2
$\begingroup$

I am trying to understand why light undergoes a phase shift when reflecting off one side of a half-silvered mirror, but not the other side.

This Wikipedia page and this answer both give the following explanation:

  • According to the Fresnel equations, a wave:
    • Undergoes a phase shift of π when reflecting off the boundary with a denser medium.
    • Does not undergo any phase shift when reflecting off the boundary with a less dense medium.
  • When light approaches the silvered side of a half-silvered mirror and reflects off the air-to-glass boundary, it will undergo a phase shift (because glass is more optically dense than air).
  • When light approaches the non-silvered side of a half-silvered mirror, enters the glass and then reflects off the glass-to-air boundary, it will not undergo a phase shift (because air is less optically dense than glass).

However, this explanation seems to completely neglect the silver coating. It seems to me that there is no air-glass boundary at all, but instead an air-silver boundary and a silver-glass boundary.

What am I missing?

$\endgroup$
  • $\begingroup$ Related Refractive index of mirror. $\endgroup$ – Farcher Sep 17 '19 at 5:49
  • $\begingroup$ physics.stackexchange.com/a/330656 there is, ideally a 180 degree shift when reflecting from a metal surface. $\endgroup$ – boyfarrell Sep 18 '19 at 8:07
  • $\begingroup$ @boyfarrell Then why is there no phase shift for light approaching from the non-silvered side of the mirror? This is what I don't understand. $\endgroup$ – Vacuous Sep 18 '19 at 10:27
  • $\begingroup$ Diagram would help this question get a good answer $\endgroup$ – boyfarrell Sep 18 '19 at 11:29
  • $\begingroup$ Who says there would be no phase shift from the non-silvered side? It's just that very little light will be reflected at the interface between air and the not-silvered-side-of-the-glass. So approaching from the non-silvered side, you'd get a small fraction of the light reflected from the air-glass interface with a phase shift of pi, most of the light would enter the glass. $\endgroup$ – Holzner Sep 19 '19 at 7:31
0
$\begingroup$

With coherent light from a laser one can assume that the path lengths of a Michelson interferometer are equal (for the purpose of your question) and due to the narrow bandwidth of the laser light one can also assume low dispersion (equal path lengths through the glass).

With everything assumed to be equal the only difference between the two paths that the light travels is:

There is a phase change for a reflection when a wave propagating in a lower-refractive index medium reflects from a higher-refractive index medium, but not in the opposite case.

Michelson interferometer fringe formation

Notice that the light travels through the glass twice, once from air to glass to mirror, and once from mirror to glass to air; one direction has increasing index and the other doesn't.

Reference:

"Phase shift between the transmitted and the reflected optical fields of a semireflecting lossless mirror is π/2" (1980), by Vittorio Degiorgio, in the American Journal of Physics 48, 81 (1980); https://doi.org/10.1119/1.12238

Assuming a lossless beamsplitter the energy conservation condition $\left | E_0 \right |^2 = \left | E_t \right |^2 + \left | E_r \right |^2$ gives $\phi^{'}_t - \phi^{''}_t = \phi^{'}_r - \phi^{'}_r + \pi$, thus:

$$\phi_R = \phi_T + \pi / 2$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.