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From this link's overview : Afshar's experiment uses a variant of Thomas Young's classic double-slit experiment to create interference patterns to investigate complementarity. Such interferometer experiments typically have two "arms" or paths a photon may take.1 One of Afshar's assertions is that, in his experiment, it is possible to check for ...

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The company TeachSpin builds a two-slit interference instrument for use in the advanced/modern physics lab. There are several descriptions and schematics at that website. They do not have a lens before or after the single slit. The single slit is physically close to the source. I've used the instrument several times and it gives beautiful results for both ...

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Are you sure that the single slit between lenses 1 and 2 is a slit and not a pinhole? If I were setting this up without a laser, I would use a pinhole below the diffraction limited spotsize of the first lens at the focus: this gives you an aberration free spherical wave at the output of the pinhole (same idea as a point diffraction interferometer / ...

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I think you have a few misconceptions here. You start by talking about the particles in the beam "not interfering with each other" so the "wave function of each particle is lambda/p". There are at least two problems with this statement. I'll take the last part first. It looks like you are confusing "wave function" with "wave length". The wave function ...

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For a measurement event to occur, it's not necessary that the particle is detected. This is outlined in detail in the Renninger negative-result experiment, which consists of a radioactive atom surrounded by two hemispherical particle detectors, which are assumed to be 100% efficient so they detect every particle coming out from the atom. Generally one of ...

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Since no one else has contributed, and the most satisfactory answer was the pdf I found, I will state that that is the answer to my question: "The Illusion of Light as Transverse Electromagnetic Waves" (bit.ly/1etSb0n)

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If the slits are on top of each other, then the light travelling through each slit goes the same distance and therefore has the same phase. In this case, the distance between the fringes is infinite. On the other hand, if the slits are very far apart, then even a small angle incurs a large path difference, so the fringes are very close together. Thus we have ...

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This calculation assumes that the other screen is very far away, that is, $y \ll D$. So redraw your diagram so that the green line is very nearly horizontal, and you'll get the conclusion.

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The "lines" of constructive interference are not really "lines", although far away from the points they approach lines. What are they? They are "The set of points P where the distance from P to the top slit, minus the distance from P to the bottom slit, is a certain fixed value". And what is that set of points? You may have learned it in high-school math ... ...

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