Designing a broadband AR coating, start to finish

A machine-vision integrator sends you this. A protective window sits in front of a camera lens on a factory line. The window is getting cleaned with solvent every shift, so it has to be hard-coated, and the customer wants the transmission loss gone.

Two of those numbers deserve a note before we start. The band and the 0.4 % average are lifted straight from a standard catalog visible AR coating, so they are what the industry actually publishes. The R_max ≤ 1.0 % is ours, not the catalog's, because catalog coatings publish an average and an average alone cannot be toleranced. Pick any worst-case limit you like; the point is that a spec without one is not a spec.

And where does 0 to 20° come from? The window sits ahead of the lens, so it sees the whole field, and on a 16 mm lens with a 2/3-inch sensor that field runs to about 19°. Round up and you get 20°. This is the constraint the catalog part cannot satisfy, because catalog visible AR coatings are specified at 0° and say nothing about what happens at 20°. That gap is the whole reason to design instead of buy.

Superstrate Air, substrate BK7, scan 400 to 800 nm, angle 0°, unpolarized. Five settings, and one of them is a decision.

The scan range is the first decision, and it costs you one sentence of thought. The tool opens at 300 to 1300 nm, and on this material set that range fires the material validity advisory, because our Ta₂O₅ dataset stops at 350 nm and our SiO₂ dataset stops at 1250 nm. Set 400 to 800 and the advisory clears. A coating you cannot compute across your own band is not a design yet.

Run the bare substrate first. N-BK7 has n = 1.518522 at 550 nm, and the bare surface reflects:

A whole 3 mm window with both faces bare transmits 91.8496 % averaged over the band. You are being asked to recover almost 8 points of transmission, and the coating has to earn them on both faces.

The number that sets the difficulty is the ideal single-layer index: the square root of the substrate index, 1.2323. No durable coating material comes close to that. MgF₂, the lowest-index film anyone deposits routinely, sits near 1.38. So a single layer physically cannot reach 0.4 %, and the job is multilayer from the first minute.

One sentence of physics: reflection at a single interface is set by the index mismatch alone, which is why the substrate is fixed before any coating exists. Full treatment: Fresnel Equations.

This is the step that separates a design from a guess, and it is three eliminations, each with a number attached.

Bulk crystal data is not film data. Our built-in TiO₂ is bulk rutile with n = 2.6479 at 550 nm. No TiO₂ film is 2.65: the six thin-film TiO₂ datasets in the catalog that are valid at 550 nm run from 2.1644 to 2.5166, and which one you get depends on whether the film came out anatase, amorphous or something in between. Our built-in SiO₂ is Malitson bulk fused silica at n = 1.459911, while the ion-assisted e-beam SiO₂ film is 1.479093, higher by 0.019. That is not a rounding error, it is a design error waiting to happen.

The deposition method picks the dataset. Since the shop runs ion-assisted e-beam, the honest choice is a low/high pair characterized on that process, in one lab, by one method: main/SiO2/Gao and main/Ta2O5/Gao, both ion-assisted e-beam monolayers reverse-engineered by Gao, Lemarchand and Lequime. Their mutual consistency matters more than either value being the "best" number in the literature.

And the process you have costs you layers. If the shop ran magnetron sputtering instead, the matched pair would be main/SiO2/Lemarchand with main/Nb2O5/Lemarchand, and the high index would be 2.360317 instead of 2.157262. Higher contrast means fewer layers for the same performance. We are staying on the ion-assisted path because that is the shop we have, and the cost of that choice shows up in step 6 as extra layers.

One sentence of physics: a coating's performance comes from the index contrast between its layers, and the contrast you get is a property of the deposition process, not of the chemical formula. Full treatment: Refractive Index and Sellmeier and Docs: optical constants dominate.

Start with the textbook move: a quarter-wave of the low-index material at 550 nm. For SiO₂/Gao that is 92.9624 nm, and it gives R = 3.2601 % at 550 nm, 3.3144 % averaged over the band. You have removed less than a quarter of the reflection, because 1.479 is nowhere near the ideal 1.2323.

