--- title: "Fresnel formulae" date: "`r format(Sys.time(), '%d %B, %Y')`" author: "baptiste Auguié" output: rmarkdown::html_vignette: toc: true toc_depth: 2 fig_width: 7 fig_height: 4 fig_caption: true vignette: > %\VignetteIndexEntry{fresnel} %\VignetteEngine{knitr::rmarkdown} \usepackage[utf8]{inputenc} --- ## Single interface Let us consider the situation depicted below, ![Schematic of reflection and transmission by a single interface. Note that the orientation of the vectors has been chosen so as to have a consistent picture at normal incidence.](interface.png) where we define the Fresnel coefficients as the ratio of the complex _amplitude_ of the _electric_ fields, $r= E_r/E_i$ for the reflection coefficient, and $t= E_t/E_i$ for the transmission coefficient. The Fresnel coefficients can be obtained by considering the continuity of the tangential component of the $E$ and $H$ fields at the interface. The continuity of the normal components of $D$ and $B$ does not yield supplementary conditions and will not be considered. It is worth noting that the continuity of $H^y$ is only valid for materials that do not sustain a surface current resulting from the addition of external charges. ### TE-polarised light For TE-polarised light, the continuity of $E^y$ reads, \[ E_i + E_r = E_t \] which yields, $$ 1+r =t. $$ The continuity of $H^x$ can be written as, $$ H_i\cos\theta_i - H_r\cos\theta_i = H_t\cos\theta_t $$ It is convenient to express the angles in terms of the normal component of the $k$-vectors, $$ k_z = n k_0 \cos\theta. $$ for the incident, reflected, and refracted fields, so that $\cos\theta_t / \cos\theta_i=\dfrac{k_{z2}n_1}{k_{z1}n_2}$. The complex magnitude of the magnetic and electric fields are linked in each medium by the optical impedance. Its expression can be obtained from the induction law in a homogeneous medium, \[ \nabla\times\mathbf{E} = -\partial_t \mathbf{B} \] expressed in Fourier space\footnote{assuming here a homogeneous plane wave} as, \[ \mathbf{k}\times \mathbf{E} = i\omega \mathbf{B} \] The ratio $E/H$ is the impedance, written as, $$ \frac{E}{H}= c \mu \mu_0 / n= \sqrt{\frac{\mu\mu_0}{\varepsilon\varepsilon_0}}=Z\cdot Z_0. $$ We can define the admittance $Y$ as the inverse impedance, $Y=\dfrac{1}{Z\cdot Z_0}$ The continuity of $H^y$ can be expressed as, \[ Y_1 (1 - r) = Y_2 \frac{n_1k_{z2}}{n_2k_{z1}}t. \] After rearranging, we obtain, $$ 1 - r = \frac{\mu_1k_{z2}}{\mu_2k_{z1}}t. $$ We can summarize the two continuity relations in the following system, $$\left\{ \begin{aligned} 1 + r &= t\\ 1 - r &= \frac{\mu_1 k_{z2}}{\mu_2 k_{z1}}t \end{aligned}\right. $$ Solving for $r$ and $t$ yields the result, $$ t_{12}^s=\frac{2\mu_2 k_{z1}}{\mu_2 k_{z1}+\mu_1k_{z2}},\qquad r_{12}^s=\frac{\mu_2 k_{z1}-\mu_1k_{z2}}{\mu_2 k_{z1}+\mu_1k_{z2}}. $$ ### TM-polarised light For TM-polarised light, it is generally easier to consider the Fresnel coefficients for the magnetic field, denoted $\rho$ and $\tau$. The continuity of $H^y$ reads, \[ 1 - \rho = \tau \] The continuity of $E^y$ can be written, \[ \frac{k_{z1}}{\varepsilon_1} + \rho \frac{k_{z1}}{\varepsilon_1} = \tau \frac{k_{z2}}{\varepsilon_2} \] We can summarize the two continuity relations in the following system, $$\left\{ \begin{aligned} 1 - \rho &=\tau\\ (1 + \rho) \frac{k_{z1}}{\varepsilon_1}&=\frac{k_{z2}}{\varepsilon_2}\tau \end{aligned}\right. $$ Solving for $\rho$ and $\tau$ yields the result, $$ \tau_{12}^p=\frac{2\varepsilon_2 k_{z1}}{\varepsilon_2 k_{z1}+\varepsilon_1k_{z2}},\qquad \rho_{12}^p=\frac{\varepsilon_2 k_{z1}-\varepsilon_1k_{z2}}{\varepsilon_2 k_{z1}+\varepsilon_1k_{z2}}. $$ To summarize, for a single interface from 1 to 2 with normal along the $z$ direction, the Fresnel coefficients read, $$ \begin{aligned} \rho_{12}^p&=\frac{\varepsilon_2 k_{z1}-\varepsilon_1k_{z2}}{\varepsilon_2 k_{z1}+\varepsilon_1k_{z2}},&{}& r_{12}^s=\frac{\mu_2 k_{z1}-\mu_1k_{z2}}{\mu_2 k_{z1}+\mu_1k_{z2}}\\ \tau_{12}^p&=\frac{2\varepsilon_2 k_{z1}}{\varepsilon_2 