Thin-Film Interference

Where a bubble gets its colour

A soap bubble is colourless stuff. Water, a little detergent, a wall maybe two hundred nanometres thick. Nothing in it is red or green. Yet hold one up in white light and it is banded with colour, and the bands move as the film drains and thins. An oil slick on a wet road does the same thing, and so does the purple sheen on a camera lens.

The double slit gave you interference between two beams that left the same source and travelled different distances. A thin film does exactly that, with one difference worth the whole lesson: here the two beams are made by the film itself. Light hits the top surface, and part of it bounces straight back. The rest goes in, crosses the film, bounces off the bottom surface, and comes back out travelling alongside the first beam. Two waves, same source, slightly different paths. They interfere. Which colours survive depends on the thickness, and that is why the colour changes as the film drains.

The extra path, measured in wavelengths

Take the light coming in straight down, at normal incidence, on a film of thickness $t$ and refractive index $n$. Call the wavelength in vacuum $\lambda$.

The second beam crosses the film twice, down and back, so it travels an extra distance $2t$. What matters for interference is not the distance but how many wavelengths fit into it, and inside the film the wavelength is not $\lambda$. The frequency of light cannot change when it enters glass (the atoms at the surface are driven at the frequency they are hit with) but the speed drops to $v = c/n$. Since $\lambda_n = v/f$ and $\lambda = c/f$,

$\lambda_n = \frac{\lambda}{n}$

Light of 600 nm in vacuum is 400 nm long inside a film of index 1.50. So the number of wavelengths crammed into that extra $2t$ is

$N = \frac{2t}{\lambda_n} = \frac{2nt}{\lambda}$

The product $2nt$ is called the optical path difference: the geometric path multiplied by the index, which converts it into "distance measured in vacuum wavelengths". Almost every mistake in this topic is dropping the $n$ from that product, or putting it in twice.

The flip

If that were the whole story, a film would be bright whenever $2nt$ was a whole number of wavelengths. It is not the whole story, and the reason is something you have already met on a rope.

Send a pulse down a rope tied to a wall. The pulse comes back upside down. Send it down a rope that is tied to a much lighter string, and it comes back the right way up. The reflection inverts when the wave is meeting something harder to move than what it came from.

Light does the same at a boundary, and "harder to move" means higher refractive index. A reflection off a medium of higher index inverts the wave. A reflection off a medium of lower index does not. An inversion is a phase shift of half a cycle, so it is worth exactly $\lambda/2$ of extra path. People call it the half-wave loss; this lesson will call it a flip.

Now there are two reflections to check, and each one either flips or it does not. Write the three indices down in order (above the film, the film, below the film) and compare across each boundary.

If the two reflections both flip, or neither does, the flips cancel and the only thing left is the path. Constructive reflection then needs a whole number of wavelengths:

$2nt = m\lambda \qquad (\text{bright, even number of flips})$

with $m = 0, 1, 2, \dots$. If exactly one of them flips, half a cycle is added on top of the path, and the two conditions trade places:

$2nt = \left(m + \tfrac{1}{2}\right)\lambda \qquad (\text{bright, one flip})$

Dark is whichever one is left over. There is no formula to memorise beyond these two lines, and memorising which is which for soap will hurt you the first time a problem puts the film on glass instead of air. Count the flips every time.

Soap in air is air-soap-air: $1.00 \to 1.33$ flips, $1.33 \to 1.00$ does not. One flip. Oil on water is $1.00 \to 1.46 \to 1.33$: still one flip. A magnesium fluoride coating on glass is $1.00 \to 1.38 \to 1.52$, rising all the way: two flips.

Two films you can do in your head

Example 1. The thinnest bright soap film. A soap film of index 1.33 hangs in air. Green light of 550 nm hits it head on. What is the smallest nonzero thickness that reflects it most strongly?

Air-soap-air is one flip, so bright means $2nt = (m + \frac12)\lambda$. The smallest nonzero thickness comes from the smallest order, $m = 0$:

$2(1.33)t = \tfrac{1}{2}(550\text{ nm}) = 275\text{ nm}$

so $t = 275/2.66 = 103.4$ nm. Round to 103 nm. Notice the shape of the answer: $t = \lambda/(4n)$. A quarter of the in-film wavelength, because the light crosses it twice and a half-wave round trip is what the single flip needs.

