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 \(\small{\sqrt{\mu_o\varepsilon_o}}\)
