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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.

fig. 1. Two interface transmission.

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

[I]

1. \label{eqn:twoInterfaceNormal:20}
\begin{aligned}
A e^{-j k_1 z} \\
A r e^{j k_1 z} \\
\end{aligned}
2. \label{eqn:twoInterfaceNormal:40}
C e^{-j k_2 z} \\
D e^{j k_2 z} \\
3. \label{eqn:twoInterfaceNormal:60}
A t e^{-j k_3 (z-d)}

Matching at $$z = 0$$ gives
\label{eqn:twoInterfaceNormal:80}
\begin{aligned}
A t_{12} + r_{21} D &= C \\
A r &= A r_{12} + D t_{21},
\end{aligned}

whereas matching at $$z = d$$ gives

\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}

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

\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}

Solving for $$C’, D’$$ we get

\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}

so

\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}

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

\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}
}

### A slab

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

\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}.

so the reflection and transmission coefficients are

\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}

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

\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}

or

\label{eqn:twoInterfaceNormal:260}
\inv{\eta_1} \cos\theta_1 = \inv{\eta_2} \cos\theta_2,

so the matching condition for normal incidence is just

\label{eqn:twoInterfaceNormal:280}
\eta_1 = \eta_2.

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

\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}

so the matching condition yields
\label{eqn:twoInterfaceNormal:320}
\begin{aligned}
t
&=
t_{12} t_{21} e^{j\phi} \\
&=
e^{j\phi} \\
&=
e^{-j k_2 d}.
\end{aligned}

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