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In the double slit experiment, light passes through two slits and the waves interfere and form an interference pattern. A single slit is required for diffraction.

So, I was thinking about whether the light in the interference setup undergoes diffraction first at the two slits?

I think that's the case, because light waves spread due to diffraction at the two slits. The two waves interfere to form the interference pattern. Is my thinking correct?

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2 Answers 2

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According to the Huygens–Fresnel principle, every point of the wavefront is a new spherical wave source. Of course, you don't see infinite individual waves; what you see is the result of summing (interference) infinite waves.

enter image description here

This means there is always interference, even if there are no obstacles. Diffraction would be a consequence of blocking part of the wavefront, so the waves which are left interfere in some fancy way. This principle can be used to describe refection, refraction and diffraction.

For a single slit several times bigger than the wavelength (the dots are the wave sources):

enter image description here

If the slit is as big as the wavelength you see a single spherical wave (I wouldn't be sure to consider this diffraction at all):

enter image description here

There is something similar to the Huygens–Fresnel principle in quantum electrodynamics. The path integral formulation says that when light (and any other particle) travels to a point $A$ to a point $B$, you have to sum every possible trajectory. Each trajectory has the same probability, they only differ in phase.

So for the two slit, if you compute each possible path you would get the classical result.

enter image description here

So I would say that diffraction is a particular case of interference where some part of the wavefront has been blocked.

But the difference between interference and diffraction is not clear. As Feynman said: "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".

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    $\begingroup$ Sir, but what's the mistake in thinking the situation this way? Light will anyway undergo diffraction when it encounters a narrow slit(as in the diffraction setup). So, the light should also diffract before interfering at the two slits in the interference setup. $\endgroup$ Oct 4, 2013 at 17:23
  • $\begingroup$ Even in the single slit there is interference. More precisely, since you block the rest of the wave, the wave in the middle of the slit cannot interfere whith the rest. So you observe what would correspond to an spherical wavefront, instead of a plane wave. I'll try to add some images when I get to a pc. $\endgroup$
    – jinawee
    Oct 4, 2013 at 17:39
  • $\begingroup$ @RajathKrishnaR I've updated the answer. $\endgroup$
    – jinawee
    Oct 7, 2013 at 14:21
  • $\begingroup$ So that means that in you very 1st diagram where you have shown the wavefront at t and t+ dt , the secondary waves at t somehow interfere constructively and the wavefront at t+dt is the constructive wavefront. But where does destructive interference occur because in between the will be trough $\endgroup$
    – Shashaank
    Mar 12, 2017 at 11:17
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Diffraction at a slit is due to the interference of the Hyugence wavelets. But then the interference of the waves from the two slits is due to this diffraction at the slits. When the distance between the slits is great and diffraction is not too strong (slits not too pointlike) the waves after the slits would not bend strong and superimpose on each other and not form the interference picture. But one can say also it is due to the interference of the wavelets from the diffracted waves. For me diffraction is equivalent to bending of light. Interference is practically when we observe max min interference patterns.

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