Show that the plane of incidence is perpendicular to the surface of reflection Is it possible to derive from the boundary conditions of the Maxwell equations for E and H, that the plane of incidence for an EM wave is perpendicular to the reflection surface? How? If not, what additional condition is needed?
 A: Let's assume the plane of incidence is the plane containing the normal to the surface and the wavevector of the incident wave. We have to prove that the wavevectors of the reflected and transmitted waves lay in the same plane.
If we assume plane and monochromatic waves we have that our fields can be expressed as follows:
$$
\vec{E}_{in} = \vec{E}_{0,in} \exp\, [i(\vec{k}_{in}\cdot \vec{r} - \omega_{in}t)]
$$
$$
\vec{E}_{r} = \vec{E}_{0,r} \exp\, [i(\vec{k}_{r}\cdot \vec{r} - \omega_{r}t)]
$$
$$
\vec{E}_{t} = \vec{E}_{0,t} \exp\, [i(\vec{k}_{t}\cdot \vec{r} - \omega_{t}t)]
$$
The boundary conditions impose that:
$$
\vec{n}\times\vec{E}_{in} + \vec{n}\times\vec{E}_{r} = \vec{n}\times\vec{E}_{t} 
$$
For this to be true you need that the phases match at the boundary:
$$
(\vec{k}_{in}\cdot \vec{r} - \omega_{in}t) = (\vec{k}_{r}\cdot \vec{r} - \omega_{r}t) = (\vec{k}_{t}\cdot \vec{r} - \omega_{t}t)
$$
Since this has to be true for all points along the surface, it must hold:
$$
\omega_{in} = \omega_{r} = \omega_{t}
$$
Therefore for all times it has to be:
$$
\vec{k}_{in}\cdot \vec{r} = \vec{k}_{r}\cdot \vec{r} = \vec{k}_{t}\cdot \vec{r} = const
$$
But the last expression is the equation of a plane orthogonal to all three wavevectors, thus all the wavevectors have to lay in the same plane, which is the incidence plane.
A: By DEFINITION, the plane of incidence contains the NORMAL to the surface of reflection.  Any plane containing the normal of a surface is perpendicular to the surface itself.  This has nothing to do with electromagnetism.  It is a purely geometrical property.
