The boundary conditions of an electromagnetic wave hitting a surface When I try to solve the Fresnel equations, I don't understand the condition:
E(i)+E(r)=E(t)
where E(i) is the incident wave, E(r) is the reflected wave and E(t) is the transmitted wave. 
So the question is how it can be that the E(t) equal both fields while the intuitive equation is that E(i)=E(r)+E(t)...
 A: The boundary condition here is derived from the Faraday-Maxwell law, and says that the component of the electric field parallel to the boundary is continuous. That is, the electric field parallel to the boundary is the same either side of the boundary.
Since solutions of Maxwell's equations can superpose, that means if there are multiple electric fields on one side of the boundary (in this case, the components of the fields due to the incident and reflected waves), then these must be added, in a vectorial way, and their sum must be the same either side of the boundary.
Your intuitive approach is perhaps confusing electric field with energy.
Energy is conserved at the boundary, but this says that the flux in equals the flux out
$$\vec{N}_i\cdot d\vec{s} = -\vec{N}_r\cdot d\vec{s} + \vec{N}_t\cdot d\vec{s},$$
where the terms represent the Poynting vectors integrated over the beam area, and the minus sign before the reflected term is because the Poynting vector of the reflected wave and $d\vec{s}$ are in opposite directions.
A: It's not true that $\textbf{E}(i)+\textbf{E}(r)=\textbf{E}(t)$. This holds for the components parallel to the surface only.

So the question is how it can be that the E(t) equal both fields while the intuitive equation is that E(i)=E(r)+E(t)...

Your intuition is presumably about energy conservation, but that's a different thing. The equation you're asking about just says that the tangential component of the field is continuous, which follows directly from Maxwell's equations.
