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### Motivation

In class an outline of normal transmission through a slab was presented. Let’s go through the details.

### Normal incidence

The geometry of a two interface configuration is sketched in fig. 1.

Given a normal incident ray with magnitude \( A \), the respective forward and backwards rays in each the mediums can be written as

- [I]
- \begin{equation}\label{eqn:twoInterfaceNormal:20}

\begin{aligned}

A e^{-j k_1 z} \\

A r e^{j k_1 z} \\

\end{aligned}

\end{equation} - \begin{equation}\label{eqn:twoInterfaceNormal:40}

C e^{-j k_2 z} \\

D e^{j k_2 z} \\

\end{equation} - \begin{equation}\label{eqn:twoInterfaceNormal:60}

A t e^{-j k_3 (z-d)}

\end{equation}

Matching at \( z = 0 \) gives

\begin{equation}\label{eqn:twoInterfaceNormal:80}

\begin{aligned}

A t_{12} + r_{21} D &= C \\

A r &= A r_{12} + D t_{21},

\end{aligned}

\end{equation}

whereas matching at \( z = d \) gives

\begin{equation}\label{eqn:twoInterfaceNormal:100}

\begin{aligned}

A t &= C e^{-j k_2 d} t_{23} \\

D e^{j k_2 d} &= C e^{-j k_2 d} r_{23}

\end{aligned}

\end{equation}

We have four linear equations in four unknowns \( r, t, C, D \), but only care about solving for \( r, t \). Let’s write \(

\gamma = e^{ j k_2 d }, C’ = C/A, D’ = D/A \), for

\begin{equation}\label{eqn:twoInterfaceNormal:120}

\begin{aligned}

t_{12} + r_{21} D’ &= C’ \\

r &= r_{12} + D’ t_{21} \\

t \gamma &= C’ t_{23} \\

D’ \gamma^2 &= C’ r_{23}

\end{aligned}

\end{equation}

Solving for \( C’, D’ \) we get

\begin{equation}\label{eqn:twoInterfaceNormal:140}

\begin{aligned}

D’ \lr{ \gamma^2 – r_{21} r_{23} } &= t_{12} r_{23} \\

C’ \lr{ \gamma^2 – r_{21} r_{23} } &= t_{12} \gamma^2,

\end{aligned}

\end{equation}

so

\begin{equation}\label{eqn:twoInterfaceNormal:160}

\begin{aligned}

r &= r_{12} + \frac{t_{12} t_{21} r_{23} }{\gamma^2 – r_{21} r_{23} } \\

t &= t_{23} \frac{ t_{12} \gamma }{\gamma^2 – r_{21} r_{23} }.

\end{aligned}

\end{equation}

With \( \phi = -j k_2 d \), or \( \gamma = e^{-j\phi} \), we have

\begin{equation}\label{eqn:twoInterfaceNormal:180}

\boxed{

\begin{aligned}

r &= r_{12} + \frac{t_{12} t_{21} r_{23} e^{2 j \phi} }{1 – r_{21} r_{23} e^{2 j \phi}} \\

t &= \frac{ t_{12} t_{23} e^{j\phi}}{1 – r_{21} r_{23} e^{2 j \phi}}.

\end{aligned}

}

\end{equation}

### A slab

When the materials in region I, and III are equal, then \( r_{12} = r_{32} \). For a TE mode, we have

\begin{equation}\label{eqn:twoInterfaceNormal:200}

r_{12}

=

\frac{\mu_2 k_{1z} – \mu_1 k_{2z}}{\mu_2 k_{1z} + \mu_1 k_{2z}}

= -r_{21}.

\end{equation}

so the reflection and transmission coefficients are

\begin{equation}\label{eqn:twoInterfaceNormal:220}

\begin{aligned}

r^{\textrm{TE}} &= r_{12} \lr{ 1 – \frac{t_{12} t_{21} e^{2 j \phi} }{1 – r_{21}^2 e^{2 j \phi}} } \\

t^{\textrm{TE}} &= \frac{ t_{12} t_{21} e^{j\phi}}{1 – r_{21}^2 e^{2 j \phi}}.

\end{aligned}

\end{equation}

It’s possible to produce a matched condition for which \( r_{12} = r_{21} = 0 \), by selecting

\begin{equation}\label{eqn:twoInterfaceNormal:240}

\begin{aligned}

0

&= \mu_2 k_{1z} – \mu_1 k_{2z} \\

&= \mu_1 \mu_2 \lr{ \inv{\mu_1} k_{1z} – \inv{\mu_2} k_{2z} } \\

&= \mu_1 \mu_2 \omega \lr{ \frac{1}{v_1 \mu_1} \theta_1 – \frac{1}{v_2 \mu_2} \theta_2 },

\end{aligned}

\end{equation}

or

\begin{equation}\label{eqn:twoInterfaceNormal:260}

\inv{\eta_1} \cos\theta_1 = \inv{\eta_2} \cos\theta_2,

\end{equation}

so the matching condition for normal incidence is just

\begin{equation}\label{eqn:twoInterfaceNormal:280}

\eta_1 = \eta_2.

\end{equation}

Given this matched condition, the transmission coefficient for the 1,2 interface is

\begin{equation}\label{eqn:twoInterfaceNormal:300}

\begin{aligned}

t_{12}

&= \frac{2 \mu_2 k_{1z}}{\mu_2 k_{1z} + \mu_1 k_{2z}} \\

&= \frac{2 \mu_2 k_{1z}}{2 \mu_2 k_{1z} } \\

&= 1,

\end{aligned}

\end{equation}

so the matching condition yields

\begin{equation}\label{eqn:twoInterfaceNormal:320}

\begin{aligned}

t

&=

t_{12} t_{21} e^{j\phi} \\

&=

e^{j\phi} \\

&=

e^{-j k_2 d}.

\end{aligned}

\end{equation}

Normal transmission through a matched slab only introduces a phase delay.