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Looking at this diagram from Wikipedia:

enter image description here

I was trying to make sense of the sentence

the interference is constructive at C and destructive at D

Let's take a look at the superposition at C.

  1. Say that the photon arrives at phase $0$.
  2. There are 50% that it will go through the lower route and 50% that it will take to upper route.
  3. If it goes through the lower route: the first half-silvered plane mirror will keep it's phase at $0$, the second mirror turn it to $\pi$ and the last half-silvered plane mirror will keep it's phase at $\pi$.
  4. If it goes through the upper route: the first half-silvered plane mirror will turn it's phase at $\pi$, the second mirror turn it to $0$ and the last half-silvered plane mirror will keep it's phase at $0$.

Since the outcome of the first route is $\pi$ and the outcome of the second route is $0$ there should be a destructive interference at C.

Under the same assumptions, one can show that there should be a constructive interference at D.

What am I missing, why did I get the opposite result?

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  • $\begingroup$ To get to $C$ the upper beam is reflected three times and the lower beam is reflect once, so their phases are the same. To get to $D$ the upper beam is reflected twice and the lower beam twice, so actually I get the same phase at $D$ as well. Hmm. $\endgroup$ Aug 19, 2015 at 14:58
  • $\begingroup$ @JohnRennie Just below the diagram in Wikipedia, you can see that 'Mirrors in the lower left and upper right corners are half-silvered'. Looking at the phase diagram of the half-silvered mirror, you can see that on the upper route, the last reflection does not change the phase of the photon. So it actually changes it's phase only twice. $\endgroup$
    – Michael
    Aug 19, 2015 at 15:50
  • $\begingroup$ @Michael If the phase diagram is correct, it seems the wikipedia page had it backwards. No big deal $\endgroup$
    – Jim
    Aug 19, 2015 at 16:37

1 Answer 1

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In step 4, LAST half silvered plane acts as a mirror not as a window, therefore it changes the phase back to $\pi$ and not keeping it at 0.

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  • $\begingroup$ If you look at the phase diagram of the half-silvered mirror you can see that going through the upper path, the reflection should not change the phase. $\endgroup$
    – Michael
    Aug 19, 2015 at 16:04
  • $\begingroup$ On a closer look I now see that they assume the reflection from the silver paint. Why is that important? the phase and amplitude of a reflected\transmitted light is derived from fresnel equation and is dependent on the refraction indexes on both sides of the reflection plane. You can see for the derivation here(link in next comment) that Er/Ei~n1*cos(theta_i)-n2*cos(theta_t). Therfore it can give -\+ dependent on the ratio between n1/n2. So in one case the light incidate from air to glass you will get a pi shift, other case n1 & n2 switch places and you get no phase shift. $\endgroup$ Aug 19, 2015 at 20:00
  • $\begingroup$ google.co.il/… $\endgroup$ Aug 19, 2015 at 20:00
  • $\begingroup$ So D and C are now opposite, which one is constructive and which is destructive will be decided by the phase added in 'B'. $\endgroup$ Aug 19, 2015 at 20:13

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