Thin-Film Interference and Anti-Reflection Coatings
Light reflected from a thin film's interfaces interferes. Thickness and reflection phase changes determine the reflected intensity. A quarter-wave coating can reduce reflection at a chosen wavelength; zero reflection also requires a suitable refractive-index match.
What a learner can do afterwards
- Works out the condition for constructive interference in a film, including the reflection phase change
- Explains the colours of an oil film and of a soap bubble as it drains
- Designs the thickness of an anti-reflection coating for a stated wavelength
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As in phasor addition, reflected waves interfere according to their relative phase. Here light strikes the film normally, and the materials are transparent, nonabsorbing dielectrics. Reflection toward a higher refractive index flips the electric-field phase by half a cycle. Reflection toward a lower index has no phase flip. These normal-incidence rules do not describe the phase changes in total internal reflection.
Let t be thickness, n the film index and lambda0 the vacuum wavelength. The extra round-trip optical path is 2nt. A soap film in air has one reflection flip. Reflection maxima satisfy 2nt = (m + 1/2)lambda0; minima satisfy 2nt = m lambda0, with m = 0, 1, 2, ... . When nt is much smaller than every visible wavelength, reflection is weak across the visible band. It approaches zero as thickness approaches zero.
For a soap film in air, m = 0 in the maximum rule gives the smallest bright thickness t = lambda0/(4n). The first nonzero minimum uses m = 1 and gives t = lambda0/(2n). With n = 1.25 and lambda0 = 620 nm, these are 124 nm and 248 nm. A given thickness can favour one wavelength while suppressing another.
The calculations compare the smallest bright thickness with the first nonzero dark thickness.
White light contains many wavelengths. As a soap film drains, its changing thickness changes which wavelengths reflect strongly, altering the colour. An oil film can have air above and water below. For oil index 1.46 and water index 1.33, only the front reflection flips. At a fixed viewing angle, thickness variations across one oil film can produce colour bands. Each patch can reflect a mixture of wavelengths.
For a coating with n_air < n_film < n_glass, both reflections flip. Their relative reflection phase change is zero. A reflection minimum needs half a cycle from propagation: 2n_film t = lambda0/2. The smallest positive thickness is t = lambda0/(4n_film), a quarter of the wavelength inside the film. This design assumes normal incidence and one chosen wavelength.
Design a coating for 600 nm light with film index 1.50, between air of index 1.00 and glass of index 2.25. Both reflections flip. Use t = 600/(4 × 1.50) = 100 nm. This calculation sets the quarter-wave phase condition. Other wavelengths have different round-trip phases.
This shows the thickness calculation, not a scaled drawing of the coating.
The two-ray picture identifies phase minima. Exact coating reflection includes repeated internal reflections. A uniform lossless quarter-wave layer at normal incidence has zero reflection only when n_film² = n_air n_glass. Magnesium fluoride of index 1.38 on glass of index 1.52 in air does not meet that condition, so its minimum remains nonzero. A single quarter-wave layer does not cancel every visible wavelength.
At normal incidence, count reflection flips and the round-trip phase. Quarter-wave thickness can reduce reflection. Zero reflection also requires the correct index match.
The lesson, practice questions and flashcards were written for this dataset.
The reading and flashcards were written for this dataset. Linked videos credited to Khan Academy.
More to read
Checking phase and index in a coating
A reflected wave can come from the front of a thin film or return after travelling inside it. Their relative phase depends on both propagation and reflection. For normal incidence on transparent, nonabsorbing dielectrics, reflection toward a higher index adds half a cycle. Reflection toward a lower index adds no phase flip.
Write the film thickness as t, its index as n and the vacuum wavelength as lambda0. The round-trip optical path is 2nt. A soap film surrounded by air has one reflection flip. Its reflection maxima satisfy 2nt = (m + 1/2)lambda0, while minima satisfy 2nt = m lambda0, with nonnegative integer m. A finite film much thinner than visible wavelengths reflects weakly across the visible band. Its reflection approaches zero in the zero-thickness limit.
Under white light, changing thickness changes which wavelengths reflect strongly. This accounts for changing soap-film colours and can produce bands across an oil film. Check the media on both sides: a different index order can change the reflection flips.
For air, a coating and glass with increasing indices, both reflections flip. Propagation must then add half a cycle for a reflection minimum. The smallest thickness is lambda0/(4n), one quarter of the wavelength inside the film.
A phase minimum need not be zero reflection. Exact reflection includes repeated internal reflections. At normal incidence, a uniform lossless quarter-wave layer reaches zero reflection at its design wavelength only when n² equals the incident index times the substrate index. Without that match, some reflection remains. One quarter-wave layer does not cancel the entire visible band.
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