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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?

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That sounds like something either explained in Jackson's or in Griffith's book – A friendly helper Apr 16 '13 at 21:34
What is the definition of the plane of incidence? I think it involves the normal vector of the local surface. – DarenW Apr 16 '13 at 22:53
up vote 1 down vote accepted

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.

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