The layer that makes a reflection vanish

Every window reflects about eight per cent of the light that falls on it, four from each surface, and every one of those reflections is light that did not get through. A camera lens with ten air-glass surfaces would lose a third of its light and fill the image with a haze of stray reflections, and until the 1930s that is exactly what such a lens did. The cure is a film about a tenth of a micrometre thick, and it does not reduce the reflection so much as arrange for it not to happen.

What the step costs, and what it depends on #

A wave meeting a change of medium divides. How much goes back is decided entirely by the ratio of the two impedances, and by nothing else:

$$r = \frac{Z_1 - Z_2}{Z_1 + Z_2}, \qquad R = r^2.$$

Plot the reflected and transmitted amplitudes against the impedance ratio and the one place nothing comes back is where the ratio is one — meaning there is no boundary at all. That is what makes an antireflection coating a genuinely clever object rather than an obvious one: it does not remove a boundary, it adds a second one, and arranges for the two returns to cancel each other.

For light, the impedance of a transparent medium is inversely proportional to its refractive index, so the ratio is a ratio of indices and $R = ((n_1-n_2)/(n_1+n_2))^2$. Glass at 1.52 in air gives 4.3 per cent, which sounds small and is not: ten surfaces at 4.3 per cent transmit $0.957^{10} = 64$ per cent.

Two reflections, arranged to cancel #

Insert a layer of some third medium. There are now two boundaries, and light reflects at both. Those two reflected waves travel back into the first medium together and superpose, and superposition is not a matter of adding intensities.

Two waves of equal amplitude half a cycle apart annihilate, and that is the whole mechanism — not less light coming back, but two returns cancelling. The condition is therefore two conditions: the two reflections must be equal in size, which fixes the index of the coating, and opposite in phase, which fixes its thickness. Neither alone suffices, and a coating that gets one right and the other wrong reflects more than a bare surface.

The phase condition is the thickness. A wave that enters the layer, crosses it, reflects, and crosses back has travelled twice the thickness; making the thickness a quarter of a wavelength in the layer makes that round trip half a wavelength, so the second reflection returns half a cycle behind the first. The amplitude condition is the impedance. The two reflection coefficients are $(Z_0 - Z)/(Z_0 + Z)$ and $(Z - Z_s)/(Z + Z_s)$, and setting them equal gives

$$Z = \sqrt{Z_0 Z_s},$$

the geometric mean. In optics that reads $n = \sqrt{n_0 n_s}$, and for glass in air it is $\sqrt{1.52} = 1.233$.

That the cancellation survives arbitrarily large mismatches is the strongest statement in the subject and it deserves its own sentence. The layer does not soften the step; it splits it into two equal steps whose reflections destroy each other. Two steps of ratio $\sqrt{k}$ in place of one of ratio $k$ — and the geometric mean is precisely the number that makes them equal.

It is worth checking the amplitude condition rather than accepting it, because the check shows what the approximation is. The first reflection has amplitude $r_1 = (n_0 - n)/(n_0 + n)$ and the second $r_2 = (n - n_s)/(n + n_s)$, and setting $r_1 = r_2$ and clearing denominators gives $n^2 = n_0 n_s$ exactly. No approximation has been made. But the net reflected amplitude is not simply $r_1 - r_2$: light reflecting off the second surface can reflect again off the first from inside, and again, and the true answer is the sum of that infinite series. Summing it — which is what the matrix method does in closed form — gives

$$r = \frac{r_1 + r_2 e^{2i\delta}}{1 + r_1 r_2 e^{2i\delta}},$$

and at $\delta = \pi/2$ the exponential is $-1$, so the numerator is $r_1 - r_2$ and vanishes when the two are equal, whatever the denominator does. The multiple reflections change the transmitted phase and cannot resurrect the reflection, which is why the geometric-mean condition is exact rather than a first-order result.

What it costs: bandwidth #

The thickness is a quarter of a wavelength at one wavelength only. Away from it the round-trip phase is not half a cycle, the two reflections no longer cancel completely, and the reflectance climbs.

This is not a defect of the design but the design’s other half: a device that works by interference works over a band, and the narrower the feature the wider the band. A single quarter-wave layer is the shallowest possible interference structure and therefore has the widest useful band — the coated surface in the first figure stays under a quarter of the bare reflectance across most of the visible. Stack more layers to do better at the design wavelength and the band contracts. The way out of that trade is to stop repeating and start tapering, which buys a band with no design wavelength in it at all.

The stop band is the interesting object here, because it is a property of the periodicity rather than of the number of periods. A structure that repeats with a period comparable to the wavelength reflects totally over an interval of wavelengths, and the interval’s width is set by how strong the repetition is. That statement is not about optics. It is the same statement as an electron in a crystal having forbidden bands whose width is set by the strength of the lattice potential, and the two are the same calculation with different names for the impedance.

