The colors of a soap film

The reflected spectrum is a cosine

Consider light of wavelength $\lambda$ that travels normal to the film, perpendicular to its two surfaces. Two beams return toward the eye. One is reflected by the front surface. The other crosses the film, is reflected by the back surface, and crosses the film again.

Light is an electromagnetic wave: an electric field and a magnetic field that oscillate together as the wave travels. The amplitude of a beam is the size of its electric field’s oscillation, and the brightness of the beam is proportional to the square of its amplitude. In the figure, each beam is labeled with its amplitude as a multiple of the incoming beam’s amplitude.

All the beams in the figure travel along the same line, normal to the film. They are drawn side by side so that each one can be labeled. The right half shows the two returning waves at one instant. The back-surface wave is the front-surface wave turned upside down, and it trails the front-surface wave by a distance $\tau$, derived below. The light that reaches the eye is the sum of the two waves.

How much each surface reflects. When light travels normal to a surface, its electric and magnetic fields both lie parallel to the surface. In a medium of refractive index $n$, the amplitude of the magnetic field is proportional to $n$ times the amplitude of the electric field. Suppose a beam of amplitude $1$ crosses a boundary from a medium of index $n_1$ into a medium of index $n_2$. It splits into a reflected beam of amplitude $r$ and a transmitted beam of amplitude $t$. Maxwell’s equations require each field to take the same value on both sides of the boundary. For the electric field, this condition reads $1 + r = t$. The reflected beam travels in the opposite direction, so its magnetic field points the opposite way, and the condition for the magnetic field reads $n_1(1 - r) = n_2\,t$. Solving these two equations gives the Fresnel coefficients for normal incidence,

$r = \frac{n_1 - n_2}{n_1 + n_2}, \qquad t = \frac{2n_1}{n_1 + n_2}.$

In the soap film, the index is $1$ for air and $n$ for water. At the front surface, light passes from air into water, so $r = (1-n)/(1+n)$ and $t = 2/(1+n)$. For $n = 1.33$, $r = -0.142$. At the back surface, light passes from water into air. The two indices trade places, and the reflection coefficient becomes $(n-1)/(n+1) = -r$. The two reflections therefore have opposite signs. On its way out, the back-surface beam passes through the front surface from water into air, with transmission coefficient $t' = 2n/(1+n)$. This beam is transmitted, reflected and transmitted again, so it returns with amplitude

$t\,(-r)\,t' \;=\; -r\cdot\frac{4n}{(1+n)^2} \;=\; -r\,(1 - r^2).$

The last step uses $(1+n)^2 - (1-n)^2 = 4n$. For water, $1 - r^2 = 0.98$. From here on I approximate $0.98 \approx 1$, which changes the back-surface beam’s amplitude by $2\%$, so the two returning beams have amplitudes $r$ and $-r$.

How far the second beam trails the first. The back-surface beam travels an extra distance $2d$ inside the water, where the wavelength shortens to $\lambda/n$. It therefore falls behind the front-surface beam by $2d/(\lambda/n) = \tau/\lambda$ wavelengths, where

$\tau = 2nd$

is the optical path difference between the two beams. In air, where the wavelength is $\lambda$, this is a distance $\tau$, as in the figure. Each wavelength of delay is one full cycle of the oscillation, so the back-surface wave is shifted in its cycle by the angle $\varphi = 2\pi\tau/\lambda$. This angle is called the phase difference.

Two waves with different phases are added by writing each one as a complex number whose magnitude is the amplitude and whose angle is the phase, as in the Euler’s identity post. A shift by the angle $\varphi$ multiplies the complex number by $e^{i\varphi}$. The light reflected by the film therefore has amplitude

$r - r\,e^{i\varphi} \;=\; r\,\big(1 - e^{i\varphi}\big), \qquad \varphi = \frac{2\pi\tau}{\lambda}.$

The reflectance $R$ is the fraction of the incoming light that the film reflects. Brightness is proportional to the square of the amplitude, so $R$ is the squared magnitude of this complex number. Since $|1 - e^{i\varphi}|^2 = (1 - \cos\varphi)^2 + \sin^2\varphi = 2 - 2\cos\varphi$,

$R(\lambda) \;=\; 2r^2\Big(1 - \cos\frac{2\pi\tau}{\lambda}\Big) \;=\; 4r^2 \sin^2!\Big(\frac{\pi\tau}{\lambda}\Big).$

Taking $n = 1.33$ constant across the visible band, $r^2 = 0.0201$, so $R$ lies between $0$ and $4r^2 = 8.0\%$. In the wavenumber $\sigma = 1/\lambda$, $R = 2r^2(1 - \cos 2\pi\sigma\tau)$ is a single cosine with period $1/\tau$. Light can also bounce back and forth inside the film several times before it leaves. Summing all of these reflections gives the exact Airy formula (Born and Wolf, §7.6). Its higher harmonics $\cos(2\pi k\sigma\tau)$, for integers $k \ge 2$, shrink by a factor $r^2 \approx 0.02$ from each $k$ to the next, and the two-beam formula stays within about $4\%$ of it.

