How Does a Half-Silvered Lens Affect the Young's Two Slit Experiment?

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I don't have the equipment to carry out this experiment (and maybe it's in the literature somewhere..) however, here goes.

Elsewhere on this website I read of an experiment involving setting up the Young's two slit experiment, then replacing the screen with a large thin lens to focus the incident interference pattern onto the same screen pushed further back. But, surprise, the fringes disappear to be replaced on the screen by the image of the two slits.

I propose the following. Conduct the same experiment, but this time the thin lens is half silvered, permitting some light to pass through, the rest reflected. Would the reflected light be the interference pattern? I would suggest that it must be. If it were anything else I think the plot would thicken considerably.

Would some kind person like to try this out and tell me what was visible?
 
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The interference pattern remains as long as we don't measure the wave at any point. I don't understand what you mean by splitting up the interference pattern, light is not a either or thing, it is both according to how we chose to measure it or not. In that case my guess would be the reflected light would form a decohered distribution. This experiment requires expensive equipment, I'm not sure anyone could set it up without incurring some cost.
 
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I see you have modified your post removing the reference to Michelson and Morely's experiment of 1887 when we are considering Thomas Young's experiment of 1804. Good, at least you got that right. We are not looking for luminiferous aether.

Look, the point here is not your opinion on the matter nor mine. I am proposing an experiment. until it is conducted no one knows the answer. Probability is that this has been done long ago. If not I hope that someone will be interested enough to find out the answer.

In fact you may well be right in your conclusion. The simple fact of focussing a light onto a screen gives one result and placing a mirror there gives another, we see the source of the light reflected.

Yes, on reflection, the game is not worth the candle...is it?
 
Actually I removed the bit about using a Michelson-Morely interferometer, as basic equipment will do. But I still think the answer is obvious. Anyway I was just theorising, I have not done the experiment, call it an experimental guess.

Michelson_Interferometer_Green_Laser_Interference.jpg

Interference pattern produced with a Michelson interferometer. Bright bands are the result of constructive interference while the dark bands are the result of destructive interference.

http://en.wikipedia.org/wiki/Interference
 
I am no expert, but for this experiment can't we consider the set of cases in which the photons are reflected independently of the set of cases in which it passes through the lens -- no interference effects between those 2 cases? If so, then isn't the former case just the same as putting a mirror at the screen of the normal double slit?
 
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