# Phase difference in reflection coefficient between S and P polarisation at normal incidence?

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 $$n$$)

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.