Thin-Film Interference: Soap Bubbles and Lens Coatings

Colours Without Dyes

A soap bubble swirls with bright colours, although soapy water is colourless. A thin layer of oil on a wet road shows rainbow bands. Camera lenses and spectacles look faintly purple or green when you tilt them. None of these colours come from pigments. They come from thin-film interference: light reflecting from the top and bottom surfaces of a very thin layer, and the two reflections adding up or cancelling out depending on the wavelength. This tutorial explains the path difference, the phase change on reflection, the conditions for bright and dark reflections, and applications including anti-reflection coatings, oil films and air wedges, with worked examples. It builds on the tutorial on interference and diffraction of light.

Two Reflections from One Film

When light falls on a thin transparent film (such as a soap film in air), part of it reflects from the top surface. The rest enters the film, and part of that reflects from the bottom surface and comes back out. These two reflected waves overlap and interfere.

If they are in phase (crest meets crest), they reinforce: constructive interference, a bright reflection.
If they are exactly out of phase (crest meets trough), they cancel: destructive interference, a dark (weak) reflection.

Whether they are in phase depends on two things: the extra distance travelled by the second wave, and any phase change on reflection.

Path Difference Inside the Film

For light falling almost straight on (normal incidence) on a film of thickness t, the second wave crosses the film twice, an extra distance of 2t. Inside the film the wavelength is shorter, λ/n, where n is the film's refractive index. So what matters is the optical path difference:

optical path difference = 2nt.

Phase Change on Reflection

The second key fact: when light reflects from a boundary with a medium of higher refractive index (for example, air to glass), it undergoes a phase change of 180° (π radians), equivalent to an extra half wavelength. When it reflects from a boundary with a lower refractive index (for example, glass to air), there is no phase change. This is like a wave on a rope: reflecting from a fixed end it flips over, reflecting from a free end it doesn't.

So you must count the phase changes:

One phase change (a soap film in air: change at the top surface, none at the bottom; or an oil film on water, where n(air) < n(oil) > n(water)): bright when 2nt = (m + ½)λ, dark when 2nt = mλ.
Zero or two phase changes (a coating whose index lies between air and glass, n(air) < n(coating) < n(glass), so both reflections have a phase change): bright when 2nt = mλ, dark when 2nt = (m + ½)λ.

Here m = 0, 1, 2, … and λ is the wavelength in air.

Thin-Film Interference in Soap Bubbles

Example 1. A soap film (n = 1.33) is 400 nm thick. Which visible wavelengths are strongly reflected?

There is one phase change, so bright reflection needs 2nt = (m + ½)λ, giving λ = 2nt/(m + ½) = 1064/(m + ½) nm.

m = 0: 2128 nm (infrared).
m = 1: 709 nm (deep red).
m = 2: 426 nm (violet).
m = 3: 304 nm (ultraviolet).

Destructive reflection needs 2nt = mλ, so λ = 1064/m nm: m = 2 gives 532 nm (green). With red and violet reinforced and green cancelled, this part of the film looks magenta-purple. As the film thins and thickens, different colours are reinforced, which is why a bubble's colours swirl as the liquid drains.

The black film. Just before a soap bubble bursts, the top often turns black. When the film is much thinner than the wavelength, 2nt is almost zero, so the only difference between the two reflections is the 180° phase change. They cancel for every colour, and the film reflects almost nothing.

Oil on Water

Example 2. An oil film (n = 1.45) floats on water (n = 1.33). Where it is 300 nm thick, what colour is strongly reflected?

At the air–oil boundary there is a phase change (1.00 to 1.45, a higher index); at the oil–water boundary there is none (1.45 to 1.33, a lower index). One phase change, so bright needs 2nt = (m + ½)λ: λ = 870/(m + ½) nm. m = 1 gives 580 nm, yellow. Because the oil spreads to different thicknesses, the film shows bands of different colours, and the bands also change with the viewing angle.

Anti-Reflection Coatings

Ordinary glass reflects about 4% of the light at each surface. A camera lens with many glass elements could lose a large fraction of its light and suffer from flare and ghost images. The fix is a thin coating designed to make the two reflections cancel.

Magnesium fluoride (n = 1.38) is a common coating on glass (n ≈ 1.52). Both reflections have a phase change (air to coating, and coating to glass, are both to higher indices), so they cancel when 2nt = (m + ½)λ. The thinnest coating uses m = 0:

t = λ / (4n), a "quarter-wave" coating.

Example 3. Find the minimum thickness of magnesium fluoride to minimise reflection of 550 nm light (the middle of the visible spectrum, where the eye is most sensitive).

t = 550 / (4 × 1.38) ≈ 99.6 nm, about a tenth of a micrometre.

The coating works perfectly only at 550 nm. Red and blue light are reflected a little, which is why coated lenses look faintly purple. Better lenses use several layers of different materials to cancel reflection across the whole visible range. Where does the light go? Energy is conserved: light that is not reflected is transmitted, so a coated lens passes more light.

The same idea is used on solar panels, whose dark blue or black appearance comes from anti-reflection coatings that let more sunlight in, and on spectacles to reduce glare.

Air Wedges

Place two flat glass slides on top of each other, touching along one edge and separated at the other by a thin object such as a hair. The air between them forms a wedge whose thickness increases steadily. Under light of one wavelength you see a series of equally spaced bright and dark bands (fringes) parallel to the line of contact.

Moving from one dark fringe to the next, the air thickness changes by λ/2 (because the path difference 2t changes by λ). If the wedge has length L and the spacer has diameter D, the fringe spacing is

spacing = λL / (2D).

Example 4. Two slides 10 cm long are separated at one end by a wire 0.050 mm thick. With sodium light (λ = 589 nm), find the fringe spacing.

spacing = 589 × 10⁻⁹ × 0.10 / (2 × 0.050 × 10⁻³) = 5.89 × 10⁻⁴ m ≈ 0.59 mm.

Reversing this, by counting fringes you can measure the thickness of a hair or a thin sheet to a fraction of a micrometre. At the line of contact the thickness is zero, and the fringe there is dark because of the single phase change.

A related arrangement, a curved lens resting on a flat plate, gives circular fringes called Newton's rings. Optical workshops use such fringe patterns to check whether lenses and mirrors are shaped accurately: any distortion of the fringes shows an error smaller than a wavelength.

Summary

Thin-film interference comes from reflections at the top and bottom of a film; the optical path difference at normal incidence is 2nt.
Reflection from a higher-index medium adds a 180° phase change; reflection from a lower-index medium does not.
With one phase change, bright reflection needs 2nt = (m + ½)λ; with zero or two, it needs 2nt = mλ.
Anti-reflection coatings are a quarter wavelength thick (in the coating): t = λ/(4n).
Air wedges and Newton's rings let us measure tiny thicknesses and test optical surfaces.