= Solution
In SI <units>, the source-free <Maxwell equations> in vacuum are
$$
\nabla\cdot E=0,\qquad \nabla\cdot B=0,\qquad
\nabla\times E=-\partial_tB,\qquad
\nabla\times B=\mu_0\epsilon_0\partial_tE.
$$
Use complex amplitudes with phase $e^{i(\omega t-k\cdot x)}$ and take real parts at the end. The electric divergence equation gives $k\cdot E_0=0$. Faraday's equation gives $k\times E_0=\omega B_0$, so
$$
\boxed{B(x,t)=\Re\left[\frac{k\times E_0}{\omega}e^{i(\omega t-k\cdot x)}\right].}
$$
The magnetic divergence equation follows because $k\cdot(k\times E_0)=0$. Ampere's equation becomes $k\times B_0=-\mu_0\epsilon_0\omega E_0$. Since $k\times(k\times E_0)=-|k|^2E_0$, a nonzero wave satisfies
$$
\boxed{\omega^2=c^2|k|^2,\qquad c=(\mu_0\epsilon_0)^{-1/2}.}
$$
Thus both <fields> are transverse, mutually perpendicular for linear polarization, and obey the vacuum plane-wave dispersion relation.
For the interface let $n=e_z$ point from the minus side to the plus side and write $[E]=E_+-E_-$, similarly for $B$. Applying the integral <Maxwell equations> to a thin pillbox and a thin loop gives the <electromagnetic boundary conditions>
$$
\boxed{n\cdot[E]=\sigma/\epsilon_0,\quad n\times[E]=0,\quad
n\cdot[B]=0,\quad n\times[B]=\mu_0 j.}
$$
Here $j$ is the tangential surface current. The jump notation fixes the signs, including the direction of the normal.
For the <perfect conductor>, take the time-varying interior <fields> to vanish. Choose positive $k$ with $\omega=ck$. The incident <electric field> has complex form $E_0e_xe^{i\omega t+ikz}$. To cancel its tangential value at the surface, use the reflected <field> $-E_0e_xe^{i\omega t-ikz}$, whose wavevector points upward. The total vacuum <fields> are therefore
$$
\begin{aligned}
E&=\Re[2iE_0e^{i\omega t}\sin(kz)]e_x,\\
B&=\Re[-(2E_0/c)e^{i\omega t}\cos(kz)]e_y.
\end{aligned}
$$
The sign of each magnetic contribution follows from $B_0=k\times E_0/\omega$. At $z=0$, the <electric field> is zero and the <magnetic field> is tangential. The jump conditions yield
$$
\boxed{\sigma=0,\qquad
j(t)=\frac{2}{\mu_0c}\Re(E_0e^{i\omega t})e_x.}
$$
For a real incident amplitude this is $j=(2E_0/\mu_0c)\cos(\omega t)e_x$. It supplies the reflected wave while satisfying the normal-field and tangential-electric conditions. Any independently imposed static interior <field> is separate from this oscillatory response.
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