We present a study of a transverse magnetic mode excitation in a multi-layer atomic mirror. We give the continuity relationships at the interfaces of the setting.¹ We deduce the transverse and parallel components of the electric field through the atomic mirror and then the total component. We show that there is no enhancement of the electric field in the vaccum contrarily to the transverse electric field excitation.²

We introduce a detailed study of a transverse magnetic mode excitation in an atomic mirror. It is easy to demonstrate that in this geometry, we have: $\frac{\partial }{\partial y}=0$ . There are two independant solutions of the Maxwell equations.¹ One introduces a TE excitation (E_y, B_x and B_z).  The other one gives a TM excitation (B_y, E_x and E_z) coupled between them.

Figure 1 Resonant dielectric structure: the waveguide (Ti0₂, refraction index n₃=2.387 and thickness 2d=67.9nm) is separated from the glass prism (LaSFN18, refraction index n₁=1.893) by a layer of low index, the gap, (SiO₂, thickness a=700nm, refraction index n₂=1.49). The dielectric layers are deposited by a laser beam at $\lambda$ =780nm. The fourth medium is vacuum (refraction index n₄=1).

Maxwell equations:

The Maxwell equations [1, 3] give:

$\overrightarrow{\nabla }\wedge \overrightarrow{E}=-\frac{1}{C}\frac{\partial \overrightarrow{B}}{\partial t}$  

$\overrightarrow{\nabla }\wedge \overrightarrow{H}=\frac{4\pi }{C}\overrightarrow{J}+\frac{1}{C}\frac{\partial \overrightarrow{D}}{\partial t}$  

$\overrightarrow{\nabla }.\overrightarrow{E}=4\pi \rho$  

$\overrightarrow{\nabla }.\overrightarrow{B}=0$  

In order to introduce the right relationships, we present the electric field in the multi-layer system for a TE excitation.

$E_y^{(1)}=Aexp(-i{\gamma }_{1}z)+A^{\backslash }exp(i{\gamma }_{1}z)$  

$E_y^{(2)}=Bexp({\beta }_{2}z)+B^{\backslash }exp(-{\beta }_{2}z)$  

$E_y^{(3)}=Cexp(-i{\gamma }_{3}z)+C^{\backslash }exp(i{\gamma }_{3}z)$  

$E_y^{(4)}=Dexp({\beta }_{4}z)$  

For a transverse magnetic mode excitation, we give the two components of the electric field deduced from the Maxwell equations. The magnetic field B_y for the TM excitation has the same form as the perpendicular electric field E_y for the TE excitation.¹ See the above equations.

${\overrightarrow{E}}_T=E_x\overrightarrow{i}+E_z\overrightarrow{k}$  

$\overrightarrow{\nabla }\wedge \overrightarrow{E}=|\begin{array}{cc}\begin{array}{cc}\begin{array}{l}i\\ \frac{\partial }{\partial x}\\ E_x\end{array}& \begin{array}{l}j\\ \frac{\partial }{\partial y}\\ 0\end{array}\end{array}& \begin{array}{l}k\\ \frac{\partial }{\partial z}\\ E_z\end{array}|\end{array}=i(\frac{\partial E_z}{\partial y})-j(\frac{\partial E_z}{\partial x}-\frac{\partial E_x}{\partial z})+k(-\frac{\partial E_x}{\partial y})=i{K}_0\overrightarrow{B}$  

$E_x^{(1)}(z)=\frac{{\gamma }_1}{K_0{n}_1^2}(-Aexp(-i{\gamma }_{1}z)+A^{\backslash }exp(i{\gamma }_{1}z))$  

$E_x^{(2)}(z)=\frac{{\beta }_2}{i{K}_0{n}_2^2}(Bexp({\beta }_{2}z)-B^{\backslash }exp(-{\beta }_{2}z))$  

$E_x^{(3)}(z)=\frac{{\gamma }_3}{K_0{n}_3^2}(-Cexp(-i{\gamma }_{3}z)+C^{\backslash }exp(i{\gamma }_{3}z))$  

$E_x^{(4)}(z)=\frac{{\beta }_4}{i{K}_0{n}_4^2}Dexp({\beta }_{4}z)$  

The polarisation of the transmitted wave in the second medium is obtained from the Snell-Descartes law.

