E. The Wave Nature of Light · The Wave Nature of LightThin-Film Interference: Soap Bubbles, Oil Slicks and Anti-Reflective Coatings
E1.1 E2.1 E2.4 E3.2 E3.3 A1.12 A1.13About 75 minutes
Learning goals- I can explain thin-film colours as interference between light reflected from the top and bottom of a film.
- I can decide whether each reflection causes a half-cycle phase change, using the indices of refraction.
- I can find the thicknesses that give bright or dark reflections, using the wavelength inside the film.
- I can explain how anti-reflective coatings work and why soap films turn black just before they pop.
Words to know
thin film
A layer of transparent material whose thickness is comparable to the wavelength of light, such as a soap film or an oil slick.
phase change on reflection
When light reflects from a medium with a higher index of refraction, the reflected wave is inverted (shifted by half a cycle); reflection from a lower index causes no shift.
wavelength in the film (λ_f)
λ_f = λ/n_f, the wavelength of light inside a film of index n_f.
anti-reflective coating
A thin transparent layer, often about a quarter of a wavelength thick in the coating, that makes the light reflected from its two surfaces cancel.
iridescence
Colours that change with viewing angle, produced by interference rather than by pigments.
Colours from nothing
A soap bubble is made of clear, colourless soapy water, yet it swirls with magenta, gold and turquoise. A thin film of oil on a wet driveway after a rain shows rainbow bands. No dye makes these colours. They are made by interference between two reflections of the same light, one from the top of the film and one from the bottom.
The same physics puts the faint purple or green sheen on eyeglass lenses and camera lenses: their anti-reflective coatings are thin films designed to make reflections cancel.
Two reflections
When light hits a thin film, part reflects from the top surface (ray 1) and part enters the film, reflects from the bottom surface and comes back out (ray 2). The two rays overlap and interfere. Whether they add or cancel depends on two things:
- The extra path. For light arriving nearly straight on, ray 2 travels about 2t farther, all of it inside the film. Inside the film the wavelength is λ_f = λ/n_f, so what matters is how many film-wavelengths fit in 2t.
- Phase changes on reflection. You saw in Grade 11 (Waves and Sound) that a pulse reflecting from a fixed end of a rope comes back inverted, and from a free end comes back upright. Light behaves the same way:
- Reflection off a medium with a higher index of refraction: inverted (a half-cycle shift, like losing ½λ).
- Reflection off a medium with a lower index: no shift.
The rule: count the inversions. If exactly one of the two reflections is inverted, the rays already start half a cycle apart. If both or neither is inverted, the inversions do not change their relationship.
| Inversions | Example | Bright reflection (constructive) | Dark reflection (destructive) |
|---|
| One | soap film in air; oil (n = 1.45) on water (n = 1.33) | 2t = (m + ½)λ_f; thinnest t = λ_f/4 | 2t = mλ_f; t = 0, λ_f/2, λ_f, … |
| Zero or two | MgF₂ coating (n = 1.38) on glass (n = 1.52) | 2t = mλ_f; t = λ_f/2, λ_f, … | 2t = (m + ½)λ_f; thinnest t = λ_f/4 |
Worked exampleExample 1: A soap film
A soap film (n = 1.33) in air is lit by white light. What is the minimum thickness that strongly reflects green light of wavelength 550 nm?
- Top: air → soap (higher n): inverted. Bottom: soap → air (lower n): not inverted. One inversion.
- With one inversion, bright reflection needs 2t = (m + ½)λ_f. The thinnest is m = 0: t = λ_f/4.
- λ_f = 550 ÷ 1.33 = 413.5 nm, so t = 413.5 ÷ 4 = 103 nm.
t ≈ 103 nm (about a ten-thousandth of a millimetre).
Worked exampleExample 2: Designing an anti-reflective coating
A camera lens (n = 1.52) is coated with magnesium fluoride (n = 1.38). What is the minimum coating thickness that cancels the reflection of 550 nm light (the middle of the visible spectrum)?
- Top: air → MgF₂ (higher): inverted. Bottom: MgF₂ → glass (higher): inverted. Two inversions, which cancel out.
- So destructive interference needs 2t = (m + ½)λ_f. The thinnest is t = λ_f/4.
