In a Michelson interferometer (image from Optics by E. Hecht ) .

To quote from the same book:

As the figure shows, the optical path difference for these rays is nearly $2d \cos 0$. There is an additional phase term arising from the fact that the wave traversing the arm $OM2$ is internally reflected in the beamsplitter, whereas the $OM1$-wave is externally reflected at $O$. If the beamsplitter is simply an uncoated glass plate, the relative phase shift resulting from the two reflections will be $\pi$ radians.

Consider a beam from $B$ from $S$ towards $O$. At $O$ it splits into two beams:

Beam B1:

  1. Results from refraction of $B$ at the beam splitter $O$ towards mirror $M1$.
  2. Is reflected in the opposite direction at $M1$.
  3. Is reflected towards $D$ at $O$. Is it reflected before entering $O$ or after entering $O$ and encountering the air at the other side?

Beam B2:

  1. Results from reflection of $B$ at the beam splitter $O$ towards mirror $M2$. Should there be a phase shift here? This is should be an air/(glass/metal coating) interface.
  2. Is reflected in the opposite direction at $M2$.
  3. Goes through $O$ towards detector $D$.

Question: Could someone tell me at which of these steps a phase shift occurs? It seems to me that it could happen at steps $B1-2$, $B1-3$, $B2-1$ and $B2-2$, but that is probably not right.

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    $\begingroup$ A small advice - You have mixed up 1 and 2 in your notations for the beams. It would be nicer if B1 corresponded to M1 and likewise for the other. :) $\endgroup$ – 299792458 Mar 5 '17 at 18:32
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    $\begingroup$ Good idea, I have not noticed that. The notation now corresponds with the one which you have used :). $\endgroup$ – pseudomarvin Mar 5 '17 at 19:08

One minor detail, which is extremely important in this context, which you perhaps missed is that the beam-splitter is partially silvered at the lower surface, which implies that the appropriate location of the point O is at the lower surface of the beam-splitter (at the glass-air interface).

If we take this detail into account, the explanation is very simple:

  • There are $\pi$ phase changes at steps B1-2 and B2-2, but these are common for both beams, and hence do not contribute to any net relative phase difference. (These reflections are depicted as such in the edited diagram)

enter image description here

  • The only relative phase difference arises due to complete reflection at B1-3 (reflection at the outer, lower surface of the beamsplitter.) This occurs for only one of the two beams, not for both. Hence, net relative phase difference is $\pi$ radians. (This is represented in the diagram for the red colored beam. Note that this is not true for the other beam, shown in blue.)

  • (IMPORTANT) There is no phase change of $\pi$ in the reflected beam, arising at the original division of amplitude point O, originally. This is because the division took place at the lower interface, where the interface was glass-air and not air-glass. If we invoke the Stokes' relation, you have a phase change of $\pi$ on a reflection at a rare-dense interface. This doesn't fit the description. (Hence, no phase shift of $\pi$ for the blue beam in the figure.)

Hence, the total relative phase difference between the two coherent beams, when they recombine, is only $\pi$ radians.

(Original image edited on suggestion by Floris).

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    $\begingroup$ If you added a diagram this could be a very good answer. $\endgroup$ – Floris Mar 5 '17 at 18:07
  • $\begingroup$ @Floris - Thanks for the suggestion, sir. Done. :) $\endgroup$ – 299792458 Mar 5 '17 at 18:29
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    $\begingroup$ @TheDarkSide: Thanks for the detailed answer, it has helped me a lot. I think that B2-3 in your previous to last bullet point should be changed to B1-3. I accept full responsiblity for the notation related confusion :). $\endgroup$ – pseudomarvin Mar 5 '17 at 19:18
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    $\begingroup$ @pseudomarvin - Yes, with our consistent notations, it should be B1-3. I'll fix that part. Also, I'm glad it helped. :) $\endgroup$ – 299792458 Mar 5 '17 at 19:23
  • $\begingroup$ @TheDarkSide : can you clear my other confusion ? If there is a phase shift of π then the central fringe should be a dark back, no ? Then why in these examples we see central bright band ? pages.physics.cornell.edu/p510/O-2_Michelson_Interferometer and en.wikipedia.org/wiki/… $\endgroup$ – Spectra Nov 26 '17 at 21:52

There is an automatic change in the phase when the wave goes ..... bounces off the mirror. Then after all that you start to play with the path lengths to introduce more phase changes.

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    $\begingroup$ No - phase shift occurs on reflection only, not at transmission. $\endgroup$ – Floris Mar 5 '17 at 18:06
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    $\begingroup$ Slightly less wrong... still not quite right. Phase shift occurs during reflection when you go from low to high index - so on the external reflection, but not at the internal reflection. $\endgroup$ – Floris Mar 7 '17 at 20:12

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