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Thin Films & the Phase Shift on Reflection

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The rule that decides the whole problem

Light reflecting off a boundary picks up a half-wavelength phase shift if it is going from a lower index into a higher index medium. Reflecting off a lower-index medium produces no shift. That single rule determines whether the constructive condition is 2t = mλ or 2t = (m + ½)λ, and getting it wrong inverts every answer in the problem.

Thin film conditions
wavelength in the film: λ_film = λ_vacuum / n · path difference from two traversals: 2t · ONE net phase shift → constructive when 2t = (m + ½)λ_film · ZERO or TWO shifts → constructive when 2t = mλ_film
Count the shifts first, then choose the condition. Two shifts cancel each other, which is why zero and two give the same rule.

The counting procedure

Work through it in order. One: does the top reflection have a shift? Going from air into the film means yes if n_film > 1, which it always is. Two: does the bottom reflection have a shift? That depends on whether the material below the film has a higher or lower index than the film. Three: count the total. One net shift means the conditions swap; zero or two means they do not. Writing this count down explicitly is what keeps the problem tractable.

Why soap films are colored

A soap film in air has a higher index than the air on both sides. So the top reflection shifts and the bottom does not — one net shift. A film of varying thickness therefore satisfies the constructive condition for different wavelengths at different thicknesses, and different colors appear in bands. Where the film is vanishingly thin, 2t ≈ 0 and the one shift makes it destructive for all wavelengths — which is why the very thinnest part of a soap film looks black just before it bursts.

Anti-reflective coatings

A lens coating is designed to make reflection destructive so that more light is transmitted. Coatings are usually chosen with an index between air and glass, so both reflections have a phase shift — two shifts, which cancel — and destructive interference then requires 2t = (m + ½)λ_film. The thinnest coating that works is a quarter of the wavelength in the film. Because the condition depends on wavelength, the cancellation is exact for only one color, which is why coated lenses show a faint purple or green residual reflection.

Worked example

A soap film of index 1.33 in air appears bright green (λ = 510 nm in vacuum) in reflection. Find the thinnest film thickness that produces this.

  1. 1.Wavelength in the film: λ_film = 510/1.33 = 383 nm.
  2. 2.Count shifts: air (1.00) into film (1.33) is low to high, so the top reflection shifts. Film (1.33) into air (1.00) is high to low, so the bottom does not. ONE net shift.
  3. 3.One shift means constructive requires 2t = (m + ½)λ_film.
  4. 4.Thinnest is m = 0: 2t = 0.5 × 383 = 191 nm.
  5. 5.t = 96 nm.
Answer: About 96 nm. Note that the wavelength inside the film, not in vacuum, is what enters the condition — using 510 nm directly would overstate the thickness by a third.
Watch out

Use the wavelength in the film, λ/n, not the vacuum wavelength. The light traverses the film, so it is the film wavelength that must fit. This is the second most common error after miscounting the phase shifts.

Checkpoint

Light reflects off a boundary going from a lower-index medium into a higher-index medium. The reflected wave:

Checkpoint

The very thinnest part of a soap film in air appears black in reflected light because:

Checkpoint

Light of vacuum wavelength 600 nm enters a film of index 1.50. Its wavelength inside the film is:

On the exam

Write the phase-shift count explicitly — "top: shift; bottom: no shift; one net shift, so constructive is 2t = (m + ½)λ_film". That line is often its own rubric point and it prevents the inverted answer that otherwise looks fully worked.

Answer the 3 checkpoints as you read.

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