Three numbers from a spectrum

For the CIE 1931 standard observer, light with spectral power $P(\lambda)$ is described by three tristimulus values

$X = \int P\,\bar x\,d\lambda, \qquad Y = \int P\,\bar y\,d\lambda, \qquad Z = \int P\,\bar z\,d\lambda,$

where $\bar x, \bar y, \bar z$ are the tabulated color-matching functions. Two lights with the same $(X, Y, Z)$ look identical to the standard observer, and since $\bar y$ is the eye’s sensitivity to brightness, $Y$ is the luminance. Light reflected by the film has $P = SR$ for an illuminant with spectrum $S$. I use D65, the CIE’s standard representation of average daylight (Newton viewed his bubbles “by Reflexion of the Skies when of a white Colour”), normalized so that a perfect reflector has the white point $W = (0.950,\ 1,\ 1.089)$. The sRGB standard specifies the $3 \times 3$ matrix and the encoding curve that take $(X, Y, Z)$ to the color a screen displays.

The color is a cosine transform

Substitute $R$ into the three integrals and change variables to $\sigma = 1/\lambda$. Define three weight curves on the wavenumber axis by

$w_x(\sigma)\,d\sigma \;=\; \frac{S(\lambda)\,\bar x(\lambda)\,d\lambda}{\int S\,\bar y\,d\lambda},$

and $w_y$, $w_z$ likewise, so that their integrals are the three components of $W$. With $w = (w_x, w_y, w_z)$, the tristimulus values of the film are

$\begin{pmatrix} X \ Y \ Z \end{pmatrix}!(d) \;=\; 2r^2\big(W - \hat c(\tau)\big), \qquad \hat c(\tau) = \int w(\sigma)\cos(2\pi\sigma\tau)\,d\sigma, \qquad \tau = 2nd.$

The vector $\hat c(\tau)$ is the cosine transform (the real part of the Fourier transform) of the three weight curves, evaluated at the path difference.

The black film. At $d = 0$ the cosine equals $1$ at every wavelength, so $\hat c(0) = W$ and the film reflects nothing: the two reflections cancel exactly. For small $d$, $R(\lambda) \approx 4r^2(\pi\tau/\lambda)^2$ is proportional to $1/\lambda^2$, so the first light reflected by a thin film is weighted toward short wavelengths. Its chromaticity, in the CIE coordinates $(x, y) = (X, Y)/(X + Y + Z)$, is $(0.266, 0.281)$, bluer than the white point at $(0.313, 0.329)$. This bluish gray is the faint blue that Newton lists right after black. At $d = 10$ nm the film reflects $0.18\%$ of the light, less than a tenth of the $2\%$ reflected by a single water surface but not zero, and Newton could see “the Images of some things (as of a Candle or the Sun) very faintly reflected” from the black spot.

The orders. The luminance $Y = 2r^2\big(1 - \hat c_y(\tau)\big)$ is smallest when the wavelengths at the center of $w_y$ are extinguished, that is, when $\tau$ is a multiple of $\bar\lambda = 554$ nm, the reciprocal of the mean wavenumber of $w_y$. The computed minima are at $d = 207$, $416$, $626$ and $837$ nm, spaced by $\bar\lambda/2n = 208$ nm. With the middle of the spectrum removed, red and blue light remain, and these purple bands separate the orders: the second and third orders run through blue, green, yellow and red, and from the fourth order on, greens and pinks alternate and grow paler.

The widget evaluates the formula from the CIE tables at $5$ nm steps: the reflectance with the products $S\bar x$, $S\bar y$, $S\bar z$ dashed, the chromaticity diagram with the film’s path up to the current $d$, and the color for every $d$ up to $2\ \mu$m.

film thickness *d*

*d* = 250 nm2nd = 665 nmcolor

Newton: 2nd-order indigo, 244 nm

reflectance *R*(λ) weights *S*x̄, *S*ȳ, *S*z̄ Newton's table (water)

Colors are drawn at 1/(4*r*²) times the true reflectance, so the brightest film shows as full white. The most saturated blues and greens lie outside sRGB and are clipped.

In the first order the peak of $R$, at $\lambda = 2\tau$, moves from the ultraviolet into the visible band. At $d = 104$ nm it sits at $553$ nm, $R$ stays above half its maximum across the band, and the film is white. The zero of $R$ at $\lambda = \tau$ enters the band at $d = 143$ nm, which turns the film yellow and then red, and reaches the green at $d = 208$ nm. The chromaticity path goes once around the white point in each of the first three orders. After that, it swings between the green and pink sides of the white point, at a distance that shrinks in every order after the second.

