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I have already seen the following question: Why does wavelength affect diffraction?

But I still can't seem to understand the relationship. I understand that if the wavelength is the same size as the slit, the wave pattern will look like circles on the other side of the slit, which is maximum diffraction, because the slit will "function" as a point source.

But I don't understand why, if we reduce the wavelength, there's gonna be like more point sources in the same slit. What's the reason?

I hope I could make myself clear enough.

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  • $\begingroup$ Someone correct me if I'm wrong, but I think if the slit is too narrow, too much of the wave gets reflected; and if it's too wide, you get single slit interference: en.wikipedia.org/wiki/Diffraction#Single-slit_diffraction $\endgroup$ – Wood Nov 25 '16 at 22:40
  • $\begingroup$ That's correct. But in this case of single slit interference, there's approximately 5 point sources in the space of the slit - but, technically, if you increase the wavelength, the number of point sources will reduce. The reason for that correlation is what I don't understand. $\endgroup$ – Roberto Valente Nov 25 '16 at 22:50
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    $\begingroup$ "But in this case of single slit interference, there's approximately 5 point sources ..." In a particular visual representation someone drew five point sources. In reality this is a mathematical continuum. There are no discrete points, there is a whole wavefront. $\endgroup$ – dmckee Nov 26 '16 at 2:23
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Forget about the "number of points". There are infinite number of points in the slit.

The explanation has to do with time. Each point of the slit corresponds to a different delay that light takes to reach a given target from that point.

For instance, if the target is sideways from the slit, some points of the slit will be closer (less delay) and some points will be further (more delay). The smaller the wavelength, the more oscillations that delay will contain. More oscillations mean more cancelling between them, because averaging over many oscillations adds up to zero. To summarize this, the smaller the wavelength, the more cancelling out of waves, thus the less light. This explains why a smaller wavelength tends to have less diffraction to the side.

The more you take a target to the side of the slit, the more delay you will add, thus less and less light.

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Wavelength much larger than slit:

The slit is filled with infinitely many point sources, each one of which emits circular waves, as point sources always do. Those waves interfere constructively everywhere.

Wavelength much smaller than slit:

The slit is filled with infinitely many point sources, each one of which emits circular waves, as point sources always do. Those waves interfere constructively and destructively everywhere.

Any place receives waves from all the point sources, and those waves are not in phase, because of the different distances to different point sources. When some place seems to receive almost no waves, that's because the result of all the constructive and destructive interferences is almost zero there.

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  • $\begingroup$ thanks, I think I understand it better now. I have also found another link which I found helpful if anyone is interested here $\endgroup$ – Roberto Valente Nov 30 '16 at 1:02

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