The angle at which reflection picks a side

Light falling on a sheet of glass at 56.31° reflects, as light does. Put a polarising filter in the reflected beam and turn it, and the beam does not merely dim — at one orientation of the filter it goes out completely. Change the angle of incidence by a couple of degrees and the extinction is spoiled. There is exactly one angle at which the reflection loses a whole polarisation, and the algebra that predicts it does not explain it.

Two polarisations, because a surface breaks the symmetry #

An electromagnetic wave in free space has its electric field perpendicular to its direction of travel, and nothing distinguishes one perpendicular direction from another. A surface breaks that. Once there is a boundary, the plane containing the incoming ray and the normal — the plane of incidence — is a definite plane, and the field can be resolved into a component in it and a component along the surface.

A travelling wave has two fields perpendicular to each other and to the direction of travel, and away from any boundary the choice of which perpendicular direction to call the polarisation is arbitrary. The only thing that is not arbitrary is that the field has no component along the ray. A surface supplies the missing second axis — its own normal — and the whole of what follows is that the two resulting cases obey different boundary conditions at it.

The two are called $s$ (from senkrecht, perpendicular — the field lying in the surface) and $p$ (parallel to the plane of incidence). They obey different boundary conditions, so they reflect differently, and the whole of the Fresnel analysis is that difference.

Matching the tangential components of $\mathbf{E}$ and $\mathbf{H}$ across the boundary gives the amplitude coefficients:

$$r_s = \frac{n_1\cos\theta_i - n_2\cos\theta_t}{n_1\cos\theta_i + n_2\cos\theta_t}, \qquad r_p = \frac{n_2\cos\theta_i - n_1\cos\theta_t}{n_2\cos\theta_i + n_1\cos\theta_t},$$

with $\theta_t$ fixed by Snell’s law, which itself follows from requiring the phase to match along the surface and is therefore already a boundary condition in disguise. The reflectance is $R = r^2$ in each case. Note the asymmetry: the indices swap places between the two expressions, and that swap is the whole story.

At normal incidence both cosines are one and the two agree,

$$R(0) = \left(\frac{n_2 - n_1}{n_2 + n_1}\right)^2,$$

which for air and glass is 4%, and which they must, since at normal incidence there is no plane of incidence and no way to tell the two apart.

The angle where one of them vanishes #

Setting the numerator of $r_p$ to zero and using Snell’s law to eliminate $\theta_t$ gives

$$\tan\theta_B = \frac{n_2}{n_1},$$

Brewster’s angle, 56.31° for air into glass and 53.06° for air into water. At that angle the reflection is completely $s$-polarised, whatever the incoming light was.

The algebra says it happens. It does not say why, and the why is not in the boundary conditions at all.

The last step is the one worth stating slowly. A reflected wave is not light bouncing. It is light re-radiated by the charges in the second medium, which the incoming wave has set shaking. Those charges shake along the direction of the transmitted electric field, so for $p$-polarised light they shake in the plane of incidence, perpendicular to the refracted ray.

And an oscillating charge radiates nothing along the axis it oscillates on. The power goes as $\sin^2\theta$ from that axis, with a clean zero along it.

What a shaking charge sends where settles the rest. The pattern is $\sin^2\theta$ about the direction of the acceleration, so the emission is maximum broadside and along the axis of the shaking it is not small but exactly zero. That zero is the whole of Brewster’s angle: when the geometry puts the reflected direction along the axis the surface charges are shaking on, there is nothing for them to send that way, and the reflection for that polarisation disappears.

So the condition is geometric. The reflection vanishes when the reflected ray is perpendicular to the refracted ray, which is $\theta_i + \theta_t = 90°$, and combining with Snell’s law gives $n_1\sin\theta_i = n_2\sin(90° - \theta_i) = n_2\cos\theta_i$, that is $\tan\theta_i = n_2/n_1$. The same answer as the algebra, arrived at without any boundary conditions.

Away from the angle, and the other way through #

Change the angle and the extinction is spoiled quickly. The degree of polarisation of the reflected beam is $(R_s - R_p)/(R_s + R_p)$, and it is a sharp peak.

That curve is why polarising sunglasses do what they do and no more. Glare from a wet road seen at a shallow angle arrives near the Brewster angle and is nearly completely polarised horizontally, so a vertically transmitting filter removes almost all of it. Glare from a puddle two metres ahead arrives at a much steeper angle and is only partly polarised, and the same glasses remove only part of it.

The direction matters as much as the degree. The reflected field is the component lying in the surface, which for a horizontal surface is horizontal — so the axis of a pair of sunglasses is vertical, and the design works for water and roads and fails for a shop window, whose surface is vertical and whose glare is polarised the wrong way round. That is not a defect in the glasses. It is the plane of incidence being where the surface put it.

Sending the light the other way — from glass into air — changes the picture in a way that is worth drawing, because a second angle appears and takes over.

