After a monsoon shower, a few drops of engine oil from a scooter spread over a puddle on the road. The puddle shines with bands of colour: violet, blue, green, gold and red, in curving stripes. There is no dye in the oil, and a spoonful of it in a bottle looks plain brown. Walk round the puddle, and the colours slide to new places.
Two rays from one film
Let light fall on a film of thickness \(t\) and refractive index \(\mu\), with air on both sides. It meets the front surface at A at the angle of incidence \(i\). Part of it reflects there as ray 1. The rest refracts at the angle \(r\), reflects from the back surface at B, and leaves at C as ray 2, parallel to ray 1.
Draw CD at right angles to the two rays, as in Fig 2.1. It is a wavefront of the reflected light, so beyond it both rays travel the same distance to the eye. The difference arises before it. Ray 2 travels AB and BC inside the film, an optical path \(\mu(AB + BC)\). Ray 1 travels AD in air.
\[ AB = BC = \dfrac{t}{\cos r} \]
Why. each is the slant side of a right triangle whose height is \(t\) and whose angle at the top is \(r\).
\[ \mu(AB + BC) = \dfrac{2\mu t}{\cos r} \]
Why. the optical path is \(\mu\) times the distance inside the film.
\[ AC = 2t\tan r, \qquad AD = AC\sin i = 2t\tan r\sin i \]
Why. AC is two half-bases \(t\tan r\); in the right triangle ADC the angle at C equals \(i\).
\[ AD = 2t\tan r \cdot \mu\sin r = \dfrac{2\mu t\sin^2 r}{\cos r} \]
Why. Snell's law at A, \(\sin i = \mu\sin r\).
\[ \mu(AB + BC) - AD = \dfrac{2\mu t}{\cos r}\left(1 - \sin^2 r\right) = 2\mu t\cos r \]
Read it. subtract step 4 from step 2 and use \(1 - \sin^2 r = \cos^2 r\).
The path difference is \(2\mu t\cos r\). Seen straight on, \(r = 0\) and it is simply \(2\mu t\), twice the film's optical thickness. One more thing must be added. In a film with air on both sides, the front reflection gains a half wave and the back one does not. So the two reflected waves differ by \(2\mu t\cos r + \lambda/2\).
White light falls almost normally on a film of thickness \(400\,\text{nm}\) and \(\mu = 1.5\). What is \(2\mu t\), the extra path of the wave reflected from the back surface, in nm?
Bright and dark
The two reflected waves add when their path difference is a whole number of wavelengths. They cancel when it is a whole number plus a half. With the half wave included, bright means \(2\mu t\cos r + \lambda/2 = n\lambda\). So \(2\mu t\cos r = (n - \tfrac12)\lambda\). Dark means \(2\mu t\cos r + \lambda/2 = (n + \tfrac12)\lambda\), so \(2\mu t\cos r = n\lambda\).
For a thin film in which exactly one reflection is from a denser medium, seen in reflected light:
If both reflections, or neither, are from a denser medium, the two conditions swap.
Seen straight on, \(\cos r = 1\). The thinnest film that looks bright has \(n = 1\): \(2\mu t = \lambda/2\), so \(t = \lambda/4\mu\). The thinnest film that is dark, apart from \(t = 0\), has \(2\mu t = \lambda\), so \(t = \lambda/2\mu\). Counted in its own wavelength, a film is bright at a quarter wave thick and dark at a half wave thick.
Light of \(600\,\text{nm}\) falls normally on a film of \(\mu = 1.5\) in air. Find the least thickness for which the film looks bright in reflected light, and the least non-zero thickness for which it looks dark.
Bright needs \(2\mu t = (n - \tfrac12)\lambda\). The least is \(n = 1\): \(t = \lambda/4\mu = 600/6 = 100\,\text{nm}\).
Dark needs \(2\mu t = n\lambda\). The least non-zero is \(n = 1\): \(t = \lambda/2\mu = 600/3 = 200\,\text{nm}\).
Check: at \(100\,\text{nm}\), \(2\mu t = 300\,\text{nm} = \lambda/2\); with the half wave the total is \(\lambda\), so the waves add.
Green light of \(540\,\text{nm}\) falls normally on a film of \(\mu = 1.5\) in air. What is the least thickness, in nm, for which the film looks bright in reflected light?
Which colour a film shows
White light holds every wavelength from about \(400\) to \(700\,\text{nm}\). A film of one thickness cannot satisfy the bright condition for all of them at once. Some wavelengths are reflected strongly, some weakly, and a few not at all. The reflected light is therefore coloured, even though the film itself has no colour.
Take a film \(400\,\text{nm}\) thick of \(\mu = 1.5\), seen straight on. Here \(2\mu t = 1200\,\text{nm}\). Bright needs \(\lambda = 2400/(2n - 1)\): \(2400\), \(800\), \(480\), \(343\,\text{nm}\), and only \(480\,\text{nm}\) is visible. Dark needs \(\lambda = 1200/n\): \(1200\), \(600\), \(400\), \(300\,\text{nm}\). So the film reflects blue light of \(480\,\text{nm}\) most strongly, and \(600\,\text{nm}\) and \(400\,\text{nm}\) not at all.
