Thin-Film Interference

Start from the macroscopic Maxwell equations for a linear, isotropic, nonmagnetic medium with no free charge. Fields vary as $\exp(-i\omega t)$, units are SI, $\varepsilon(\omega)$ is the relative permittivity, $k_0=\omega/c$ is the vacuum wave number and $Y_0=\sqrt{\varepsilon_0/\mu_0}$ is the vacuum admittance.

In a uniform region $\varepsilon$ is constant, so $\nabla\cdot\mathbf E=0$. Taking the curl of Faraday's law, substituting the Ampère–Maxwell law and using $\nabla\times\nabla\times\mathbf E=\nabla(\nabla\cdot\mathbf E)-\nabla^2\mathbf E$ gives the Helmholtz equation, with $n^2=\varepsilon$:

Put the ambient medium at $z<0$, the film at $0<z<d$ and the substrate at $z>d$. The structure is uniform along $x$, so every field shares one $x$-dependence, $e^{ik_0n_0\sin\theta\,x}$, where $\theta$ is the incidence angle in the ambient. Putting $e^{ik_0(n_0\sin\theta\,x+qz)}$ into the Helmholtz equation gives each medium its normal wave number $q=\sqrt{\varepsilon-n_0^2\sin^2\theta}$. The admittance is the ratio of tangential $\mathbf H$ to tangential $\mathbf E$ for a forward wave, in units of $Y_0$: $H_y/E_x$ for TM and $-H_x/E_y$ for TE. That gives $Y=q$ (TE) or $Y=\varepsilon/q$ (TM, with $\varepsilon=n^2$). Light at an Interface derives these. Label ambient, film and substrate 0, 1, 2 in reflection subscripts and a, f, s on admittances ($Y_0$ is already the vacuum value). So $r_{01}$ is the interface formula $r=(Y_1-Y_2)/(Y_1+Y_2)$ from Light at an Interface with $Y_1\to Y_\mathrm{a}$, $Y_2\to Y_\mathrm{f}$, and $r_{12}$ uses $Y_\mathrm{f}$, $Y_\mathrm{s}$. Without a subscript, $q$ means the film's value.

Inside the film the tangential electric field is a forward plus a backward wave. Faraday's law (TE) or Ampère's law (TM) then gives the tangential magnetic field:

Here $E_\parallel$ is $E_y$ (TE) or $E_x$ (TM), and $H_\parallel$ is $-H_x$ (TE) or $H_y$ (TM). Above the film the same form holds with $A=1$, $B=r$, $q_\mathrm{a}=n_0\cos\theta$ and $Y_\mathrm{a}$; below it there is only $t\,e^{iq_\mathrm{s}k_0(z-d)}$ with $Y_\mathrm{s}$. Both tangential components are continuous at each interface.

At $z=d$, with $p=e^{i\delta}$ and $\delta=qk_0d$, continuity gives

Dividing the second by the first gives $Y_\mathrm{f}(1-x)/(1+x)=Y_\mathrm{s}$ with $x=B/(Ap^2)$, so

$\Gamma$ is the substrate reflection seen from the top of the film, delayed by one round trip. The solver stores the backward amplitude at the film's exit instead, $B\,e^{-i\delta}=A\,r_{12}\,e^{i\delta}$, so no growing exponential appears in an absorbing film. At $z=0$:

Eliminate $A$, solve for $r$, and divide the numerator and denominator by $Y_\mathrm{a}+Y_\mathrm{f}$, with $r_{01}=(Y_\mathrm{a}-Y_\mathrm{f})/(Y_\mathrm{a}+Y_\mathrm{f})$:

That is the formula from Adding up every bounce. To recover the bounces, use $t_{01}=1+r_{01}$, $t_{10}=1-r_{01}$ and $r_{10}=-r_{01}$, and expand the denominator wherever $|r_{01}\Gamma|<1$:

The first term is the top reflection. Term $m$ of the sum is a wave that enters, makes $m+1$ round trips, reflecting $m$ times off the underside of the top surface ($r_{10}$), and leaves; $r_{10}r_{12}e^{2i\delta}=-r_{01}r_{12}e^{2i\delta}$ is the round-trip factor from the lecture. The condition holds for every transparent film with $d>0$. It can fail for a thin metal film in TM when the field in the substrate is evanescent, near a surface-plasmon resonance: the series then diverges, but the closed form still holds. The same conditions give $t=Ap(1+r_{12})$ and $A=t_{01}/(1+r_{01}\Gamma)$, so

The powers come from the time-averaged normal Poynting flux $S_z=\tfrac12\operatorname{Re}(E_\parallel H_\parallel^{*})$. In the transparent ambient the incident–reflected cross terms are imaginary and drop out, leaving $\tfrac12 Y_0Y_\mathrm{a}(1-|r|^2)$; the substrate carries $\tfrac12 Y_0\operatorname{Re}Y_\mathrm{s}\,|t|^2$. What is left, $1-R-T$, is absorbed in the film.

Two notes on conventions. In TM, $r$ and $t$ are ratios of $E_x$, as in the solver. This differs from textbook $r_p$, $t_p$ written for the full field, with

at each interface: here $r=-r_p$ and $t=t_p\cos\theta_\mathrm{s}/\cos\theta$, where $\theta_\mathrm{s}$ is the angle in the substrate. $R$ and $T$ are unchanged. For an absorbing film $q$ and $\delta$ are complex and $e^{2i\delta}$ includes the decay; nothing else changes.

Two special cases. A free-standing slab has $Y_\mathrm{s}=Y_\mathrm{a}$, so $r_{12}=-r_{01}$ and $r=r_{01}(1-e^{2i\delta})/(1-r_{01}^2e^{2i\delta})$, which vanishes for a lossless film when $2qd=m\lambda$, $m=1,2,\ldots$. At normal incidence $Y=n$, so $r_{01}=r_{12}$ exactly when $n_\mathrm{f}^2=n_0n_\mathrm{s}$; a quarter-wave film, $n_\mathrm{f}d=\lambda/4$, makes $e^{2i\delta}=-1$, the numerator becomes $r_{01}-r_{12}=0$, and reflection vanishes.