Thin-Film Interference

A soap film drains, thin at the top and thick at the bottom, against a black card. Every colour you see is computed from the light the film reflects. Beside it is the ghost: the same two surfaces added as ray optics adds them, with no interference. Click the film, or type a thickness, to see the spectrum it reflects.

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Lesson

The theory — Thin-Film Interference

Thin-film interference: light reflected from the two surfaces of a layer comes back as two waves, the second delayed by δ = 4πn1d/λ for a layer of index n1 and thickness d at normal incidence. Each surface reflects an amplitude r = (na − nb)/(na + nb), negative where light goes into a denser material. Adding the waves, with every bounce inside the film, gives the reflectance R = (r01² + r1s² + 2r01r1s cos δ)/(1 + r01²r1s² + 2r01r1s cos δ).

What each symbol means

d
the film's thickness, in nm
n_1
the film's refractive index at the wavelength in question; water is 1.3347 at 550 nm
δ
how far the light from the bottom surface falls behind, in radians: 2π is one whole wavelength
Y
the luminous reflectance in white light: the reflected spectrum weighted by the eye's sensitivity, 100% for a perfect white reflector
Assumes
Light arriving head-on, a film of even thickness over the patch being looked at, and no absorption. A soap film is treated as water; the soap layers on each face are not modelled. White light is CIE illuminant D65.

The thinnest soap film is the darkest 🖖

Press About to burst. The top of the film is now 10 nm of water, and it is black. One water surface on its own reflects 2.055% of 550 nm light, and ray optics, adding the film's two surfaces, expects 4.027%. The film reflects 0.197%, under a tenth of a single surface. The two reflections are almost the same size, but the one from the top, where light goes from air into water, comes back turned over (r = −0.1434), and the one from the bottom, water into air, does not (+0.1434). In a film much thinner than the wavelength they return almost in step and cancel. Thicken the film and the light from the bottom falls behind by δ = 4πn1d/λ; at a quarter wave, 103.0 nm, the two add and the film reflects 7.892%, nearly twice the ghost. Just before black the film is a dim blue-grey, because the shortest wavelengths, with the largest δ, are the last to cancel. Newton watched it happen on a soap bubble under a glass cover: a black spot grew at the top where, he wrote, "At first I thought there had been no Light reflected from the Water in that place", and then, looking closer, saw a candle faintly reflected in it.

A thick film is white again, and still interferes 🖖

Press A 3 µm film. At 550 nm this film reflects 7.635%, almost the most any thickness can. Yet in white light it reflects 4.013%, against 4.025% for the ghost, and it looks the same grey. The spectrum shows why: 11 reflection peaks between 380 and 780 nm, so each of the three colour-matching functions the page integrates spans several bright fringes and dark ones, and the sum comes out as the ghost. Now press Green cancelled. At 206 nm there is a single trough in the visible, centred on 550 nm, which removes green and leaves a violet 0.256 from white. The most colourful film anywhere on the scale is within a few nanometres of this one. Colour needs a gap in the spectrum as wide as a colour, which only a thin film makes, and from about 2 µm up to the 5 µm the page reaches, the film never again comes more than 0.005 from white, against 0.256 for this violet. The second book of Newton's Opticks opens with it, as something others had already seen: thin plates show colours "according to their various thinness, altho' at a greater thickness they appear very clear and colourless". Ray optics gets a thick film right only because white light averages the fringes away. Light of a single wavelength never does.

A coating twice as thick undoes itself 🖖

Press Coated lens. 100 nm of magnesium fluoride on crown glass cuts the reflection at 550 nm from 4.239% to 1.247%. Some of that is the gentler steps in index: adding the two surfaces as ray optics does gives 2.754%. The rest is cancellation. Both reflections turn the wave over, since each goes into a denser material, and a quarter-wave layer delays the second by half a turn (δ = 0.5013 turn), so they arrive out of step. Now press Coating twice as thick. At 200 nm the delay is a whole turn, the reflections are back in step, and the face reflects 4.239% at 550 nm, the same as bare glass; in white light it reflects 4.082%, against 4.24% bare. More coating is no better; the right thickness is. Because a quarter wave fits only one wavelength, the coated face reflects 1.634% at 450 nm and 1.418% at 650 nm, more at both ends of the spectrum than in the middle, which is why the coated face in the picture is faintly violet. Full cancellation needs a layer of index √1.5185 = 1.2323: type it as the film index, set d to the quarter wave the results give, 111.6 nm, and the 550 nm reflection falls to 0.000%. Carl Zeiss applied on 1 November 1935 for a patent on a layer evaporated onto the glass in a vacuum. It names as its forerunner a British patent of Taylor's from 1904, prompted by old lenses whose weathered patches looked more transparent than the polished glass beside them.

Problem solved in full

  1. How many glass surfaces halve the light? 3 steps

    A camera lens has many glass surfaces in a row. How many uncoated N-BK7 surfaces does it take for half the light to be lost to reflection, and how many coated ones? Count only the light that passes every surface without being reflected. Press Coated lens: the page gives the bare surface's 550 nm reflectance and the coated one's.

    Open this problem in the tool

    1. Each surface passes 1 − R of what reaches it, so k surfaces pass (1 − R)k. Set that to one half:

    2. For bare glass, with R = 4.239%:

    3. For the coated surface, with R = 1.247%:

    Answer

    Seventeen bare surfaces, or 56 coated ones. The working gives k = 16.00 and 55.24, and surfaces come whole: sixteen bare ones still pass 50.01%, so it takes nine uncoated lens elements in air, not eight, to lose half of the 550 nm light. Ten surfaces pass 64.8% bare against 88.2% coated. The light that is lost does not vanish: it bounces between the surfaces and some of it reaches the picture as haze. Zeiss's 1935 patent gives that fogging of the negative as its second reason for the coating.

Learning path

Wave optics

References (3)