electromagnetic-radiationpolarizationreflection
When a light beam reaches a dielectric surface, the incident and reflected beams have different intensities depending on polarization. For the so-called Brewster's angle, the reflected light is lineary polarized.
My question is: how does this law work in case of mirror-like surface, when (ideally) all the light is reflected?
To calculate the transmitted and scattered wave state at any angle from a plane interface, you need to resolve the incident field into s- and p-linear polarised wave amplitudes and multiply the two complex amplitudes by the amplitude Fresnel transmission and reflexion co-efficients. By "amplitude Fresnel co-efficient I mean the ratio of complex wave amplitudes rather than modulus squared of the Fresnel co-efficients, which gives the fraction of power reflected / transmitted and which are also sometimes called the transmission and reflexion co-efficients.
There is nothing "special" about the polarisation of the throughgoing wave calculated by this approach at the Brewster angle. It is simply a general elliptical polarised state which must be represented by two complex amplitudes $\vec{a}$ of the chosen basis polarisation states, or as the Stokes parameters $s_j = \vec{a}^\dagger \sigma_j \vec{a}$ where $\sigma_j$ are the Pauli spin matrices. The latter approach only represents the field to within a factor of $\pm 1$ (both $\vec{a}$ and $-\vec{a}$ have the same Stokes paramters).
Upon reflection at the Brewster angle the component polarised in the plane of incidence will be zero. So you are left with light polarised perpendicular to the plane of incidence.
But this doesn't mean you can say the transmitted wave is also linearly polarised. The reason is you have to satisfy the boundary condition that the component of E-field tangential to the boundary is continuous.
The only simplification here is that the reflected wave tangential component is just the reflected amplitude of that "half" of the incoming wave that had its polarisation perpendicular to the plane of incidence. You have of course the additional constraints at the Brewster angle (for light going into a medium of refractive index $n$ from vacuum).
$$ n = \sin \theta_i /\cos \theta_i\ \ \ \ \cos \theta_i = \sin \theta_t $$
I think this leads to an s-polarisation transmission (power) coefficient of
$$T_s = 1 - \left(\frac{\cos \theta_i - \tan \theta_i \cos (\sin^{-1}(\cos\theta_i))}{\cos\theta_i + \tan \theta_i \cos(\sin^{-1}(\cos\theta_i))}\right)^2 $$ and a p-polarisation transmission coefficient of 1 (because it is at the Brewster angle).
Thus the polarisation state of the transmitted wave will still depend on the angle of incidence (the value of the Brewster angle), but the p-polarised component will be stronger than the s-polarised component. The transmitted light would be unpolarised if the Brewster angle was $\pi/4$ (when $n=1$), falling monotonically to completely polarised when the Brewster angle is $\pi/2$ (when $n=\infty$). The plots below show the s-polarisation transmission coefficient and then the degree of p-polarisation at the Brewster angle as a function of refractive index. The degree of polarisation is defined as $(T_p -T_s)/(T_p+T_s) = (1-T_s)/(1+T_s)$.
Since most conventional dielectric materials have refractive indices in the range $1<n<3$ for visible light, then the transmitted light never has a degree of linear (p-)polarisation of more than about 50%. There are some materials with much higher refractive indices at microwave/radio frequencies (e.g. water).
Best Answer
The easy answer is to say that Brewster's law only applies to reflection from the interface with a transparent medium, and a mirror isn't transparent. Indeed for an ideal perfect mirror, all light of both polarizations is reflected perfectly, so there is nothing to say.
For an actual real-world mirror, the metal mirror surface will have a finite skin depth, and can be considered a dielectric medium with a very large, complex index of refraction. This does lead to a small polarization dependence of the reflection coefficient for near grazing incidence angles. The analogue of Brewster's angle occurs at an angle given by 2*pi*(skin depth) / wavelength above grazing, where the parallel polarized reflection coefficient will reach a minimum ( but not zero, still close to 1). For more details, see for example Landau and Lifshitz, volume 8, section 87.