If you shine monochromatic light on a reflective metal surface, say polished silver, does the reflected light undergo a $180^{\circ}$ phase change like it does when it reflects off a medium with a higher refractive index?
[Physics] Does light change phase when reflecting off a metal surface
opticsreflectionvisible-light
Related Solutions
The key word here is continuity.
The continuity boundary conditions for the electromagnetic field vectors sets these phenomenons. The tangential components of $\vec{E}$ and $\vec{H}$ must be continuous across an interface - the only way that they can not be is if there is a surface current flowing (which cannot happen in dielectrics). Likewise, the normal components of the displacement and induction must be continuous across the boundary (the only way that they cannot is if nett charge gathers on the surface - again - not possible in a dielectric.
On the incidence side there are two fields - the incident and reflected one. The sum of these two must have the same phase as the transmitted wave, by the continuity conditions. If you write down boundary conditions for tangential and normal components, then you find that the only way that continuity can be upheld is if the incident and transmitted waves are in phase, whilst the reflected wave may be either exactly in or out of phase, depending on whether the waves are crossing an increase or decrease in refractive index.
Continuity of boundary conditions holds for different physical reasons for different waves, but the conditions are often similar. We don't often find fields in nature with a sudden, discontinuous change in them.
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.
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:
$$\phi_R = \phi_T + \pi / 2$$
Best Answer
Yes, if the metal has infinite conductivity so the induced currents in it are in anti-phase with the electric field of the primary wave.
In practice, metals have finite conductivity, the oscillations of charges on the surface will be slightly off the anti-phase so the reflected wave will have phase shift lower than 180 degrees.