$\begin{array}{l}de(21)E_x=\frac{1}{i{K}_0{n}^2}\frac{\partial B_y}{\partial z}\\ de(20)i{K}_0{B}_y=(\frac{\partial E_x}{\partial z}-\frac{\partial E_z}{\partial x})\end{array}$  

$\begin{array}{l}i\mu \epsilon K_0{E}_z=-i{K}_x{B}_y\\ E_z=-\frac{K_x}{K_0{n}^2}{B}_y\end{array}$  

$B_y^{(1)}=Aexp(-i{\gamma }_{1}z)+A^{\backslash }exp(i{\gamma }_{1}z)$  

$B_y^{(2)}=Bexp({\beta }_{2}z)+B^{\backslash }exp(-{\beta }_{2}z)$  

$B_y^{(3)}=Cexp(-i{\gamma }_{3}z)+C^{\backslash }exp(i{\gamma }_{3}z)$  

$B_y^{(4)}=Dexp({\beta }_{4}z)$  

$y_1^2=n_1^2{k}_0^2{\cos }^2\theta$  

${\beta }_2^2=(n_1^2\sin \theta -n_2)K_0^2$  

$y_3^2=(n_3^2-n_1^2\text{si}{\text{n}}^2\theta )K_0^2$  

${\beta }_4^2=(n_1^2{\sin }^2\theta -n_4^2)K_0^2$  

$E_x=\frac{1}{i{K}_0{n}^2}\frac{\partial B_y}{\partial z}$  

$E_x^{(1)}(z)=\frac{{\gamma }_1}{K_0{n}_1^2}(-Aexp(-i{\gamma }_{1}z)+A^{\backslash }exp(i{\gamma }_{1}z))$  

$E_x^{(2)}(z)=\frac{{\beta }_2}{i{K}_0{n}_2^2}(Bexp({\beta }_{2}z)-B^{\backslash }exp(-{\beta }_{2}z))$  

$E_x^{(3)}(z)=\frac{{\gamma }_3}{K_0{n}_3^2}(-Cexp(-i{\gamma }_{3}z)+C^{\backslash }exp(i{\gamma }_{3}z))$  

$E_x^{(4)}(z)=\frac{{\beta }_4}{i{K}_0{n}_4^2}Dexp({\beta }_{4}z)$  

We give the continuity equations in the multi-layer system:

$\overrightarrow{s.}({\overrightarrow{D}}_2-{\overrightarrow{D}}_1)=0$  

$n_i^2{E}_i^1(z)=n_{i+1}^2{E}_{i+1}^1(z)$  

$\overrightarrow{s}\times ({\overrightarrow{E}}_2-{\overrightarrow{E}}_1)=\overrightarrow{0}$  

$E_x^{}(i)=E_x^{}(i+1)i=1;2;3$   at each interface.

$\overrightarrow{s}\times ({\overrightarrow{H}}_2-{\overrightarrow{H}}_1)=\overrightarrow{0}$   

$B_y^1(d+a)=B_y^2(d+a)$   

$E_x^1(d+a)=E_x^2(d+a)$   

$B_y^2(d)=B_y^3(d)$   

$E_x^2(d)=E_x^3(d)$   

$B_y^3(-d)=B_y^4(-d)$   

$E_x^3(-d)=E_x^4(-d)$   

$n_i^2{E}_z^{}(i)=n_{i+1}^2{E}_z^{}(i+1)$   

$Aexp(-i{\gamma }_1(d+a))=-A{}^{‵}exp(-i{\gamma }_1(d+a))+Bexp({\beta }_2(d+a))+B{}^{‵}exp(-{\beta }_2(d+a))$   

$\frac{{\gamma }_1}{n_1^2{K}_0}Aexp(-i{\gamma }_1(d+a))=\frac{{\gamma }_1}{n_1^2{K}_0}-A{}^{‵}exp(-i{\gamma }_1(d+a))-\frac{{\beta }_2}{i{n}_1^2{K}_0}[Bexp({\beta }_2(d+a))-B{}^{‵}exp(-{\beta }_2(d+a))$   

