Thin-film interference

Light striking a film partly reflects at the front surface; the rest enters, reflects at the back, and comes back out alongside the first beam. The two are copies of one wavefront, divided by amplitude rather than by a pair of slits, so they hold a fixed phase relation and interfere whatever the source — sunlight on a bubble does it as well as a laser. They emerge a few hundred nanometres apart, far below what the eye resolves, so they arrive as a single beam. At normal incidence the second wave simply crosses the film twice: an extra 2t of geometry. Off normal the crossing is longer, 2t/cos θₜ, but the first beam also gets a head start along the surface, and the two effects combine to leave an optical path difference of 2 n t cos θₜ, with θₜ fixed by Snell's law, sin θᵢ = n sin θₜ. Tilt a bubble and cos θₜ falls, so every bright wavelength shifts towards the blue — which is why the colours run as you turn your head.

Inside the film the wave slows to c/n while its frequency is unchanged, so its wavelength shrinks to λₙ = λ/n. Green light of vacuum wavelength 550 nm is only 550/1.33 = 414 nm long inside a soap film, and 550/1.38 = 399 nm inside magnesium fluoride. Interference does not care about distance as such; it cares how many wavelengths fit into the extra trip. Multiplying the geometric distance by n converts it into the equivalent distance in vacuum, which is what optical path means. Crossing a film of index 1.33 and thickness 300 nm twice is 600 nm of geometry but 2nt = 2 × 1.33 × 300 nm = 798 nm of optical path, and 798/550 = 1.45 wavelengths of green. Writing the condition with 2nt on one side and a vacuum λ on the other keeps every length in the same currency, and it explains why making the film thicker and making it denser do the same thing to the colour.

Not all of the phase difference is bought with distance. A wave reflecting from a medium of higher refractive index — optically denser, slower — comes back inverted, a phase change of π, worth half a wavelength of extra path. Reflecting from a lower index gives no shift at all. The mechanical analogue is exact: a pulse on a light string meeting a heavy one returns upside down, while a pulse meeting a lighter string returns the right way up. Apply the rule at each surface and then take the difference. Soap in air runs 1.00 → 1.33 → 1.00: the front flips, the back does not, so the two reflections start half a wavelength apart before any path is counted. Magnesium fluoride on glass runs 1.00 → 1.38 → 1.52: both flip, the flips cancel, and the boundaries contribute nothing. Only the difference matters, which is why no layer in a stack can be judged without looking at what lies on either side of it.

Now put the two pieces together. A bright reflection means the waves leave in step. With one net flip that needs 2nt = (m + ½)λ; with no net flip it needs 2nt = mλ. The dark condition is the other line of each pair. Nothing is memorised except the flip rule. Work an antireflection coating with it: uncoated crown glass of index 1.52 reflects [(1.52 − 1.00)/(1.52 + 1.00)]² = 4.3% of the light at normal incidence. Coat it with magnesium fluoride, n = 1.38. Both surfaces flip, so cancellation needs the path to supply the whole half wavelength: 2nt = (m + ½)λ, thinnest at t = 550 nm ÷ (4 × 1.38) = 99.6 nm. That drops the reflected intensity at 550 nm to a little over 1%. It would fall to zero only if the two reflected amplitudes matched, which requires n_c = √1.52 = 1.23 — lower than any hard, durable coating material reaches.

Let the film be a wedge of air trapped between two glass plates that touch along one edge. The reflections now come from glass into air at the top of the gap, no flip, and from air into glass at the bottom, one flip. One net flip again, so with n = 1 in the gap the dark fringes fall where 2t = mλ. The line of contact has t = 0 and is therefore dark; a bright edge means a flip has been miscounted, and it is the fastest check that two plates are truly touching. If the gap opens to d over a length L then t(x) = dx/L, the wedge angle is θ ≈ d/L, and successive dark fringes sit Δx = λL/(2d) = λ/(2θ) apart. Trap a human hair 70 µm thick at the far edge of plates 10.0 cm long and θ = 7.0 × 10⁻⁴ rad, so 546 nm mercury light rules the gap with fringes 0.390 mm apart; the 256 of them between the contact line and the hair fix its diameter to within half a wavelength. Curve one plate instead of tilting it and the straight fringes close into Newton's rings, dark at the centre for the same reason.

White light supplies every wavelength at once, and one thickness satisfies the bright condition for only a few of them. Take a soap film whose optical path is 2nt = 900 nm, with one net flip: it is bright where λ = 900/(m + ½), which lands on 1800 nm and 360 nm outside the visible band and 600 nm inside it, while the dark condition λ = 900/m puts a null at 450 nm. Blue is removed, orange survives, and that patch looks orange-red. A slightly thicker patch shifts every one of those wavelengths, which is what draws the bands on a bubble. Now drain the bubble: its top thins until 2nt ≪ λ, the path contributes nothing, the lone half-wave flip is all that remains, and every visible wavelength cancels — the film goes black a few seconds before it bursts. Push the other way to a millimetre of window glass and the orders crowd so closely that every colour is bright somewhere inside the eye's resolution, while the two reflections are separated by far more than daylight's coherence length. The colour disappears and glass looks like glass.