Try the classic quarter-half-quarter, L-2H-L, at 92.96 / 127.48 / 92.96 nm. It averages 2.5286 % and peaks at 4.2992 %, which is worse at its worst point than bare glass.

For reference, the best single layer anyone actually deposits, an MgF₂ quarter-wave, reaches 1.3317 % at 550 nm. That is the number in our AR tutorial, and it is still three times your spec.

Hand design stops here. Every stack you can reason your way to by hand is worse than the target. That is the argument for the optimizer.

One sentence of physics: a single layer cancels the two reflections only when its index is the geometric mean of its neighbours, and no durable oxide is that low. Full treatment: How Anti-Reflection Coatings Work.

Set up the merit function the obvious way: target R = 0 at fifty wavelengths across 425 to 675 nm, all weights equal, normal incidence. The merit is the single number the optimizer drives down. Then refine a four-layer stack.

The optimizer is a Pro feature on Photizon. Every result it produced is printed below, and forward simulation is free, so you can reproduce every curve in this guide without an account.

The four-layer result looks great:

Design A (air side first): SiO₂ 82.78 / Ta₂O₅ 104.63 / SiO₂ 23.28 / Ta₂O₅ 14.48 nm.

It passes at normal incidence and fails the worst-case limit at 20°. Not by much, and that is the point: you would never have caught it, because you never asked.

Why does tilting hurt? The optical path through every layer shortens as the beam tilts, so the whole design slides toward the blue. For this design the 0° reflectance minimum sits at 614.56 nm and the 20° minimum at 604.72 nm, a shift of 9.84 nm. You can estimate the direction and the scale of that shift in advance, but not the number, because it is a property of the whole stack and not of any one layer.

One sentence of physics: tilting the beam shortens the optical path through every layer, which blue-shifts the entire design, by an amount set by the stack rather than by any single film. Full treatment: Docs: angle sweep.

The fix is to put the angle into the merit function. Add a second target section at 20°, same fifty wavelengths, same unit weights, and re-run.

Design B: SiO₂ 83.86 / Ta₂O₅ 106.19 / SiO₂ 24.12 / Ta₂O₅ 14.31 nm.

Now it passes. With 0.008 percentage points of margin, which is not margin, it is luck. Step 6 turns that into a real design.

One thing worth knowing about the merit number. When every target is R = 0 with unit weight, the merit function is the RMS reflectance over the target grid, checked here against an independent calculation and agreeing to all 18 digits. So "MF = 3.79 × 10⁻³" means "0.379 % RMS reflectance", and you can set tolerance thresholds in units your customer understands.

Design B has a 14.31 nm layer. Ask a coating house to hold 14.31 nm and they will ask what your tolerance is, because shutter timing, rate stability and the crystal monitor all get worse in relative terms as the layer gets thinner. A practical floor for a production run is 15 nm. That is a process fact, not a numerical convenience.

Impose it and the four-layer design dies. Re-optimize four layers with a 15 nm minimum and the best you can do is R_max = 1.0903 % at 0°. The spec that Design B scraped past is now out of reach with four layers, because Design B was leaning on a layer the shop cannot make.

So add layers. Here is the whole family, every design found by the same search at the same 15 nm floor, with the same two-angle merit:

Two things in that table are worth stopping on.

The merit and the spec do not agree. Five layers has a lower RMS than six, and a higher peak. Optimizing an RMS merit does not optimize a maximum, so the ranking you get from the merit number is not the ranking your customer cares about. Hold that thought; step 7 makes it expensive.

The optimizer is a local refiner, and it gets worse with layers. Two hundred random four-layer starting points land in 87 different local minima. At eight layers, 196 of the 200 land in a minimum of their own. Blind searching stops working, which is why real designs are grown rather than found: take the good design you have, add a pair, re-refine.

And by eight layers the minima stop being distinguishable. Eight of them sit within 0.4 % of each other in merit, and across those eight the worst case at 0° runs from 0.5225 % to 0.6009 %. The eight-layer row above is one member of that valley, computed at the thicknesses this guide publishes; a different run of the same search lands on a different member with a different peak. The merit is an RMS, so it pins the average and leaves the maximum loose. That is not a rounding problem, and step 7 is where it stops being an academic one.

