# Intensity of reflexion of circular polarization

Circularly polarized light, can be caracterized as : $$E(r,t) = E_0 \cos(kz-\omega t) \hat{x} + E_0 \sin(kz-\omega t) \hat{y}$$

In my course, the Intensity of EM radiation is defined as $$I = |U|^2$$. And we defined polarized monochromatic waves as follows (i already put the phase for (right) circular): $$U(r,t) = (E_x \hat{x} + E_y \hat{y} e^{i \pi/2})e^{i(kz - \omega t)}$$ Now, i can get the first equation by taking the real part of the second one.

But now, let it reflect of some glass and let's assume the s and p polarizations coincide with the x-y axes. How do i get to $$I = I_0(R_s + R_p)/2$$. Where the $$R_s$$ and $$R_p$$ are the reflectivities. The Instensity as defined in the course doesn't seem well defined to me. Do i take the sum of the instensity along each axis (But then i miss a factor of $$1/2$$) ?

Yes, you take the sum of the intensity along each axis. But you do not miss the factor of $$1/2$$ by doing that. The reason is that the total input intensity to the reflection is $$I_0=U^2=|E_x|^2 + |E_y|^2$$ and $$|E_x|=|E_y|$$; Along each axis the input intensity is $${I_0}/{2}$$. With reflection reflectivity $$R_s$$ and $$R_p$$, your output intensity is $$\frac{I_0}{2}R_s+\frac{I_0}{2}R_p=I_0(R_s+R_p)/2.$$