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Interference in thin films

The beautiful colours in soap films and in oil floating on water are due to interference by reflection. We will first consider the interference due to a parallel-sided film as shown in Figure 1. Interference can be seen at O by division of amplitude, some light being reflected from each side of the film.

Part of the light beam travels a distance BEP = 2BE while the other travels a distance BM.
The path difference is therefore = 2nBE BM

Now the distance 2BE is in the thin film which has a refractive index n and so the equivalent path length in a vacuum is 2nBE
So path difference = 2nBE nBN = n(BE + EF - BN)
= nNF = 2nt cos r

where n is the refractive index of the film of thickness t.

Path difference in a thin film = 2nt cosr

For a soap film in air a phase change will occur at the first face and so a maximum will occur when the path difference is equal to an odd number of half wavelengths.


Path difference in a thin film = 2nt cosr = (m + )l

where m = 0, 1, 2, etc.

For an oil film on water the phase changes will depend on the refractive index of the oil.

If we now consider the soap film and normal incidence (i.e. cos r = 1) it is easy to see why thick films reflect light but do not show coloured effects and appear transparent and why very thin films appear 'black' i.e. do not reflect light. The idea of the next set of calculation is to see which wavelengths will 'fit' into the film and so give a maximum. In other words to find out which colours will be seen. Consider a soap film of refractive index 1.33 illuminated by white light.

For t =10-7m and m = 0         l = 5.32x 10-7m = 532 nm (green)
while for m=1         l = 1.77x10-7 m = 177 nm (ultraviolet)
Therefore only one colour is visible at this thickness at normal incidence.

For t = 10-6 m and m= 0         l = 5.32 x 10-6 m = 5320 nm (infrared)
m = 4         l = 5.91 x 10-7 m = 591 nm (yellow)
m = 5         l = 4.75 x 10-7 m = 475 nm (blue)
All higher values of m give wavelengths in the ultraviolet region, so only two colours are visible.

For t= 10-5 m and with m= 0         l = 5.32 x 10-5m (infrared)
with m=36         l = 7.28x10-7m (red)
with m=59         l= 4.52x10- 7m (violet)

All higher values of m give wavelengths in the ultraviolet region. Therefore values of m between 36 and 59 will "fit", there are 24 wavelengths that give maxima and so the appearance is very nearly white.

For thicker films still many more wavelengths can 'fit in', and so all thicker films appear to reflect white light.

For very thin films the distance travelled inside the film is insignificant and so the two reflected waves are almost exactly out of phase with each other (due to the phase change at one surface); they interfere destructively and the film appears 'black'. For this reason when a soap film goes black it is about to burst.

Example problem
Example problem A film of oil 0.0005 mm thick and of refractive index 1.42 lies on a pool of water.
Which colour will be missing from the spectrum when a point on the film is viewed at 400 to the vertical?

Destructive interference occurs: ml = 2ntcos r ml = 2 x 1.42 x 5 x 10-7 x cos r
Therefore: ml = 2 x 1.42 x 5 x 10-7 x 0.89
m = 1   1.26 x 10-6 m (infrared),    m = 2    6.33 x 10-7 m (orange),    m = 3   4.22 x 10-7 m (ultraviolet).
Thus the only colour missing from the visible part of the spectrum will be an orange line of wavelength 6.33 x 10-7 m.

The colours in a soap film can be observed clearly by projecting them on to a screen. A wire ring with a soap film formed across it should be mounted vertically in the beam of a high-intensity light source in such a way that the light is focused just behind the film. A further lens may then be used to project the colours formed on to a screen. As the water runs from the film it gets progressively narrower at the top and turns black when it is about to break. Beautiful effects may be obtained by blowing gently on the film.


Student investigation
It has been suggested that the gun turrets on a ship (Figure 3) may be made more stable by using inter-ference damping. This means feeding an oscillation into the mounting of the same type and amplitude as that causing the vibration, but out of phase with it by half a period. Test this in the laboratory using two vibrators and a signal generator and a method for delaying the output of one vibrator
© Keith Gibbs 2011