So I'm reading this textbook on modern optics by Fowel and it's saying that the magnitudes of H and E are related by
$$ H = \dfrac{\epsilon \omega E} {k} $$
Then it says because u=w/k (u is the velocity in the medium) we get
$$H= \epsilon u E $$
That's all fine until it says the refractive index is n=c/u which gives
$$H=\dfrac{nE} {Z_0} $$ where $Z_0=( \mu / \epsilon )^{1/2} $

My question is how does the refractive index, n, appear on the top? Am I missing a step somewhere that's not just direct substitution?

Question

So I'm reading this textbook on modern optics by Fowel and it's saying that the magnitudes of H and E are related by 
$$ H = \dfrac{\epsilon \omega E} {k} $$ 
Then it says because u=w/k (u is the velocity in the medium) we get
$$H= \epsilon u E $$ 
That's all fine until it says the refractive index is n=c/u which gives 
$$H=\dfrac{nE} {Z_0} $$  where $Z_0=( \mu / \epsilon )^{1/2} $ 

My question is how does the refractive index,  n,  appear on the top? Am I missing a step somewhere that's not just direct substitution?

athecool · Answer

Anyone?

raffle_snaffle · Answer

I need to see the text book? Epsilon and mu must equal n/Z