Needle synthesis automates the same idea, and it is worth running. Started from Design B it reaches a six-layer stack at RMS 3.055 × 10⁻³, better than the six-layer search found, but its thinnest layer is 9.08 nm, below the process floor. Force a 10 nm floor and you get seven layers with two of them at 10 to 11 nm. Needle optimizes the physics, not the factory. Read its answer, then re-impose your floor.

One sentence of physics: each added layer is one more interference term available to flatten the curve, with diminishing returns and rising sensitivity, and the optimizer only ever finds the minimum nearest where you started. Full treatment: Thin-Film Interference and Docs: the optimizer is local.

Nominal performance is a fiction. Real runs miss thickness. Ask your coating house what they hold; a common answer for a production oxide process is 1 % of thickness or 0.5 nm, whichever is larger, with run-to-run index repeatability of about ±0.005. Put those numbers into Monte Carlo.

Run it against the merit function first, with a yield threshold of 4.0 × 10⁻³, which by the identity above means "RMS reflectance stays under 0.40 %". Any trial that clears that threshold has its reflectance, averaged over the band and over both angles, below 0.40 %. Be precise about what it does not say. It bounds the two-angle average, not each angle separately, and it is evaluated on the fifty-point merit grid rather than the 1 nm reporting grid.

But the worst-case limit is a different question, and the merit function does not answer it. So ask it directly: put a single target at the wavelength and angle where the design comes closest to failing, and set the yield threshold to 1.0 %. Every design here peaks on a band edge, at 425 nm for 0° and 675 nm for 20°, so there are two candidate points, and you ask both and keep the worse answer.

Look at the five-layer row. It passes the merit-based yield 99.2 % of the time and holds the actual worst-case limit 50.6 % of the time. If you had stopped at the merit number you would have quoted a design that fails incoming inspection on one part in two.

And closest to failing is not the same as highest. The blue edge at 0° wins every row in that table, and it wins for some designs even when the taller nominal peak is at the other edge, because the spread matters as much as the height. The final design in step 9 peaks at 675 nm and 20°, at 0.4360 %, but the run-to-run standard deviation there is only 0.110 percentage points; at 425 nm and 0° the peak is lower, 0.3979 %, and the spread is 0.245 points. That is 5.1 standard deviations of headroom at the tall edge against 2.5 at the short one, so the tall peak holds 1.0 % essentially always and the short one 96.6 % of the time. Rank your candidate points by margin divided by spread, not by height.

The fix is to tell the optimizer what you actually care about. The peaks are at the band edges, so weight the band edges. Give every target outside 450 to 650 nm a weight of 3 and re-refine. Nothing else changes.

Weighting the edges costs a little average reflectance and buys a lot of peak. Six layers with weighted edges is a legitimate answer if cost dominates. Eight layers with weighted edges is the one to quote, because it holds the average limit with 45 % of the allowance unused and the worst-case limit with 56 % unused, at every angle in the cone.

Two more things Monte Carlo tells you. Thickness error dominates: with index error switched off the spread at the binding point (425 nm, 0°) drops only from 0.2453 % to 0.2335 %, so thickness control accounts for about 90 % of the variance and index repeatability for the rest. And the sensitivity ranking names layer 8, the 15 nm Ta₂O₅ layer against the substrate: 9.5 times more sensitive than the next layer, 40 times the least sensitive one. That single line belongs on the RFQ as a monitoring note.

One honest limitation. The confidence band drawn on the R(λ) plot is computed at normal incidence, unpolarized, always, whatever angles your merit function contains. The yield number does use every angle, because it comes from the merit function. So read the yield from the number and read the band as "normal incidence".

One sentence of physics: the merit is smooth in every thickness, so Gaussian thickness errors map to a near-Gaussian merit distribution whose width is set by the gradient, which is why the sensitivity ranking and the yield are the same fact seen twice. Full treatment: Docs: optimization.

You have been designing reflectance of one surface. Nobody measures that. They put a part in a spectrophotometer and read transmission, and the part has two surfaces and 3 mm of glass between them, so the back surface contributes a reflection of its own.