Notice that $\tan\theta_B = n_2/n_1$ and $\sin\theta_c = n_2/n_1$ share the same ratio, so the Brewster angle is always the smaller of the two when it exists in the denser medium. A ray polarised in the plane of incidence therefore always meets its vanishing angle before it meets the angle at which it cannot escape.

What the surface does with the rest of it #

A reflectance is only half of an accounting, and the other half is worth doing explicitly because it is where the arithmetic can be caught out.

The transmitted fraction is not $1 - R$ by definition; it is computed from the transmission coefficients, and then compared with $1 - R$. The comparison has a subtlety: the transmitted beam travels in a different medium at a different angle, so the power crossing a given area of the boundary involves both the impedance of the second medium and the projection of the beam onto the surface. Putting both in,

$$T_s = \frac{n_2\cos\theta_t}{n_1\cos\theta_i}\,t_s^2, \qquad t_s = \frac{2n_1\cos\theta_i}{n_1\cos\theta_i + n_2\cos\theta_t},$$

and the sum $R_s + T_s$ comes to one at every angle — to about four parts in $10^{16}$, which is double-precision arithmetic and not physics. Getting that projection factor wrong is the standard error, and it produces a transmittance that is wrong by tens of per cent at large angles while looking perfectly reasonable at small ones.

The projection factor in the transmittance comes from geometry rather than from the fields. A beam of a given width in air becomes a beam of a different width in the glass, because the angles differ, so the same power now crosses a different area — and intensity and power are therefore not the same accounting. Snell’s law fixes the two angles, and it is the ratio of their cosines, times the ratio of the indices, that appears in front of $|t|^2$.

At the Brewster angle the $p$-polarised beam is transmitted with no loss at all — $T_p = 1$ exactly, since $R_p = 0$ — which is a stronger statement than it looks. A window admits a p-polarised beam perfectly, with no anti-reflection coating and no thin film, purely by being tilted.

Making light polarised on purpose #

Reflection at the Brewster angle is a polariser that throws away 85% of the light it is given. Transmission through a pile of plates is a polariser that keeps almost all of it and works far less well.

The Brewster surface earns its place elsewhere, where the loss is the point rather than the cost.

Historically it earned something else. Étienne-Louis Malus discovered polarisation by reflection in 1808 by looking at the windows of the Luxembourg Palace through a calcite crystal at sunset and finding that one of the two images vanished as he turned it. He had been working on double refraction and expected a property of the crystal; what he had found was a property of the glass. David Brewster measured the angle for a range of materials over the following decade and found the tangent law empirically — before anybody had a wave theory to derive it from, and long before there was any suggestion that light was made by shaking charges. The geometric explanation given above had to wait for that suggestion, and it is a rare case of a law being known exactly for sixty years while its reason was unavailable in principle.

The Brewster surface earns its keep in one place where the loss is the point rather than the cost. A laser cavity with its windows cut at the Brewster angle transmits one polarisation with no reflection loss at all and taxes the other by a few per cent per pass; over hundreds of passes that difference is enough to make the laser run polarised, and the window contributes nothing to the loss of the surviving mode. It is a polariser built out of the absence of an interface rather than out of an absorber.

The sky, and the photographer’s rule #

The same radiation pattern that empties the reflection at one angle also polarises the sky, and the two are the same argument with the second medium replaced by single molecules.

Sunlight sets a molecule’s electrons shaking along the direction of the field, which is perpendicular to the direction the sunlight came from. Look at that molecule from ninety degrees away and only one of the two possible shaking directions can radiate toward the observer — the other one points straight at them, and a charge sends nothing along the axis it shakes on. So light scattered at a right angle to the Sun is polarised, and light scattered forward or backward is not.

Which gives a rule anybody can check. The sky’s polarisation is strongest along a band ninety degrees from the Sun, running through the zenith when the Sun is on the horizon, and it is zero toward and away from the Sun. A polarising filter turned in front of that band darkens it dramatically and does nothing to the sky near the Sun — which is the whole of what a photographer’s polariser does to a landscape, and why its effect depends on which way the camera is pointing rather than on the filter.

The polarisation reaches about three quarters rather than one, and the shortfall is informative. Some of it is multiple scattering: light that has bounced twice arrives from a direction unrelated to the Sun’s. Some of it is that molecules are not spheres, so the induced dipole is not exactly parallel to the field and the zero is not exact — the same anisotropy correction that spoils the agreement between the sky’s brightness and air’s refractive index.

Several insects navigate by that pattern, reading the direction of polarisation across a patch of sky and recovering the Sun’s position from it when the Sun itself is behind cloud.

Two ways to stop a reflection #

A Brewster window and an anti-reflection coating both deliver a surface that does not reflect, and the comparison between them is the sharpest way to see what each is doing.

A coating works by interference: a quarter-wave layer of an intermediate index sends back two reflections of equal size and opposite phase, and they cancel. That makes it exact at one wavelength and one angle of incidence, and approximate on either side of both; a good multilayer stack extends it across the visible, at the cost of many layers and a design that is a compromise everywhere. It works for both polarisations, and it works at normal incidence, which is where most optics is used.

