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In Young's double slits experiment, there are points on the screen where the probability of an impact is null. I drill a hole at such a point and place a second screen E2 behind E1. I guess that there will be no impact on the second screen. Wheeler said that looking to one of the slit will reveal particle properties. To look in the direction of one slit I take a long thin tube between E1 and E2. One of its ends is at the hole and the other one on E2. Will there be impacts on E2 through this tube?

I am not sure of my answer to this problem: before piercing the screen at point p where there is a dark fringe, all the paths that arrive there stop with annihilation of the photon. There are mainly two categories of paths that play an important role, these are those who join each of the slots at p in as approximate straight line. the others contribute very little to the resulting amplitude. and each of the two groups provide a phase 1 and 2 which vanishes at p. let's go directly to the case with a tube behind the hole pointing towards one of the slots. we still have the group of paths from the target slot that now extend into the tube with an increasing phase. the other group must deviate from the straight line to get into the tube; They will not contribute coherently to the final result of the detection at the bottom of the tube. this group will not be able to provide a phase opposite to the identical one provided by each of the paths in a straight line. So we will have a possible detection on E2 at the bottom of the tube. There is another argument remove the screens and just keep the telescope pointing to one of the sources. no photon from the other sourc will cross it. and some from the observed source will do it. so we have an device that gives partial wich path information and therefore the probability can no longer be zero at p as we supposed.

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  • $\begingroup$ It would reveal particle properties of a wave that's zero in the region, so... $\endgroup$
    – The Vee
    Mar 1, 2018 at 9:57
  • $\begingroup$ We have three devices. A screen then a screen with a hole and then a screen with a tube. One cannot deduce anything if the observables do not commute.To be zero in a region is not an absolute property. If i remove the screen it is no more zero. $\endgroup$
    – Naima
    Mar 4, 2018 at 9:28
  • $\begingroup$ We can't observe energy that doesn't exist, so it's not going to do much. The screen is simply an absorber of energy. You know what is also acting as an absorber of energy? Free space. Light simply keeps going and never returns from it. That is exactly what a screen does, so your modification amounts to, literally, nothing as far as the actual double slit is concerned. $\endgroup$ Jun 2, 2023 at 17:00

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Here is a schematic of the double slit experiment

dblslit

Here is a double slit experiment one photon at a time.

enter image description here

Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

If you make a hole on the screen where there is an accumulation of photons, you would be generating a new coherent light source, like the one in the schematic on the left. If you make a hole in a dark line, there will be no light behind it.

What you are doing between E1 and E2 is irrelevant to the quantum mechanical problem with boundary condition " photons impinging in double slits at fixed distance apart and fixed width" . You are not interfering with the boundary conditions of the problem, the way you would if you put a tube from E1 to the slits, where then one would expect the interference pattern would be lost.

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It is an easy experiment to do; you should do it.

I think you intend the tube to point at one of the slits and thereby serve as a directional filter. This is equivalent to blocking the other slit, and will enable photons to enter and traverse the tube.

It is important to note that when the intensity at an interference null is zero, it doesn't mean that nothing is going on there. In fact, in a simple interference pattern formed by two collimated beams crossing each other at an angle, light is necessarily passing through each zero-intensity region to get to the next high- intensity region.

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