Why Is My AR Coating Purple?

A coated lens held up to the light shows a color in reflection — often a faint purple, blue, or green — while what you see through it looks clear. That color is not a coating on top of the coating. It is the reflectance spectrum itself, turned into a color by your eye.

This tutorial takes three designs that are already published here — bare glass,
a two-layer V-coat, and a four-layer broadband AR coating — and asks one question: what color does each one show, in reflection and in transmission?

The conditions

Color is not a property of a coating alone. It depends on the light source and on whose eye is looking, so those have to be fixed before any number means anything. Throughout this article:

Illuminant: CIE standard D65 (average daylight)
Observer: CIE 1931 2° standard observer
Angle of incidence: 0° (normal)
Polarization: unpolarized (s/p average)
Reflected color is computed from one coated front surface (R), the same single-surface convention as the design articles.
Transmitted color is computed from the whole 1 mm plate (T), including both surfaces and the substrate, evaluated incoherently.
Spectra are sampled over 380–780 nm; the color integral uses a 5 nm step.

Change the illuminant or the observer and the numbers move. D65 and the 2° observer are the common default and the one used here.

From a spectrum to a color

The eye has three cone types, so any spectrum collapses to three numbers. The CIE formalizes this with the tristimulus integrals:

X = k\int S(\lambda)R(\lambda)\bar x(\lambda)d\lambda,\quad Y = k\int S(\lambda)R(\lambda)\bar y(\lambda)d\lambda,\quad Z = k\int S(\lambda)R(\lambda)\bar z(\lambda)d\lambda

Here $S(\lambda)$ is the illuminant, $R(\lambda)$ is the coating's reflectance (or transmittance) as a fraction, and $\bar x,\bar y,\bar z$ are the color-matching functions. The constant $k$ is chosen so that $Y = 100$ for a
perfect diffuser, which makes:

$Y$ the luminance factor — how bright the reflection or transmission is, in percent. It is essentially the luminous (eye-weighted) average of the spectrum.
The chromaticity $(x, y)$ , derived from $X, Y, Z$ , the hue and saturation — the color independent of brightness.

Two further numbers make the color easier to name:

Dominant wavelength — the pure spectral color the eye reads the hue as.
Excitation purity — how far the color sits from white, as a percentage. Low purity is near-grey; high purity is vivid.

A design can have a very low $Y$ (a dim reflection) and still have high purity (a saturated hue). That combination is exactly what a strong AR coating produces — and, as the next section shows, a hue can be so dim that it reads as black until something bright is reflected in it.

Bare glass: a neutral grey reflection

Let's check the color characteristics of 1 mm thick BK7 Glass plate. Uncoated BK7 reflects about 4.2% at every visible wavelength.

Because the spectrum is nearly flat, the reflected light keeps the color of the source: a dim, essentially neutral grey.

Quantity	Reflected (one surface)	Transmitted (total)
Luminance factor $Y$	4.24%	91.85%
Chromaticity $x, y$	0.3104, 0.3267	0.3129, 0.3292
CIE $L^a^b^*$	24.441, −0.016, −0.466	96.759, 0.002, 0.116
Dominant wavelength	481.5 nm	583.8 nm
Excitation purity	1.07%	0.12%

The D65 white point is at $x, y = 0.3127, 0.3290$ . Both of bare glass's colors sit almost exactly on it: purity is near zero. Reflection is a dark grey only because $Y$ is low, not because it is colored. Transmission is a bright, essentially colorless white.

The broadband AR coating: a saturated violet, too dark to see

The four-layer coating holds front-surface reflectance below about 0.3% across 450–650 nm, but the residual reflection is not flat. It climbs steeply toward the blue and, less sharply, toward the red:

Wavelength	Coated surface R
400 nm	6.34%
450 nm	0.29%
500 nm	0.08%
550 nm	0.10%
600 nm	0.01%
650 nm	0.26%
700 nm	0.97%

The eye integrates this — but weighted by its own sensitivity, which peaks in the green near 555 nm and is nearly blind in the deep blue. So the 6.34% spike at 400 nm adds almost nothing to the luminance $Y$ ; what it does is set the hue. It pushes the chromaticity far out toward the violet corner of the spectral locus — a dominant wavelength of 449 nm and 73% purity — while the luminance stays near zero.

Quantity	Reflected (one surface)	Transmitted (total, one surface coated)
Luminance factor $Y$	0.085%	95.65%
Chromaticity $x, y$	0.2045, 0.0962	0.3131, 0.3298
CIE $L^a^b^*$	0.768, 4.097, −7.52	98.299, −0.166, 0.368
Dominant wavelength	449.1 nm	571 nm
Excitation purity	73.12%	0.33%

Here is the catch, and it is the whole point of keeping hue and brightness apart. The chromaticity is a vivid violet, but the luminance is only $Y = 0.085\%$ — so the reflected-color swatch above is essentially black. There is almost no reflected light to carry the hue, and against a neutral background the eye sees near-black, not violet. Compared with bare glass the coating is about 50× dimmer in reflection ( $Y$ 0.085% versus 4.24%) yet far more saturated (73% versus 1.1% purity). Making the AR coating better drives the reflected brightness toward zero, and the color you could name goes with it.

