I am trying to figure out what is happening to the charged particles in the reflective surface in the case of Brewster's angle. When this angle is different than zero or Brewster's angle, then the reflected light is only "partially" polarised. Why is the oscillation of the particles such that the reflected wave is polarized parallel to the reflecting surface when the angle is Brewster's angle?

I understand the explanation behind separating the electric fields into their components, yet I don't understand what causes this polarization to be observed? How does the vector component of the electric field emerge from the reflective surface while the other component remains as a part of the refracted wave?

By "parallel" I mean oscillating to the left and right- represented by dots (not into the page) in the diagram below. There are two planes I have talked about, the first one being the plane of the page(plane of incidence) and the plane perpendicular (the plane parallel to the reflecting surface, which is also the plane of polarization)

enter image description here

  • $\begingroup$ What do you mean by "vertical", and by "x" and "y" directions? Those terms are meaningless if you don't set up the conventions correctly. $\endgroup$ Commented Dec 28, 2020 at 11:27
  • $\begingroup$ @EmilioPisanty I have edited the question $\endgroup$
    – ten1o
    Commented Dec 28, 2020 at 11:35
  • 4
    $\begingroup$ Wikipedia explains it quite clearly (under Explanation, paragraph starting "the physical mechanism for this..."). Is that explanation insufficient? If so, what aspects of it are unclear? $\endgroup$ Commented Dec 28, 2020 at 12:04
  • $\begingroup$ It's not clear to me why one component of the electric field is reflected while the other is refracted, more specifically, why the reflected ray has no component in the plane of incidence. The 90 deg between the reflected ray and refracted ray signifies that the reflected ray is a component of the refracted ray, not the incident ray. @EmilioPisanty $\endgroup$
    – ten1o
    Commented Dec 28, 2020 at 12:18

1 Answer 1


An unpolarized wave can be represented by the sum of two waves with perpendicular polarisations and equal amplitude. The unpolarised incident wave can be considered to be of this nature, with one polarisation in the plane of incidence (p-polarised) and another at right angles to that (s-polarised), and with the electric fields of both being perpendicular to the direction of incident wave motion.

When the electric field of the incident wave is incident upon the interface, it sets up an electric field in the medium. Because of the continuity conditions for the electric field (any components tangential to the surface are continuous) and the requirement for a fixed phase relationship between the incident, reflected and transmitted waves, we obtain the law of reflection and Snell's law of refraction.

One way to think about how the fields of the reflected and transmitted waves are produced, is to imagine that oscillations are set up in small electric dipoles in the medium. The oscillations are driven by the electric field in the medium, i.e. the electric fields of the transmitted wave. These oscillating dipoles then re-radiate electromagnetic waves. Since the incident wave consists of both p- and s-polarised waves then the dipole oscillations will also have components in the plane of incidence and at right angles to that, but must also be perpendicular to the direction of the transmitted wave.

The key property of electric dipole radiation here is that no radiation is emitted along the oscillation axis of the dipole. This means that, at the Brewster angle, the electric field of the reflected travelling wave contains no contribution from the dipoles oscillating in the same direction as the reflected wave is travelling, but only from the dipole oscillations at right angles to that. As a result the p-polarised incident light is not reflected at all and the reflected wave is solely s-polarised.

At other angles the dipole oscillations in the medium are not exactly aligned with the reflected wave direction and so the p-polarised light is reflected to some extent; resulting in a partially polarised reflected wave. Note that the s-polarisation is not affected by these considerations since the dipole oscillations due to this polarisation are always at right angles to the reflected wave direction. Note also that because the oscillating dipoles would in general have different amplitudes in each direction, then the transmitted wave is always partially polarised for light incident at the Brewster angle (see https://physics.stackexchange.com/a/294528/43351 ).

  • $\begingroup$ By "dipoles oscillating in the same direction of the reflected wave", do you mean perpendicular to some common axis of the refracted and reflected rays? $\endgroup$
    – ten1o
    Commented Dec 30, 2020 at 12:01
  • $\begingroup$ @ten1o In the direction of the reflected wave is a unique direction. I mean that the axis of oscillation is in the direction that the reflected wave travels. $\endgroup$
    – ProfRob
    Commented Dec 30, 2020 at 12:04
  • $\begingroup$ I think it would be more clear if you could talk about paragraph 2 in relation to the last paragraph, as in the part "The dipole oscillations are not exactly in the same direction as the reflected wave; resulting in a partially polarised reflected wave. ", it is understood as if the reflected wave is produced by some other process than dipole oscillations. $\endgroup$
    – ten1o
    Commented Dec 30, 2020 at 12:32

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