How does interference occur in a thin film of uniform thickness, and what is the path difference?

In one line

Light reflected from the top and bottom surfaces of a thin film comes from the same wave (division of amplitude), so the two reflected beams interfere, with an effective path difference \(2\mu t\cos r - \lambda/2\).

Explained simply

The rainbow colours on a soap bubble or an oil patch on a wet road are thin-film interference. Part of the light bounces off the top surface and part goes in, bounces off the bottom and comes back out. The two reflected beams overlap; depending on the film thickness and wavelength, they add (bright colour) or cancel (dark).

Technically, this is interference by division of amplitude: one incident wave is split into two by partial reflection, so the two beams are coherent.

How it works

1. A ray hits a film of thickness \(t\) and refractive index \(\mu\) at angle \(i\). Part reflects at A (ray 1); part refracts at angle \(r\), reflects from the bottom at B, and emerges at C (ray 2).
2. Geometric path difference \(= \mu(AB + BC) - AN\), where AN is the path ray 1 travels in air while ray 2 is inside the film.
3. From geometry, \(AB = BC = t/\cos r\) and \(AN = AC\sin i = 2t\tan r\sin i\). Using \(\sin i = \mu\sin r\): \(AN = 2\mu t\sin^2 r/\cos r\).
4. So path difference \(= \dfrac{2\mu t}{\cos r} - \dfrac{2\mu t\sin^2 r}{\cos r} = 2\mu t\cos r\).
5. Ray 1 reflects from a denser medium (air to film), so it suffers a phase change of \(\pi\), equal to an extra path \(\lambda/2\) (Stokes' law). Ray 2 reflects from film to air at the bottom, with no phase change.
6. Effective path difference \(\Delta = 2\mu t\cos r - \lambda/2\).

The film must be thin: if it is thick, the path difference exceeds the coherence length, and with white light the many overlapping colours wash out, so no pattern is seen.

Key formula

\(\Delta = 2\mu t\cos r - \dfrac{\lambda}{2}\). \(\mu\): refractive index of film, \(t\): thickness (m), \(r\): angle of refraction inside the film, \(\lambda\): wavelength in air (m).

Quick example

A soap film (\(\mu = 1.33\), \(t = 500\) nm) is seen in white light at normal incidence (\(\cos r = 1\)). \(2\mu t = 1330\) nm. Bright reflection when \(2\mu t = (2n-1)\lambda/2\): \(\lambda = 2660/(2n-1)\) gives 2660, 887, 532, 380 nm. Only 532 nm (green) is well inside the visible range, so the film looks greenish.

Common mistakes

• Forgetting the \(\lambda/2\) phase change on reflection at the denser surface.
• Writing \(2\mu t\cos i\) instead of \(2\mu t\cos r\).
• Calling this division of wavefront; thin films use division of amplitude.

How MU asks it

A standard 5-mark derivation: "Derive the conditions for maxima and minima for a thin film of uniform thickness in reflected light". Draw the ray diagram with A, B, C, N marked, derive \(2\mu t\cos r\), explain the \(\lambda/2\) term, then write the conditions.