k_{z1}+\varepsilon_1k_{z2}},&{}& t_{12}^s=\frac{2\mu_2 k_{z1}}{\mu_2 k_{z1}+\mu_1k_{z2}}. \end{aligned} $$ Note that, $$ r_{ij}=-r_{ji}. $$ To verify the conservation of energy, one must consider the irradiance defined in terms of the electric field as $I=\frac{1}{2}Y|E|^2$, yielding, $$ R=\frac{I_r}{I_i}=|r|^2, \qquad T=\frac{I_t}{I_i}=\frac{Y_1}{Y_2}|t|^2. $$ and we can verify that, indeed, $R+T=1$ for a single interface. ## Reflectivity of a layer From the viewpoint of ray optics, a thin layer will support an infinite number of internal reflections (absorption and irregularities will however reduce the intensity in a physical situation). The infinite series of reflected orders can be expressed in the form a geometric sum, leading to a closed form formula as shown below. ![Schematic of reflection and transmission by a multilayer stack. A few reflected orders are noted `A`, `B`, `C` and `D` for the first film.](stack.png) An incident plane wave with amplitude $A$ impinges on the first interface. It can be reflected, $B=r_{01}A$, or transmitted. The total response of the slab can be obtained by following each order of reflection inside the slab (`C`, `D`, $\dots$). Upon transmission at the first interface, the wave amplitude is $t_{01}A$. Application of Fermat's principle yields a phase change $\Delta \phi= k_{z1}d$ when the wave hits the second interface. The reflection coefficient at this interface is $r_{12}$. The partial wave reflected from this path, noted `C`, is therefore $C=t_{10}t_{01}r_{12}\exp(2i k_{z1}d)A$. Similarly, in `D`, \[ D=t_{10}t_{01}r_{10}r_{12}^2\exp(4i k_{z1}d)A \] And, for the ${j^\text{th}}$ partial wave, \[ t_{10}t_{01}r_{12}^{j}r_{10}^{{j-1}}\exp(2i {j}k_{z1}d)A \] The wave reflected by the slab is the sum of these contributions, \[ r_{\text{slab}}A=B+C+D+\dots=\left[r_{01}+t_{10}t_{01}r_{12}\sum_{j=0}^{\infty}r_{12}^jr_{10}^{j}\exp(2ji k_{z1} d)\right]A \] For clarity, I write $\beta=r_{12}r_{10}\exp(2i k_{z1}d)$. The summation of all partial waves is thereby expressed as a geometrical sum, \[ r_{\text{slab}}=r_{01} + \left(t_{10}t_{01}r_{12}\exp(2i k_{z1}d)\right)\sum_{j=0}^{\infty}\beta^j \] Recalling that the sum of a geometric series of common ratio $q$ is $\frac{1}{1-q}$, we can write, \[ r_{\text{slab}}=r_{01} + \frac{t_{10}t_{01}r_{12}\exp(2i k_{z1}d)}{1-r_{12}r_{10}\exp(2i k_{z1}d)} \] We note that for either polarisation we have the following identity, $$ t^s_{ij}t^s_{ji} = (1-r^s_{ij})(1+r^s_{ij})= 1-(r^s_{ij})^2 ,\qquad t^p_{ij}t^p_{ji} =\frac{Y_i}{Y_j}(1-r^p_{ij})\frac{Y_j}{Y_i}(1+r^p_{ij}) = 1-(r^p_{ij})^2 $$ Using this equation and the substitution $r_{10} = -r_{01}$ we finally obtain, $$ r_{\text{slab}}=\frac{r_{01}+r_{12}\exp(2i k_{z1}d)}{1+r_{01}r_{12}\exp(2i k_{z1}d)}. $$ For the transmission, one obtains, $$ t_{\text{slab}}=\frac{t_{01}t_{12}\exp(i k_{z1}d)}{1+r_{01}r_{12}\exp(2i k_{z1}d)}. $$ ## Reflectivity of a multilayer structure When N layers are stacked together, the reflection coefficient of the structure can be found by applying recursively the preceding formula for a single layer. This amounts to considering one of the reflection coefficients to be the effective reflection accounting for all the layers behind. Let us consider explicitly the case of a 3 layer system. The procedure is as follows, * Compute and store for later use all single-interface coefficients, namely $r_{01}$, $r_{12}$, $r_{23}$, $r_{34}$ * Combine the previous coefficients to compute the reflectivity of the last layer, i.e. $r_{24} = \frac{r_{23}+r_{34}\exp(2i k_{z3}d_3)}{1+r_{23}r_{34}\exp(2i k_{z3}d_3)}$. This value can be stored in a temporary variable $r_{\text{tmp}}$ * Go up, combining one interface at a time, $r_{\text{tmp}}=r_{14} = \frac{r_{12}+r_{24}\exp(2i k_{z2}d_2)}{1+r_{12}r_{24}\exp(2i k_{z2}d_2)}$ * the combined reflectivity of the multilayer is obtained when one has reached the top interface, $r_{\text{tmp}} = r_{04} = \frac{r_{01}+r_{14}\exp(2i k_{z1}d_1)}{1+r_{01}r_{14}\exp(2i k_{z1}d_1)}$