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 $1/2$) ?


1 Answer 1


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 $\frac{I_0}{2}R_s+\frac{I_0}{2}R_p=I_0(R_s+R_p)/2.$


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