A lens with a high permeability In real life materials with high $\mu$ values are not realistic, so lenses are made with high $\epsilon$ materials. But what would be the impact of achieving the same refractive index by increasing $\mu_r$ too.
For instance, if we made a lens with $\epsilon_r=\mu_r$ the wave impedance of free space would be the same as the lens, right? So would the lens just work with much much much less reflections? How big of an impact would this make?
Or what if we used a lens with $\epsilon_r=1,\mu_r>1$ , how different would different ways of achieving the same refractive index be? Which would be the best pick?
 A: The refractive index is $\nu = \sqrt{\epsilon_r \mu_r}$ because the speed of light is $c/\nu$ in that medium. At this level of idealization it does not matter what the relative permittivity or relative permeability is, only their product matters when it comes to ideal propagation in a homogeneous medium.
If the two media have different permittivities, $\epsilon_r, \epsilon'_r$, and permeabilites, $\mu_r, \mu'_r$ , then the Fresnel reflection formulas will also change from the commonly known ones as in https://en.wikipedia.org/wiki/Fresnel_equations because you now have to match both the E and H fields at the interface. For example, in the case of the incident E field parallel with the interface the reflectivity $\mathcal R$ becomes, see Jackson 7.39:
$$\mathcal R = \frac{2\nu cos {\mathcal i}}{\nu cos i + \frac{\mu_r}{\mu'_r}\sqrt{\nu'^2-\nu^2 sin^2 i}}$$
If now you assume that the angle of incidence is $i=0$ and also the special case $\mu_r/\epsilon_r=\mu_r/\epsilon_r$ you do get $\mathcal R=0$, as expected.
