Find the direction of the parallel polarized E field reflecting off the boundary.

  • Thread starter yungman
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  • #1
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Main Question or Discussion Point

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:

[tex]\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 [/tex]

I know

[tex] \hat {E_r} \;=\; \hat x cos \theta_i + \hat z sin \theta_i [/tex]

My question is how can I derive the direction of [itex] \vec {E_r}[/itex] 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
 

Answers and Replies

  • #2
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You know from reflections that
[tex]\hat{n}_r=\hat{x}\sin(\theta_i)-\hat{z}\cos(\theta_i)[/tex]
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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