A Brewster surface works by geometry: there is nothing to radiate in that direction, at any wavelength. So it is exact across the whole spectrum — up to the small movement of the angle with dispersion — and exact for one polarisation only, and only at one angle. It needs no coating, cannot be damaged, and survives any power level the substrate survives.

The two therefore occupy opposite corners. A camera lens, used near normal incidence with unpolarised light across the visible, must be coated and cannot be tilted. A high-power laser cavity, working at one wavelength, at one angle, with a polarisation it is happy to select, uses the tilt and needs no coating at all — which matters because a coating is the part of an optic that fails first under intense light.

The phase the intensity hides #

The note that $r_p$ changes sign at the Brewster angle deserves more than a caveat, because it is the basis of an entire measurement technique.

The amplitude coefficients are signed, and away from normal incidence they are in general complex — for an absorbing medium, or beyond the critical angle, the reflection imposes a phase shift as well as an attenuation. The two polarisations acquire different phase shifts, so light that arrives linearly polarised at some angle to the plane of incidence leaves elliptically polarised, with an ellipse whose shape encodes the ratio of the two complex coefficients.

Measuring that ellipse is ellipsometry, and it is remarkably sensitive: the ratio is a ratio of two quantities measured on the same beam at the same instant, so almost everything that could go wrong — source fluctuations, detector drift, absorption on the way in and out — cancels out of it. What survives is a pair of numbers, at each angle and wavelength, from which the index and the thickness of a surface layer can be recovered.

The sensitivity is what makes it useful. A layer a single molecule thick changes the ellipse detectably, so the technique measures oxide growth, adsorbed films and the thickness of coatings in real time and without touching them. All of it lives in the phase, and every figure in this essay has squared it away.

Where the model stops #

The media are transparent and non-magnetic. The coefficients above assume real refractive indices. A metal has a complex index, both reflectances stay large at every angle, and the $p$ curve has a minimum rather than a zero — the pseudo-Brewster angle — which never reaches the axis. Extinction by reflection off metal is not available.

The surface is smooth and sharp. A boundary that is graded over a fraction of a wavelength, or rough on that scale, scatters rather than reflecting, and the polarisation is spoiled. A wet road works because water is smooth; a dry one does not, and the difference is the same one that separates a mirror from a white wall.

Both media are isotropic. In a birefringent crystal the index depends on polarisation to begin with, and there is no single $n_2$ to put in the formula.

One wavelength at a time. The index depends on colour, so dispersion moves the Brewster angle: for crown glass it shifts by about 0.4° between red and blue. A single angle cannot extinguish the whole spectrum, which is a small effect and a real one — the extinction at a Brewster window is best in the middle of the band and imperfect at its edges.

The beam is a plane wave. A real beam is a bundle of directions of some angular width, and the extinction it achieves is the average of the reflectance over that width. Since $R_p$ is quadratic in the departure from $\theta_B$ — it has a zero of second order, not first — a beam of half-width $\delta$ reflects an average of order $\delta^2$ rather than zero. For a beam converging at $f/8$, that is about one part in ten thousand: excellent, and not nothing, and it puts a floor under every measurement made with a Brewster window that the plane-wave formula says has no floor.

What the pictures cannot show #

Every reflectance drawn is an intensity, and the amplitude coefficients carry signs. Crossing the Brewster angle, $r_p$ passes through zero and changes sign — a phase flip of 180° — which is invisible on a squared quantity and is measurable in an interferometer. The figures plot $R$ and lose it.

The geometry figure draws the charges’ shaking direction as a double arrow at one point on the refracted ray. There are of course charges everywhere in the medium, all shaking, and the reflected wave is their coherent sum; the single arrow stands for all of them and cannot show that the cancellation along the axis is a property of each one separately rather than of the arrangement.

And nothing here shows the refracted beam’s polarisation, which is partially polarised in the opposite sense to the reflected one — necessarily, since the two have to add back up to what arrived.

Where the ladder goes next #

This rung has taken the fact that light has a direction of shaking and asked what a boundary does with it. The step was to stop treating reflection as a thing surfaces do and start treating it as radiation by the charges in the second medium — at which point the answer came from a radiation pattern rather than from a boundary condition.

The rungs above are the ones where the two descriptions stop agreeing so comfortably: reflection from an absorbing medium, where the phase between the two polarisations becomes the measurable quantity and ellipsometry is built on it; reflection from a stack of thin layers, where the interference between many surfaces is the whole design; and scattering by small particles, where the same sin²θ pattern makes the sky polarised at right angles to the sun by exactly the argument used here.

The transferable habit: when an expression gives a clean zero, look for a geometry rather than a cancellation. A quantity that vanishes because two terms happen to be equal usually vanishes only approximately and only at one place; a quantity that vanishes because something is pointing along an axis vanishes exactly, and stays vanished when the numbers change.