= Solution
For a <plane electromagnetic wave> with <linear polarization>, write $\mathbf E=E_0\widehat{\mathbf x}\cos(kz-\omega t)$ and $\mathbf H=E_0\widehat{\mathbf y}\cos(kz-\omega t)/(\mu_0c)$. The <Poynting vector> is instantaneous <energy> flux:
$$
\mathbf S=\frac{E_0^2}{\mu_0c}\cos^2(kz-\omega t)\widehat{\mathbf z}.
$$
The <light intensity> is its cycle average, not its peak:
$$
\boxed{I=\langle|\mathbf S|\rangle=\frac{E_0^2}{2\mu_0c}=\frac12\epsilon_0cE_0^2.}
$$
An incident <electromagnetic wave> transports longitudinal <momentum> $\mathcal E/c$ per <energy> $\mathcal E$. Perfect reflection back along the incident direction reverses that <momentum>, transferring $2\mathcal E/c$ to the mirror. Dividing by illuminated <area> and time gives the mean <radiation pressure>
$$
\boxed{p=\frac{2I}{c}=\frac{2\langle|\mathbf S|\rangle}{c}.}
$$
Here $\mathbf S$ denotes the \b[incident travelling wave]. The average net <Poynting vector> of the incident-plus-reflected <standing wave> is zero, although its <radiation pressure> is nonzero.
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