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Thanks

marlon

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- Thread starter marlon
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Thanks

marlon

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Meir Achuz

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E'=E_{incident}[(n_1-n_2)/(n_1+n_2)] for normal incidence.

This shows the phase change. The FRs are derived in most junior level EM or optics texts. They follow from the BCs on E and H at the interface of two dielectrics.

The transverse components of each are continuous.

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Daniel.

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So my question still stands...Anyone who knows the answer ???

regards

marlon

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[tex] E_{reflected}=E_{incident}\frac{n_{1}-n_{2}}{n_{1}+n_{2}} [/tex]

If [itex] n_{1} < n_{2} [/itex],then [itex] E_{reflected}= - k E_{incident} [/itex] (1)

,where [itex] k=:\frac{n_{2}-n_{1}}{n_{1}+n_{2}}>0 [/itex] (2)

Okay.Now,u write,following (1) & (2)

[tex]E_{reflected}=k E_{incident}e^{i\pi} [/tex] (3)

Is it any clear?Guess not.I introduce the phases and the polarization vectors,okay,then

[tex] \vec{E}_{reflected}=\vec{e_{p}}E_{reflected}e^{i\left(\vec{k}_{reflected}\cdot\vec{r}-\omega t\right)} [/tex] (4)

Using (3),u see where that phase change comes from.

http://scienceworld.wolfram.com/physics/FresnelEquations.html

(For the graph).

Here's a nice course

http://www.ece.rutgers.edu/~orfanidi/ewa/

BTW,it's PHASE.

Daniel.

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thanks dexter

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