# Decomposing reflectance for a wave whose vibration is at an angle with incidence plane

I am solving exercises where I occasionally need to decompose the radiance of an EM wave in its components, but I don't understand how it's done. In this context, reflectance ($$R$$) and transmissivity ($$T$$), defined as power ratios derived from Fresnel's laws, should always verify that $$R+T=1$$ due to the conservation of energy. In the case I'm having trouble with, where the electric field is neither $$p-$$ nor $$s-$$polarized, but at an angle $$\theta$$ with the incidence plane of a wave that interacts with an interphase between two dielectric materials, my professor decomposes R as:

$$R = R_s \sin^2 \alpha + R_p \cos^2 \alpha$$

If $$R_s$$ and $$R_p$$ could be seen as components of a vector (which I guess is not possible), there would need to be a $$^2$$ exponent in $$R_s$$, $$R_p$$ and $$R$$. But this doesn't seem to be the case, and I am confused as to why this happens. Does this have anything to do with the requirement that $$R+T = 1$$, or is this a mistake on my professor's behalf?

## 1 Answer

$$R_s$$ and $$R_p$$ are indeed the squared moduli of the reflection coefficients of the fields $$r_s$$ and $$r_p$$. In fact, by calculating the reflected field first using $$r_s$$ and $$r_p$$ first and afterwards calculating the reflected power (squaring $$\vec E$$) and using $$R_s = \vert r_s\vert^2$$ and $$R_p = \vert r_p\vert^2$$ you arrive at $$E_{refl}^2=E^2 (R_s \sin^2\theta + R_p\cos^2\theta)$$ what leads directly to the formula you cited.

• Thank you very much! My book didn't point at the relation between $r$ and $R$, and I therefore didn't think they were so obviously related. – mar Dec 26 '19 at 18:43