THIN FILM
Soap-film colors: 2nt = (m−½)λ — and a hidden half-flip makes a vanishing film go black, not bright.
Use the simulation above to change the variables and play through the guided stages. The explanation below describes the default starting values; the simulation updates its explanation as you experiment.
Setup
A transparent film — a soap bubble, an oil slick, a lens coating — only a few hundred nanometres thick, lit by white light. Light reflects off both surfaces; the two returns overlap and interfere. Which color bounces back depends on a number you cannot see: the thickness.
Two beams
Two interface amplitudes return with Fresnel coefficients r₀₁ and r₁₂; a negative coefficient carries the π flip. The lower ray disappears entirely when film and substrate indices match. The round-trip phase locates extrema, while the amplitudes set their real contrast.
Solve
The round-trip phase is δ = 4πnt/λ. Phase tells where the returns align, but perfect anti-reflection also requires equal interface amplitudes: at normal incidence that means n² = n_sub as well as quarter-wave thickness.
Colors
The spectrum now plots the exact lossless multilayer reflectance R = |(r₀₁+r₁₂e^{iδ})/(1+r₀₁r₁₂e^{iδ})|². Slide the thickness and its physical Fresnel peaks march through the visible band; they are not normalized to artificial unit height.
Audit
Audited: at zero thickness the stack collapses to the direct air–substrate Fresnel reflectance; an unmatched quarter-wave coating keeps a nonzero residual; an index-matched quarter-wave coating reaches zero; and n = n_sub removes the lower reflected ray entirely.