Old Friends

If $\mu$ is to $B$ as $\varepsilon$ is to $E$, what then is $n$ to?  Why, in the first place photons reduce velocity?

$v^2=\cfrac{1}{\mu\varepsilon}$

$\cfrac{v^2_1}{v^2_2}=\cfrac{\mu_2\varepsilon_2}{\mu_1\varepsilon_1}$

Since,

$c^2=\cfrac{1}{\mu_o\varepsilon_o}$

If we define,

$n_o=\sqrt{\mu_o\varepsilon_o}$

then,

$n_r=\cfrac{n}{n_o}=\sqrt{\cfrac{\mu\varepsilon}{\mu_o\varepsilon_o}}$

$n_r=\cfrac{1}{v\sqrt{\mu_o\varepsilon_o}}$

$n_r.v=c$

$n$ is the result of both $\mu$ on $B$ and $\varepsilon$ on $E$.  $n_r$ is defined relative to $c$, or $\small{\sqrt{\mu_o\varepsilon_o}}$