at the bottom of this page.

The overall result looks like this:

The top shows the graph of the total intensity (in red) "squashed" by the one-slit diffraction pattern (in blue). Below that is shown what the pattern actually would look like on the screen, and below that a schematic of the laser beam approaching the slits.

Workout: Interference from two wide slits

Joe Redish 4/28/12 and 7/8/19

Last Modified: July 9, 2019