How wide the band is can be got from the same expression without solving anything. Near the design wavelength write $\delta = (\pi/2)(1 + \epsilon)$ with $\epsilon$ the fractional detuning. Then $e^{2i\delta} \approx -1 + i\pi\epsilon$, the numerator becomes $r_1 - r_2 + i\pi\epsilon r_2$, and with the amplitudes matched the reflectance grows as $\epsilon^2$. So the reflectance rises quadratically from its zero rather than linearly — the cancellation is not fragile, and a coating a few per cent off thickness is still a good coating. That quadratic is why one layer buys such a wide useful band, and it is the reason the purple of a coated lens is a broad residual rather than a narrow line: the reflectance is small across the middle of the visible and rises at both ends together.

A tarnish, noticed and then made deliberately #

The history is unusually short and unusually clear about the difference between observing an effect and understanding it.

Lord Rayleigh recorded in 1886 that old, tarnished glass often transmitted more light than freshly polished glass of the same kind — a result that seems absurd until the tarnish is recognised as a thin surface layer of altered composition and lower index. Harold Dennis Taylor, working on photographic lenses at Cooke, made the same observation in 1892 and went further: he found that deliberately aged lenses performed better and wrote about how to produce the effect on purpose by chemical attack. Neither drew the design condition. Both had noticed a layer, and neither asked what index it ought to have.

The condition arrived when the calculation was done. Alexander Smakula at Zeiss patented evaporated coatings in 1935, and the war made them universal: coated optics gave gunsights and periscopes a visible advantage in dim light, and the technique was classified for the duration. The whole of what separates Taylor’s tarnish from Smakula’s coating is the geometric mean — the difference between a layer that helps and a layer designed so that two amplitudes are equal.

The identical arithmetic, in sound #

Nothing above used any property of light. Replace the refractive index by an acoustic impedance and every line survives.

Every ultrasound probe has such a layer, and the bandwidth trade above is the reason a probe cannot be both efficient and broadband. The same construction appears wherever two impedances have to be joined: the horn on a loudspeaker, the taper on a microwave waveguide, and — in a form so old nobody thinks of it as physics — the flare of a trumpet, which matches the small impedance of a tube to the large impedance of open air over a band rather than at a point.

An impedance is the square root of tension times linear density, which is where the whole idea comes from in its simplest case: two strings of different mass per length under the same tension reflect a pulse at their join, and a third string of intermediate impedance placed between them can cancel that reflection. The optical case is the same arithmetic with the impedance being an index, and the acoustic one is a matching layer on an ultrasound transducer.

How the calculation is actually done #

Two boundaries can be handled by adding two amplitudes. Twenty cannot, because each internal reflection reflects again, and the infinite series has to be summed. The general method replaces the series with a product of matrices, one per layer:

$$M_j = \begin{pmatrix} \cos\delta_j & i\sin\delta_j/\eta_j \\ i\eta_j\sin\delta_j & \cos\delta_j\end{pmatrix}, \qquad \delta_j = \frac{2\pi n_j d_j}{\lambda}.$$

Multiply them in the order the wave meets them, apply the result to the substrate’s impedance, and read off the reflection coefficient. Every figure in this essay is that product, evaluated with different arguments — which is why the single layer, the residual of a real coating, the mirror’s stop band and the ultrasound match are not four calculations but one.

From inside the layer, a quarter of a wavelength is the distance from a node to an antinode — so the coating holds exactly the piece of a standing wave that inverts the phase on a round trip. That is the geometrical way to see the thickness condition, and it explains why it is a quarter and not a half: the wave crosses the layer twice, and two quarters make the half-cycle that produces cancellation.

Why the residual is purple, and what that says #

The colour of a coated lens is worth taking seriously as a measurement, because it is the residual made visible.

A single magnesium-fluoride layer designed for 550 nm is at its best in the green and rises at both ends of the visible, as the first figure shows. What reflects is therefore blue plus red, in roughly equal measure, and blue plus red is purple. The hue is not a property of the material — magnesium fluoride is colourless — but of where the design wavelength was put, and a coating designed for the red end reflects blue-green and looks distinctly blue instead.

Multilayer coatings, which can hold the reflectance low across a wider band, have a fainter and greener residual, and the change in the colour of camera lenses over the last fifty years is a direct record of how many layers the coating has. It is a rare case of an optical specification being legible by eye from across a room.

The same reasoning runs the other way for the stack that maximises reflection. A dielectric mirror designed at 550 nm reflects green totally and transmits the rest, so it looks green in reflection and magenta in transmission — and the two are exactly complementary, because nothing is absorbed and every photon does one or the other. A metal mirror is grey in both, which is the visible signature of absorption, and is the reason a dielectric mirror is used wherever the light being thrown away matters.

Why no material has the index the arithmetic wants #

The design condition for glass in air asks for 1.233, and the coating actually used has 1.38. That gap is the whole difference between a removal and a reduction, and it is not a failure of chemistry that better searching would fix.

Refractive index in a transparent solid tracks two things: how densely the atoms are packed and how easily their electron clouds distort. Both have floors. The least distortable anion available is fluorine, which is why every low-index coating material is a fluoride — magnesium fluoride at 1.38, lithium fluoride at 1.39, cryolite at 1.35 — and why the list stops there. A solid whose index is much below 1.35 is a solid whose atoms are barely interacting, which is to say not much of a solid.