So why do real coated lenses so clearly look purple? Because you usually catch them reflecting a bright source — a window, a lamp, the sky. Even 0.1% of a bright light is enough for the eye to read the violet tint, and a real lens reflects from both of its surfaces at once. The hue in the swatch and the purple glint on a lens are the same chromaticity; what changes is how much light sits behind it.

The transmitted color, meanwhile, is a slightly brighter white than bare glass — the coating sends more light through without tinting it.

Reveal the hue: the exposure control

The swatch is black because Color Evaluation scales it against a perfect, 100%-reflecting white — a 0.085% reflection simply has no brightness to show. To see the hue anyway, raise the Exposure control next to the swatch (×10 … ×1000) or choose Fit hue. It rescales the swatch only — the $Y$ , $x, y$ , and $L^*a^*b^*$ readout stays exactly the same — and at full brightness the patch becomes a vivid blue-violet, rgb(131, 33, 255).

Turning exposure up is the on-screen version of the coating catching a brighter source. It is why the same reflection is invisible against a dim wall but glints violet when it reflects a lamp, a window, or the sky — the black swatch and the purple glint on a real lens are the one chromaticity seen at two exposures.

The V-coat: color of a deep single-wavelength minimum

The V-coat drives reflectance to nearly zero at 550 nm and rises on both sides. Its residual reflection is therefore missing its green and strongest at the blue and red ends — the recipe for a magenta or
purple reflection, more saturated than the broadband coating because the minimum is narrower.

Quantity	Reflected (one surface)	Transmitted (total, one surface coated)
Luminance factor $Y$	fill from app	fill from app
Chromaticity $x, y$	fill from app	fill from app
CIE $L^a^b^*$	fill from app	fill from app
Dominant wavelength	fill from app	fill from app
Excitation purity	fill from app	fill from app

Why the averages do not tell you the color

It is tempting to summarize a coating with one number, such as average reflectance. But two of these designs make nearly the same scalar impression and look completely different:

Bare glass and the AR coating both transmit a near-white with luminance in the low-to-mid 90s of a percent. In transmission they are hard to tell apart by color.
In reflection they are opposites in hue: a neutral grey versus a strongly violet-biased reflection — even though the AR coating reflects far less total light, and is in fact so dim it reads as black.

A luminous average throws away where in the spectrum the light sits. Color keeps it. That is why "average R is low" and "the reflection is strongly colored in hue" are both true of the same coating at the same time.

Doing this in TFStudio

The Color Evaluation window takes the same TMM spectrum used everywhere else and applies the integral above.

Open one of the designs.
In Color Evaluation, set Characteristic to R for reflected color or T for

transmitted color. 3. Set Illuminant to D65, Observer to 2°, AOI to 0°, and

Polarization to average. 4. For reflected color use the Front evaluation mode (one coated surface);

for transmitted color use Total (the whole plate). 5. Read the swatch and the $xyY$ , $L^*a^*b^*$ , dominant wavelength, and purity

values from the panel.

The chromaticity diagram plots where the color sits relative to the D65 white point and the spectral locus, which makes the difference between "near white" and "strongly colored" visible at a glance.

What this does and does not tell you

The color is stated for D65 and the 2° observer at normal incidence. Under a different source — an incandescent lamp, a phone screen — or at an angle, the color shifts. Tilt in particular moves the AR minimum and changes the reflected hue, as covered in What Angle and Polarization Do to an AR Coating.
Reflected color here is one surface. A real plate in air also reflects from its back surface, which adds a second, weaker contribution.
A low luminance $Y$ in reflection means little light is reflected, not that the surface is dark to look at; against a bright background the colored reflection is what you notice.
Color says nothing about absorption on its own. These dielectric coatings are essentially lossless in the visible; a colored reflection from an absorbing film would need the same analysis plus an energy balance.

Recap

Color is the reflectance or transmittance spectrum weighted by the eye and a chosen light source; fix the illuminant and observer first.
Bare glass reflects a near-neutral grey (purity ~1%) and transmits a clear white.
The broadband AR coating reflects a strongly violet hue that is so dim (50× less light than bare glass) it reads as near-black; it shows that violet only when reflecting a bright source, and it transmits a clean white.
Color needs both a hue and enough light to see it: a similar luminous average, or a vivid chromaticity at near-zero luminance, can hide what a surface actually looks like.