Perpendicular to Plane of Incidence

**Figure 11.2:** Polarization component of the incident (and reflected and refracted) waves perpendicular to the plane of incidence.
$\begin{figure} \centerline{\epsfbox{figures/plane_perp_inc.eps}} \end{figure}$

The electric field in this case is perforce parallel to the surface and hence $\Vec{E}\cdot\Hat{n} = 0$ and $\vert\Vec{E}\times\Hat{n}\vert = 1$ (for incident, reflected and refracted waves). Only two of the four equations above are thus useful. The $\Vec{E}$ equation is trivial. The $\Vec{B}$ equation requires us to determine the magnitude of the cross product of $\Vec{B}$ of each wave with $\Hat{n}$ . Let's do one component as an example.

**Figure:** Geometry of $\Vec{B}_0 \times \Hat{n}$ .
$\begin{figure} \centerline{\epsfbox{figures/B_cross_n.eps}} \end{figure}$

Examining the triangle formed between $\Vec{B}_0$ and $\Hat{n}$ for the incident waves (where $\theta_i$ is the angle of incidence), we note that $B_\perp = B_0 \cos(\theta_i)$ and thus: 1&mu#mu;| $\Vec{B}$ _0 × $\Hat{n}$ | & = & 1&mu#mu; B_0 (&thetas#theta;_i)
& = & &mu#mu;&epsi#epsilon;&mu#mu; E_0 (&thetas#theta;_i)
& = & &epsi#epsilon;&mu#mu; E_0 (&thetas#theta;_i).

Repeating this for the other two waves and collecting the results, we obtain: E_0 + E_0'' & = & E_0'
&epsi#epsilon;&mu#mu; (E_0 - E_0'')(&thetas#theta;_i) & = & &epsi#epsilon;'&mu#mu;' E_0'(&thetas#theta;_r) This is two equations with two unknowns. Solving it is a bit tedious. We need: (&thetas#theta;_r) & = & 1 - ^2(&thetas#theta;_r)
& = & 1 - n^2n'^2^2(&thetas#theta;_i)
& = & n'^2 - n^2^2(&thetas#theta;_i)n' Then we (say) eliminate

using the first equation: &epsi#epsilon;&mu#mu; (E_0 - E_0'')(&thetas#theta;_i) = &epsi#epsilon;'&mu#mu;' (E_0 + E_0'') n'^2 - n^2^2(&thetas#theta;_i)n' Collect all the terms: E_0 ( &epsi#epsilon;&mu#mu; (&thetas#theta;_i) - &epsi#epsilon;'&mu#mu;' n'^2 - n^2^2(&thetas#theta;_i)n' ) =
E_0''(&epsi#epsilon;'&mu#mu;'n'^2 - n^2^2(&thetas#theta;_i)n' + &epsi#epsilon;&mu#mu; (&thetas#theta;_i) ) Solve for

: E_0'' = E_0 ( &epsi#epsilon;&mu#mu; (&thetas#theta;_i) - &epsi#epsilon;'&mu#mu;' n'^2 - n^2^2(&thetas#theta;_i) n' ) ( &epsi#epsilon;&mu#mu; (&thetas#theta;_i) + &epsi#epsilon;'&mu#mu;' n'^2 - n^2^2(&thetas#theta;_i) n' )

This expression can be simplified after some tedious cancellations involving nn' = &mu#mu;&epsi#epsilon;&mu#mu;'&epsi#epsilon;' and either repeating the process or back-substituting to obtain : E_0'' & = &E_0 ( n(&thetas#theta;_i) - &mu#mu;&mu#mu;' n'^2 - n^2^2(&thetas#theta;_i) ) ( n(&thetas#theta;_i) + &mu#mu;&mu#mu;'n'^2 - n^2^2(&thetas#theta;_i) )
E_0' & = & E_0 2n(&thetas#theta;_i) ( n(&thetas#theta;_i) + &mu#mu;&mu#mu;'n'^2 - n^2^2(&thetas#theta;_i) )