Physical meaning of boundary conditions at interface for electric field

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 $$E_{i}=E_{t}+E_{r}$$?

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 $$E_i + E_r = E_t$$.

• I can see why $E_{i}+E_{r}=E_{t}$ must hold true for there to be continuity at the boundary. However, wouldn't that imply that $E_{t}>E_{i}$ if part of the incident wave is reflected? Oct 19, 2022 at 23:06
• that depends on the sign of the product $E_iE_r$ and at any rate these are vectors. The energy of the wave must be conserved and the wave amplitude is not the wave energy. Oct 20, 2022 at 1:52

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.

• I think "by conservation of energy" was an inappropriate argument on my part. What I am trying to get at is that if $E_{i}+E_{r}=E_{t}$, then shouldn't $E_{t}>E_{i}$ if part of the incident wave is reflected? Wouldn't this essentially create an electric field out of nothing, merely because some of the incident wave was reflected? Oct 19, 2022 at 23:12
• @BrunoNowak and as I said, electric field can be negative. e.g. $E_r$ is negative if $n_t > n_i$. Oct 19, 2022 at 23:14
• I think I understand now. Although the electric fields of the reflected and incident waves may point in opposite directions at the interface, their sum must be such that it points in the same direction and has the same magnitude as the electric field of the transmitted wave to ensure continuity across the boundary. If that is correct, thank you for help! If not, please let me know. Oct 19, 2022 at 23:20
• I realized that your answer makes sense when $n_{t}>n_{i}$, but what happens in the case where $n_{t}<n_{i}$ and some of the incident wave is reflected? Doesn't that lead to the contradiction that I mentioned earlier since both $E_{i}$ and $E_{r}$ have the same sign? Oct 20, 2022 at 0:05
• @BrunoNowak The continuity equation is satisfied and energy conservation is satisfied. Why shouldn't $E$ be larger in the medium with lower refractive index? Oct 20, 2022 at 7:51