When unpolarized light is incident at Brewster's angle, what I understand is that the reflected wave is S-Polarized. What is the polarization state of the transmitted (refracted) part of the wave?


A p-polarized wave is completely transmitted into an interface at the Brewster's angle. A s-polarized wave is both reflected and refracted at the same angle.

An unpolarized beam having both s- and p- components will give rise to a completely s-polarized reflected wave (since none of the p-component reflected) and a slightly p-polarized transmitted wave (since both components were transmitted at least partially).

EDIT: As Rob's answer shows, the degree of polarization in the transmitted wave depends on the difference in refractive index of the two media forming the interface. My comments only concern typical situations of glass and air, water and glass, etc. See Robs answer for a more complete answer.

  • $\begingroup$ The degree of polarisation depends entirely on what the Brewster angle is. It can vary from zero to 1. $\endgroup$ – Rob Jeffries Nov 24 '16 at 9:23
  • $\begingroup$ @Rob I've incorporated your comments. $\endgroup$ – Mark Hannel Nov 24 '16 at 19:00

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).

s-polarisation power transmission as a function of refractive index

Degree of p-polarisation

  • $\begingroup$ This is the more complete answer. Nice job! I'd like to see a plot of s-transmission vs refractive index of the second media. The highest Ive encountered is titantia at ~2.5. $\endgroup$ – Mark Hannel Nov 24 '16 at 18:59
  • 1
    $\begingroup$ @MarkHannel Plots added. I agree that conventional dielectrics would never lead to a high degree of linear polarisation in the transmitted visible light. $\endgroup$ – Rob Jeffries Nov 24 '16 at 19:50

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