Thin-film interference

Thin-film interference occurs when light reflects from both surfaces of a transparent layer a few hundred nanometers to a few micrometers thick. The two reflected waves overlap and interfere: where they arrive in phase the film reflects strongly, and where they arrive half a wave apart it reflects weakly. Because the phase difference depends on the film's thickness, refractive index and the wavelength, a film of nonuniform thickness shows colored bands in white light, as on a soap bubble or an oil slick, and a film of controlled thickness acts as a coating: a quarter-wave layer of magnesium fluoride, 99.6 nm thick, lowers the reflectance of glass at 550 nm from 4.3% to 1.3%.

Phase difference and reflection phase

For a film of index $n$ and thickness $d$, the wave reflected from the lower surface travels an extra optical path of $2nd\cos\theta_t$, where $\theta_t$ is the angle of refraction inside the film. The phase difference between the two reflections is

$$\delta = \frac{2\pi}{\lambda}\,2nd\cos\theta_t + \Delta\phi_r,$$

where $\Delta\phi_r$ accounts for reflection phases. By the Fresnel equations, reflection from a higher-index medium adds a phase of $\pi$ and reflection from a lower-index medium adds none. A soap film in air ($n \approx 1.33$) has one $\pi$ shift between its two reflections, so reflection maxima occur at $2nd\cos\theta_t = (m + \tfrac{1}{2})\lambda$; a magnesium fluoride layer on glass has a $\pi$ shift at both surfaces, which cancel, so its minima occur at $2nd = (m + \tfrac{1}{2})\lambda$.

The film must be thinner than the coherence length of the light for the reflections to interfere. White light has a coherence length of the order of a micrometer, which is why interference colors appear only in films up to roughly a micrometer thick; a millimeter-thick glass window shows fringes only with narrow-linewidth light such as a laser, where it acts as an etalon.

Quarter-wave anti-reflection layer

A single layer cancels reflection completely at the design wavelength when it is a quarter wave thick, $d = \lambda/(4n_c)$, and its index is $n_c = \sqrt{n_s}$ on a substrate of index $n_s$ in air. With a non-ideal index the residual reflectance at normal incidence is

$$R = \left(\frac{n_s - n_c^2}{n_s + n_c^2}\right)^2.$$

For MgF₂ ($n_c$ = 1.38) on BK7 ($n_s$ = 1.52) at 550 nm, $d$ = 99.6 nm and $R$ = 1.3%, against 4.3% for bare glass; the ideal index would be 1.23. The anti-reflection coating entry covers multilayer designs. At oblique incidence $\cos\theta_t$ shortens the effective path: at 30° in air, $\cos\theta_t$ = 0.932 inside MgF₂, so the reflectance minimum moves from 550 nm to 513 nm. Every thin-film component shifts toward shorter wavelengths with angle in this way.

Soap films, oil slicks and Newton's rings

A soap film draining under gravity thins from the bottom up, producing horizontal color bands. The thinnest film that reflects 550 nm strongly is $\lambda/(4n)$ = 103 nm thick. As the film thins further its reflectance falls at all wavelengths, and below about 30 nm the two reflections, separated by the single $\pi$ shift, nearly cancel across the visible, so the film looks black just before it breaks. An oil layer on water behaves the same way when its index exceeds that of water.

Newton's rings are thin-film fringes in the wedge of air between a convex lens and a flat plate. In reflection the center is dark, and the $m$th dark ring has radius $r_m = \sqrt{m\lambda R}$: for a surface of radius $R$ = 1 m in 589 nm light, 0.77 mm for the first ring and 1.09 mm for the second. The pattern is a classical test of surface shape.

Multilayer stacks

Stacking alternating high- and low-index quarter-wave layers makes all the reflections add in phase, giving a dielectric mirror or distributed Bragg reflector that reflects strongly within a stop band: a TiO₂/SiO₂ mirror designed for 550 nm reflects from about 480 to 644 nm. Combinations of such stacks with spacer layers give bandpass filters, edge filters and dichroic mirrors. Their reflectance at any wavelength, angle and polarization is computed with the characteristic matrix method.

Measuring film thickness

The reflectance spectrum of a film several wavelengths thick shows fringes. For adjacent maxima at $\lambda_1 < \lambda_2$ at normal incidence, neglecting dispersion,

$$d = \frac{\lambda_1\lambda_2}{2n(\lambda_2 - \lambda_1)}.$$

A silicon dioxide film ($n$ = 1.46) with adjacent maxima at 520 and 560 nm is 2.49 µm thick. Reflection phases cancel in this difference, but dispersion of $n$ does not, and a silica film thinner than about 0.32 µm shows less than one full fringe between 400 and 700 nm, so its thickness must be found by fitting a model spectrum. A reflection from the back of a transparent substrate adds an incoherent background that reduces fringe contrast.

Common questions

Why do soap bubbles show colors?

Each thickness reinforces some wavelengths and suppresses others, so white light reflected from a film a few hundred nanometers thick is colored. The color changes as the film drains and thins.

Why does a very thin film look black?

When the film is much thinner than a quarter wavelength, the two reflections differ in phase by almost exactly the $\pi$ reflection shift and cancel for every visible wavelength.

How is thin-film interference related to a Fabry-Perot etalon?

Both are the same multiple-reflection interference in a layer with two surfaces. The term thin-film interference is used for layers thinner than the coherence length of white light; an etalon is usually thick and highly reflective, giving narrow transmission peaks.

References: E. Hecht, Optics, 5th ed. (Pearson, 2017); M. Born and E. Wolf, Principles of Optics, 7th ed. (Cambridge University Press, 1999).