Condition I: ${\bf t}\cdot [{\bf E}_1 - {\bf E}_2 ] = 0$

From Eq.46 the tangential component of the electric field become

$$E_x + E_{x}' = E_{x}'' \ \ \ \ \ {\rm at} \ z = 0$$ (57)

$$E_p \cos\theta e^{i(k\sin\theta x - \omega t)} + R_p \cos\theta' e^{i(k'\sin\theta' x - \omega' t)} = T_p \cos\theta'' e^{i(k'' \sin\theta'' x - \omega'' t)}$$ (58)

$$\ \ \ \ \ \ \ \ \ \ {\rm for \ any} \ x \ {\rm at} \ z = 0, \ {\rm then}$$

$$\omega = \omega' = \omega''$$ (59)

$$k\sin\theta = k'\sin\theta' = k'' \sin\theta''$$ (60)

The magnitude of the wavevector is given by Eq.21

$$\sin\theta = \sin\theta'=\sin(\pi-\theta'), \ \ \ {\rm because}\ \ k=k'$$ (61)

The incident angle equals to the reflection angle, and

$$k \sin\theta = \tilde{n}_1 \frac{\omega}{c} \sin\theta = k'' \sin\theta'' = \tilde{n}_2 \frac{\omega'' }{c} \sin\theta''$$ (62)

So we can get Snell's law because $\omega = \omega''$

$$\tilde{n}_1 \sin\theta = \tilde{n}_2 \sin\theta''$$ (63)

The Eq.58 becomes

$$(E_p - R_p ) \cos\theta = T_p \cos\theta''$$ (64)

From above we can get

$${\bf E}_0 = \left( \begin{array}{c} E_p \cos\theta \\ E_s \\... ...p \cos\theta'' \\ T_s \\ -T_p \sin\theta'' \end{array}\right)$$ (65)

There is another condition

$$E_y + E_{y}' = E_{y}'' \ \ \ \ \ {\rm at} \ z = 0$$ (66)

$$(E_s + R_s)e^{i(k\sin\theta x -\omega t)} = T_s e^{i(k'' \sin\theta'' x -\omega t)}$$ (67)

$$(E_s + R_s) = T_s$$ (68)