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In the diagram below I've drawn Wheeler's delayed choice experiment, in the case where a 2nd half-silvered mirror has been inserted near the detectors.

---->\-------->\
     |         |
     v         v
     \-------->\--> D2 (0%)
               |
               v
               D1 (100%)

Every summary of this experiment I've found says that in this case destructive interference will occur at D2 and constructive interference will occur at D1.

I understand why the two paths interfere constructively at D1:

  • One path has 1 reflection => phase change of 180$^\circ$
  • The other path has 3 reflections => phase change of 3 x 180$^\circ$ which equals 180$^\circ$ modulo 360$^\circ$

I don't understand why the two paths interfere destructively at D2:

  • One path has 2 reflections
  • The other path also has 2 reflections

Can anyone explain why interference is supposed to be destructive at D2?

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    $\begingroup$ I believe you drew the diagram wrong. D1 is getting some paths with one reflection and some with 3, so D1 should have the destructive interference and be 0% $\endgroup$
    – TKoL
    Feb 10, 2021 at 17:19
  • $\begingroup$ Yes, given the answer you appear to be right. I copied the diagram and the intensity split from quantamagazine.org/… $\endgroup$ Feb 11, 2021 at 9:21

1 Answer 1

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For beam-splitters there is a phase difference of +- Pi/2 between transmitted an reflected beam. If you take that into account and the direction of the beam vs. splitter, everything seems fine.Look here: https://arxiv.org/ftp/arxiv/papers/1509/1509.00393.pdf

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