Topic 14.9: Thin-Film Interference

What Topic 14.9 requires

Topic 14.9 carries one learning objective, 14.9.A, describe the behavior of light that interacts with a thin film. Five essential knowledge statements sit under it, three of them with sub-statements, and unlike 14.7 and 14.8 this topic does print a boundary statement.

• 14.9.A.1 When light travels from one medium to another, some of the light is transmitted, some is reflected, and some is absorbed.
• 14.9.A.2 The phase change of a reflected ray depends on the relative indices of refraction of the materials with which the ray interacts. Sub-statements: 14.9.A.2.i (a phase change of 180 degrees occurs when a light ray is reflected from a medium with a greater index of refraction than the medium through which the ray is traveling), 14.9.A.2.ii (no phase change occurs when a light ray is reflected from a medium with a lower index of refraction than the medium through which the ray is traveling).
• 14.9.A.3 The phase of a wave does not change when it is refracted as it passes from one medium into another.
• 14.9.A.4 Thin-film interference occurs when light interacts with a medium whose thickness is comparable to the light's wavelength. Sub-statements: 14.9.A.4.i (the interactions between the initial reflected light and the light exiting the thin film after being reflected from the second interface exhibit wave interference behavior, resulting in a single wave that is the sum of the two interacting waves), 14.9.A.4.ii (the amount of constructive or destructive interference between the two reflected waves depends on the relationship between the thickness of the film, the wavelength of light, any phase shifts, and the angle at which the incident light strikes the film).
• 14.9.A.5 Practical examples of thin-film interference include the color variations seen in soap bubbles and oil films, as well as antireflection coatings. Sub-statements: 14.9.A.5.i (the spectrum of colors observed in oil films and soap bubbles arises from differences in the thickness of the film), 14.9.A.5.ii (antireflection coatings eliminate reflected light by applying the relationships between indices of refraction, phase shift, and wave interference to create destructive interference of the light reflected from the two surfaces of the coating), 14.9.A.5.iii (the simplest antireflection coating has a thickness equal to one-quarter of the wavelength of the light in the coating, and the index of refraction of the coating is greater than that of air and less than that of the surface upon which the coating is applied, which assumes incident light is normal to the surface).

Boundary statement, in full: "Quantitative analysis of thin-film interference is limited to waves that are normal to the incident surface." That is the whole statement as printed, with no exception clause. Take it literally: every number you are asked to produce here assumes light arriving straight on. Off-axis light is describable but not calculable.

The suggested skills are 1.A (create diagrams, tables, charts, or schematics to represent physical situations), 2.B (calculate or estimate an unknown quantity with units from known quantities), 2.C (compare physical quantities between two or more scenarios or at different times and locations in a single scenario) and 3.B (apply an appropriate law, definition, theoretical relationship, or model to make a claim). Skill 1.A being listed first is a hint about method: draw the film, mark the two reflections, and label the indices before you compute anything.

Unit 14 is weighted at 12 to 15 percent of the multiple-choice section across a suggested 14 to 23 class periods.

The two waves that interfere, and the one thing that can flip

14.9.A.1 sets the stage: when light travels from one medium to another, some of the light is transmitted, some is reflected, and some is absorbed. A boundary is never all or nothing, and that partial reflection is what makes thin films work.

Picture a film of thickness $t$ and index $n_{\text{film}}$ sitting on something else, with light arriving from above and straight on, as the boundary statement requires. Two waves come back toward your eye.

• Wave 1 reflects off the top surface of the film and never enters it.
• Wave 2 enters the film, crosses it, reflects off the bottom surface, crosses back, and exits through the top.

14.9.A.4.i names that pair: the interactions between the initial reflected light and the light exiting the thin film after being reflected from the second interface exhibit wave interference behavior, resulting in a single wave that is the sum of the two interacting waves. What you see is that sum.

Two things make the waves differ in phase, and only two.

One: the extra distance. Wave 2 travels down and back through the film. At normal incidence that is exactly $2t$ of extra path, which is why the boundary statement restricts quantitative work to normal incidence: tilt the light and the extra path is no longer $2t$.