Example 2. The coating on a camera lens. Glass of index 1.52 is coated with magnesium fluoride, index 1.38, to kill the reflection at 550 nm. Find the minimum coating thickness.

Here the indices rise at both surfaces, so two flips, and the plain condition $2nt = m\lambda$ is the bright one. Dark is therefore $2nt = (m + \frac12)\lambda$, and the minimum thickness is again the quarter wave:

$t = \frac{\lambda}{4n} = \frac{550\text{ nm}}{4(1.38)} = 99.6\text{ nm}$

Same arithmetic, opposite meaning. That is the whole reason to count flips rather than memorise.

A quarter-wave coating always sets the two reflected waves half a cycle apart, but they only cancel completely if they have the same amplitude. At normal incidence the reflected amplitude ratio at a boundary is $r = (n_1 - n_2)/(n_1 + n_2)$. Setting the size of $r$ equal at both surfaces, $(1 - n_c)/(1 + n_c) = (n_c - n_g)/(n_c + n_g)$, and multiplying out gives $n_c = \sqrt{n_g}$. For $n_g = 1.52$ the ideal coating index is 1.23. No cheap durable solid has that index, which is why real coatings use 1.38 and leave a faint purple reflection behind.

White light: several colours at once

Shine white light on a film of fixed thickness and every wavelength gets the same $2nt$ but a different number of wavelengths out of it. Some colours land on the bright condition, some on the dark one, most in between. The reflected light is what is left.

At the very thin end something clean happens. Let $t \to 0$ for a soap film in air. Then $2nt \to 0$, the path contributes nothing, and the single flip is all that is left. The two beams are half a cycle apart for every wavelength at once, so the whole spectrum cancels and the film looks black. That black band at the top of a draining film is the last thing you see before it bursts.

The most common hard problem gives you two wavelengths that are both missing, or both enhanced, with none in between, and asks for the thickness. Two conditions, two unknowns: the thickness and the order.

Example 3. Oil on water, two colours missing. An oil film of index 1.50 floats on water, air above. In reflection, 500 nm and 600 nm are missing and nothing between them is. Find $t$.

Air-oil-water is $1.00 \to 1.50 \to 1.33$: one flip. So dark is $2nt = m\lambda$. The longer wavelength needs fewer wavelengths to fill the same path, so it takes the smaller order. "No missing wavelength between them" means the orders are consecutive. Call them $m$ for 600 nm and $m+1$ for 500 nm:

$2nt = m(600) = (m+1)(500)$

Expand: $600m = 500m + 500$, so $100m = 500$ and $m = 5$. Then $2nt = 5(600) = 3000$ nm, and

$t = \frac{3000\text{ nm}}{2(1.50)} = 1000\text{ nm}$

Check it: $2nt/\lambda$ is 5 at 600 nm and 6 at 500 nm, both whole numbers, and between them $3000/5.5 = 545$ nm is not a dark wavelength. Good.

The same setup with a wavelength allowed in between just changes the counting. If 500 nm and 700 nm are missing with exactly one other dark wavelength between them, the orders differ by 2: $m(700) = (m+2)(500)$ gives $m = 5$, so $2nt = 3500$ nm. For oil of index 1.46 that is $t = 3500/2.92 = 1199$ nm.

Wedges and rings

Nothing above assumed the thickness was the same everywhere. If $t$ varies slowly across the film, apply the condition point by point: each place where $2nt$ hits the right value is a fringe, and the fringes map out the shape of the gap.

Two flat plates touching at one edge and held apart at the other make an air wedge, $n = 1$. Air between glass is one flip (no flip on the way in at glass-air, a flip at air-glass below), so dark means $2t = m\lambda$. Adjacent dark fringes differ by one order, so the gap changes by $\lambda/2$ between them. If the plates are separated by a wire of diameter $50.0\ \mu$m, the number of dark fringes from the contact line to the wire is

$\frac{2t_{\max}}{\lambda} = \frac{2(50.0\times10^{-6})}{589\times10^{-9}} = 170$

And if you know the fringe spacing along the plate, you know the wedge angle: the gap climbs by $\lambda/2$ over one spacing, so $\theta \approx \lambda/(2 \Delta x)$ for a small angle in air, or $\lambda/(2n\Delta x)$ if the wedge is filled.

