In the book Principles of Lasers by Orazio Svelto, at chapter 4.3 "Wave Reflection and Transmission at a Dielectric Interface", the author tells that

If the wave is initially in the medium of refractive index n_1 and it is normally incident on the surface, the electric field reflectivity is $r_{12}=\frac {n_1-n_2} {n_1+n_2}$

Then the author also tells:

For non-normal incidence, the expressions for electric field reflectivity and transmission are more complicated and depend also on the field polarization

So the author is saying that waves with different polarization have same reflectivity if the beam is normally incident on the surface. However i know that Fresnel equations for normally incident beam tells: $$r_p=\frac {n_2-n_1} {n_1+n_2}=-r_s$$ So the two polarization have a reflectivity coefficient that differs by a minus sign. How is that possible?


1 Answer 1


$S$ and $p$ polarizations are defined with the assumption of non-normal incidence, and they converge as angle of incidence $\theta \to 0$. A “surface” implicitly has continuous rotational symmetry about the surface normal (unless otherwise stated). Thus, there can be no difference between light polarized in the $x$ direction vs $y$ direction for normal incidence in the $z$ direction.

Your last equation is wrong. What you’re probably thinking of is light incident from opposite sides of an interface, from $n_1 \to n_2$ vs $n_2 \to n_1$. Those indeed differ only by a sign.


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