I designed the following interferometer:
If the light beam -theoretically- is very narrow , will light be diffracted away from point B from QED's perspective?
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1$\begingroup$ This config will not work as an interferometer. It needs another BS to recombine beams before the screen. Interference effects are only observed if you provide a path between reflective/transmissive surfaces ... by varying path length you will either allow or deny light to travel. $\endgroup$– PhysicsDaveCommented May 29 at 18:46
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$\begingroup$ @PhysicsDave how come it will not work as an interferometer? If the beam splitter were, say, a transparent dielectric of width $h$ and refractive index $n$, then the transmitted beam would have a $(n-1)h$ path difference and there would be interference, wouldn't there? $\endgroup$– LagrangianoCommented May 29 at 19:13
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$\begingroup$ You can take a dielectric (or even a well made piece of glass) and it will have a strong transmission/reflectance property on its own that depends on the thickness/n. So you may have no interference at all if your source gets 100% reflected! Having 2 elements is the same as the front surface and back surface of the glass. By saying you want to make an interferometer you are really needing 2 arms where can vary path length. $\endgroup$– PhysicsDaveCommented May 29 at 21:49
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$\begingroup$ The term interference classically/high school implies that photons are cancelling at the screen, but this is incorrect. What is really happening is that the EM field (the photon is dumb) guides everything ... the EM field just won't resonate if the path length and frequency don't match ... i.e. no transmission. $\endgroup$– PhysicsDaveCommented May 29 at 21:53
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$\begingroup$ The photon interferes with itself! ... per Dirac. The real meaning is somehow the photons know where to go .... how .... the EM field is very active .... all electrons in the apparatus contribute .... especially an excited electron in and atom waiting to release a photon. $\endgroup$– PhysicsDaveCommented May 29 at 21:57
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1 Answer
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The interferometer in the figure is almost the same as a Mach-Zehnder interferometer; it only lacks the last beamsplitter. Why is this last beamsplitter important? Without it you won't see the interference fringes. They are there, but they are too small to see. The reason is that the wavelength of visible light is extremely small, less than a mikron. With the last beamsplitter, one can align the two beams so that angle between them becomes very small, thus giving larger fringes.
As far as QED is concerned, it only reproduces what you would see classically in this scenario. In other words, regardless of whether the beams are very small or not, one can only have interference in the region where they overlap. In terms of QED, you can say that paths of photons outside of the beam are weighed by a very small probability amplitude.
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$\begingroup$ @filippiefanus thanks for the clear answer. So, for the above setup, a dark fringe will be in the center of the region -where the 2 beams overlap- then bright fringes will follow and etc. Thanks all for the valuable comments. I might ask another question separately. $\endgroup$ Commented Jun 2 at 20:23