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Intensity

Without wanting to get all tedious about it, you should be able to compute the transmission coefficient and reflection coefficient for all of these waves from these results. These are basically the fraction of the energy (per unit area per unit time) in the incident wave that is transmitted vs being reflected by the surface.

This is a simple idea, but it is a bit tricky to actually compute for a couple of reasons. One is that we only care about energy that makes it through the surface. The directed intensity of the wave (energy per unit area per unit time) is the Poynting vector $ \Vec{S}$ . In equation 11.35 above, we found the time-average Poynting vector in terms of the $ E$ -field strength and direction of propagation: = 12&epsi#epsilon;&mu#mu;|E|^2(k) (where we have written the direction of propagation in terms of $ \hk =
\vk/k$ to avoid confusion with the normal to the surface $ \hn$ , which we recall is $ \hz$ , not $ \hk$ ).

We only care about the energy flux through the plane surface and thus must form $ \Vec{S}\cdot\Hat{n}$ for each wave: I_0 = S_n & = & 12&epsi#epsilon;&mu#mu;|E_0|^2 (&thetas#theta;_i)
I_0' = S_n' & = & 12&epsi#epsilon;'&mu#mu;'|E_0'|^2 (&thetas#theta;_r)
I_0'' = S_n'' & = & 12&epsi#epsilon;&mu#mu;|E_0''|^2 (&thetas#theta;_i)
This is ``easy''11.8 only if the waves are incident $ \perp$ to the surface, in which case one gets: T & = & I_0'I_0 = &epsi#epsilon;'&mu#mu;&epsi#epsilon;&mu#mu;' |E_0'|^2|E_0|^2
& = & 4nn'(n'+n)^2 R & = & I_0''I_0 = |E_0''|^2|E_0|^2
& = & (n' - n)^2(n'+n)^2 As a mini-exercise, verify that $ T + R = 1$ (as it must). Seriously, it takes only three or four lines.


next up previous contents
Next: Polarization Revisited: The Brewster Up: Dynamics and Reflection/Refraction Previous: Parallel to Plane of   Contents
Robert G. Brown 2017-07-11