= Solution
Reflected and transmitted waves arise because the fields must satisfy the electromagnetic <boundary conditions>. With no free surface charge or current, tangential $E,H$ and normal $D,B$ are continuous. At normal incidence between lossless media, define the real <wave impedance> $Z_j=\sqrt{\mu_j/\epsilon_j}$. Writing $r=E_r/E_i$ and $t=E_t/E_i$ in a common transverse electric-field direction gives
$$
1+r=t,\qquad\frac{1-r}{Z_1}=\frac{t}{Z_2},\qquad\boxed{r=\frac{Z_2-Z_1}{Z_2+Z_1},\quad t=\frac{2Z_2}{Z_1+Z_2}.}
$$
For the <power normalization of Fresnel amplitudes>, power fractions are \b[$R=|r|^2$] and \b[$T=(Z_1/Z_2)|t|^2$], so $R+T=1$. An electric-field amplitude coefficient alone is not a transmitted power fraction.
At oblique incidence, phase matching along the interface gives <Snell's law>, $n_1\sin\theta_i=n_2\sin\theta_t$. Applying the same <boundary conditions> separately to the two linear <polarizations> gives the <Fresnel equations>. For nonmagnetic dielectrics the parallel-polarized reflection vanishes at $\tan\theta_B=n_2/n_1$. Above the critical angle into a lower-index lossless medium, the transmitted field is evanescent and reflection is total. A perfect conductor instead has zero tangential <electric field> and unit reflected power; finite conductivity introduces penetration and absorption.<Electromagnetism by David Tong>[Chapter 7: reflection and transmission at dielectric interfaces]{ref}
Back to article page
