Power carried by a light wave: intensity and brightness

We are generally interested in the optical power per unit area of a light beam, as illustrated below:

As you might imagine, a modest amount of optical power illuminating a very small area can cause significant damage.  There is some lack of uniformity in the usage of terminology in optics, but the power per area is commonly referred to as the intensity ($latex W / m^2$).  If we have the electromagnetic wave fields $latex \vec{E}$ and $latex \vec{H}$, we can calculate the intensity from the Poynting vector,

$latex \displaystyle \vec{P} = \vec{E} \times \vec{H} $

The direction of the Poynting vector is the direction of the power flow, which for the plane waves we’ve written will be the same as the direction of propagation.  The magnitude of the Poynting vector is the instantaneous intensity.  For time-harmonic propagating waves, both $latex \vec{E}$ and $latex \vec{H}$ will each oscillate in time as $latex \cos{\omega t}$.  Therefore, the instantaneous intensity will oscillate in time as $latex \cos^2{\omega t}$.  Using the cosine double-angle formula, this equals $latex (1/2) (1 + \cos{2 \omega t})$.   You can see there’s one non-fluctuating term and one term fluctuating at twice the optical frequency.   As a reminder, optical frequencies are on the order of $latex 10^{15} Hz$.  Your eye and many optical detection devices can’t detect the rapidly oscillating term in the instantaneous intensity, but instead detect the intensity averaged over many cycles of oscillation (time-averaged intensity).  The averaging will wipe out the term oscillating at double frequency, leaving behind the DC term.  In terms of the phasor electromagnetic fields, we can write the time-average intensity as

$latex \displaystyle \vec{P}_{avg} = \frac{1}{2} Re (\vec{E}_{phasor} \times \vec{H}^*_{phasor} ) $

which, for a plane wave, becomes

$latex \displaystyle \vec{P}_{avg} = \hat{a}_n \frac{ \sqrt{\epsilon} |A|^2} {2 \sqrt{\mu}} $

where $latex A$ is the amplitude of the electric field of the plane wave.  As promised, the intensity is proportional to the square of the electric field.