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What is the impact of obstacles on the interference pattern if the obstacles are arranged to be located in the destructive interference locations as follow:

enter image description here

According to my understanding to classical physics, there will an impact for these obstacles. According to my understanding to QED, there will be no impact on the interference pattern. Please advise.

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Judging from your picture, there will be almost no influence. But such a picture usually does not show all necessary components of the wave.

  1. For sound waves, for instance, the colors might represent the pressure as a wave function in space, but there is also the velocity (described by 3 other functions in space, or 2 if you remain in 2D). The influence of the obstacles can only be small if the velocity normal to their surface is already zero everywhere before you bring them in. After you bring them in you of course force the situation to be like that, so what your wave picture colors show should be the initial situation!
  2. For EM waves, there are even 6 components (3 each for $E$ and $B$). And then the $B$ field component normal to the obstacle surface should be zero and also the $E$ field parallel to it should be zero.

So there could exist situations where indeed the influence is small, but it is difficult to visualize the field in such a way that you can immediately see whether that is the case.

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  • $\begingroup$ Thank you for your answer. I'm just trying to understand the meaning of destructive interference in Young's double slit experiment. So, according to classical physics, each slit will generate a spherical waves that interfere destructively with each other wave in certain points. However, the 2 waves propagate normally after interference. So, i assume that any obstruction even in the destructive interferance location will have an impact on thesperical wave. $\endgroup$
    – Wael Khatib
    Commented Apr 28 at 18:38
  • $\begingroup$ On the other hand, according to Quantum Electrodynamics, the possibility of having photon in the destructive interference points equals zero, which means photon will never reach to that location (destructive interference points); hence, obstructions in these locations will never affect the propagation of wave. $\endgroup$
    – Wael Khatib
    Commented Apr 28 at 18:45
  • $\begingroup$ I don't know what you mean by "propagate normally after interference". In fact, if the observation screen with the interference pattern is relatively far from the two slits, all waves travel more or less in the same direction (radially out from the location of the two slits). And in that case a drawing like you present will work. But of course you have to know in advance what the pattern is to know where all the required field components are zero, as explained, but if you do that correctly, it would certainly look like in the drawing. (And it will be similar for the quantum system.) $\endgroup$
    – Jos Bergervoet
    Commented Apr 28 at 18:45

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