In the figure above, a thin film has one broad hump: it reflects one band of colours. As \(t\) grows, the hump moves to longer wavelengths and a second hump enters from the violet end. By \(700\,\text{nm}\) two peaks share the visible range, at \(467\) and \(600\,\text{nm}\), and three wavelengths are missing. The colours of a thicker film are mixtures, paler than those of a thin one.
A film of \(\mu = 1.5\) and \(t = 500\,\text{nm}\) in air is viewed so that the angle of refraction inside it is \(53°\) (take \(\cos 53° = 0.6\)). Find the path difference \(2\mu t\cos r\), the visible wavelength reflected most strongly, and the visible wavelength that is missing.
One reflection gains a half wave, so bright needs \(2\mu t\cos r = (n - \tfrac12)\lambda\) and dark needs \(2\mu t\cos r = n\lambda\).
\(2\mu t\cos r = 2 \times 1.5 \times 500 \times 0.6 = 900\,\text{nm}\).
Bright: \(\lambda = 1800/(2n - 1) = 1800, 600, 360\,\text{nm}\). Dark: \(\lambda = 900/n = 900, 450, 300\,\text{nm}\).
Seen at this slant the film reflects \(600\,\text{nm}\) most strongly, and \(450\,\text{nm}\) is missing.
Why do the colours of the puddle move when you walk round it? A slanting view makes \(r\) larger and \(\cos r\) smaller. So the same film has a shorter path difference, and it behaves like a thinner film seen straight on. Each colour band slides to where the oil is a little thicker.
A film of \(\mu = 1.5\) and \(t = 400\,\text{nm}\) in air is lit normally by white light. Which visible wavelengths are missing from the reflected light?
A soap film (\(\mu = \tfrac43\)) seen straight on first shows a bright green of about \(540\,\text{nm}\) where it is thinnest. Roughly how thick is the film there, in nm?
A very thin film looks black
Let the film become far thinner than a quarter of any wavelength. Then \(2\mu t\) is almost zero, and the only difference left between the two reflected waves is the half wave. That is the dark condition with \(n = 0\), and it holds for every wavelength at once. So the film reflects no colour at all and looks black. In the figure above, a film \(10\,\text{nm}\) thick reflects at most about a twentieth of what a bright film would.
What happens to the light that is not reflected? It goes through. The light a film transmits is made of a direct wave and a wave reflected twice inside the film. Both inner reflections are from the film towards air, a rarer medium, so neither gains a half wave. The transmitted waves differ by \(2\mu t\cos r\) alone. Where the reflected light is dark, the transmitted light is bright, and the other way round.
The two patterns are complementary: together they hold all the light, because the film absorbs almost nothing. The colours of the transmitted light are pale, though. Bare glass reflects about \(4\,\%\) at each face, so even the brightest reflection of a glass-like film is only about \(15\,\%\). The transmitted light therefore changes only between about \(85\) and \(100\,\%\).
Not every rainbow on a surface comes from a thin film. The colours of a CD or a DVD come from its closely spaced tracks, which spread white light into its colours like a grating. That is diffraction, the subject of the chapter on diffraction. A thin film has two surfaces and no tracks.
A part of a soap film is black in reflected light because it is far thinner than a wavelength. How does the same part look in transmitted light?
An oil film on a puddle shows its colours in bands, and each band follows a curve. Explain why.
The oil spreads unevenly, so its thickness changes from place to place. At each place, the wavelength reflected most strongly depends on the thickness through \(2\mu t\cos r = (n - \tfrac12)\lambda\). All the points where the oil has the same thickness reflect the same colour. Those points lie along a curve, like a contour line of the oil's thickness. So each colour forms a band that follows a curve of equal thickness.
A peacock's feather owes its blue and green to no pigment. Its tiny barbs carry thin, regular layers, a fraction of a micrometre apart, and light reflected from them interferes, just as in a film. That is why the feather's colour shifts as the bird turns: a slanting view changes the path difference. Soap bubbles at a mela, the sheen on a beetle's wing and the oil on a puddle all colour themselves the same way.
Now return to the monsoon puddle. The oil is a film with a half wave at its front surface only. Where it is thin enough to hold one bright wavelength, it glows in that colour, and each colour band traces a line of equal thickness. Where it is a few micrometres thick, many wavelengths are missing at once, spread across the spectrum, and the oil looks plain. A slanting view shortens the path, so the bands slide as you walk.
- Forgetting the half wave, and calling a film bright when \(2\mu t = n\lambda\).Count the denser reflections first. With one of them, \(2\mu t = n\lambda\) is dark and \(2\mu t = (n - \tfrac12)\lambda\) is bright.
- Using the angle of incidence \(i\) in \(2\mu t\cos r\).The angle is \(r\), inside the film. Find it from \(\sin i = \mu\sin r\).
- Using the wavelength inside the film in the conditions.The factor \(\mu\) in \(2\mu t\) already turns the thickness into a path in air. Use the wavelength in air.
- Expecting a very thin film to look brightest, because its path difference is nearly zero.A nearly zero path leaves only the half wave, so the reflections cancel. A very thin film looks black.