The solution to the wave equation and Faraday's law in a uniform layer, the field is a forward wave plus a backward wave described by the following $ \begin{align} E_{x}(z) &= E_{+}e^{-jknz}+E_{-}e^{+jknz} \\ H_{y}(z)&= \frac{n}{\eta_0}\Big(E_+e^{-jk_0nz}-E_-e^{+jk_0nz}\Big) \end{align} $ where $n$ is the refractive index and $\eta_{0}=\sqrt{ \mu_{0}/\epsilon_{0} }$ Tangential $E$ and $H$ must be continuous, so at an interface between materials with refractive indices $n$ and $n'$ we have $ \begin{align} E_+'+E_-'&=E_++E_- \\ n'\,(E_+'-E_-')&=n\,(E_+-E_-) \end{align} $ This can be rewritten in matrix form as $ \begin{bmatrix} E'_{+} \\ E'_{-} \end{bmatrix}=\underbrace{ \frac{1}{2n'}\begin{bmatrix}n'+n & n'-n \\ n'-n & n'+n \end{bmatrix} }_{ V }\begin{bmatrix} E_{+} \\ E_{-} \end{bmatrix} $ When light moves through a layer with thickness $d$, it gains phase defined by phase thickness $\delta\equiv k_{0}nd$, where $k_{0}$ is the free-space wavenumber $k_{0}=\frac{\omega}{c}$ which doesn't change between layers with varying $n$. This phase shift gives us the propagation matrix $P$: $ \begin{bmatrix} E'_{+} \\ E'_{-} \end{bmatrix}=\underbrace{ \begin{bmatrix}e^{-j\delta} & 0 \\ 0 & e^{+j\delta} \end{bmatrix} }_{ P }\begin{bmatrix} E_{+} \\ E_{-} \end{bmatrix} $ Note that - Quarter wave layers: $d_i=\dfrac{\lambda_0}{4n_i}$, $\delta=\frac{\pi}{2}$, anti-reflection - Half wave layers: $\delta=\pi$, $P=-I$, transparent Therefore we can calculate the steady state solution by chaining these matrices corresponding to mediums from left to right: $ \mathbf M=\mathbf V_{\text{last}}\,\mathbf P_N\,\mathbf V_{N}\cdots\mathbf P_1\,\mathbf V_{1} $ Then we normalize the incident wave to 1 and solve for the reflective coefficient: $ \begin{pmatrix}T\\0\end{pmatrix}=\mathbf M\begin{pmatrix}1\\\Gamma\end{pmatrix} $ $\Gamma=-\frac{M_{21}}{M_{22}}=$ reflective coefficient $R=\lvert \Gamma \rvert^{2}=$ reflected power $T_{\text{power}}=\frac{n_{\text{exit}}}{n_{\text{in}}}\lvert T \rvert^{2}=$ transmitted power