= Solution
Define the dual tensor by
$$
\widetilde F^{\mu\nu}
=\frac12\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}.
$$
For metric signature $(-,+,+,+)$ and $\epsilon^{0123}=+1$, two independent quadratic <electromagnetic field invariants> are
$$
\boxed{
F_{\mu\nu}F^{\mu\nu}
=2\left(B^2-\frac{E^2}{c^2}\right)}
$$
and
$$
\boxed{
F_{\mu\nu}\widetilde F^{\mu\nu}
=-\frac4c\,\mathbf E\cdot\mathbf B}.
$$
Changing orientation or the sign convention for $F$ can change the displayed overall signs but not their vanishing.
For a vacuum <plane electromagnetic wave> with wavevector $\mathbf k$,
$$
\mathbf k\cdot\mathbf E=0,
\qquad
\mathbf B=\frac{\mathbf k\times\mathbf E}{\omega},
\qquad
\omega=c|\mathbf k|.
$$
Hence
$$
\mathbf E\cdot\mathbf B=0,
\qquad
B^2=\frac{E^2}{c^2},
$$
for every transverse polarization. Both Lorentz scalars therefore vanish.
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