$\Vec{E}$ Parallel to Plane of Incidence

Now the magnetic field is parallel to the surface so $\Vec{B}\cdot\Hat{n} = 0$ and $\vert\Vec{B} \times \Hat{n}\vert = 1$ . This time three equations survive, but they cannot all be independent as we have only two unknowns (given Snell's law above for the reflected/refracted waves). We might as well use the simplest possible forms, which are clearly the ones where we've already worked out the geometry, e.g. $\Vec{E}_0 \times \Hat{n} = E_0\cos(\theta_i)$ (as before for $\Vec{B}_0$ ). The two simplest ones are clearly: (E_0 - E_0'')(&thetas#theta;_i) & = & E_0'(&thetas#theta;_r)
&epsi#epsilon;&mu#mu;(E_0 + E_0'') & = & &epsi#epsilon;'&mu#mu;' E_0' (from the second matching equations for both $\Vec{E}$ and $\Vec{B}$ above).

It is left as a moderately tedious exercise to repeat the reasoning process for these two equations - eliminate either $E_0'$ or $E_0''$ and solve/simplify for the other, repeat or backsubstitute to obtain the originally eliminated one (or use your own favorite way of algebraically solving simultaneous equations) to obtain: E_0' & = & E_0 2nn'(&thetas#theta;_i) &mu#mu;&mu#mu;' n'^2(&thetas#theta;_i) + nn'^2 - n^2^2(&thetas#theta;_i)
E_0'' & = & E_0 &mu#mu;&mu#mu;' n'^2(&thetas#theta;_i) - nn'^2 - n^2^2(&thetas#theta;_i) &mu#mu;&mu#mu;' n'^2(&thetas#theta;_i) + nn'^2 - n^2^2(&thetas#theta;_i)

The last result that one should note before moving on is the important case of normal incidence (where $\cos{\theta_i} = 1$ and $\sin(\theta_i) = 0$ ). Now there should only be perpendicular solutions. Interestingly, either the parallel or perpendicular solutions above simplify with obvious cancellations and tedious eliminations to: E_0' & = & E_0 2n n' + n
E_0'' & = & E_0n' - nn' + n

Note well that the reflected wave changes phase (is negative relative to the incident wave in the plane of scattering) if $n>n'$ . This of course makes sense - there are many intuitive reasons to expect a wave to invert its phase when reflecting from a ``heavier'' medium^11.7.