Thin films and changes in the fringes

A soap bubble

A soap bubble shows bands of colour that change as the film thins. There is no coloured light in the soap. The colours come from interference. Light is reflected from the front surface of the film and from the back surface. The two reflected waves come from the same wave, so they are coherent, and they interfere.

Interference in a thin film

A film of refractive index $n$ and thickness $t$ lies in air. Light falls on it straight on, or nearly. The second reflected wave goes through the film twice, so the path difference is $2t$ in the film, and the optical path difference is $2nt$.

There is one more thing. A wave that is reflected at the surface of a denser medium (one with a higher refractive index) is turned over. It gains a phase change of $\pi$, which is the same as half a wavelength. A wave that is reflected at the surface of a rarer medium has no such change. For a film in air, the reflection at the front surface (air to film) has the change. The reflection at the back surface (film to air) has none. So the two reflected waves have an extra half wavelength between them:

• Bright in reflection: $2nt = \left(m + \tfrac12\right)\lambda$, with $m = 0, 1, 2, \dots$
• Dark in reflection: $2nt = m\lambda$, with $m = 0, 1, 2, \dots$

Here $\lambda$ is the wavelength in air. For light that comes at an angle, the path becomes $2nt\cos r$, with $r$ the angle of refraction in the film.

Look at the top of a vertical soap film. It is very thin, so $2nt \approx 0$, and the film looks dark. The two reflected waves have the half-wavelength difference and nothing else, so they cancel. With white light, each thickness reflects some colours strongly and others weakly. This is why a film shows bands of colour.

A worked example. A soap film, of refractive index $1.33$, is $290$ nm thick, in air. Which wavelengths in the visible range, $400$ nm to $700$ nm, does it reflect strongly, and which are missing?

1. The optical path. $2nt = 2 \times 1.33 \times 290\ \text{nm} = 771.4\ \text{nm}$.
2. Bright. $\lambda = \dfrac{2nt}{m + 1/2}$. For $m = 1$: $\lambda = \dfrac{771.4\ \text{nm}}{1.5} = 514.3$ nm. For $m = 0$ it is $1543$ nm (infrared), and for $m = 2$ it is $309$ nm (ultraviolet).
3. Dark. $\lambda = \dfrac{2nt}{m}$: for $m = 1$ it is $771$ nm (infrared), and for $m = 2$ it is $386$ nm, below $400$ nm.
4. The result. Only $514.3$ nm is in the visible range and strongly reflected. The film looks green.

The anti-reflection coating

A camera lens or a pair of spectacles often looks slightly purple. A thin coating on the glass cuts down the reflection. Take a glass of index $1.52$ and a coating of magnesium fluoride, $n = 1.38$. The coating is between air and glass. The reflection at both surfaces is from a denser medium, so both have the half-wavelength change, and it cancels. The two reflected waves are out of step if $2nt = \lambda/2$, so the least thickness is

For light at the middle of the visible range, $\lambda = 550$ nm, $t = \dfrac{550\ \text{nm}}{4 \times 1.38} = 99.6$ nm. A film one quarter of a wavelength thick, in the film, is called a quarter-wave coating. Light of other wavelengths is not cancelled as well, and a little red and violet comes back, which gives the purple look.

Changes to the fringes of Young's experiment

In course one we studied Young's double slit, with a fringe width $\beta = \lambda D/d$. Here are the standard changes. The first two you know already.

• The whole setup in a liquid of index $n$: $\lambda \to \lambda/n$, so $\beta \to \beta/n$.
• A thin transparent sheet over one slit. Let the sheet have thickness $t$ and index $\mu$. Light that goes through it has an extra optical path of $(\mu - 1)t$, since the sheet holds $\mu t$ of optical path in place of $t$ of air. The central bright fringe, which has zero path difference, moves to the side where this extra path is made up: towards the covered slit. The shift is

The pattern moves by $\dfrac{(\mu - 1)t}{\lambda}$ fringes. The fringe width stays the same.

• Moving the source slit. If the first slit $S_0$ is moved sideways by $s$, at a distance $L$ from the double slit, the whole pattern moves the other way by $\Delta y = \dfrac{sD}{L}$.
• Sheets of different thickness on both slits. The shift is given by the difference of the two extra paths, $(\mu_1 - 1)t_1 - (\mu_2 - 1)t_2$.

Which sheets move the central fringe by exactly one fringe width? They are the ones with $(\mu - 1)t = \lambda$. Try $t = 1.2\ \mu\text{m}$ and $\mu = 1.5$ for $\lambda = 600$ nm: the shift reads $1.00$ fringes.

A worked example. Sodium light of wavelength $589$ nm lights two slits, $0.50$ mm apart. The screen is $1.20$ m away. A sheet of index $1.50$ and thickness $4.0\ \mu\text{m}$ covers one slit. How far does the central fringe move?

1. The fringe width. $\beta = \dfrac{\lambda D}{d} = 1.41$ mm. The sheet does not change it.
2. The extra path. $(\mu - 1)t = 0.50 \times 4.0\ \mu\text{m} = 2.00\ \mu\text{m}$.
3. The number of fringes. $\dfrac{2.00\ \mu\text{m}}{589\ \text{nm}} = 3.40$ fringes.
4. The shift. $\Delta y = 3.40 \times 1.41\ \text{mm} = 4.80\ \text{mm}$, towards the covered slit.
5. Check. The direct formula, $\dfrac{(\mu - 1)tD}{d} = \dfrac{2.00 \times 10^{-6}\ \text{m} \times 1.20\ \text{m}}{0.50 \times 10^{-3}\ \text{m}} = 4.80\ \text{mm}$, agrees.

Practise with new numbers

Try it in Python

The program finds which colours a soap film reflects strongly, and which ones it does not reflect at all, for a given thickness.

Common mistakes

• Forgetting the half-wavelength change. For a film in air, the bright condition is $2nt = (m + \tfrac12)\lambda$, not $2nt = m\lambda$.
• Using $\lambda$ from the film in the condition. The condition with $2nt$ uses the wavelength in air. If you use the wavelength in the film, $\lambda/n$, the path difference is $2t$.
• Using $\mu t$ for the shift of the fringes. The extra path is $(\mu - 1)t$.
• Thinking that a sheet changes the fringe width. It does not. It only moves the whole pattern.

Remember these

• Film in air, straight on: bright in reflection when $2nt = (m + \tfrac12)\lambda$, dark when $2nt = m\lambda$.
• A very thin film looks dark in reflection.
• Quarter-wave coating: $t = \lambda/4n$ cancels the reflection of one wavelength.
• A sheet over one slit: shift $= (\mu - 1)tD/d$ towards the covered slit. The fringe width is unchanged.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Why does the top of a vertical soap film look dark, just before it breaks?

2. A sheet is put over one slit of a double slit. Does the fringe width change? In which direction does the pattern move?

3. A thin sheet of index $1.5$ and thickness $2\ \mu\text{m}$ covers one slit. Light of wavelength $500$ nm is used. By how many fringes does the pattern move?

4. The source slit $S_0$ of a double slit is moved up. Which way does the central fringe move?

References

• OpenStax, University Physics Volume 3, Chapter 3 (Interference): interference in thin films.
• NCERT Physics Class 12, Chapter 10 (Wave Optics): Young's experiment and its changes. Thin-film interference belongs to A-Level, IB and AP.