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Lets say we have a double slit setup where a plane wave hits 2 narrow slits and travels down a small channel. One of the channels has a dielectric which will slow down the light passing through it. The dielectric's refractive index is continuous so that all the light is transmitted and none is reflected. The light passing through the dielectric gets a phase delay of $2\pi n$ where n is a positive integer greater than the distance between the slits divided by the wavelength of light used plus 1. If we send through one photon at a time, will there be no interference pattern? If we sent one wave front through the double slit, the waves coming out of either slit wouldn't be able to interfere with each other due to them never overlapping due to their phase delay. I suppose it could be possible that there could be multiple photons in one wave front and that it could also be possible that one photon's wave function could be spread out over multiple wave fronts. I guess that kind of complicates things. I'm guessing that if we send a continuous plane wave through we'll see a diffraction pattern as normal. Though if we knew how many photons were entering both slits combined over time and the amount of photons hitting our screen over time, we may be able to determine how many of the photons passed through each slit. I don't know if that knowledge would interfere with a diffraction pattern's formation though. Any help would be appreciated.

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    $\begingroup$ The nature of Fourier optics means that you can't have a single oscilation of a wave with a well defined frequency (classical framing). Or that a wave-packet particle thingy (pop-sci photon) with a well defined frequency is stretched-out across many wavelength of longitudinal distance (semi-quantum framing). Which makes the notion "we delay the light on one path so it can't interfere with itself" rather tricker than it seems at first. I suspect we can construct a varient of the energy-time HUP to cover this case. $\endgroup$ – dmckee --- ex-moderator kitten Dec 21 '19 at 21:36
  • $\begingroup$ Aside: Fourier optics is almost entirely neglected in pop-sci treatments, and is given too short shrift in most undegraduate treatments. Which is a shame because it, well, illuminates many imporatnt issues. $\endgroup$ – dmckee --- ex-moderator kitten Dec 21 '19 at 21:39
  • $\begingroup$ Diffraction patterns come from the slit and spreads across the whole screen. If you have a double slit each one spreads out across the same screen. You cannot have photons diffracting at the edges of the two slits and then travel through two different mediums on their way to all parts of the same screen. $\endgroup$ – Bill Alsept Dec 21 '19 at 22:38
  • $\begingroup$ @dmckee Ah, thanks for bringing that point up. I was just thinking of the wave function like an electromagnetic plane wave. I guess that got me kind of confused. $\endgroup$ – Laff70 Dec 21 '19 at 22:44
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It appears that you're thinking of a photon as if it were a tiny particle-- but it's not. It is a wave, extended in width, height, and length. Its length depends on the coherence length of the source. Actually, the coherence length of the source is the effective length of the photons. When each individual photon goes through the double slit interferometer, it is a wave going through both slits. But when it is detected at downstream, it's found to be at a random point location with a likelihood proportional to the intensity of the wave.

In the experiment you described, you are delaying the wave at one slit by some amount. As long as that amount is less than the coherence length of the source (i.e., the length of the wave packet), it is still possible to get unterference where the two parts of the wave overlap.

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  • $\begingroup$ I consider a photon to be a very small particle, too. $\endgroup$ – my2cts Dec 22 '19 at 15:04
  • $\begingroup$ At the moment/location of detection, a photon can be considered a particle for most purposes. But between emission and detection, it is best described as a wave. $\endgroup$ – S. McGrew Dec 22 '19 at 15:28
  • $\begingroup$ I consider it to be a tiny particle all time. The wave I see as the equivalent of the wave function of the electron. $\endgroup$ – my2cts Dec 22 '19 at 16:02
  • $\begingroup$ @S.McGrew There is no proof of waves. Besides you can describe a particle but you cannot describe a light wave without resorting to billions of photons. $\endgroup$ – Bill Alsept Dec 23 '19 at 16:49

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