- λ_f = 550 ÷ 1.38 = 398.6 nm, so t = 398.6 ÷ 4 = 99.6 nm.
t ≈ 100 nm: a quarter-wavelength coating. Red and violet are not cancelled as completely, so the faint reflection that remains is purplish.
In an anti-reflective coating, the reflections from the top and bottom of the coating come back half a cycle apart and nearly cancel. The light energy that is not reflected is transmitted into the lens instead, so the lens passes more light.Worked exampleExample 3: Which colour does an oil slick reflect?
A film of oil (n = 1.45) 400 nm thick floats on water (n = 1.33). For light arriving straight on, which visible wavelengths (400 to 700 nm) are strongly reflected, and which are cancelled?
- Top: air → oil (higher): inverted. Bottom: oil → water (lower): not inverted. One inversion.
- Bright: 2t = (m + ½)λ/n, so λ = 2n t/(m + ½) = 2(1.45)(400)/(m + ½) = 1160/(m + ½) nm. m = 1: 773 nm (infrared); m = 2: 464 nm; m = 3: 331 nm (ultraviolet).
- Dark: λ = 1160/m nm. m = 2: 580 nm; m = 3: 387 nm (just past violet).
Blue light near 464 nm is strongly reflected and yellow light near 580 nm is cancelled, so this part of the slick looks blue. Where the oil is thicker or thinner, other colours win: that is why a slick shows bands.
Why bubbles turn black before they pop
A soap film held upright drains: gravity pulls the liquid down, so the film is thin at the top and thicker toward the bottom. Each thickness reflects a different colour, so you see horizontal colour bands that slide downward as the film drains.
When the top becomes much thinner than a wavelength (t → 0), the extra path 2t is almost nothing. But soap films have one inversion, so the two reflections are half a cycle apart and cancel for every colour. The top of the film looks black, and the film is about to burst.
The light that is not reflected is transmitted, so in transmitted light a thin film shows the complementary colours. Energy is never destroyed: interference only redistributes it.
Thin-film colours appear in nature too, for example on some beetles, bird feathers and butterfly wings, where layered microscopic structures act as thin films. This iridescence changes colour as you change your viewing angle, because the path difference depends on the angle.
Just for funWhy are soap bubbles such good listeners? They reflect on everything you say, and they only burst into silence when they've thinned out completely.
Hands-on activityWatch a soap film drain
You need: dish soap, water, a cup, an optional teaspoon of glycerin or sugar (makes longer-lasting films), a wire loop (a bent paper clip, a pipe cleaner or a cup without a bottom), a lamp, and a dark cloth or dark wall as a background.
- Mix about 1 part dish soap with 10 parts water (add the glycerin or sugar if you have it).
- Dip the loop and hold the film vertically in front of the dark background, with the lamp shining on it from your side so you see its reflection.
- Watch for 1 to 2 minutes. Sketch the colour bands and how they move.
- Watch the top of the film closely just before it pops.
- Try tilting the film or blowing gently on it.
Grown-up note: soap stings eyes; rinse with water if any splashes. Wipe up drips so no one slips.
What to look for: horizontal bands of colour that move downward as the film drains, with the bands crowding together near the thicker bottom. The top turns black (or nearly invisible) shortly before the film breaks, because it has become much thinner than a wavelength and the two reflections cancel.
Your turn
1When does light reflecting from a surface undergo a half-cycle phase change?
When it reflects off a medium with a higher index of refraction than the one it is travelling in.
2How many inversions occur for a soap film (n = 1.33) surrounded by air?
One (at the top surface only).
3Find the minimum thickness of a soap film (n = 1.33) that strongly reflects 600 nm light.
One inversion, so t = λ/(4n) = 600 ÷ (4 × 1.33) = 113 nm.
4Find the next thickness that strongly reflects 600 nm light in that film.
5Find the minimum (non-zero) thickness of that soap film that gives a dark reflection for 600 nm.
t = λ/(2n) = 600 ÷ 2.66 = 226 nm.
6A coating with n = 1.25 is put on glass (n = 1.52). Find the minimum thickness to cancel reflection of 633 nm light.
Two inversions, so t = λ/(4n) = 633 ÷ 5.00 = 127 nm.
7Why does the top of a draining soap film look black?
It is much thinner than a wavelength, so the path difference is nearly zero, and the one inversion makes the two reflections cancel for all colours.
