# Two-slit interference

#### Prerequisites

Let's now apply the ideas of Huygens' principle to the classic experiment that shows the wave character of a propagating signal: interference. Here's the critical idea we draw from observing interference:

If adding more of your signal can result in a smaller result, then your signal must possibly be both positive and negative so that two signals can cancel each other — at least at particular places in space. It can't cancel everywhere if it carries energy. If some places are cancelling, others must be getting stronger.

The key components of the wave model that allow us to get this result are:

• The signals (light or sound) are oscillating.
• Signals from multiple sources just add — superposition.

The second one seems strange at first sight. Superposition means that the waves don't interfere with each other — in one sense. The signals freely pass through each other. This is true both of sound (up to intensities that create near vacuums, like shock waves in explosions) and light (up to extremely high energies like gamma rays where the light has enough energy to create matter out of the vacuum). The issue that makes this a dangerous bend is that in common speech the word "interference" can have multiple meanings. So what we have here is this:

• The signals can pass through each other so they don't "interfere" with each others' propagation through space.
• At any given point, the result from multiple signals is the sum of those signals; so one signal can "interfere" with the result produced by another signal at a given place in space.

## The two source example: Where do we expect to see bright and dark spots?

Our two-source-interference examples might look like water waves passing through two openings in a jetty as shown in the figure at the left (or from two elephants walking in shallow water as shown at the far right), sound waves coming from two coordinated speakers as shown in the figure in the middle, or laser light passing through a double slit as in the figure on the right.

In the discussion of the math of Huygens' principle we showed the algebraic equations of how sinusoidal waves add at different points in space. This is necessary if we want to calculate the actual observed intensities of the combined waves. But if we are only interested in asking about where do the two waves add constructively and destructively, we can get away with a geometrical argument and we don't have to worry about adding oscillating functions and trig identities and time averaging rapidly oscillating signals and lots of other messy math.

So let's ask the question:

If we have two sources of outgoing circular/spherical waves, when we observe the superposed result of those two waves on a distant screen, where do they add constructively and where do they add destructively?

What guides us in answering the question is the general result we inferred from looking at the math:

Whether two oscillating waves add constructively or destructively at a particular point in space is determined by the phase difference between the oscillations at that point. If the phase difference is an integral multiple of 2π, they oscillate together and add constructively. If the phase difference is an odd multiple of π, they oscillate oppositely and will cancel.

If the two sources are oscillating in phase, then what determines the phase at the observation point is the distance from the two sources. The phase difference between the two waves received at the observation point is equal to $2π$ times the ratio of the difference in the distances to the two sources divided by the wavelength:

$$\Delta \phi = 2\pi \frac{\Delta r}{\lambda}$$

All we actually need from this equation are the following results:

If the extra path difference from the observation point to the farther source is an integral number of wavelengths, $\Delta r = n\lambda$, the two sources will add constructively.

If the extra path difference from the observation point to the farther source is an integral number of wavelengths plus one half a wavelength, $\Delta r = (n + ½)\lambda$, the two sources will add destructively (cancel).

This tells us where to expect bright spots and dark spots in an interference pattern. All we have to do now is the geometry to figure out what the extra distance is to the farther source.

## Two-slit interference: the geometry

For our example of a two-source interference, let's choose the laser incident on two very thin slits producing a pattern on a screen far away. There are three reasons for doing this: (1) the arrangement produces a large multiplication factor that actually lets us infer the wavelength of light from measurements we can make with a meter stick; (2) this is a "simplest" example that allows us to develop an understanding of the phenomenon that we can build into an understanding of more complex measuring devices, light diffraction gratings, that rely on similar physics; (3) this situation leads to "easily calculable" examples that often appear on standardized tests; and finally (4) provides a nice example of a place where you can in principle get by with memorizing a bunch of equations, but they are easy to confuse, and a simple two-step analysis from the basic physical meaning will allow you to easily regenerate those equations and will never lead you astray. (See the webpage an evolutionary model of memory.)

Here's the setup we will consider. We will have a laser beam incident on a mask with two slits that are very close together, a distance $a$ apart. (They are shown magnified in the blow-up circle.)  We will assume the slits are very narrow and will ignore their width, treating them each as sources of outgoing circular waves. (For simplicity, we are doing this in 2D, not worrying about the third dimension.)

The light coming from the two slits propagates to a screen a distance L away from the slits where it hits a screen. At the line from the center point between the slits (shown dotted) we expect a bright spot, since there the distance to each of the slits is the same.  $Δr$ is 0 so the waves received at that point from the two sources should add constructively.

We then want to figure out what the pattern is as we go up and down on the screen ($y$ varies from 0). We will assume

$$L >> a.$$

In a typical experiment like you might do in lab or see in a lecture demonstration, $a$ ~ a fraction of a millimeter, while $L$ might be a number of meters. So the ratio $L/a$ in this experiment is typically a factor of several thousand.