$0=-Bexp({\beta }_{2}d)-B^{\backslash }exp(-{\beta }_{2}d)+Cexp(-i{\gamma }_{3}d)+C^{\backslash }exp(i{\gamma }_{3}d)$   

$0=+\frac{{\beta }_2}{i{n}_2^2{K}_0}[Bexp({\beta }_{2}d)-B^{\backslash }exp(-{\beta }_{2}d)+\frac{{\gamma }_3}{n_3^2{K}_0}[Cexp(-i{\gamma }_{3}d)-C^{\backslash }exp(i{\gamma }_{3}d)$   

$0=-Cexp(i{\gamma }_{3}d)+C^{\backslash }exp(-i{\gamma }_{3}d)+Dexp(-{\beta }_{4}d)$   

$0=\frac{{\beta }_4}{i{K}_0}Bexp(-{\beta }_{4}d)-\frac{{\gamma }_3}{n_3^2{K}_0}[Cexp(i{\gamma }_{3}d)-C^{\backslash }exp(-i{\gamma }_{3}d)$   

We extract some factors from the matrix and call them I, II, III, IV, V, VI and VII:

$\text{I}=exp(-i{\gamma }_1(d+a))$   

$\text{II}=exp(i{\gamma }_1(d+a))$   

$\mathrm{III}=exp({\beta }_{2}d)$   

$\text{IV}=exp(-{\beta }_{2}d)$   

$\text{V}=exp(-i{\gamma }_{3}d)$   

$\text{VI}=exp(i{\gamma }_{3}d)$   

$\text{VII}=exp(-{\beta }_{4}d)$   

$\begin{array}{l}{\text{I}}_1=|\begin{array}{l}\\ \\ \\ \\ \\ \end{array}\begin{array}{cccc}1& -1& -1& 0\\ \frac{{\beta }_2}{i{n}_2^2{K}_0}& \frac{-{\gamma }_3}{n_3^2{K}_0}& \frac{{\gamma }_3}{n_3^2{K}_0}& 0\\ 0& exp(2{\gamma }_{3}d)& exp(-2{\gamma }_{3}d)& -1\\ 0& \frac{{\gamma }_3}{n_3^2{K}_0}exp(2{\gamma }_{3}d)& \frac{-{\gamma }_3}{n_3^2{K}_0}exp(-2{\gamma }_{3}d)& \frac{{\beta }_4}{i{K}_0}\end{array}\begin{array}{l}\\ \\ \\ \\ \\ \end{array}|\\ {\text{I}}_{\text{2}}=|\begin{array}{l}\\ \\ \\ \\ \\ \end{array}\begin{array}{cccc}1& -1& -1& 0\\ -\frac{{\beta }_2}{i{n}_2^2{K}_0}& \frac{-{\gamma }_3}{n_3^2{K}_0}& \frac{{\gamma }_3}{n_3^2{K}_0}& 0\\ 0& exp(2{\gamma }_{3}d)& exp(-2{\gamma }_{3}d)& -1\\ 0& \frac{{\gamma }_3}{n_3^2{K}_0}exp(2{\gamma }_{3}d)& \frac{-{\gamma }_3}{n_3^2{K}_0}exp(-2{\gamma }_{3}d)& \frac{{\beta }_4}{i{K}_0}\end{array}\begin{array}{l}\\ \\ \\ \\ \\ \end{array}|\\ \\ \end{array}$   

${\text{I}}_{\text{1}}=\frac{2i}{K_0^2}\left[\frac{{\gamma }_3}{n_3^2}(\frac{{\beta }_2}{n_2^2}+{\beta }_4)cos(2{\gamma }_{3}d)-(\frac{{\gamma }_3^2}{n_3^4}-\frac{{\beta }_2{\beta }_4}{n_2^2})sin(2{\gamma }_{3}d)\right]$   

