If a perfectly circularly polarized wave of light is incident on a dielectric medium (coming from air) at Brewster's Angle, what will the polarization state of the transmitted wave be? I am aware that the reflected wave will be linearly polarized, but will the transmitted wave be linearly polarized as well because the light was perfectly circularly polarized to begin with, and the perpendicular component was completely eliminated by the fact that the angle of incidence is Brewster's Angle?


2 Answers 2


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).

  • $\begingroup$ I recognize that "There is nothing "special" about the polarisation of the throughgoing wave calculated by this approach at the Brewster angle" in the general case, but in the case of a circularly polarized plane wave, is it not possible to know that the transmitted wave would be linearly polarized? Is it possible for the transmitted wave to be linearly polarized? Or is there no way of knowing without more information? $\endgroup$
    – weskpga
    Dec 11, 2013 at 18:18

If you split the circularly polarised light up into two perpendicular polarisations with a $\pi/2$ phase lag between them. You are free to choose which unit vectors to represent these perpendicular polarisations - so choose one in the plane of incidence and one parallel to it.

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 that (asssuming $n_{\rm air}=1$):

$$ n = \sin \theta_i /\cos \theta_i\ \ \ \ \cos \theta_i = \sin \theta_t $$

But I think this still leads to an s-polarisation transmission 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 the Brewster angle).

Thus the polarisation state of the transmitted wave will still depend on the angle of incidence. $T_s$ is a bell shaped curve peaking at $T_s=1$ for $\theta_i= \pi/4$, corresponding to $n=1$ (i.e. no interface and the transmitted light is still circularly polarised). For other values of the Brewster angle (corresponding either side of $\pi/4$ to medium to air or air to medium), the result will be elliptically polarised light.


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