Overview#

• Thin-film interference causes the colors on soap bubbles, oil slicks, and anti-reflection coatings. Light reflected from the front and back surfaces of a thin transparent film interferes.

• The phase difference between the two reflected waves depends on three factors: (1) reflection phase shifts at each interface, (2) path length difference \(2L\) (ray \(r_2\) travels through the film twice), and (3) the internal wavelength \(\lambda_n = \lambda/n\) in the film.

• Reflection can add a phase shift of \(\pi\) (half-wavelength); refraction never does.

Setup#

Light is incident on a thin film of thickness \(L\) and index \(n_2\), with media \(n_1\) and \(n_3\) on either side. Ray \(r_1\) reflects from the front surface; ray \(r_2\) reflects from the back surface and emerges. Both reach the observer. For near-normal incidence, the path length difference is \(2L\) (the extra distance \(r_2\) travels in the film).

Reflection phase shifts#

Rule: Reflection causes a phase shift of \(\pi\) rad (equivalent to half-wavelength) only when the incident light is in the medium with the smaller index of refraction (i.e., reflecting off a “denser” medium).

Refraction never causes a phase shift.

Film in air#

For a film in air: \(n_1 = n_3 = 1\), \(n_2 = n > 1\).

• \(r_1\) (front surface): air \(\to\) film, \(n_1 < n_2\) \(\Rightarrow\) phase shift 0.5\(\lambda\).

• \(r_2\) (back surface): film \(\to\) air, \(n_2 > n_3\) \(\Rightarrow\) no phase shift.

So the reflection shifts contribute 0.5\(\lambda\) to the phase difference between \(r_1\) and \(r_2\). The path adds \(2L\) in the film, i.e., \(2L/\lambda_n = 2nL/\lambda\) wavelengths.

Total phase difference (in wavelengths) \(= 0.5 + 2nL/\lambda\).

Very thin film (\(L \ll \lambda\)): \(2nL/\lambda \approx 0\), so total phase \(\approx 0.5\) \(\Rightarrow\) destructive \(\Rightarrow\) film appears dark in reflected light.

Films between other media#

When \(n_1\), \(n_2\), \(n_3\) differ (e.g., coating on glass), the bright and dark equations may interchange. Use an organizing table:

1. For \(r_1\): does reflection at the front interface cause a phase shift? (Yes if \(n_1 < n_2\).)

2. For \(r_2\): does reflection at the back interface cause a phase shift? (Yes if \(n_2 < n_3\).)

3. If the reflection shifts differ (one yes, one no): use the film-in-air formulas above.

4. If the reflection shifts are the same (both yes or both no): the path alone determines relative phase; bright and dark conditions swap.

Anti-reflection coating#

A thin layer (e.g., MgF\(_2\), \(n_2 = 1.38\)) on glass (\(n_3 = 1.50\)) can eliminate reflection at a chosen wavelength. Both interfaces give a phase shift (air\(\to\)MgF\(_2\) and MgF\(_2\)\(\to\)glass: \(n_2 < n_3\)). For destructive interference of reflected light:

Minimum thickness (\(m = 0\)): \(L = \lambda/(4n_2)\). For \(\lambda = 550\) nm: \(L \approx 100\) nm.

Wedge-shaped films#

For a wedge (thickness varies from \(L_L\) to \(L_R\)), the brightness varies along the film. Apply the same equations at each point; \(L\) changes, so different regions satisfy bright or dark conditions.

• At each end, determine whether that thickness gives bright or dark.

• Count how many times \(2nL/\lambda\) increases by \(\frac{1}{2}\) (or 1) between the ends to find the number of fringes.

Summary#

• Reflection phase shift: \(\pi\) when incident from lower \(n\) to higher \(n\); otherwise 0.

• Film in air: Bright \(2nL = (m+\frac{1}{2})\lambda\); dark \(2nL = m\lambda\). Very thin \(\Rightarrow\) dark.

• Other configurations: Use reflection rules; bright/dark may swap.

• Anti-reflection: \(L = \lambda/(4n)\) for destructive reflection.

• Wedge: Thickness varies; count fringes by applying conditions along the film.