${\text{I}}_{\text{2}}=\frac{2i}{K_0^2}\left[\frac{{\gamma }_3}{n_3^2}(-\frac{{\beta }_2}{n_2^2}+{\beta }_4)cos(2{\gamma }_{3}d)-(\frac{{\gamma }_3^2}{n_3^4}+\frac{{\beta }_2{\beta }_4}{n_2^2})sin(2{\gamma }_{3}d)\right]$   

$\text{X}=\frac{{\gamma }_3}{n_3^2}(\frac{{\beta }_2}{n_2^2}+{\beta }_4)cos2{\gamma }_{3}d-(\frac{{\gamma }_3^2}{n_3^4}-\frac{{\beta }_2{\beta }_4}{n_2^2})sin2{\gamma }_{3}d$   

$\text{Y}=\frac{{\gamma }_3}{n_3^2}(-\frac{{\beta }_2}{n_2^2}+{\beta }_4)cos2{\gamma }_{3}d-(\frac{{\gamma }_3^2}{n_3^4}+\frac{{\beta }_2{\beta }_4}{n_2^2})sin2{\gamma }_{3}d$   

$detA=\frac{2}{K_0^3}(G+iH)$  $detA=\frac{2}{K_0^3}(G+iH)$  

$detA=\frac{2}{K_0^3}(G+iH)exp(i{\gamma }_3(d+a))exp(-{\beta }_{4}d)$   

With:$\text{G=}\frac{{\beta }_2}{n_2^2}(\mathrm{X}exp({\beta }_{2}a)+\text{Y}exp(-{\beta }_{2}a))$   

$\text{H=}\frac{{\gamma }_1}{n_1^2}(-\mathrm{X}exp({\beta }_{2}a)+\text{Y}exp(-{\beta }_{2}a))$   

$\left(\begin{array}{ccccccc}-1& & -exp({\beta }_{2}a)& -exp(-{\beta }_{2}a)& 0& 0& 0\\ \frac{-{\gamma }_1}{n_1^2{K}_0}& & \frac{{\beta }_2}{i{n}_2^2{K}_0}exp({\beta }_{2}a)& \frac{-{\beta }_2}{i{n}_2^2{K}_0}exp(-{\beta }_{2}a)& 0& 0& 0\\ 0& & 1& 1& -1& -1& 0\\ 0& & \frac{-{\beta }_2}{i{n}_2^2{K}_0}& \frac{{\beta }_2}{i{n}_2^2{K}_0}& \frac{-{\gamma }_3}{n_3^2{K}_0}& \frac{{\gamma }_3}{n_3^2{K}_0}& 0\\ 0& & 0& 0& exp(2i{\gamma }_{3}d)& exp(-2i{\gamma }_{3}d)& -1\\ 0& & 0& 0& \frac{{\gamma }_3}{n_3^2{K}_0}exp(2i{\gamma }_{3}d)& -\frac{{\gamma }_3}{n_3^2{K}_0}exp(-2i{\gamma }_{3}d)& \frac{{\beta }_4}{i{K}_0}\end{array}\right)$  

$\begin{array}{l}det{A}^{\backslash }=-\frac{{\beta }_2}{i{n}_2^2{K}_0}exp({\beta }_{2}a)|\begin{array}{l}\\ \\ \\ \\ \\ \end{array}\begin{array}{cccc}1& -1& -1& 0\\ \frac{{\beta }_2}{i{n}_2^2{K}_0}& \frac{-{\gamma }_3}{n_3^2{K}_0}& \frac{{\gamma }_3}{n_3^2{K}_0}& 0\\ 0& exp(2{\gamma }_{3}d)& exp(-2{\gamma }_{3}d)& -1\\ 0& \frac{{\gamma }_3}{n_3^2{K}_0}exp(2{\gamma }_{3}d)& \frac{-{\gamma }_3}{n_3^2{K}_0}exp(-2{\gamma }_{3}d)& \frac{{\beta }_4}{i{K}_0}\end{array}\begin{array}{l}\\ \\ \\ \\ \\ \end{array}|\\ -\frac{{\beta }_2}{i{n}_2^2{K}_0}exp(-{\beta }_{2}a)|\begin{array}{l}\\ \\ \\ \\ \\ \end{array}\begin{array}{cccc}1& -1& -1& 0\\ -\frac{{\beta }_2}{i{n}_2^2{K}_0}& \frac{-{\gamma }_3}{n_3^2{K}_0}& \frac{{\gamma }_3}{n_3^2{K}_0}& 0\\ 0& exp(2{\gamma }_{3}d)& exp(-2{\gamma }_{3}d)& -1\\ 0& \frac{{\gamma }_3}{n_3^2{K}_0}exp(2{\gamma }_{3}d)& \frac{-{\gamma }_3}{n_3^2{K}_0}exp(-2{\gamma }_{3}d)& \frac{{\beta }_4}{i{K}_0}\end{array}\begin{array}{l}\\ \\ \\ \\ \\ \end{array}|\end{array}$  