Two: the reflections themselves. This has no analogue in Topic 14.8 and it is the whole difficulty of Topic 14.9. 14.9.A.2 states that the phase change of a reflected ray depends on the relative indices of refraction of the materials with which the ray interacts, and its two sub-statements say precisely when:

• 14.9.A.2.i A phase change of 180 degrees occurs when a light ray is reflected from a medium with a greater index of refraction than the medium through which the ray is traveling.
• 14.9.A.2.ii No phase change occurs when a light ray is reflected from a medium with a lower index of refraction than the medium through which the ray is traveling.

A 180 degree phase change is half a cycle, so it is equivalent to adding or subtracting half a wavelength. That is why it changes answers rather than decorating them.

14.9.A.3 closes the last gap: the phase of a wave does not change when it is refracted as it passes from one medium into another. Crossing a boundary does nothing to the phase. Only bouncing off one can. So the accounting is complete: extra path $2t$, plus zero, one or two half-wavelength flips.

Counting the shifts: the two questions to ask first

Before any arithmetic, answer two questions in order. Skill 1.A, create diagrams, tables, charts, or schematics to represent physical situations, is listed first on this topic for a reason: draw three layers and write the index in each one.

Question 1: at the top surface, is the film's index greater than the index above it? If yes, wave 1 flips. If no, it does not.

Question 2: at the bottom surface, is the index below the film greater than the film's own index? If yes, wave 2 flips. If no, it does not.

Note what is being compared each time. 14.9.A.2.i weighs the medium the ray is reflecting from against the medium the ray is travelling in, and at the top surface that means whatever is above the film while at the bottom it means the film itself. The pair changes between the two questions, and so can the answer.

Then count. Only the net number matters, and only whether it is odd or even.

Two flips cancel. That is the step people miss, and it is why "there is a shift, so add a half wavelength" is wrong as a habit. A shift only matters if the other wave does not have one too.

Three standard arrangements, run through the two questions:

• Soap bubble in air. Air ($n \approx 1.00$), soap film ($n \approx 1.33$), air below. Top: $1.33 > 1.00$, flip. Bottom: $1.00 < 1.33$, no flip. One flip, net half wavelength.
• Oil film on water, with the oil's index above the water's. Top: flip. Bottom: no flip. One flip, net half wavelength.
• Antireflection coating on glass. Air, coating, glass, with the index rising at every step, which is what 14.9.A.5.iii requires. Top: flip. Bottom: flip. Two flips, net zero.

A film with air on both sides always gives exactly one flip, whatever the film is, because the index goes up then down. A coating whose index sits between air and the substrate always gives two.

The wavelength inside the film, and where 2t comes from

One piece is still missing, and it is easy to forget: the $2t$ of extra path happens inside the film, so it must be compared with the wavelength inside the film, not the wavelength in air.

The CED prints no symbol for that wavelength, but you can build it from two printed equations plus one statement from earlier in the unit.

• The AP Physics 2 equation sheet prints $n = \dfrac{c}{v}$, so the speed inside a medium is $v = c/n$.
• The same sheet prints $\lambda = \dfrac{v}{f}$.
• Essential knowledge 14.3.A.1.iv, in Topic 14.3, states that the frequency of a wave does not change when it travels from one medium to another.

Put them together. The frequency is fixed, the speed drops by a factor of $n$, so the wavelength drops by the same factor:

where $\lambda_0$ is the wavelength in vacuum, which is what a question means when it says "light of wavelength 550 nm" without naming a medium. This is a derived result. It is not printed on the sheet and it is not in Topic 14.9, so it is a two-line derivation you should be able to produce, in the spirit of skill 2.A from the neighbouring topics.

It also gives 14.9.A.4 a precise meaning. That statement says thin-film interference occurs when light interacts with a medium whose thickness is comparable to the light's wavelength, and the wavelength to compare with is the one inside the film. A visible-light film is therefore a few hundred nanometres thick.

