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In the double slit experiment (with photons say) (see diagram below), what happens if the detector screen is moved close enough to the slits that the two light "cones" don't touch? Do the photons now cease to pass simultaneously through both slits?

enter image description here

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    $\begingroup$ Last time I checked, point sources don't send out light within a cone. Could you explain why you think the waves don't cover all of space? $\endgroup$ Dec 12, 2019 at 14:22

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No, they do not. Moreover, the interference is still there, but in a more complex form.

When the screen is close to $S_1$ and $S_2$ they are still a collection of point-like sources according to Huygens-Fresnel principle. You could integrate the contribution of each point within the slits and get the near-field interference. When you put the screen far away there is not much point in that integration, you could consider $S_1$ and $S_2$ just as two point sources.

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As noted in the other answers, the diagram is somewhat misleading. A more accurate picture is shown here:

enter image description here

The waves through the slits spread out in all directions. (As pointed out in the comments, the amplitude is much larger in the "forward" directions when the width of the slit is much larger than the wavelength of the light, which is perhaps what your original diagram was trying to convey. Even so, there is still some small wave amplitude in most directions from the slit.)

The reason we usually don't care about the waves spreading out at large angles (beyond their smaller amplitude, as noted above) is that it allows us to use the Fraunhofer approximation, which assumes that we are at a large distance from the screen (compared to the size of the pattern in the screen.) In the region you indicate, we are not necessarily at a "large" distance from the screen, and so we would need to use the Fresnel approximation instead. The mathematics required to analyse this case are rather more complicated, and so it's usually skipped in an introductory physics course.

Finally, nothing that I've said here requires the light to behave quantum-mechanically or classically. The only difference is that if we think of the light as being composed of photons, the intensity pattern should be treated as a probability distribution.

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  • $\begingroup$ I think the statement, "The reason we usually don't care about the waves spreading out at large angles (as opposed to "straight ahead") is simply that it allows us to use the Fraunhofer approximation, ..." is technically correct, but a bit misleading in this case. In fact, unless each slit is extremely narrow (on the order of a single wavelength), the light diverging from the slit is effectively confined to a "cone". If light from slit A doesn't overlap light from slit B, there is no spatial interference and no fringes are formed. It's trivial to demonstrate this. $\endgroup$
    – S. McGrew
    Dec 12, 2019 at 14:55
  • $\begingroup$ If you look at your diagram, you will notice the first "wavelets" out of each slit don't touch. The next two almost touch and then the third overlap. What if the detector is so close that just the "first" or "second" "wavelets" have got through? If the gap is wide enough can't the detector be so close that the light through the slits cannot interact given the speed of light? $\endgroup$
    – fundagain
    Dec 12, 2019 at 15:12
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    $\begingroup$ @fundagain: The "first" peak from one slit doesn't have to interfere with the "first" peak from the other slit to get constructive interference. You could equally well have interference between the "first" peak from one slit and the second, third, fourth, fifth, sixth... peak from the other slit. In fact, this is what happens in the far-field approximation as well; the nth-order peak in the interference pattern corresponds to a difference of n cycles between the waves arriving from each of the two slits. $\endgroup$ Dec 12, 2019 at 15:15
  • $\begingroup$ Tx. That's what I was missing. $\endgroup$
    – fundagain
    Dec 12, 2019 at 15:17
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    $\begingroup$ @S.McGrew: While the light is largely confined to the diffraction angle, it doesn't go to zero exactly outside those cones. If you put a detector in front of slit A and blocked slit A, you'd still get some (low-intensity) light from slit B; and when both slits are open, the amplitude of the waves from slit B would cause a small deviation from the amplitude (and intensity) from slit A alone. I've edited the answer to address your point. $\endgroup$ Dec 12, 2019 at 15:21
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That diagram is not correct.

There is no cone of light; light from each slit fills the entire space beyond the slits. That means that the "interference region" fills the entire space. There is no special region where there is interference, outside of which there is not.

Close to the screen the intensity of the light might be weak, and interference will have a more complex pattern than in the far field, but it's there nonetheless.

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If the screen is so close to the slits that the two "cones" don't overlap and land on separate areas of the screen, it is obviously easy to know which slit each photon goes through. No interference fringes will appear.

However there is still temporal interference, because there is an "anticorrelation" between photon counts in the areas where photos hit the screen. That is, if the beam illuminating the slits is so dim that it can be considered to be a stream of single photons, and if the detectors are sensitive enough and fast enough to distinguish single photon arrivals, detectors in those two areas will never detect a photon at the same time. A plot of photon counts over time for each detector will look (and be) random, but plots of the two sets of counts will be "opposite".

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