$\begin{array}{l}-\frac{{\gamma }_1}{n_1^2{K}_0}exp({\beta }_{2}a)|\begin{array}{l}\\ \\ \\ \\ \\ \end{array}\begin{array}{cccc}1& -1& -1& 0\\ \frac{{\beta }_2}{i{n}_2^2{K}_0}& \frac{-{\gamma }_3}{n_3^2{K}_0}& \frac{{\gamma }_3}{n_3^2{K}_0}& 0\\ 0& exp(2{\gamma }_{3}d)& exp(-2{\gamma }_{3}d)& -1\\ 0& \frac{{\gamma }_3}{n_3^2{K}_0}exp(2{\gamma }_{3}d)& \frac{-{\gamma }_3}{n_3^2{K}_0}exp(-2{\gamma }_{3}d)& \frac{{\beta }_4}{i{K}_0}\end{array}\begin{array}{l}\\ \\ \\ \\ \\ \end{array}|\\ +\frac{{\gamma }_1}{n_1^2{K}_0}exp(-{\beta }_{2}a)|\begin{array}{l}\\ \\ \\ \\ \\ \end{array}\begin{array}{cccc}1& -1& -1& 0\\ -\frac{{\beta }_2}{i{n}_2^2{K}_0}& \frac{-{\gamma }_3}{n_3^2{K}_0}& \frac{{\gamma }_3}{n_3^2{K}_0}& 0\\ 0& exp(2{\gamma }_{3}d)& exp(-2{\gamma }_{3}d)& -1\\ 0& \frac{{\gamma }_3}{n_3^2{K}_0}exp(2{\gamma }_{3}d)& \frac{-{\gamma }_3}{n_3^2{K}_0}exp(-2{\gamma }_{3}d)& \frac{{\beta }_4}{i{K}_0}\end{array}\begin{array}{l}\\ \\ \\ \\ \\ \end{array}|\end{array}$  

$det{A}^{\backslash }=-(\frac{{\beta }_2}{i{n}_2^2{K}_0}+\frac{{\gamma }_1}{n_1^2{K}_0})exp({\beta }_{2}a){\text{I}}_{\text{1}}-(\frac{{\beta }_2}{i{n}_2^2{K}_0}-\frac{{\gamma }_1}{n_1^2{K}_0})exp(-{\beta }_{2}a){\text{I}}_{\text{2}}$  

$det{A}^{\backslash }=-\frac{2}{K_0^3}(G-iH)exp(-i{\gamma }_3(d+a))exp(-{\beta }_{4}d)$  