Here is how good coatings get rejected. The spec says T ≥ 99 %. The witness piece is coated on one side, because that is how it came out of the chamber. It reads 95.5 %, the lot fails, and there was never anything wrong with the coating.

MIL-C-48497A anticipated this. §4.1.3(d), verbatim: "Where transmission characteristics are required the witness piece shall simulate the axial thickness of the component by either computational methods or using a witness piece that has the same thickness as the coated component." Say on the RFQ which one you are asking for.

One sentence of physics: a substrate thousands of wavelengths thick has no fixed phase relationship between its two surfaces, so their contributions add as intensities. Full treatment: Docs: coherence.

Two things about those numbers before you quote them.

They already include coating absorption, and it is not evenly spread. Reach for the obvious sanity check — two coated surfaces at R_avg = 0.2068 % should transmit 100 − 2(0.2068) = 99.59 % — and you land 0.085 percentage points above the table. That gap is real and it is Ta₂O₅. Our Ta₂O₅ dataset carries a measured extinction coefficient that climbs toward the blue: k = 2.1 × 10⁻⁵ at 550 nm, 2.29 × 10⁻⁴ at 425 nm, and exactly zero from 640 nm up. So the part absorbs 0.0869 % on average across the band and 0.4544 % at 425 nm, and two-thirds of the whole band's absorption is packed into the 50 nm at the blue edge. At the worst point the linear estimate is off by 0.44 percentage points, which is the entire difference between 99.20 % and 98.76 %. If you sanity-check a coated transmission with 1 − 2R and the answer comes out high, look for absorption before you look for an arithmetic error.

What our model leaves out: bulk internal transmittance of the glass. Our BK7 model carries no absorption at all, so the substrate contributes exactly zero loss above, and real 3 mm N-BK7 loses a little across the visible. A measured part will therefore read slightly under 99.50 %, on top of the coating absorption that is already in the number. That is a limitation of the model, not of the coating, and it is cheaper to say so here than to discover it on a spectrophotometer.

Eight layers, ion-assisted e-beam SiO₂ and Ta₂O₅ on N-BK7, air side first.

Performance, per surface, 425 to 675 nm:

R at 550 nm and normal incidence is 0.2923 %. The whole 3 mm part, both faces coated, transmits 99.5010 % on average.

Layers 3 and 8 sit exactly on the 15 nm floor. That is the optimizer pressing against your process constraint, and it means those two layers are where the design and the factory are arguing. Layer 8 is also, not coincidentally, the most sensitive layer in the tolerance analysis.

Open the final design in the Thin-Film Simulator. Simulating it is free. Saving it to your library is a Plus feature; the spec sheet below is not gated, so you can take the design with you either way.

Everything above exists to produce this. Copy it, fill in the two bracketed fields, send it.

Laser damage. Film stress. Packing density and the humidity shift that comes with it. Thickness uniformity across a 25 mm part in a real chamber. Bulk absorption in the substrate. Every one of those can move a real coating, and none of them is in a transfer-matrix calculation. What that calculation tells you is what the design is worth if the shop hits the constants and the thicknesses you gave them, and step 7 is how you find out what happens when they do not.

• H. A. Macleod, Thin-Film Optical Filters, 5th ed., CRC Press (2017).
• A. V. Tikhonravov, M. K. Trubetskov, G. W. DeBell, "Application of the needle optimization technique to the design of optical coatings," Appl. Opt. 35, 5493 (1996).
• L. Gao, F. Lemarchand, M. Lequime, Opt. Express 20, 15734 (2012); and J. Eur. Opt. Soc. Rapid Publ. 8, 13010 (2013).
• C. C. Katsidis, D. I. Siapkas, "General transfer-matrix method for optical multilayer systems with coherent, partially coherent, and incoherent interference," Appl. Opt. 41, 3978 (2002).
• MIL-C-48497A, Coating, Single or Multilayer, Interference: Durability Requirements For, 8 September 1980 (inactive).
• ISO 9211-3 and ISO 9211-4 (named, not quoted).
• RefractiveIndex.INFO, https://refractiveindex.info