An Undergraduate Optics Problem – The Brewster Angle

Recently, a lab-mate of mine asked me if there was an intuitive way to understand Brewster’s angle. After trying to remember how Brewster’s angle was explained to me from Griffiths’ E&M book, I realized that I did not have a simple picture in my mind at all! Griffiths’ E&M book uses the rather opaque Fresnel equations to obtain the Brewster angle. So I did a little bit of thinking and came up with a picture I think is quite easy to grasp.

First, let me briefly remind you what Brewster’s angle is, since many of you have probably not thought of the concept for a long time! Suppose my incident light beam has both components, s– and p-polarization. (In case you don’t remember, p-polarization is parallel to the plane of incidence, while s-polarization is perpendicular to the plane of incidence, as shown below.) If unpolarized light is incident on a medium, say water or glass, there is an angle, the Brewster angle, at which the light comes out perfectly s-polarized.

An addendum to this statement is that if the incident beam was perfectly p-polarized to begin with, there is no reflection at the Brewster angle at all! A quick example of this is shown in this YouTube video:

So after that little introduction, let me give you the “intuitive explanation” as to why these weird polarization effects happen at the Brewster angle. First of all, it is important to note one important fact: at the Brewster angle, the refracted beam and the reflected beam are at 90 degrees with respect to each other. This is shown in the image below:

Why is this important? Well, you can think of the reflected beam as light arising from the electrons jiggling in the medium (i.e. the incident light comes in, strikes the electrons in the medium and these electrons re-radiate the light).

However, radiation from an oscillating charge only gets emitted in directions perpendicular to the axis of motion. Therefore, when the light is purely p-polarized, there is no light to reflect when the reflected and refracted rays are orthogonal — the reflected beam can’t have the polarization in the same direction as the light ray! This is shown in the right image above and is what gives rise to the reflectionless beam in the YouTube video.

This visual aid enables one to use Snell’s law to obtain the celebrated Brewster angle equation:

n_1 \textrm{sin}(\theta_B) = n_2 \textrm{sin}(\theta_2)


\theta_B + \theta_2 = 90^o

to obtain:

\textrm{tan}(\theta_B) = n_2/n_1.

The equations also suggest one more thing: when the incident light has an s-polarization component, the reflected beam must come out perfectly polarized at the Brewster angle. This is because only the s-polarized light jiggles the electrons in a way that they can re-radiate in the direction of the outgoing beam. The image below shows the effect a polarizing filter can therefore have when looking at water near the Brewster angle, which is around 53 degrees for water.

To me, this is a much simpler way to think about the Brewster angle than dealing with the Fresnel equations.

6 responses to “An Undergraduate Optics Problem – The Brewster Angle

  1. For the p polarization, I get that when the angle of reflected ray and refracted ray is 90 deg, then the refracted ray has no reflected ray components and thus, no reflection. I don’t get why the incident ray turns into the refracted ray first, and then, the components are calculated for the reflected ray. In other words, why doesn’t the incident ray component directly transfer to the reflected ray (and has to undergo snell’s law first)?


    • Hi, indeed you are right. It does not turn into the refracted ray first by any means. All that happens is that the light interacts with the electrons of the medium. They will then oscillate and cannot radiate light in the direction of this oscillatory motion.


  2. Hello, Do you mind elaborating the rationale behind this statement:

    “at the Brewster angle, the refracted beam and the reflected beam are at 90 degrees with respect to each other”

    Thank you, the article has been very useful!


    • The 90 deg difference between the refracted and the reflected light is the Brewster condition. The drawing is a little poor, but you can see the 90 deg angle there.


      • Yes I could see the 90 deg in the figure, but it still felt like an axiom rather than an explanation. I dug in in other sources, and this condition could be explained using the Fresnel formula. The R_P coefficient has a tan(theta_i + theta_t) term is the denominator. If theta_i + theta_t = 90 deg, R_P goes to zero. Just putting this in here, in case other people are also wondering. Thank you!

        Liked by 1 person

  3. I still don’t understand why there is no Brewster’s angle for s polarazation?


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