The colors fade to white

As $d$ grows, $\hat c(\tau) \to 0$ by the Riemann–Lebesgue lemma (the Fourier transform of an integrable function tends to zero at infinity), so a thick film reflects the illuminant’s own color at $2r^2 \approx 4\%$, the reflectance of two water surfaces whose reflections add without interfering. The cosine in $R$ has period $1/\tau$ in $\sigma$, and the visible band holds $\tau\,(1/380 - 1/780)\ \text{nm}^{-1}$ of its periods: $1.8$ at $d = 0.5\ \mu$m and $5.4$ at $d = 1.5\ \mu$m. A weight curve that spans several periods averages the cosine to nearly zero.

The widths of the weight curves set the rate of decay. Replacing $w_y$ by a Gaussian with the same mean $\bar\sigma = 1/\bar\lambda$ and standard deviation $s$ gives

$\hat c_y(\tau) \;\approx\; \cos(2\pi\bar\sigma\tau)\; e^{-2\pi^2 s^2 \tau^2},$

the oscillation that produces the orders, under a Gaussian envelope. For the tabulated D65 and $\bar y$, $s = 0.136\ \mu\text{m}^{-1}$, and the envelope falls to $1/e$ at $\tau = 1/(\sqrt{2}\,\pi s) = 1.65\ \mu$m, a thickness of $0.62\ \mu$m. The modulus of the exact transform of the tabulated curve stays within $0.032$ of this envelope up to $d = 1.2\ \mu$m. The transform of a narrower weight curve would decay more slowly. This is the width trade-off of the Fourier transform from the Heisenberg post. Computed from the formula, the chromaticity stays within $0.03$ of the white point for every $d \ge 1.25\ \mu$m and within $0.01$ for every $d \ge 1.61\ \mu$m.

Newton’s table

Newton measured the rings of color in the thin film of air between two lenses pressed together, and compiled a table of the thickness “at which each Colour is most intense and specifick”, in millionths of an inch, for films of air, water and glass (Book II, Part II). Selected entries of his water column, converted at $25.4$ nm per millionth of an inch, next to the computed color at that thickness:

The brightest computed film, at $104$ nm, is Newton’s white of the first order at $98$ nm. His violet of the second order, at $213$ nm, sits next to the first luminance minimum at $207$ nm, and his last entry, the ruddy white of the seventh order at $1467$ nm, comes out as a pinkish white near the white point.

Newton also viewed the bubble obliquely (Observation 19). At angle $\theta$ to the normal inside the film the path difference is $2nd\cos\theta$, so a color seen at thickness $d$ at normal incidence appears where the film is $d/\cos\theta$ thick. Snell’s law gives $\sin\theta \le 1/n$, so the largest factor, at grazing incidence, is $n/\sqrt{n^2 - 1} = 1.52$ for water. Newton’s ratio for his most oblique view: “By the best of my Observations it was between 15 and 15-1/2 to 10.”

The Michel-Lévy chart

A plate of a birefringent crystal between crossed polarizers, at $45°$ to them, transmits the fraction $\sin^2(\pi\Gamma/\lambda)$ of the polarized light, where the retardation $\Gamma$ is the optical path difference between the plate’s two polarizations. That is $R/4r^2$ with $\Gamma = \tau$, so the strip in the widget is also the interference color chart that mineralogists use to identify minerals in thin section, with $2nd$ as the retardation (for the same illuminant, and with the birefringence taken constant across the band). Sørensen (2013) recomputed that chart from first principles.

References

• Isaac Newton. Opticks: or, a Treatise of the Reflections, Refractions, Inflections and Colours of Light. London, 1704. Fourth edition, 1730. Book II, Part I, Observations 17–19 on soap bubbles, and Part II, the table of thicknesses of colored plates.
• Max Born and Emil Wolf. Principles of Optics, 7th edition. Cambridge University Press, 1999. Chapter 7: two-beam interference in a plane-parallel plate (§7.5) and multiple-beam interference (§7.6).
• CIE. Colour-matching functions of CIE 1931 standard colorimetric observer. Dataset, 2019, from ISO/CIE 11664-1:2019.
• CIE. CIE standard illuminant D65. Dataset, 2019, from ISO/CIE 11664-2:2022.
• IEC 61966-2-1:1999. Multimedia systems and equipment – Colour measurement and management – Part 2-1: Colour management – Default RGB colour space – sRGB.
• Bjørn Eske Sørensen. A revised Michel-Lévy interference colour chart based on first-principles calculations. European Journal of Mineralogy, 25(1):5–10, 2013.

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Originally published at aseemrb.me.