If we split a collimated light beam by using a half-silvered mirror, then the two resulting beams (A and B) have exactly the same intensity. Since the light is made of photons, that means that half of the photons go through path A, and the other half through path B.

If we now reflect both beams by a mirror and the two beams then pass a second half-silvered mirror and enter two detectors as explained in the picture:

then we expect the A beam to be split into two beams. We will call them A1 and A2. A1 goes to dectector 1, while A2 goes to detector 2. Each one contains 50% of A-photons, that is, 25% of the photons of the original light beam:

On the other hand, we also expect the B beam to be split into two beams. We will call them B1 and B2. B1 goes to dectector 1, while B2 goes to detector 2. Each one contains 50% of B-photons, that is, 25% of the photons of the original light beam:

So the amount of photons that should arrive to detector 1 is 25% + 25% = 50%, and the same for detector 2:

Nevertheless, once we have carried out the experiment, what we found is that 100% of photons arrive to detector 2 and no photon arrives to detector 2!

Moreover, what is even more puzzling, if we obstruct channel A (or B, it does not matter), then we detect the same number of photons in detector 2 as the number detected in detector 1. Are you able to figure it out? Try it!

Please, explain your reasoning. You can post your attempted answers in the comment box below. Please, do not use Facebook or Twitter to give your answers.

Photons are behaving like waves. In output 1 both waves are synchronous and intensity is added. In output 2 one wave comes dephased 90º and the output is destroyed. See wave interference.

ReplyDeleteHot, but you still need to explain the details. The phase on reaching the second beamsplitter is simply the path length divided by the wavelength, multiplied by 2π On recombination at the beamsplitter, if the two paths are of

Deleteequal length, then the phases are equal. So both detectors should show constructive interference, shouldn't they?

Is it related to the number of times A is mirrored ?

ReplyDeleteYes, it is.

DeleteBut for detector 1, I calculate no phase difference, while for detector 2, I calculate a phase difference of 2π, which should give constructive interference.

Deletehttps://books.google.fr/books?id=y03ABAAAQBAJ&pg=PA74&dq=light+beam+mirrors+two+detectors&hl=fr&sa=X&ved=0ahUKEwjO7qK85IvXAhUC0hoKHS42Cd8Q6AEIJzAA#v=onepage&q=light%20beam%20mirrors%20two%20detectors&f=false

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