Anti-reflective coating

Why does adding another transparent layer to a lens make it reflect less light rather than more, and why is the leftover reflection purple?

Every boundary between two transparent materials sends part of the light back. A bare lens has one such boundary; coat it and there are two, air to coating and coating to glass. The obvious arithmetic says the coated lens should reflect more. Yet it plainly reflects less, and the little it returns is a faint magenta rather than white. Adding a surface subtracted a reflection. Where has the arithmetic gone wrong?

First, how big is the thing we are trying to remove? At near-normal incidence the reflected fraction at a boundary between refractive indices n1 and n2 is ((n1 - n2)/(n1 + n2))^2. For air and ordinary crown glass, 1.00 and about 1.52, that is (0.52/2.52)^2 = 0.0426 — roughly 4.3% per surface. Elsewhere in this collection that same four per cent is what turns a night window into a mirror; here it is the enemy.

Now the error. Adding "two reflections are more than one" adds intensities. Light is a wave, and two waves must be added as amplitudes, with their relative timing. Two equal amplitudes half a cycle apart sum to nothing. So the real question is not how many reflections there are, but whether they can be arranged to oppose — which needs two conditions at once: the right delay and the right sizes.

Take the delay. Let the coating have index n_c and thickness t. The second reflection makes an extra round trip inside the film, an optical path of 2 n_c t. We want that to be half a wavelength — unless something else has already shifted the phase. Something can: reflection off a higher index inverts the wave, off a lower index it does not. Air to coating is low-to-high, so the first copy flips. Coating to glass, with the coating index sitting between air's and glass's, is also low-to-high, so the second copy flips too. Both are inverted, the inversions cancel each other, and only the path is left to do the work:

2 n_c t = lambda/2, so t = lambda / (4 n_c) — the quarter-wave thickness. For n_c = 1.38 and green light at 550 nm that is 550/(4 x 1.38), close to 100 nm.

That is worth contrasting with a draining soap film, where the back reflection is high-to-low and does not flip, so the two inversions do not cancel and a vanishingly thin film goes black. The coating is the opposite case, which is exactly why its rule is a quarter wave and not zero.

Now the sizes, which is where the design really lives. Cancellation is only complete if the two amplitudes match:

(n_c - 1)/(n_c + 1) = (n_g - n_c)/(n_g + n_c)

Cross-multiply. The left side becomes n_c n_g + n_c^2 - n_g - n_c; the right becomes n_g n_c + n_g - n_c^2 - n_c. The n_c n_g and n_c terms cancel from both sides, leaving n_c^2 - n_g = n_g - n_c^2, so n_c = sqrt(n_g). For n_g = 1.52 the ideal coating index is about 1.233. Nothing was assumed to get that; it fell out. And notice what it says: the coating must sit geometrically between air and glass, not halfway.

The trade-off is the interesting part. No durable solid has an index near 1.233. The workhorse single-layer coating is magnesium fluoride, index about 1.38 — a recalled material constant, and the closest hard, adherent material available. So the sizes do not match, and cancellation is incomplete. Work it out: the first amplitude is 0.38/2.38 = 0.160, the second is (1.52 - 1.38)/(1.52 + 1.38) = 0.14/2.90 = 0.048. Opposed, they leave 0.111, and squaring gives about 1.2%. One layer takes 4.3% down to a little over 1% — a factor of three, from a film a tenth of a micron thick.

And the purple falls out of the same arithmetic. A quarter-wave thickness is exactly a quarter wave for one wavelength only, and designers centre it near 550 nm where the eye is most sensitive. Computing the residual across the visible band for a 100 nm magnesium fluoride film gives about 2.2% at 400 nm, 1.26% at its 550 nm minimum, and 1.6% at 700 nm. What comes back is therefore enriched at both the blue and the red ends and starved in the middle — and blue plus red, with the green removed, is magenta. The colour is not a dye. It is the shape of the residual.

Two honest limits. This is the single-layer story; real camera and spectacle coatings are stacks of several layers of alternating index, which flatten the dip across the whole visible band and can push reflection below a tenth of a per cent, at the cost of more delicate manufacture. And the tidy two-reflection sum ignores light bouncing repeatedly inside the film; the numbers quoted above come from the full expression including those, which is why the minimum is 1.26% rather than exactly 1.24%.

• How multilayer stacks broaden the dip, and why the same trick run in reverse makes a mirror that reflects almost everything.

Every term the collection defines is gathered in the glossary.