= Solution
An incident <electromagnetic wave> generally produces reflected and transmitted waves because the two media have different electromagnetic response. Consider a flat interface between homogeneous, linear, lossless media with <permittivities> $\epsilon_j$ and <magnetic permeabilities> $\mu_j$. Their wave speeds and <wave impedances> are
$$
v_j=\frac1{\sqrt{\mu_j\epsilon_j}},\qquad Z_j=\sqrt{\frac{\mu_j}{\epsilon_j}}.
$$
For each <plane electromagnetic wave>, $\mathbf H=\hat{\mathbf k}\times\mathbf E/Z_j$. With no <free surface> charge or current, the <interface conditions for electromagnetic fields> require continuous tangential $\mathbf E,\mathbf H$ and normal $\mathbf D,\mathbf B$. Equality of the <phase> at every point on the boundary gives equal <angular frequencies> and tangential <wavevectors>. Hence the reflection <angle> equals the incidence <angle> and <Snell's law> is
$$
n_1\sin\theta_i=n_2\sin\theta_t,
$$
where $n_j=c/v_j$ and each <angle> is measured from the normal.
At normal incidence use the same tangential axis for all electric <amplitudes>. The two independent boundary equations are
$$
E_i+E_r=E_t,\qquad \frac{E_i-E_r}{Z_1}=\frac{E_t}{Z_2}.
$$
Solving gives the <amplitude> <Fresnel equations>
$$
\boxed{r=\frac{E_r}{E_i}=\frac{Z_2-Z_1}{Z_2+Z_1},\qquad t=\frac{E_t}{E_i}=\frac{2Z_2}{Z_1+Z_2}.}
$$
For nonmagnetic media, these become $r=(n_1-n_2)/(n_1+n_2)$ and $t=2n_1/(n_1+n_2)$. Reflection into a medium of larger <refractive index> therefore reverses the electric <amplitude>'s sign at normal incidence.
The normal <Poynting vector> determines the reflected and transmitted fractions of incident <power>. At normal incidence,
$$
\mathcal R=|r|^2,\qquad \mathcal T=\frac{Z_1}{Z_2}|t|^2,\qquad
\boxed{\mathcal R+\mathcal T=1.}
$$
It is the <power> fractions, rather than the squared electric <amplitudes> alone, that add to unity.
At oblique incidence separate <transverse electric polarization> ($s$, <electric field> perpendicular to the incidence plane) from <transverse magnetic polarization> ($p$, <electric field> in that plane). For nonmagnetic media, writing $c_i=\cos\theta_i$ and $c_t=\cos\theta_t$, the <Fresnel equations> are
$$
r_s=\frac{n_1c_i-n_2c_t}{n_1c_i+n_2c_t},\qquad
t_s=\frac{2n_1c_i}{n_1c_i+n_2c_t},
$$
$$
r_p=\frac{n_2c_i-n_1c_t}{n_2c_i+n_1c_t},\qquad
t_p=\frac{2n_1c_i}{n_2c_i+n_1c_t}.
$$
Here choose a common $s$ direction and, for each ray, the $p$ direction as $\hat{\mathbf s}\times\hat{\mathbf k}$. This makes the reflected $p$ basis reverse its tangential direction at normal incidence, explaining why the displayed $r_p$ then has the opposite sign to $r_s$. Physical <reflectance> is independent of this sign convention. For either polarization with propagating transmission,
$$
\mathcal R=|r|^2,\qquad \mathcal T=\frac{n_2\cos\theta_t}{n_1\cos\theta_i}|t|^2.
$$
For $p$ polarization, zero reflection occurs at the <Brewster angle> $\tan\theta_B=n_2/n_1$. If $n_1>n_2$, incidence above the <critical angle for total internal reflection> $\sin\theta_c=n_2/n_1$ produces an <evanescent field in total internal reflection> and $\mathcal R=1$ for an isolated lossless interface. A lossy second medium requires complex response coefficients and an absorption term in the <energy> balance.
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