The extra path is $2t$ only at normal incidence, straight down and straight back. That is what the boundary statement protects. At an angle the geometry brings in a factor depending on the refracted angle inside the film, and 14.9.A.4.ii does list the angle at which the incident light strikes the film among the things the interference depends on. The boundary statement then removes that angle from anything quantitative you will be asked.

The two condition sets, and why they swap

Here is the part to get right, because there is no single pair of conditions for thin films. The constructive and destructive conditions trade places depending on how many half-wavelength flips there were. Any page that gives you one pair and calls it the rule is giving you half the topic.

Write $\lambda_{\text{film}} = \lambda_0 / n_{\text{film}}$ and let $m = 0, 1, 2, \ldots$

Case A, exactly one flip (a film with a lower-index medium on the far side, so a soap bubble in air or oil on water):

The two waves already differ by half a wavelength before the path is counted, so the path has to supply another half wavelength to bring them back in step.

Case B, zero flips or two flips (an antireflection coating between air and glass, or a film with a higher-index medium on both sides):

The two waves start in step, so the path difference alone decides.

Compare the two boxes. The right-hand sides are identical and the words attached to them are exchanged. That is the whole sign question in this topic, and 14.9.A.4.ii is the CED's version of the same point: the amount of constructive or destructive interference between the two reflected waves depends on the relationship between the thickness of the film, the wavelength of light, any phase shifts, and the angle at which the incident light strikes the film. The phase shifts sit in that list on equal footing with the thickness.

Two notes on the indexing:

• In Case A, $m = 0$ on the destructive line gives $2t = 0$. A vanishingly thin film reflects almost nothing. On a vertical soap film that drains until the top is far thinner than a wavelength, the top goes black just before it breaks, and this is why.
• In Case B, $m = 0$ on the constructive line also gives $t = 0$, which is not a film, so the useful orders start at $m = 1$. The destructive line in Case B starts usefully at $m = 0$, and that is the antireflection coating.

None of these four lines is printed anywhere. The AP Physics 2 equation sheet has 15 equations in the Waves, Sound, and Optics group and not one of them mentions a film, a thickness or a phase shift, and the CED prints no relevant equation and no derived equation under Topic 14.9. You reconstruct all of it from 14.9.A.2 and the geometry, which is what skill 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim, is asking for.

So do not memorise four lines. Memorise the accounting: count the flips, write $2t$, express everything in $\lambda_{\text{film}}$, and ask whether the path needs a whole number or a half-odd number of film wavelengths to bring the two waves into step.

The antireflection coating, and using it to check yourself

14.9.A.5.iii is a worked answer the CED hands you, and you can run it backwards to confirm you have the conditions the right way round. In full: the simplest antireflection coating has a thickness equal to one-quarter of the wavelength of the light in the coating, and the index of refraction of the coating is greater than that of air and less than that of the surface upon which the coating is applied. This assumes incident light is normal to the surface.

Two separate requirements are packed in there.

The index ordering: $n_{\text{air}} < n_{\text{coating}} < n_{\text{substrate}}$. The index rises at both surfaces, so both reflections flip. That is Case B.

The thickness: $t = \lambda_{\text{coating}} / 4$, a quarter of the wavelength in the coating, not in air.

Now check it against Case B. An antireflection coating wants the reflected light gone, so 14.9.A.5.ii says the aim is to create destructive interference of the light reflected from the two surfaces of the coating. Case B destructive is $2t = \left(m + \frac{1}{2}\right)\lambda_{\text{film}}$, and the thinnest coating is $m = 0$:

which is exactly the quarter-wavelength the CED states. That agreement is the point: if the quarter-wave coating does not fall out of the destructive line, you have the two cases swapped, and you can catch it before the answer goes down. In terms of the wavelength you are actually given, in air or vacuum, $t = \dfrac{\lambda_0}{4\,n_{\text{coating}}}$, which is just $\lambda_{\text{film}} = \lambda_0 / n$ substituted in.

