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I always learnt in school that diffraction and interference happens if the slit width is of the order of wavelength. But what if the slit width is less than that of wavelength. Does diffraction happens in such conditions? If so then is there a mathematical representation for that?enter image description here

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marked as duplicate by Ben Crowell, Jon Custer, John Rennie visible-light Dec 7 '18 at 8:58

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    $\begingroup$ Then it acts like a point source, radiating in all directions. $\endgroup$ – Pieter Dec 6 '18 at 8:57
  • $\begingroup$ @Pieter if slit width is just a little bit small then wavelength then does diffraction happen in such case? $\endgroup$ – user210956 Dec 6 '18 at 9:02
  • $\begingroup$ See also physics.stackexchange.com/questions/141562/… $\endgroup$ – Ben Crowell Dec 6 '18 at 14:53
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The angle of diffraction $\theta$ can be approximated by $$\theta \approx \frac{\lambda}{d}$$ where $\lambda$ is the wavelength and $d$ is the width of the slit.

Now in case of $d<\lambda$, i.e. $\frac{\lambda}{d} > 1$, $\theta$ continues to remain maximum, that is $\frac{\pi}{2}$. Which can be interpreted (as correctly pointed out by @Pieter) the slit will act like a point source and radiate in all directions.

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Think about what's going on. Light passes through the slit and goes in all directions from all parts of the slit. To reach any point, the light that comes from different parts of the slit has traveled different distances and is delayed different amounts. When the delay from different parts of the slit is just the right amount that you get a whole wave, the wave cancels with itself and you get a minimum.

When the slit is one wavelength wide, at the center maximum the delay can't be more than half a wavelength. So it can't cancel as effectively. You get less light at the center point than you would if a lens made sure that the light from all parts of the slit arrived at the same time, but you get plenty.

If the slit is more than one wavelength wide, there can be two minima where one whole wavelength arrives at the same time and cancels, two more minima where two whole wavelengths arrive at the same time and cancel, and so on. Between the minima are maximum points. At one of them one whole wavelength cancels leaving another half. At the next two whole wavelengths cancel leaving a half. Each of them gets dimmer.

Now look at your example where the slit is less than a wavelength wide. There can't be anywhere that a whole wavelength arrives at once and cancels, because you don't get a whole wavelength's delay anywhere. The intensity will not be the same everywhere, though. At the center the fraction of a wavelength you get at once is a minimum, and at 90 degrees it's a maximum. The intensity will fall off to the sides.

But you can't see the intensity of the light rise to another maximum, because with a slit that's one wavelength wide, the minimum comes at 90 degrees. With a slit that's less than a wavelength wide, the minimum would come at an angle that's bigger than 90 degrees, and so it hits the wall the slit is in.

This leads to a question I don't know the answer to. Suppose you could build your slit with, say, two razor blades mounted on micrometers so you could get the distance just right. And suppose you could put the razor blades at a 90 degree angle to each other, so the wall the slit is in is not flat at all. Can diffraction bend light more than 90 degrees? I'm not sure what the theory says about that. And I haven't heard that the experiment has been done.

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  • $\begingroup$ If you've got another question, you shoul not write it in your answer. Rather, ask a separate question. It'll be welcomed. $\endgroup$ – FGSUZ Dec 6 '18 at 12:34
  • $\begingroup$ Good point. It relates to this question -- the quasi-minimum will be beyond 90 degrees if it exists -- and I would have given the answer if I knew it. But I'll ask it as a separate question. $\endgroup$ – J Thomas Dec 6 '18 at 12:47

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