Brewster angle glare
A Socratic walk-through of Brewster angle glare — reasoned out one step at a time, not lectured.
The question we started with
THE QUESTION #Why does glare off water disappear through polarized glasses at one viewing angle but not at others?
Put on polarised sunglasses and look at a lake. At most angles the glare dims but stays. Then, at one particular slant — roughly halfway between looking straight down and looking along the surface — it does not dim. It goes, and you see straight to the bottom.
A filter that works completely at one angle and only partly at others is telling us something specific. It means that at that angle the reflected light is not merely mostly one thing, but entirely one thing. What could force the surface into so exact a restriction?
Reasoning it through
REASONING #We already know from the ordinary account of a window that a boundary between air and a transparent medium reflects a small fraction and transmits the rest — about two per cent for water at near-normal incidence, four for glass — and that the reflected fraction climbs steeply as the view flattens toward grazing. That account treats light as one thing. Split it in two and something new appears.
Light's electric field oscillates across its direction of travel, which leaves two independent choices for a beam hitting a surface: the field can vibrate in the plane containing the incoming ray and the surface normal, or across that plane. Sunlight arrives with both in equal measure. The question is whether the surface treats them alike.
To see why it cannot, ask what reflection physically is. The light entering the water shakes the electrons in it, and those oscillating charges radiate. What we call the reflected beam is the sum of that re-radiation heading back out. So the reflected beam is not bounced light; it is light made afresh by little oscillating dipoles, and it inherits their limitations.
Here is the limitation that matters: a dipole radiates nothing along its own axis of oscillation. Shake a charge up and down and the radiation goes out sideways, strongly in the plane perpendicular to the shaking, and exactly zero along the line it is shaking on.
Now put those two facts together. The electrons are driven along the electric field of the refracted beam inside the water. For the component vibrating in the plane of incidence, that direction is perpendicular to the refracted ray, in that plane. And the reflected beam leaves at the mirror angle. So there will be one geometry — one incidence angle — at which the direction the reflected beam would have to take lies exactly along the axis those electrons are shaking. At that angle they cannot send anything that way. That component of the reflection is not reduced; it is forbidden.
The geometry is easy to pin down. The condition is that the reflected ray and the refracted ray leave at right angles to each other. Since the reflected ray leaves at the incidence angle θ, the refracted ray must be at 90° − θ. Snell's law says sin θ = n sin(90° − θ) = n cos θ, so tan θ = n. For water with n = 1.33 that gives θ = 53°, measured from the vertical — which is a line of sight about 37° above the water. For glass, n = 1.5 and θ = 56°.
And the other component? Its electrons shake horizontally, parallel to the water's surface and perpendicular to the reflected ray — a direction along which they radiate perfectly well. So at that angle the reflection consists of that component alone. Evaluating the Fresnel formula there gives about eight per cent of what arrived in that polarisation for water: not a small glare, but a completely one-sided one. Since the surface is horizontal, that surviving light is horizontally polarised — which is exactly why sunglasses are built with a vertical transmission axis, and why turning your head sideways brings the glare back.
The analogy
THE ANALOGY #Think of someone waving a long stick to signal to friends around them. Wave it side to side and everyone off to the sides sees it clearly — except the two people standing directly in line with the stick's travel, who see only a dot growing and shrinking. Move to that spot and the signal disappears, not because the waving stopped but because you are standing along the axis of the motion.
the observer in line with the stick still sees it, only not its sweep, whereas the electrons genuinely radiate nothing along their axis — and the observer's position here is not chosen freely but forced by the reflection angle, so the extinction happens at exactly one incidence angle rather than wherever you care to stand.
Clarifying the model
THE MODEL #Some refinements, in order of how much they matter in practice.
The extinction is complete only for that polarisation and only at that angle. Ten degrees either side, the in-plane reflection has climbed back off zero, so some of the glare is now vertically polarised and passes your lenses no matter how you orient them. This is precisely the observation the question started from, and it is a clean falsification test in both directions: rotate a polarising filter while looking at glare from a range of angles, and there should be exactly one angle at which some orientation gives total extinction. If the glare could be extinguished completely at every angle, or at none, the dipole account would be wrong.
Ripples blur it. The angle is defined against the local surface, so on a rippled lake different facets meet your eye at different angles and only some are at extinction.
Metals have no true Brewster angle. Their refractive index is complex, and the in-plane reflection falls to a shallow minimum rather than to zero — the pseudo-Brewster angle. Polarised sunglasses therefore work far better on water, glass and wet tarmac than on chrome.
Finally, the reflected light being horizontally polarised depends on the reflecting surface being horizontal. Glare off a vertical shop window is polarised the other way, and vertically-axed sunglasses barely touch it — another everyday check on the whole picture.
A picture of it
THE PICTURE #How to readThis is a flow diagram of one hundred units of unpolarised sunlight arriving at still water at the Brewster angle, split first into its two polarisation components and then into what is reflected and what enters the water. Follow the upper path: of the fifty units vibrating across the plane of incidence, about 3.9 turn back as glare. Now follow the lower path, and notice what is missing — there is no branch at all from "In the plane" to "Reflected glare". That absent ribbon is the whole phenomenon: at this one angle the reflection is not merely weak in that polarisation, it does not exist, so every unit continues into the water.
What became clearer
WHAT CLEARED #Reflection is re-radiation by driven electrons, and driven electrons are silent along their own axis. That single constraint means there is one incidence angle — where reflected and refracted rays are perpendicular, so tan θ = n — at which one polarisation cannot be reflected at all. The surviving glare is then purely horizontal, which a vertical filter removes completely. At every other angle the reflection contains both polarisations, no filter orientation can catch all of it, and the glare merely dims.
Where to go next
ONWARD #- Why the same right-angle condition is used to cut windows in laser cavities at Brewster's angle so they reflect nothing at all.
- How the sky's own polarisation, produced by the same dipole rule applied to scattering rather than reflection, interacts with what you see through the lenses.
Key terms
TERMS #| Term | What it means |
|---|---|
| Brewster's angle | the incidence angle at which reflected and refracted rays are perpendicular and the in-plane polarisation is not reflected; tan θ equals the refractive index ratio. |
| Plane of incidence | the plane containing the incoming ray and the normal to the surface, against which the two polarisation components are defined. |
| Dipole radiation | the field emitted by an oscillating charge, maximal perpendicular to its motion and zero along it. |
Every term the collection defines is gathered in the glossary.