Curve one surface instead and the fringes become circles. A plano-convex lens resting on a flat plate leaves a gap that, for $r \ll R$, has thickness $t = r^2/(2R)$, where $R$ is the radius of curvature. These are Newton's rings.

Example 4. The tenth dark ring. A lens of $R = 2.00$ m sits on a flat plate, air in the gap, lit by 589 nm light. Find the radius of the tenth dark ring, with the dark centre counted as $m = 0$.

One flip, so dark is $2t = m\lambda$. Substituting $t = r^2/(2R)$ gives $r^2/R = m\lambda$, so $r = \sqrt{m\lambda R}$:

$r = \sqrt{10 \times (589\times10^{-9}\text{ m}) \times 2.00\text{ m}} = \sqrt{1.178\times10^{-5}} = 3.43\times10^{-3}\text{ m}$

That is 3.43 mm. Two things fall out of $r = \sqrt{m\lambda R}$ for free. The rings crowd together as you go out, because $r$ grows like $\sqrt{m}$. And if a liquid of index $n$ fills the gap, replace $\lambda$ by $\lambda/n$, so every ring shrinks by $\sqrt{n}$: a 3.00 mm ring in air becomes $3.00/\sqrt{1.33} = 2.60$ mm under water.

The centre is only dark if exactly one of the two reflections flips there. A lens of index 1.52 on a plate of index 1.75 with liquid of index 1.62 between them goes $1.52 \to 1.62 \to 1.75$: two flips, and the centre is bright instead. Then bright is $2nt = m\lambda$, the centre is $m = 0$, and the first bright ring away from it is $m = 1$, at $r = \sqrt{\lambda R/n}$.

Angles, and looking through instead of at

At an angle the geometry changes slightly. The beam inside the film travels along a slant, and the reflected beam that never entered gets a head start. Work the geometry through and the optical path difference becomes

$\Delta = 2nt\cos\theta_t$

where $\theta_t$ is the refracted angle inside the film, from $\sin\theta_i = n\sin\theta_t$. The cosine, not the sine, and the angle inside, not the angle you aimed at. At normal incidence $\cos\theta_t = 1$ and you recover $2nt$. Tilting always shrinks $\Delta$, so the colour of a film shifts toward the blue as you tip it, which is why a bubble's bands crawl when you move your head.

Finally, whatever is not reflected is transmitted. The transmitted beams pick up flips at different places, and the upshot is simple: the transmitted condition is always the opposite of the reflected one. A film that looks dark in reflection is at peak brightness looking through it. For a soap film in air, reflection is dark at $2nt = m\lambda$, so transmission is maximal there: $t = \lambda/(2n)$ at the smallest nonzero thickness, 188 nm for 500 nm light.

How to spot one of these

Reach for thin-film interference when a problem mentions a film, coating, slick, bubble, gap, or wedge whose thickness is comparable to a wavelength (tens to a couple of thousand nanometres) and then asks about colour, missing wavelengths, brightness, or darkness in reflected or transmitted light. Three indices in a stack is the giveaway.

Then run the same four steps every time. One: write the three indices in order and count the flips, which fixes whether $m\lambda$ or $(m+\frac12)\lambda$ is the bright side. Two: write the optical path as $2nt$, or $2nt\cos\theta_t$ off normal, with $n$ the index of the film. Three: if the geometry varies, put in the thickness law: $t = r^2/(2R)$ for rings, $t = x\theta$ for a wedge. Four: if two wavelengths are given, set the two conditions equal and solve for the order first, then the thickness. The order is always an integer, and if your algebra gives a non-integer you have the flip count or the ordering of the wavelengths backwards.

Waves have taken you as far as they go for now. Next you leave optics for the other thing light forced physicists to rewrite: how space and time themselves behave when something moves near the speed of light, in Special Relativity: Kinematics.