More practice: warm-up
1Light in air reflects off glass. Is it inverted?
Yes (glass has the higher index).
2Light inside glass reflects off air. Is it inverted?
3Write the wavelength inside a film of index n_f.
4About how much farther does ray 2 travel than ray 1 at near-normal incidence?
5Name two products that use anti-reflective coatings.
Any two: eyeglasses, camera lenses, solar panels, binoculars, some screens.
More practice: on your own
1A soap film (n = 1.33) is 500 nm thick. Which visible wavelength is most strongly reflected, and which are cancelled?
One inversion. Bright: λ = 2nt/(m + ½) = 1330/(m + ½): 887 nm (m = 1, infrared), 532 nm (m = 2, green), 380 nm (m = 3, just beyond violet). Dark: λ = 1330/m: 665 nm (red) and 443 nm (violet-blue).
2A coating (n = 1.30) on glass (n = 1.52) is designed for 500 nm. Give the two smallest thicknesses that cancel the reflection.
Two inversions, so 2t = (m + ½)λ/n: t = λ/(4n) = 96 nm and 3λ/(4n) = 288 nm.
3Why do anti-reflective coatings usually leave a faint purple or greenish reflection?
A quarter-wave coating cancels only one wavelength perfectly (usually in the middle of the spectrum). The ends of the spectrum (red and violet, which together look purplish), are only partly cancelled, so they still reflect a little.
4Explain why an oil slick shows bands of different colours.
Its thickness varies from place to place. Each thickness gives constructive interference for a different wavelength, so different colours are reflected in different places.
5A film of oil (n = 1.50) floats on water (n = 1.33). Find the minimum thickness that strongly reflects 600 nm light.
One inversion (top only), so t = λ/(4n) = 600 ÷ 6.00 = 100 nm.
Challenge
1A lens (n = 1.52) has a MgF₂ coating (n = 1.38) that is 150 nm thick. For light arriving straight on, which visible wavelength is reflected most strongly? Is this a good anti-reflective coating for visible light?
Two inversions. Bright: 2nt = mλ, so λ = 2(1.38)(150)/m = 414/m nm: 414 nm (violet) for m = 1. Dark: λ = 414/(m + ½): 828 nm (m = 0) and 276 nm (m = 1), both outside the visible range. It strongly reflects violet and cancels no visible colour, so it is a poor anti-reflective coating.
2A film of index 1.40 lies on glass of index 1.60. A student uses the soap-film rule (one inversion) to find the minimum thickness for strong reflection of 560 nm light and gets 100 nm. Find the correct answer and explain the error.
Air → film (higher) inverts; film → glass (higher) also inverts: two inversions, not one. Bright reflection needs 2t = mλ/n, so the smallest non-zero thickness is t = λ/(2n) = 560 ÷ 2.80 = 200 nm. (100 nm is λ/(4n), which here gives destructive interference.)
3An anti-reflective coating cancels reflected light. Use conservation of energy to explain what happens to that light, and why this matters for solar panels.
Energy cannot vanish. Destructive interference in the reflected direction means the energy goes into the transmitted light instead. For solar panels, more light enters the cell instead of reflecting away, so the panel can produce more electrical energy.
Quiz
1. Light travelling in air reflects off a soap film. The reflected wave is…
2. A soap film (n = 1.33) in air: the minimum thickness that strongly reflects 532 nm light is…
3. An anti-reflective coating with n between air and glass should have a minimum thickness of…
4. Why does a soap film look black just before it pops?
5. A thin film shows different colours in different places mainly because…
6. What happens to the light energy that an anti-reflective coating keeps from being reflected?
Big idea: thin-film interference (E2.4, E3.2), separation of white light into colours (E3.3) and a key technology, anti-reflective coatings (E1.1). The decisive step in every problem is counting phase inversions before choosing a formula.
How to help: the draining soap film is a lovely demonstration. Ask your teen to explain why the bands move downward and why the top goes black.
Common difficulties:
- Forgetting to use λ_f = λ/n inside the film.
- Applying the soap-film rule to a coating on glass (two inversions, not one).
- Using t = λ/4 when the question wants t = λ/2 (or the reverse); write down "one inversion" or "two inversions" first.
- Thinking destructive interference destroys energy. It sends the energy elsewhere (into the transmitted light).