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I am tutoring a high school student in physics, and struggling to find a good explanation as to why the interference pattern from a diffraction grating has thinner maxima compared to a double slit.

Does anyone have an explanation of this that an average high school physics student could understand? The content they have covered just includes the basics of Young's double slit experiment - no single slit interference or mathematical derivations of the interference equation ($d \sin \theta = n \lambda$). An ideal explanation would include very little maths (ideally none), and be somewhat intuitive to understand. Cheers

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  • $\begingroup$ Did you prove that the slit width and the distance between the slits are the same in the double slit obstacle and in the diffraction grating? $\endgroup$ – HolgerFiedler Nov 18 '17 at 5:04
  • $\begingroup$ Yes, but that is part of what makes the whole thing so hard to explain: that equation suggests no differences if $d$ is kept constant $\endgroup$ – Jordan Nov 18 '17 at 5:22
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A diffraction grating is effectively many small slits next to one another. At a maximum, all of the phases of the light from the individual slits line up perfectly, so the point is bright.

At an intermediate point, the phases from each slit are each effectively random, so the phases from any two slits are as likely to cancel out as to add up. The net result as you have a lot of slits is that rather than a continuum of constructive to partial to destructive interference, you get either constructive or basically destructive interference. As you get more slits, you transistion more quickly from "constructive," where the phases line up, to "random," where the phases don't tend to line up.

Another method: you can state without proof that for $n$ slits, there are $n-1$ points of destructive interference between any two points of destructive interference. (For your own purposes: you can show this graphically with phasors and drawing regular polygons with said phasors). If we think of a diffraction grating as infinity slits next to each other, this implies that there are infinite points of destructive interference between any two points of constructive interference and so intuitively every point that's not constructive is destructive.

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  • $\begingroup$ Nice, I think appealing to the laws of probability is quite a good way to approach this. So there are only specific points where constructive interference can occur, and anywhere else is subject to "random" phases interfering, which are on average destructive. $\endgroup$ – Jordan Nov 18 '17 at 5:27
  • $\begingroup$ Unfortunately, this could suggest that interference is governed by randomness, rather than deterministic, but I suppose it's harder to explain deterministically without going through a more complex proof/explanation $\endgroup$ – Jordan Nov 18 '17 at 5:29
  • $\begingroup$ @Jordan Pretty much. The simplest explanation I can think of that's rigorous involves phasors, so it's a bit much for a high schooler, probably. $\endgroup$ – Chris♦ Nov 18 '17 at 5:31
  • $\begingroup$ @Jordan I've added some info about that method to my answer. $\endgroup$ – Chris♦ Nov 18 '17 at 5:38

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