In our study of interference between waves generated from two sources we assumed that each of our slits were points. We did not consider the possibility that the distance between the slit and the observation point might vary significantly. What does "significantly" mean here? When can we really ignore the width of the slits? Since what we were doing in the two source analysis was to compare the distances from the source to the observation point — on scales of the order of a wavelength — we might reasonably assume that the following works:

A source may be treated as a point for purposes of considering interference as long as the path difference from different parts of the source doesn't change more than a small fraction of a wavelength.

Note that this does NOT say "as long as the slit is small compared to the size of a wavelength." (We've put a strikethrough this to emphasize that that's not what matters — and to discourage anyone from cutting and pasting the statement and forgetting to include the "NOT".) It might be just fine to ignore the size of a source that is many wavelengths large if the observation point is far enough away and in the right place.

This tells us why for a 20-cm-wide speaker putting out sound at middle C (262 Hz → $λ = v_0/f$ = [(343 m/s)/(262 1/s)] = 1.3 m) at 2 meters away it is OK to ignore the width of the speaker, but for a laser putting  light through a 1/4 mm slit at $λ$ = 600 nm, it is NOT OK to ignore the width of the slit at 2 meters away.

For this webpage, we consider the effect of the width of the source that does effect an interference pattern. We begin by showing that a single slit can all by itself produce an interference patters — as a result of wavelets coming from different parts of the slit interfering with each other.

What is "diffraction"?

Sorry! This is another one of those "sloppy terminology" issues that creeps up in science all the time. The term diffraction was invented to describe the observation (made in the 17th century) that light didn't appear to travel exactly in straight lines. Near the edge of a shadow was a (typically small) blurring of the edge. Introductory physics texts tend to use the term "diffraction" to refer to the interference pattern coming from a single source and "interference" to refer to the pattern coming from a multiple source. But what is "single" and what is "multiple" is in the eye of the beholder and is not absolute. Is the interference of x-rays scattered from two different molecules in a crystal "interference" (because it is from two molecules) or "diffraction" (because it is from one crystal)? In his famous introductory physics text,  physicist Richard Feynman says (quoted in Wikipedia)

"no-one has ever been able to define the difference between interference and diffraction satisfactorily. It is just a question of usage, and there is no specific, important physical difference between them."

and the Wikipedia article on diffraction refers to what we have called two-slit interference as "interference pattern from two-slit diffraction". In these pages we will mostly use the term "interference" when we mean the combination of two or more waves from sharply identified sources, and "diffraction" when we mean the interference pattern produced by an extended source. But don't get hung up trying to define these terms too precisely.

Internal interference from a single slit conceptually

To consider the effect of the non-0 size of a source of waves, we'll use the example used in our webpage, Two-slit interference: a laser beam incident on a narrow slit producing a pattern on a screen that is much farther away than the size of the slit — or than the wavelength of the wave. 

In Huygens' model of wave propagation, each point on a wavefront acts as a source of outgoing spherical waves or wavelets; and the resulting wavefront is the integral of all these wavelets. To see what this implies for a slit of non-negligible width, take a look at the figure at the right. We show a plane wave incident on the slit. In the middle of the slit, each point on the plane wave creates outgoing spherical wavelets. One way to think about this is to use the water wave analogy. If a series of parallel straight water waves are approaching a straight jetty with a break in it, in each point in the break, the incoming waves drive the water in the middle of the break up and down and these oscillating bits of water are the source of outgoing circular waves.

To get the actual result on the screen, we have to do an integral. This is pretty messy and we won't do it here; but we can figure out qualitatively what is happening by a geometrical analysis similar to what we did in our discussion of two slit interference.

The point is, every point on the wavefront across the slit is serving as a source of outgoing waves as shown in the figure at the right. When we reach the observation point, y, we actually have wavelets arriving from every point. At the central point, $y = 0$, we expect most of them to add up to give a bright spot. But as y moves up the screen from the center, the rays coming from the lower points on the slit have to travel farther. (As in our discussion of two-slit interference, for this situation we will assume that $L >> d >> λ$ so the small angle approximation holds for trig functions of $θ$.)


Something interesting happens when the wavelet from the very top of the slit and the mid-point of the slit have an extra 1/2 wavelength (as shown in the figure at the right). The extra distance traveled by the wave from the mid-point is $Δr = λ/2$. The contributions of these two wavelets will be out of phase and cancel each other.

So what? Those are only two wavelets out of many. But if you now start to move down from the top wavelet, you will see, that every source of wavelets in the top half of the slit is matched by a wavelet from the bottom half of the slit that is 180o out of phase with it!  As a result, every wavelet has a wavelet that cancels it. The total result at this angle is 0 intensity — a dark spot. (You can see how this works by imagining sliding the two arrows from the slit to the screen and the associated dashed lines downward rigidly. The little triangle that gives us the extra distance $Δr$ remains the same.)

I refer to this point in the patters (where $Δr =  λ/2$) as the squash point-- because the amplitude of the light from the slit is "squashed down" to 0 at this position. (This becomes more relevant when we have an additional pattern — such as the one produced by a pair of slits.)

As the angle increases, there is still a cancellation but not a perfect one and some intensity comes back. The total pattern that one gets looks like the figure on the left below: a strong central bright area with a weakly bright line on both sides of the dark.  (This comes approximately from dividing the slit in thirds so the top third cancels the second leaving the final third bright. This also suggests — correctly — that the intensity of the first bright spot past the first dark spot has an intensity of about 1/9 of the central spot.)

In the figure above, the $\Delta r$'s refer to the path difference between a wavelet from the top of the slit and the center of the slit. 

Internal interference from a single slit — the math

Adding up the result from all the sources across the slit is much harder than adding up the two sine waves for two-source interference: you have to do a very messy integral. The result is fairly straightforward however. What you get is something like this:*

$$I = A^2 \frac{sin^2{z}}{z^2}\quad \mathrm{where}\quad z = \frac{\pi d}{\lambda L}y$$

If you plot this in the Desmos graphing calculator, it gives a graph that looks like this: (Of course in Desmos you need to rewrite it as $y = B \sin^2{Cx}/(Cx)^2$ since it demands x as the horizontal and y as the vertical axes.)

where again the $\Delta r$'s refer to the path difference between a wavelet from the top of the slit and the center of the slit. 


For a more extended object, diffraction creates oscillations in the intensity at the edge of the sharp shadow.  A graph of this phenomenon is shown in the figure at the right. If your shadow is created by an extended source, these patterns from each point on the source will be shifted slightly from each other and you won't see the dark rings at the edge of the shadow.

One place where you can see this is if you wear glasses and go walking in the rain on a dark night. A distant streetlight viewed through a droplet of water on your glasses will show dark diffraction lines at the edge of the bright drop.

The next question we have to consider is: What happens when we have two slits and the width of the slits is not small enough to ignore? The pattern gets more complex — and realistic. We treat this in the follow-on.

This is a funny function. If you look at $\sin^2{z} / z^2$ for $z=0$, it looks like 0/0 — undefined. But if you use the small angles approximation for small $z$, $\sin{z} \approx z$ so this function near 0 looks like $z^2/z^2 = 1$. Fortunately, Desmos knows how to handle this.


Article 717
Last Modified: July 8, 2019