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According to the Fresnel equations, the effective reflectivity of an arbitrarily polarized electromagnetic wave is given as:

$$ \Gamma = \frac{\Gamma_p + \Gamma_s}{2} , $$

where \$\Gamma_p\$ and \$\Gamma_s\$ are the reflectivities for p and s polarization, respectively.

I assume the common scenario of an electromagnetic wave traveling through empty space (\$n_1=1; \epsilon_1=\epsilon_0, \mu_1=\mu_0\$) hitting a non-magnetic surface (\$\mu_2=\mu_0\$) with dielectric \$\epsilon_2=\epsilon_0 \epsilon_2'\$ (\$n_2=\sqrt{\epsilon_2'}\$) at a straight angle.

If I plug these into the Fresnel equations with an angle of incidence \$\theta_i=0\$, I obtain:

$$ \Gamma_p = \frac{\epsilon_2'-\sqrt{\epsilon_2'}}{\epsilon_2'+\sqrt{\epsilon_2'}} , \\ \Gamma_s = \frac{1-\sqrt{\epsilon_2'}}{1+\sqrt{\epsilon_2'}} . $$

Then it is easy to see that \$\Gamma=(\Gamma_p+\Gamma_s)/2=0\$ ! This suggests that any circularly polarized wave hitting a surface at a perpendicular angle would be perfectly aborbed. That must be nonsense.

Where is the mistake?

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  • \$\begingroup\$ THis is easy to answer: what do the p and s in your polarization names stand for? \$\endgroup\$ – Marcus Müller Jan 7 at 8:23
  • \$\begingroup\$ senkrecht und parallel ;-) How does this answer? (I see how the wave-fronts are reaching the surface for both cases: de.wikipedia.org/wiki/Fresnelsche_Formeln#/media/… and de.wikipedia.org/wiki/Fresnelsche_Formeln#/media/…) \$\endgroup\$ – divB Jan 7 at 17:46
  • \$\begingroup\$ senkrcht means "normal to the surface"; and your wave is hitting the surface in a straight angle, and since there's no longitudinal field components in electromagnetic waves in free space... \$\endgroup\$ – Marcus Müller Jan 7 at 19:36
  • \$\begingroup\$ Sorry, I really don't understand. Both, \$\Gamma_p \neq 0\$ and \$\Gamma_s \neq 0\$! For example, for \$\epsilon_2'=9\$, \$\Gamma_p = 0.5\$ and \$\Gamma_s = -0.5\$. This makes sense. In both cases things are reflected. My question is why does in the circularly polarized case nothing seem to be reflected? \$\endgroup\$ – divB Jan 7 at 19:45
  • \$\begingroup\$ maybe I'm misinterpreting what "in a straight angle" means \$\endgroup\$ – Marcus Müller Jan 7 at 19:51

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