What matters is the difference in the distances from the point on the screen to the two slits. Since $L$ is so large compared to $a$, the angles we will be dealing with are very small. When we look in the magnifier the paths to the point on the screen are almost parallel. This makes the geometry reasonably simple. Our angles are small, so we can use the small angle approximations, $\sin{θ} \approx θ, \tan{θ} \approx θ$ (with $\theta$ in radians).

If we go up by a distance $y$ from the central point on the screen, the line to the slits will be up by an angle we'll call $\theta$. In the small angle approximation,

$$θ \approx y/L.$$

When we look in our magnifying glass, we see that the line measuring the distance to the lower slit has been pulled out a bit longer — an extra amount $Δr$. Since $L$ is so far away compared to $A$ we can treat it as nearly infinity, so the lines to the point on the screen are nearly parallel. This means that the angle pulled out next to a also has to be $\theta$. (What else could it be? There are no other angles in the diagram except $90-\theta$ and that's too big.) So our extra distance is

$$Δr = a \sin{θ} = aθ.$$

We can then put these last two equations together to get

$$Δr = \frac{ay}{L}.$$

This is the key equation that allows us to make sense of the pattern. It says if you are not at the center of the pattern, one of the paths is longer.

## Interference conceptually

For the two waves coming along each path, if we freeze an instant in time, we see that more wiggles fit on the longer path. Therefore, even if the waves started out oscillating together at the source slits, at our observation point they won't be oscillating together any more. This is shown schematically in the figure at the right. (The slit separation is magnified so it's easier to see what's happening.) What are the implications?

When $y$ is 0, the extra distance is 0 so the two sources add coherently at that point and produce a bright spot. As $y$ moves up, the extra distance, $\Delta r$, gets larger. When it is equal to 1/2 a wavelength, the two sources are out of phase and cancel each other. We get a dark spot. As $y$ gets larger, when it gets to a whole wavelength, we get a bright spot again, and so on.

Therefore, putting $Δr$ in as some integral number of wavelengths, $nλ$ (for a bright spot), or putting $Δr$ in as some integral number of wavelengths plus 1/2 a wavelength, $Δr = (n+½) λ$, (for a dark spot) and solving for $y$ we get

bright spots when:   $y = nλ \frac{L}{a}$

dark spots when:  $y = (n+½) λ \frac{L}{a}.$

where $n$ stands for any integer.

It's tricky remembering which of these is which, and they can be switched in various situation (like using a quarter-wave plate in front of one of the slits). It's much more reliable to remember the geometry and be able to reconstruct the equation $Δr = ay/L$ for yourself. The bright and dark situations follow from this immediately and in a physically sensible way.

The separation between the bright spots are therefore basically the wavelength of light magnified by the factor $L/a$. The fact that $L/a$ is large means that this experiment has been arranged so as to magnify $λ$ by a factor of many thousands. This allows us to measure the wavelength of light with a meter stick!

Another thing to notice is that the position of the spots in the pattern on the screen depends inversely on the separation of the slits. Making the slits closer together spreads the pattern further apart and vice versa. This is a common phenomenon in waves and interference patterns — making the source more compact makes the pattern produced more spread out.

The resulting pattern looks like a series of bright lines. (In the technical literature, these are referred to as fringes.)

The fading out of the pattern to the left and right is caused by another phenomenon — diffraction. The interference of the light from different parts of the slit with itself. This is considered in the follow-ons.

## Interference mathematically

The conceptual ideas bring together many complex ideas. While it's worth disentangling them and getting a good picture of what's happening, in some sense it's easier to "just do the math". The basic idea is that we are adding together two sine waves that have come to a particular point in space traveling different distances. We are then squaring the result since the intensity, $I$, is proportional to the square of the E field, and then time averaging over the very fast oscillations:

$$E = A\sin{(kr_1 - \omega t)} + A\sin{(kr_2 - \omega t)}$$

$$I = \langle E^2 \rangle_t$$

where the brackets-sub-t means time average.

This is straightforward, but a fair mess. We're not going to go through it here (though we may provide a "technical" follow-on in the future). The result is this (modulo some constant factors to get the units right which we will not pay attention to here):

$$I = A^2 \cos^2{(k \Delta r)} = A^2 \cos^2{(2\pi\frac{ay}{\lambda L})}.$$

Let's read the meaning into this math. We know that cosine goes through a full oscillation when its argument changes by $2\pi$. So when $y$ changes by an amount $\lambda L/a$ (dimensionality L), we get a full oscillation.

If we plot a graph of this as a function of y (on Desmos graphing calculator) you get the result below. (Careful! On Desmos you are always plotting "y vs x" so in our equation, $y$ is Desmos' x and our $I$ is Desmos' y. So you have to tell it to plot $y = B\cos^2{Cx}.$ )

This is the mathematical representation of the red fringes in the red on black photo above.

The qualitative features are good. There is a peak at the center, and as y increases or decreases, the intensity oscillates between the maximum value and zero — light and dark places. Almost. As mentioned in the conceptual section, the "fading out" of the fringes on the side is not described by this function. We'll see how it is represented in the follow-ons.

Joe Redish 4/26/12 and 7/7/19

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