$\text{j}=\frac{2}{K_0^3}exp(-i{\gamma }_3(d+a))exp(-{\beta }_{4}d)$  

$detA=\text{j(G+iH)}$   

:  j^* is the conjugate of j

$det{A}^{\backslash }=-{\text{j}}^{\ast }\text{(G-iH)}$   

$\left(\begin{array}{ccccccc}& 1& -1& -exp(-{\beta }_{2}a)& 0& 0& 0\\ & \frac{-{\gamma }_1}{n_1^2{K}_0}& \frac{-{\gamma }_1}{n_1^2{K}_0}& \frac{-{\beta }_2}{i{n}_2^2{K}_0}exp(-{\beta }_{2}a)& 0& 0& 0\\ & 0& 0& 1& -1& -1& 0\\ & 0& 0& \frac{{\beta }_2}{i{n}_2^2{K}_0}& \frac{-{\gamma }_3}{n_3^2{K}_0}& \frac{{\gamma }_3}{n_3^2{K}_0}& 0\\ & 0& 0& 0& exp(2i{\gamma }_{3}d)& exp(-2i{\gamma }_{3}d)& -1\\ & 0& 0& 0& \frac{{\gamma }_3}{n_3^2{K}_0}exp(2i{\gamma }_{3}d)& -\frac{{\gamma }_3}{n_3^2{K}_0}exp(-2i{\gamma }_{3}d)& \frac{{\beta }_4}{i{K}_0}\end{array}\right)$   

$detB=-\frac{2{\gamma }_1}{n_1^2{K}_0}{\text{I}}_{\text{1}}$   with ${\text{I}}_{\text{1}}=\frac{2i}{K_0^2}\text{X}$

$detB=-4\frac{i{\gamma }_1}{n_1^2{K}_0^3}\text{X }exp\text{(-}d\text{(}{\beta }_2+{\beta }_4))$   

$\left(\begin{array}{ccccccc}& 1& -exp({\beta }_{2}a)& -1& 0& 0& 0\\ & \frac{-{\gamma }_1}{n_1^2{K}_0}& \frac{{\beta }_2}{i{n}_2^2{K}_0}exp({\beta }_{2}a)& \frac{-{\gamma }_1}{n_1^2{K}_0}& 0& 0& 0\\ & 0& 1& 0& -1& -1& 0\\ & 0& \frac{-{\beta }_2}{i{n}_2^2{K}_0}& 0& \frac{-{\gamma }_3}{n_3^2{K}_0}& \frac{{\gamma }_3}{n_3^2{K}_0}& 0\\ & 0& 0& 0& exp(2i{\gamma }_{3}d)& exp(-2i{\gamma }_{3}d)& -1\\ & 0& 0& 0& \frac{{\gamma }_3}{n_3^2{K}_0}exp(2i{\gamma }_{3}d)& -\frac{{\gamma }_3}{n_3^2{K}_0}exp(-2i{\gamma }_{3}d)& \frac{{\beta }_4}{i{K}_0}\end{array}\right)$   

$det{B}^{\backslash }=\frac{2{\gamma }_1}{n_1^2{K}_0}{\text{I}}_2$   with ${\text{I}}_{\text{2}}=\frac{2i}{K_0^2}\text{Y}$

$det{B}^{\backslash }=4\frac{i{\gamma }_1}{n_1^2{K}_0^3}\text{Y }exp\text{(}d\text{(}{\beta }_2-{\beta }_4))$  

$\left(\begin{array}{ccccccc}& 1& -exp({\beta }_{2}a)& -exp(-{\beta }_{2}a)& -1& 0& 0\\ & \frac{-{\gamma }_1}{n_1^2{K}_0}& \frac{{\beta }_2}{i{n}_2^2{K}_0}exp({\beta }_{2}a)& \frac{-{\beta }_2}{i{n}_2^2{K}_0}exp(-{\beta }_{2}a)& \frac{-{\gamma }_1}{n_1^2{K}_0}& 0& 0\\ & 0& 1& 1& 0& -1& 0\\ & 0& \frac{-{\beta }_2}{i{n}_2^2{K}_0}& \frac{{\beta }_2}{i{n}_2^2{K}_0}& 0& \frac{{\gamma }_3}{n_3^2{K}_0}& 0\\ & 0& 0& 0& 0& exp(-2i{\gamma }_{3}d)& -1\\ & 0& 0& 0& 0& -\frac{{\gamma }_3}{n_3^2{K}_0}exp(-2i{\gamma }_{3}d)& \frac{{\beta }_4}{i{K}_0}\end{array}\right)$   

$detC=\frac{2{\beta }_2{\gamma }_1}{i{n}_2^2{n}_1^2{K}_0^2}|\begin{array}{ccc}-\frac{{\gamma }_1}{K_0}& 0& 0\\ 0& exp(-2{\gamma }_{3}d)& -1\\ 0& -\frac{{\gamma }_3}{K_0}exp(-2{\gamma }_{3}d)& \frac{{\beta }_4}{i{K}_0}\end{array}|$  

