Condition II: ${\bf n}\cdot ({\bf B}_1 - {\bf B}_2 )=0$

For the boundary consdition in Eq.41

$${\bf k} \times {\bf E}_0 = \left\vert \begin{array}{ccc} {\bf i} & {\bf j} & {\bf k} \\ k\s... ...& k\cos\theta \\ E_p \cos\theta & E_s & -E_p \sin\theta \end{array}\right\vert$$

$$= k[ -{\bf i}E_s \cos\theta +{\bf j} (E_p\cos^2\theta + E_p \sin^2\theta ) + {\bf k}E_s \sin\theta ]$$ (69)

$$\omega{\bf B} = {\bf k} \times {\bf E}_0 = \left( \begin{array}{c} -kE_s \cos\theta \\ kE_p \\ kE_s \sin\theta \end{array}\right)$$ (70)

$${\bf k}' \times {\bf R}_0 = \left\vert \begin{array}{ccc} {\bf i} & {\bf j} & {\bf k} \\ k\s... ...-k\cos\theta \\ -R_p \cos\theta & R_s & -R_p \sin\theta \end{array}\right\vert$$

$$= k[ {\bf i}R_s \cos\theta +{\bf j} (R_p\cos^2\theta + R_p \sin^2\theta ) + {\bf k}R_s \sin\theta ]$$ (71)

$$\omega{\bf B}' = {\bf k}' \times {\bf R}_0 = \left( \begin{array}{c} kR_s \cos\theta \\ kR_p \\ kR_s \sin\theta \end{array}\right)$$ (72)

$${\bf k}'' \times {\bf T}_0 = \left\vert \begin{array}{ccc} {\bf i} & {\bf j} & {\bf k} \\ k''... ...s\theta'' \\ T_p \cos\theta'' & T_s & -T_p \sin\theta'' \end{array}\right\vert$$

$$= k'' [ -{\bf i}T_s \cos\theta'' +{\bf j} (T_p\cos^2\theta + T_p \sin^2\theta'' ) + {\bf k}T_s \sin\theta'' ]$$ (73)

$$\omega{\bf B}'' = {\bf k}'' \times {\bf T}_0 = \left( \begin{array}{c} -k'' T_s \cos\theta'' \\ k'' T_p \\ k'' T_s \sin\theta'' \end{array}\right)$$ (74)

The boundary condition $B_z + B'_z = B''_z$ becomes

$$\frac{k\sin\theta}{\omega} ( E_s + R_s) e^{i(k\sin\theta x - \omega t)} = \frac{k'' \sin\theta'' }{\omega} T_s e^{i(k'' \sin\theta'' x - \omega t)}$$ (75)

$$\tilde{n_1}(E_s + R_s ) \sin\theta = \tilde{n_2}T_s \sin\theta''$$ (76)

$$E_s + R_s = T_s$$ (77)

We used the Snell's law in the last equation. This equation is the same as Eq.68.