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I'm reading a paper which says

The beam that reflects from a Fabry–Perot cavity is actually the coherent sum of two different beams: the 'promptly reflected beam', which bounces off the first mirror and never enters the cavity; and a 'leakage beam', which is the small part of the standing wave inside the cavity that leaks back through the first mirror.

I understand this. However, the author then writes

If the cavity is resonating perfectly, then the promptly reflected beam and the leakage beam are exactly 180° out of phase. In this case the two beams interfere destructively, and the total reflected beam vanishes.

I don't understand why these two beams would be 180° out of phase.

This is my (flawed?) understanding: enter image description here

The light beam comes in as beam 1. Beam 2, the part that is reflected, get a 180° phase shift w.r.t. beam 1. Beam 3 continues without any phase shift w.r.t. beam 1. Part of beam 3 is reflected as beam 5. So, beam 5 has a 180° phase shift w.r.t. beam 3. The part of beam 5 that goes back through the first mirror is called beam 6, and this has the same phase as beam 5. So, both 2 and 6 are both shifted by 180°, each. So, they are in phase, and would constructively interfere... right?

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    $\begingroup$ Note the requirement on the cavity resonating perfectly. This means that any attempt at explaining this with a finite number of in-cavity reflections (let alone just one) are doomed to failure. $\endgroup$ – Emilio Pisanty Sep 12 '13 at 15:04
  • $\begingroup$ @EmilioPisanty So... how would you explain it then? $\endgroup$ – User 17670 Sep 12 '13 at 15:07
  • $\begingroup$ The way I've always seen it done is using transfer matrices. From what I understand, it handles all the reflections in each material by just looking at the resultant E and B fields and matching boundary conditions, very similar to the quantum tunneling problem through a finite barrier. $\endgroup$ – YungHummmma Sep 12 '13 at 15:37
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in an F-P cavity such as this...the cavity thickness is quarter wavelength thick, such that 3-5 travels a distance of half a wavelength (180 degrees shift even before accounting for the phase shift due to reflection). when that is accounted for, 1 and 6 are actually in phase. therefore 2 and 6 are still antiphase--antireflection achieved.

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  • $\begingroup$ As I understand it (and also mentioned in the above paper), the thickness of the cavity must be an integer multiple of half-wavelengths in order to have resonance. $\endgroup$ – User 17670 Sep 12 '13 at 15:59
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    $\begingroup$ that is true for resonance, but not necessarily for destructive intereference at the first reflector. if however we know the reflecting surfaces are half-wavelength apart, and yet destructive interference occurred at the first reflector, the logical and correct conclusion is that there was no phase change resulting from reflection at the second reflector. this is possible if the refractive index between the two reflecting surfaces is higher than that behind the second reflector (aka an etalon). $\endgroup$ – gregsan Sep 12 '13 at 18:02
  • $\begingroup$ my original answer is in line with your assumption that phase change occurred at both surfaces of reflection, which is typically the case for antireflective coatings where light travels from air, into coating material and then into glass, in order of increasing refractive index. $\endgroup$ – gregsan Sep 12 '13 at 18:05
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    $\begingroup$ Right, I see where I went wrong now. As you suggest, the FP 'cavity' must be an etalon. So, there is no phase change on the second reflection. Thanks for helping me work it out. $\endgroup$ – User 17670 Sep 15 '13 at 13:41

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