Below we have the Fresnel Equation at an interface for S and P polarisation: \begin{align} r^s &= \frac{n_0 \cos{\theta_0}-n_1\cos{\theta_1}}{n_0\cos{\theta_0} + n_1\cos{\theta_1}} \quad\quad t^s = \frac{2n_1\cos{\theta_0}}{n_0\cos{\theta_0} + n_1\cos{\theta_1}}\\ r^p&= \frac{n_1 \cos{\theta_0}-n_0\cos{\theta_1}}{n_1\cos{\theta_0} + n_0\cos{\theta_1}}\quad\quad t^p = \frac{2n_1\cos{\theta_0}}{n_1\cos{\theta_0} + n_0\cos{\theta_1}} \end{align} When light approaches the interface at normal incidence these equations simplify to: \begin{align} r^s &= \frac{n_0 -n_1}{n_0 + n_1} \quad\quad t^s = \frac{2n_1}{n_0 + n_1}\\ r^p&= \frac{n_1 -n_0}{n_1 + n_0}\quad\quad t^p = \frac{2n_1}{n_1 + n_0} \end{align}

When we apply this to an example where we have an aluminium mirror with refractive index, for example, of 0.93878 + i6.4195.

This would lead for S polarisation to a reflection coefficient -0.9138 - 0.2855i, leading to reflectivity of 95% but a phase shift of -0.90$\pi$. However, for P polarisation a reflection coefficient 0.9138 + 0.2855i is found, leading to the same reflectivity but a phase shift of 0.09$\pi$.

Since we come in under normal incidence, why would the orientation of the polarisation matter? The polarisation S and P are both parallel with the surface, so this would mean if I rotate the mirror I would change the phase of the reflected beam? How does this make sense? What am I missing?

Edit: this also the case for dielectric matter (i.e. real $n$)


This is explained fairly well in the relevant Wikipedia article. Your paradox just arises from the need for a sign convention when deriving the Fresnel conditions, namely that a positive reflection coefficient means the transverse wave component keeps the same direction. For s-polarised light that is the E-field, but for p-polarised light it is the H-field.

Hence a positive $r$ for p-polarised light means no phase change for the H-field, but since the reflected wave travels opposite to the incident wave, the E-field must flip.

  • $\begingroup$ So the reflection coeff. from the fresnel equations only apply on the transverse wave component? $\endgroup$
    – SjonTeflon
    Jul 2 '19 at 17:54
  • $\begingroup$ @SjonTeflon they are applied to the E-field. The H-field follows from the properties of EM waves. However, there has to be a definition of what a positive r means. $\endgroup$
    – ProfRob
    Jul 2 '19 at 21:45

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