Physical meaning of boundary conditions at interface for electric field I haven't taken an upper-level undergraduate electricity and magnetism course yet, so please keep this mind while reading or answering this question.
I am trying to understand Fresnel's reflection and transmission coefficients, which are based on the assumption that $E_{i}+E_{r}=E_{t}$, where $E_{i}$ is the amplitude of a wave incident upon a boundary of some medium, $E_{r}$ is the amplitude of a wave reflected from the boundary, and $E_{t}$ is the amplitude of the wave transmitted through the boundary into the medium. I am not sure how to interpret $E_{i}$, $E_{r}$, and $E_{t}$. If $E_{r}$ were the portion of the amplitude of the incident wave that is reflected from the boundary and $E_{t}$ were the portion of the amplitude of the incident wave that was transmitted through the boundary, then shouldn't $E_{i}=E_{t}+E_{r}$?
 A: At a dielectric interface where the permittivity has a jump the component of the $E$ field that is parallel with the interface must be continuous. So a little bit infinitesimally off on side of the interface from which the wave is incident the amplitude is $E_i + E_r$ because there you have both incident and reflected waves. On the other side you only have the transmitted wave $E_t$, therefore continuity demands that $E_i + E_r = E_t$.
A: Conservation of energy does apply, but you cannot simply equate electric field with energy. For one thing, electric field is a vector and can be negative!
The relevant energy conservation equation involves the Poynting vector, the magnitude of which is proportional to $nE^2$ in the case of the dielectric materials you are considering, where $n$ is a refractive index. Thus at normal incidence one can say
$$ n_iE_i^2 = n_t E_t^2 + n_i E_r^2\ . $$
Your confusion arises from not thinking of electric field as a vector. This is important because the continuity condition is that the vector sum of the tangential field at the boundary is continuous, not the sum of their magnitudes. When $n_t > n_i$ then indeed $E_t < E_i$ because $E_r$ is negative.
