Thin-film interference

Thin-film interference combines coherent light contributions reflected or transmitted at the two surfaces of a layer. For a film of refractive index $n$, thickness $d$ and internal angle ${\theta }_t$, the round-trip optical path difference is $2nd\cos {\theta }_t$. The relative reflection phase shift must also be included. For a nonabsorbing higher-index film surrounded by air, one of the two leading reflected contributions has a $\pi$ phase shift and the other does not. Reflection maxima therefore satisfy $2nd\cos {\theta }_t=(m+1/2){\lambda }_0$. A smooth reflection spectrum filters many visible wavelengths at once; varying thickness and angle produce spatially varying mixed colours.

For a film of refractive index $n$ and thickness $d$ in air, its round-trip propagation phase is $\delta =(4\pi d/{\lambda }_0)\sqrt{n^2-{\sin }^2{\theta }_i}$. Two neighbouring same-type interference fringes differ by $2\pi$. At small angular spacing in radians,The relation assumes slowly varying interface reflection coefficients and no rapidly varying reflection phase. At normal incidence the first derivative vanishes, so a local linear angular-spacing formula must be replaced by a quadratic expansion.

A curved transparent surface resting on a plane traps a thin film whose thickness increases radially, producing circular interference fringes. For near-normal reflected light and one relative $\pi$ reflection wave phase change, contact is dark and bright rings obey $2nt=(m+1/2){\lambda }_0$ for $m=0,1,\dots$. Here $n$ is the film's refractive index and ${\lambda }_0$ the vacuum wavelength. For spherical curvature radius $R$, $t\simeq r^2/(2R)$ gives $r_m^2=R{\lambda }_0(m+1/2)/n$. Consequently $D_{m+s}^2-D_m^2=4sR{\lambda }_0/n$ for ring diameters $D_m=2r_m$. A contact gap, altered reflection phase, or significant oblique incidence changes the formula.