Three details a question can turn on:

• The index ordering is a design requirement, not a coincidence. If the coating's index were above the substrate's, the bottom reflection would not flip, the arrangement would move to Case A, and a quarter-wave thickness would give the brightest possible reflection instead of the dimmest.
• A coating cancels one wavelength, not all of them. The thickness is chosen for one wavelength, usually in the middle of the visible range, so the extremes are only partly suppressed. That is why coated lenses look faintly purple in reflection.
• 14.9.A.5.ii says "eliminate reflected light", and the energy has to go somewhere. Light taken out of the reflection is transmitted instead, which is the whole point for a camera lens or a pair of glasses.

Soap bubbles and oil films: why the colours change across the surface

14.9.A.5 lists the practical examples: colour variations seen in soap bubbles and oil films, as well as antireflection coatings. 14.9.A.5.i gives the cause of the colours in one line: the spectrum of colors observed in oil films and soap bubbles arises from differences in the thickness of the film.

That sentence rules out the answers students reach for first. The colours are not the material's colour, and they are not light being split by refraction the way a prism splits it. One film, one index, one incident white light, and the only thing that varies from point to point is $t$.

A soap film in air is Case A. Where the film has thickness $t$, the reflection is brightest for whichever wavelengths satisfy $2t = \left(m + \frac{1}{2}\right)\lambda_0 / n$ and darkest for those satisfying $2t = m\lambda_0 / n$, so the strongly reflected vacuum wavelengths there are

Move to a point where the film is thicker and every one of those wavelengths shifts, so a different colour dominates. A vertical soap film drains under gravity and ends up thicker at the bottom, which is why it shows horizontal bands of colour that drift downward as it thins. An oil film on a puddle is uneven for its own reasons and shows irregular patches.

This is the unit's own essential question. The Unit 14 opener asks: why does it look like a rainbow when you see a puddle of water with oil in it at a gas station? The answer is 14.9.A.5.i. Different thicknesses across the puddle send different wavelengths back to your eye.

It is also a clean target for skill 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario. "At different locations in a single scenario" describes a film of varying thickness exactly. Worked examples 2 and 3 below are the two-scenario version: the same film, the same thickness, one change of substrate, and an answer that swaps.

What the equation sheet gives you, and how 14.9 is tested

The AP Physics 2 equation sheet prints 129 equations in seven groups, 15 of them in Waves, Sound, and Optics. None of them is a thin-film equation. Counting through that group: the wave relationship $\lambda = v/f$, the index $n = c/v$, Snell's law, the thin-lens equation, the magnification equation, the five interference and diffraction lines ($\Delta D = m\lambda$, $\Delta D = a\sin\theta$, $a(y_{\text{min}}/L) \approx m\lambda$, $\Delta D = d\sin\theta$, $d(y_{\text{max}}/L) \approx m\lambda$), the wave speed on a string, $T = 1/f$, the two sinusoidal wave forms, and the beat frequency. Fifteen entries, no film, no thickness, no phase shift.

The CED does not fill the gap. Topic 14.9 prints no relevant equation and no derived equation, so everything you compute here you assemble yourself, from the phase-shift rules in 14.9.A.2, the geometry $2t$, and the printed $n = c/v$ and $\lambda = v/f$.

What you can lift straight off the sheet and use:

The boundary statement decides what can be asked. "Quantitative analysis of thin-film interference is limited to waves that are normal to the incident surface." A numerical question is therefore always at normal incidence and the $2t$ path is always exact. Angle dependence is real and 14.9.A.4.ii names it, so you may be asked to describe it, for instance that a soap bubble changes colour as you tilt your head. You will not be asked to calculate it.

How it shows up. Unit 14's 12 to 15 percent weighting applies to the multiple-choice section of the exam, which is 42 questions in 85 minutes for half the score, with 4 free-response questions in 95 minutes for the other half and a calculator allowed throughout.

The four suggested skills give four shapes of question. 1.A wants the diagram: three layers, indices labelled, two reflected rays drawn, each marked flip or no flip. 2.B is the calculation, usually a minimum thickness or the wavelength most strongly reflected. 2.C is the comparison, the same film in two settings or two points on one film. 3.B is the claim: given an arrangement, say whether the reflection is enhanced or suppressed and justify it from 14.9.A.2 and the path difference.

The failure that costs most is answering with one memorised pair of conditions. Count the flips first, every time.