PX284 - S2a - fresnel relations

\begin{matrix} \hat{n} \times ({\vec{E}}_{2} - {\vec{E}}_{1}) = 0 \\ \hat{n} \times ({\vec{H}}_{2} - {\vec{H}}_{1}) = 0 \\ ({\vec{j}}_{f} = 0 in dielectrics) \end{matrix}

it is chosen to orient $\vec{E}$ such that ${\vec{E}}_{i}$ , ${\vec{E}}_{r}$ and ${\vec{E}}_{t}$ point in the same direction where $i = r = t$ , so for the first figure, $\vec{E}$ always points outwards, and for the second, it points to the right
these are called the s- and the p-components, which stand for sentrecht and parallel (perpendicular and parallel in german)
these refer to the orientation of the electric field with respect to the plane of the paper
for the s-component, the electric field is continuous at the boundary

E_{i} + E_{r} = E_{t}

⟹ 1 + r_{s} = t_{s}

H_{i} \cos i - H_{r} \cos r = H_{t} \cos t

Z = \frac{E}{H} = \frac{Z_{0}}{n}

\begin{matrix} ⟹ \frac{E_{i}}{Z_{0} / n_{1}} \cos i - \frac{E_{r}}{Z_{0} / n_{1}} \cos r = \frac{E_{t}}{Z_{0} / n_{2}} \cos t \\ n_{1} \cos i (E_{i} - E_{r}) = n_{2} \cos t E_{t} \\ n_{1} \cos i (1 - r_{s}) = n_{2} \cos t t_{s} = n_{2} \cos t (1 + r_{s}) \\ ∴ r_{s} = \frac{n_{1} \cos i - n_{2} \cos t}{n_{1} \cos i + n_{2} \cos t} \\ t_{s} = \frac{Z n_{1} \cos i}{n_{1} \cos i + n_{2} \cos t} \end{matrix}

\begin{matrix} H_{i} - H_{r} = H_{t} \\ E_{i} \cos i + E_{r} \cos r = E_{t} \cos t \end{matrix}

\begin{matrix} \frac{E_{i}}{Z_{1}} - \frac{E_{r}}{Z_{1}} = \frac{E_{t}}{Z_{2}} \\ ⟹ n_{1} (1 - r_{p}) = n_{2} t_{p} \\ (1 + r_{p}) \cos i = t_{p} \cos t \\ ∴ r_{p} = \frac{n_{1} \cos t - n_{2} \cos i}{n_{1} \cos t + n_{2} \cos i} \\ t_{p} = \frac{2 n_{1} \cos i}{n_{1} \cos t + n_{2} \cos i} \end{matrix}