Find Parallel Polarized E Field Reflection Direction

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The discussion focuses on deriving the direction of the reflected electric field vector, \vec{E_r}, for a parallel polarized electromagnetic wave incident at an oblique angle on a conductor boundary. The incident direction and electric field are defined using vector notation, with the z-axis perpendicular to the boundary. The reflection vector, \hat{n}_r, is established as dependent on the incident angle, and the relationship between the electric field in free space and the outgoing wave vector is emphasized. The correct direction of \vec{E_r} is determined by the relative index of reflection, referencing the Fresnel equations for parallel plane incidence. The thread seeks mathematical clarification on these relationships.
yungman
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For a parallel polarization EM hitting the conductor boundary in an oblique angle. z axis is perpendicular to the boundary and point into the conductor. y-axis it out of the page which give x pointing up. Let the boundary surface by xy plane. With this:

The direction of the incident is:

\hat n_i \;=\; \hat x sin \theta_i + \hat z cos \theta_i \;\hbox { and direction of }\; \hat {E_i} \;= \hat x cos \theta_i - \hat z sin \theta_i

I know

\hat {E_r} \;=\; \hat x cos \theta_i + \hat z sin \theta_i

My question is how can I derive the direction of \vec {E_r} by using formulas? I got this by looking at the reflection as I move the incident E towards the boundary...by drawing. I want to find this mathametically. Please help.

Thanks

Alan
 
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You know from reflections that
\hat{n}_r=\hat{x}\sin(\theta_i)-\hat{z}\cos(\theta_i)
If your E-field is in free space, it must be normal to this outgoing wave vector. Since you had a parallel incident E-field, there are only two possibilities and one is just the negative of the other. Which one is correct depends on the relative index of reflection; see Fresnel equations for parallel plane incidence to show which.
 

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