$detC=\frac{4{\beta }_2{\gamma }_1}{n_1^2{n}_2^2{K}_0^3}({\beta }_4-i\frac{{\gamma }_3}{n_3^2})exp(-2i{\gamma }_{3}d)$  

$detC=\frac{4{\beta }_2{\gamma }_1}{n_1^2{n}_2^2{K}_0^3}({\beta }_4-i\frac{{\gamma }_3}{n_3^2})exp(-2i{\gamma }_{3}d)exp(-{\beta }_{4}d)$  

$\left(\begin{array}{ccccccc}& 1& -exp({\beta }_{2}a)& -exp(-{\beta }_{2}a)& 0& -1& 0\\ & \frac{-{\gamma }_1}{n_1^1{K}_0}& \frac{{\beta }_2}{i{n}_2^2{K}_0}exp({\beta }_{2}a)& \frac{-{\beta }_2}{i{n}_2^2{K}_0}exp(-{\beta }_{2}a)& 0& \frac{-{\gamma }_1}{n_1^2{K}_0}& 0\\ & 0& 1& 1& -1& 0& 0\\ & 0& \frac{-{\beta }_2}{i{n}_2^2{K}_0}& \frac{{\beta }_2}{i{n}_2^2{K}_0}& \frac{-{\gamma }_3}{n_3^2{K}_0}& 0& 0\\ & 0& 0& 0& exp(2i{\gamma }_{3}d)& 0& -1\\ & 0& 0& 0& \frac{{\gamma }_3}{n_3^2{K}_0}exp(2i{\gamma }_{3}d)& 0& \frac{{\beta }_4}{i{K}_0}\end{array}\right)$  

$det{C}^{\backslash }=-\frac{4{\beta }_2{\gamma }_1}{i{n}_1^2{n}_2^2{K}_0^3}({\beta }_4+i\frac{{\gamma }_3}{n_3^2})exp(2i{\gamma }_{3}d)$  

$det{C}^{\backslash }=-\frac{4{\beta }_2{\gamma }_1}{i{n}_1^2{n}_2^2{K}_0^3}({\beta }_4+i\frac{{\gamma }_3}{n_3^2})exp(2i{\gamma }_{3}d)exp(-{\beta }_{4}d)$  

$\left(\begin{array}{ccccccc}& 1& -exp({\beta }_{2}a)& -exp(-{\beta }_{2}a)& 0& 0& -1\\ & \frac{-{\gamma }_1}{K_0}& \frac{{\beta }_2}{i{K}_0}exp({\beta }_{2}a)& \frac{-{\beta }_2}{i{K}_0}exp(-{\beta }_{2}a)& 0& 0& \frac{-{\gamma }_1}{K_0}\\ & 0& 1& 1& -1& -1& 0\\ & 0& \frac{-{\beta }_2}{i{K}_0}& \frac{{\beta }_2}{i{K}_0}& \frac{-{\gamma }_3}{K_0}& \frac{{\gamma }_3}{K_0}& 0\\ & 0& 0& 0& exp(2i{\gamma }_{3}d)& exp(-2i{\gamma }_{3}d)& 0\\ & 0& 0& 0& \frac{{\gamma }_3}{K_0}exp(2i{\gamma }_{3}d)& -\frac{{\gamma }_3}{K_0}exp(-2i{\gamma }_{3}d)& 0\end{array}\right)$  

$detD=\frac{-8i{\gamma }_1{\gamma }_3{\beta }_2}{K_0^3}$  

The coefficients $A,A^{\backslash },B,B^{\backslash },C,C^{\backslash },D$ are given by the following relationship:

$\frac{A}{detA}=\frac{A^{\backslash }}{det{A}^{\backslash }}=\frac{B}{detB}=\frac{B^{\backslash }}{det{B}^{\backslash }}=\frac{C}{detC}=\frac{C^{\backslash }}{det{C}^{\backslash }}=\frac{D}{detD}$  

