1
$\begingroup$

When incident light hits an interface of a dielectric surface at the Brewster angle, I understand that it will result in the generation of a perfectly polarized(horizontally) reflective ray and a partially polarized refracted ray. I do not understand, however, what happens at the interface that causes the reflective ray to become horizontally polarized.

I know the basic principle that it is the oscillation of charged particles that creates the reflected beam. However, I do not understand how the "light" reemitted by these electrons results in the formation of refracted and reflected rays. I have read in several places that the reflected wave doesn't have electric field vectors parallel to the refracted wave. I am hoping that someone can elucidate what is meant by this and why this occurs.

Also, it is my understanding that the dots in the diagram represent oscillations that are "into" the page, while the arrows are parallel.

enter image description here

$\endgroup$

1 Answer 1

1
$\begingroup$

Let's take two parts and put them together:

  1. If light is propagating along the $z$ direction, its wave vector $\vec k$ is parallel to $\vec e_z$.
  2. The direction of the electric field $\vec E$ is perpendicular to $\vec k$. Thus, if $\vec k = k \vec e_z$, than $$ \vec E = E_x \vec e_x + E_y \vec e_y $$

Now let's consider an incident field $\vec E_0 %= E_x\vec e_x + E_y \vec e_y + E_z \vec e_z $, which is reflected from a substrate in the Brewster configuration. What the incident field does is to excite the electrons inside the substrate. However, the excitation is done in such a way that the oscillation of the electrons is perpendicular to the the transmitted wave vector $\vec k_t$. Since we are free to choose our coordinate system, we choose $\vec k_t = k_t \vec e_z$. However, this implies that electrons do not oscillate in the $z$ direction. Hence, neither the transmitted nor the reflected beam have a $E_z$ field component.

enter image description here

Now, in the Brewster configuration the reflected and the transmitted wave vector are perpendicular to each another. Hence, the reflected wave vector $\vec k_r$ is parallel to $\vec e_y$. Thus, using the fact that the electric field must be perpendicular to the wave vector, we conclude that also the $y$ component of the reflected electric field must be zero. Hence, the reflected electric field if polarised in $x$ direction.

$\endgroup$
2
  • $\begingroup$ This is not the Brewster configuration. The Brewster configuration is when the refracted beam is orthogonal to the reflected one. $\endgroup$
    – A. P.
    Dec 27, 2020 at 18:02
  • $\begingroup$ Sorry, but this is still not what the Brewster effect is about. It describes the polarization of the reflected beam in terms of p- and s-polarization, relative to the surface, not some arbitrarily chosen coordinate system. See wikipedia. $\endgroup$
    – A. P.
    Dec 28, 2020 at 1:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.