Friday, March 27, 2009

The Double-Slit Experiment (quantitative) - Part 2

(click on the picture for a larger view)

Now that we know what a sine wave is, we can understand the double-slit experiment. I need to start with a few definitions that I probably should have put in the last post: the wavelength is the length between two successive peaks in the wave (often represented by λ) and the amplitude is the height of the wave (we'll call it A). There is a symmetry property of the wave; if you shifted the wave to the right or left by its wavelength, it would look exactly the same. In fact, you could shift the wave by any integer times the wavelength, and you wouldn't be able to tell the difference. This will be important later on.

To understand the double slit experiment, we need to ask what happens when two waves overlap. The answer depends on their "phase," or where each wave is in its oscillation relative to the other. For example, suppose two waves are perfectly "in phase," so that when one wave is peaking, so is the other. When you add these two waves together, you'll get a wave that is twice as big in amplitude.

What about when the waves are "out of phase" so that one is all the way up when the other is all the way down? In that case, the waves destructively interfere so that the addition contains no wave at all.

This interference is the key to the double-slit experiment and allows us to predict the shape of the light pattern on the screen. When the light impinges on the slit, the waves that come come out the other side are initially in phase. If you look at the the screen directly across from the slit, you would see a dark spot. However, that bright spot will be banded by bright spots, which will in turn be banded by two more dark spots in a fringe pattern. To understand why, let's zoom in on the slit right where the light passes through (on this lovely diagram I stole from here).

Here, the distance between the slit and the wall is L and the slit separation is d. Where on the wall do you expect to see bright or dark spots? If we want there to be a bright spot on the wall, then we know the two waves must interfere constructively (the first case discussed above) or be in phase. A dark spot will appear when the waves are out of phase and interfere destructively. Let θ (again, angles are always θs) be the angle between the horizontal and the position of a given fringe on the wall). If we look high on the wall, the light that came out of the top slit doesn't have to go as far as the light that came out of the bottom slit (in other words, r1 is bigger than r2). This means that the bottom ray of light will have more time to trace out its wavelength and will drop out of phase with the top ray of light. The extra distance traveled by the bottom ray is equal to d*sin θ (remember that the sine function also related the sides of a triangle and notice that the light paths r1 and r2 form a triangle with the slit). Now, remember the symmetry of the wave - a wave that is shifted by its wavelength looks the same. So if the extra distance traveled by the bottom wave equaled exactly its wavelength, it would look identical to the top wave, and the waves would interefere constructively - a bright spot would appear on the wall. If, on the other hand, the extra distance traveled was exactly half a wavelength, so that the bottom wave had just enough time to get out of phase, the two waves would interfere destructively and a dark spot would appear on the wall.

This is exactly what happens - bright spots appear if d*sin θ = λ or some multiple of λ, while dark spots appear if d*sin θ = λ/2. Wave properties predict exactly the patterns that appear in the double-slit experiment, confirming that light is like a wave.


  1. Eva thinks this is really sweet.

    Dan thinks you should get back to work :)

    (actually this is very nicely done)

  2. I like the idea behind your blog; but does your "mom" still understand the material in this style of writing? My mom hasn't been in school in 30+ years, hasn't done trigonometric math in at least that long (if ever), and would freeze up at any sight of greek symbols or subscript notation. She's a smart woman, but would still find your current explanations rather intimidating.
    I will, however, recommend this to some H.S. physics teachers I know. I think their students would find it quite handy.

  3. Anonymous,
    You do have a point. I think my mom (no need for quotes, she actually is reading) does freeze up a bit at symbols and subscripts (she made a comment in one of the earlier posts to that effect). However, she suggests to me that she's gamely trying to keep up. In the end, I imagine the math posts will go largely unenjoyed by her, but as I wrote before, I'm very reluctant to completely decouple the physics from the underlying math. And it's conceivable (if unlikely) that maybe she can help me be clearer in my writing (she's an actual writer), and I can help her get over any lingering abhorrence to abstract math.

    Thanks for reading.

  4. Very well explained. Excellent.

  5. There is a famous thought experiment in special relativity where an observer on a train watches a light in the center of the train flash and lights at either end of the train flash in response to detecting it. He sees the lights on the end of the train flash simultaneously. An observer on the platform watching the train pass, sees them flash at different times.

    What happens if the observer on the train is performing two double slit experiments using the center light as the source and having the pairs of double slits near it. The screens are at the ends of the train. Both observers should see the same interference pattern? But the one on the platform thinks photons make it to the rear screen faster than to the front screen. Can the observed wavelengths of the light somehow explain why the expects the two patterns should be identical?

  6. I meant to say "why he expects the two patterns should be identical?".

    I'm trying to discuss this "moving double slit experiment" in a forum (, and moving against the wind, so to speak. One thought that has come up is whether an experimenter can perform the double slit experiment, as usually described! By the Uncertainty Principle, the experimenter could establish the exact wavelength of the light, but then he could not measure the exact location and time of impact of a photon on the screen. He could choose to measure the exact location and time and be uncertain about the wavelengths.

    I suppose the usual way to visualize the experiment is that we look at a gross pattern of impacts, so we could say we are not measuring their time. This would let us, theoretically, say that we know the wavelength of the photons. However, this gets into the question of whether a "measurement" involves human knowledge. Does the mere existence of the screen imply that the position and time of the photon impact was measured?

  7. Very well explained. Thank you from Mexico!