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In [1] are some sum and angle difference formulations for the Fresnel formulas given a \( \mu_1 = \mu_2 \) constraint. The proof of these trig Fresnel equations is left to an exercise, and will be derived here.

We need a couple trig identities to start with.

\begin{equation}\label{eqn:fresnelSumAndDifferenceAngleFormulas:20}

\begin{aligned}

\sin(a + b)

&=

\textrm{Im}\lr{ e^{j(a + b)} } \\

&=

\textrm{Im}\lr{

e^{ja} e^{+ jb}

} \\

&=

\textrm{Im}\lr{

(\cos a + j \sin a) (\cos b + j \sin b)

} \\

&=

\sin a \cos b + \cos a \sin b.

\end{aligned}

\end{equation}

Allowing for both signs we have

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

\begin{aligned}

\sin(a + b) &= \sin a \cos b + \cos a \sin b \\

\sin(a – b) &= \sin a \cos b – \cos a \sin b.

\end{aligned}

\end{equation}

The mixed sine and cosine product can be expressed as a sum of sines

\begin{equation}\label{eqn:fresnelSumAndDifferenceAngleFormulas:40}

2 \sin a \cos b = \sin(a + b) + \sin(a – b).

\end{equation}

With \( 2 x = a + b, 2 y = a – b \), or \( a = x + y, b = x – y \), we find

\begin{equation}\label{eqn:fresnelSumAndDifferenceAngleFormulas:60}

\begin{aligned}

2 \sin(x + y) \cos (x – y) &= \sin( 2 x ) + \sin( 2 y ) \\

2 \sin(x – y) \cos (x + y) &= \sin( 2 x ) – \sin( 2 y ).

\end{aligned}

\end{equation}

Returning to the problem. When \( \mu_1 = \mu_2 \) the Fresnel equations were found to be

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

\begin{aligned}

r^{\textrm{TE}} &= \frac { n_1 \cos\theta_i – n_2 \cos\theta_t } { n_1 \cos\theta_i + n_2 \cos\theta_t } \\

r^{\textrm{TM}} &= \frac{n_2 \cos\theta_i – n_1 \cos\theta_t }{ n_2 \cos\theta_i + n_1 \cos\theta_t } \\

t^{\textrm{TE}} &= \frac{ 2 n_1 \cos\theta_i } { n_1 \cos\theta_i + n_2 \cos\theta_t } \\

t^{\textrm{TM}} &= \frac{2 n_1 \cos\theta_i }{ n_2 \cos\theta_i + n_1 \cos\theta_t }.

\end{aligned}

\end{equation}

Using Snell’s law, one of \( n_1, n_2 \) can be eliminated, for example

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

n_1 = n_2 \frac{\sin \theta_t}{\sin\theta_i}.

\end{equation}

Inserting this and proceeding with the application of the trig identities above, we have

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

\begin{aligned}

r^{\textrm{TE}}

&= \frac { n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_i – n_2 \cos\theta_t } { n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_i + n_2 \cos\theta_t } \\

&=

\frac {

\sin\theta_t \cos\theta_i – \cos\theta_t \sin\theta_i

} {

\sin\theta_t \cos\theta_i + \cos\theta_t \sin\theta_i

} \\

&=

\frac {

\sin( \theta_t – \theta_i )

} {

\sin( \theta_t + \theta_i )

}

\end{aligned}

\end{equation}

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

\begin{aligned}

r^{\textrm{TM}}

&= \frac{n_2 \cos\theta_i – n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_t }{ n_2 \cos\theta_i + n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_t } \\

&= \frac{

\sin\theta_i \cos\theta_i – \sin\theta_t \cos\theta_t

}{

\sin\theta_i \cos\theta_i + \sin\theta_t \cos\theta_t

} \\

&= \frac{\inv{2} \sin(2 \theta_i) – \inv{2} \sin(2 \theta_t) }{ \inv{2} \sin(2 \theta_i) + \inv{2} \sin(2 \theta_t) } \\

&= \frac

{\sin(\theta_i – \theta_t)\cos(\theta_i + \theta_t) }

{\sin(\theta_i + \theta_t)\cos(\theta_i – \theta_t) } \\

&=

\frac

{\tan(\theta_i -\theta_t)}

{\tan(\theta_i +\theta_t)}

\end{aligned}

\end{equation}

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

\begin{aligned}

t^{\textrm{TE}}

&= \frac{ 2 n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_i } { n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_i + n_2 \cos\theta_t } \\

&= \frac{ 2 \sin\theta_t \cos\theta_i } { \sin\theta_t \cos\theta_i + \cos\theta_t \sin\theta_i } \\

&= \frac{ 2 \sin\theta_t \cos\theta_i }

{ \sin(\theta_i + \theta_t) }

\end{aligned}

\end{equation}

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

\begin{aligned}

t^{\textrm{TM}}

&= \frac{2 n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_i }{ n_2 \cos\theta_i + n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_t } \\

&= \frac{2 \sin\theta_t \cos\theta_i }{ \sin\theta_i \cos\theta_i + \sin\theta_t \cos\theta_t } \\

&= \frac{2 \sin\theta_t \cos\theta_i }

{ \inv{2} \sin(2 \theta_i) + \inv{2} \sin(2 \theta_t) } \\

&= \frac{2 \sin\theta_t \cos\theta_i }

{ \sin(\theta_i + \theta_t) \cos(\theta_i – \theta_t) }

\end{aligned}

\end{equation}

# References

[1] E. Hecht. *Optics*. 1998.