The way past the floor is to stop using a solid. Films deposited as a sponge of silica with half their volume as air behave optically like a mixture, and indices of 1.22 to 1.25 are routinely made that way. They meet the condition exactly and they are useless on a camera lens, because a porous film wipes off, and absorbs water out of the air, and takes its index with it when it does. They are used where the surface never gets touched and the last fraction of a per cent matters — the optics inside high-power laser systems, where a per cent of reflected light is a per cent that comes back down the beam line.

There is one more escape and it is the reason magnesium fluoride is used at all rather than merely tolerated. The required index is the geometric mean, so it rises with the substrate. On a high-index glass of 1.90 the condition asks for $\sqrt{1.90} = 1.378$ — which is magnesium fluoride, to three figures. The standard coating is not a compromise on every glass; it is exact on the dense flints and progressively wrong as the glass gets lighter. That is a pleasant inversion of the usual situation, where the harder problem is the one with the bigger numbers in it.

How a quarter wave is measured while it is being made #

The thickness has to be right to a few nanometres, and the tolerance holds across a whole lens surface inside a vacuum chamber with an evaporating source in it. How that is actually achieved is a nice piece of measurement design and it uses nothing but the physics above.

The obvious approach is to measure the mass deposited, with a quartz crystal whose resonant frequency falls as material lands on it. That works and it measures the wrong quantity: what matters is the optical thickness, index times physical thickness, and the index of an evaporated film depends on how fast it was deposited and how hot the substrate was. A mass monitor has to be calibrated against an assumed density and an assumed index, and both drift.

The better approach measures the thing itself. Shine light of the design wavelength onto the part being coated and watch its reflectance as the film grows. The reflectance is the matrix expression above with the thickness increasing, so it oscillates — falling as the layer approaches a quarter wave, reaching an extremum exactly there, and rising again toward a half wave, where the layer is optically absent and the reflectance returns to the bare value.

The extremum is what makes this work, because an extremum is a place where the derivative vanishes. Close the shutter when the reflectance stops changing and the thickness is right regardless of how fast the material was arriving, how the index came out, or what the starting reflectance was. The measurement is self-correcting in exactly the way a null method is, and for the same reason: nothing has to be calibrated if the answer is read off a turning point rather than off a slope.

Both methods are used together in practice, the mass monitor giving a rate and a rough position and the optical monitor giving the stopping point. And the residual purple of a coated lens is, among other things, a record of how well that shutter was timed.

Where the model stops #

Normal incidence. Everything here assumes the wave arrives perpendicular. At an angle the optical path through the layer lengthens, the design wavelength shifts to the blue, and the two polarisations behave differently — which is why a coated lens photographed at a steep angle looks a different colour and why a coating optimised for a fast lens is designed for a cone of angles rather than for one.

A bare interface behaves quite differently as the angle changes: the two polarisations separate completely, with one of them vanishing at Brewster’s angle. A coating designed at normal incidence inherits all of that, so its performance is a function of angle as well as of wavelength — which is why a coated lens shows colour at a glancing view and why the specification of a coating always names both.

Non-absorbing, non-dispersive layers. The indices used here are real and constant. Real coating materials absorb a little and disperse, so the geometric-mean condition can be met exactly at one wavelength and only approximately at its neighbours even when the thickness is right.

Sharp boundaries. The matrix treats each interface as a step. A gradual transition over many wavelengths reflects almost nothing at all without needing any interference — which is what a moth’s eye does with a forest of sub-wavelength cones, and it is a genuinely different mechanism with a genuinely wider band.

What the pictures cannot show #

The reflectance curves give a fraction of power and say nothing about phase, which is what actually matters when several coated surfaces sit in a row and their residuals can add or cancel.

Nor do they show the light going the other way. Every figure here is drawn for a wave arriving from the low-index side; the reflectance is the same in either direction, but the phases are not, which is what makes a beamsplitter’s two outputs differ.

And none of them shows the layer being made. The thickness is a quarter of a wavelength to within a few per cent over the whole surface of a lens, which is a tolerance of a few nanometres across a hundred millimetres, and that manufacturing fact — not the physics — is why the technique dates from the 1930s rather than the 1830s.

Where the ladder goes next #

Impedance began as the number that decides what a boundary does to a wave, and has become something to be engineered rather than suffered. Two rungs follow directly. One is the multilayer design problem proper — given a target reflectance across a band, find the thicknesses — which is where the subject becomes numerical optimisation and stops being closed form. The other is the resonant cavity: two of these mirrors facing each other, with the transmission peaking sharply at the wavelengths that fit between them, which is the same matrix product with two stacks in it and is how a laser selects its frequency.

The habit worth carrying away is the geometric mean. It turns up here as the impedance that splits a step into two equal steps, and it turns up in the same role wherever a ratio has to be halved rather than a difference — in a gear train’s intermediate ratio, in an optimal transformer’s turns, in the ideal intermediate temperature of a two-stage engine. When something is best done in two equal stages, the intermediate value is the geometric mean and not the arithmetic one, and the reason is always that what compounds is a ratio.