v 4.13

\frac{E_{r 0}^{∥}}{E_{i 0}^{∥}} = - \frac{\tan (α - γ)}{\tan (α + γ)}

${\frac{E_{r0}^\parallel}{E_{i0}^\parallel} = - \frac{\tan (\alpha - \gamma)}{\tan (\alpha + \gamma)}}$