$A^{\backslash }=-A\frac{{\text{j}}^{*}(G-iH)}{\text{j}(G+iH)}$  

$B=-A\frac{4i{\gamma }_1\text{X}}{K_0^3\text{J}(G+iH)}exp(-d({\beta }_2+{\beta }_4))$  

$B^{\backslash }=A\frac{4i{\gamma }_1\text{Y}}{K_0^3\text{J}(G+iH)}exp(-d({\beta }_2-{\beta }_4))$  

$C=A\frac{4{\beta }_2{\gamma }_1}{K_0^3\text{J}(G+iH)}({\beta }_4-i{\gamma }_3)exp(-d{\beta }_4)exp(-i{\gamma }_{3}d)$  

$C^{\backslash }=-A\frac{4{\beta }_2{\gamma }_1}{K_0^3\text{J}(G+iH)}({\beta }_4+i{\gamma }_3)exp(-d{\beta }_4)exp(i{\gamma }_{3}d)$  

$D=-A\frac{8i{\gamma }_1{\gamma }_3{\beta }_2}{K_0^3\text{j}(G+iH)}$  

Results:

We take an incidence angle close to the critical angle (optimum of transmission in the TE mode). See reference.¹  

${\theta }_{inc}$   =52⁰

$\{\begin{array}{l}{E}_x=\frac{1}{i{K}_0{n}^2}\frac{\partial B_y}{\partial z}\\ E_z=-\frac{K_x}{K_0{n}^2}{B}_y\end{array}$   

${\overrightarrow{E}}_T=E_x\overrightarrow{i}+E_z\overrightarrow{k}$   

${\overrightarrow{E}}_T(4)=E_x(4)\overrightarrow{i}+E_z(4)\overrightarrow{k}$  

${\overrightarrow{E}}_T=\frac{{\beta }_4}{i{K}_0{n}^4}Dexp({\beta }_{4}z)\overrightarrow{i}-\frac{K_x}{K_0{n}^4}Dexp({\beta }_{4}z)\overrightarrow{k}$   

${\overrightarrow{E}}_T(4).conj({\overrightarrow{E}}_T(4))={|D|}^2\exp (2{\beta }_{4}z)$   

${\overrightarrow{E}}_T(1)=\frac{{\gamma }_1}{K_0{n}_1^2}(-Aexp(-i{\gamma }_{1}z)+A`exp(i{\gamma }_{1}z))\overrightarrow{i}-\frac{K_x}{K_0{n}_1^2}(Aexp(-i{\gamma }_{1}z)+A`exp(i{\gamma }_{1}z))\overrightarrow{k}$  

Figure 2 The perpendicular component EZ of the electric field (V/m).

Figure 3 The parallel component EX of the electric field (V/m).

$\{\begin{array}{l}{E}_x=\frac{1}{i{K}_0{n}^2}\frac{\partial B_y}{\partial z}\\ E_z=-\frac{K_x}{K_0{n}^2}{B}_y\end{array}$  

${\overrightarrow{E}}_T=E_x\overrightarrow{i}+E_z\overrightarrow{k}$  

${\overrightarrow{E}}_T(4)=E_x(4)\overrightarrow{i}+E_z(4)\overrightarrow{k}$  

${\overrightarrow{E}}_T=\frac{{\beta }_4}{i{K}_0{n}^4}Dexp({\beta }_{4}z)\overrightarrow{i}-\frac{K_x}{K_0{n}^4}Dexp({\beta }_{4}z)\overrightarrow{k}$  

${\overrightarrow{E}}_T(4).conj({\overrightarrow{E}}_T(4))={|D|}^2\exp (2{\beta }_{4}z)$  

${\overrightarrow{E}}_T(1)=\frac{{\gamma }_1}{K_0{n}_1^2}(-Aexp(-i{\gamma }_{1}z)+A`exp(i{\gamma }_{1}z))\overrightarrow{i}-\frac{K_x}{K_0{n}_1^2}(Aexp(-i{\gamma }_{1}z)+A`exp(i{\gamma }_{1}z))\overrightarrow{k}$