# reflection coefficient derivation for normal incidence emr

Starting with the boundary conditions for parallel E and B fields for emr normal to an interface, i am trying to derive reflection coefficient with refractive indices.
I have got as far as

$E_1=E_2\qquad \qquad$ (parallel $\vec E$ fields are equal)

$E_{0i}+E_{0r}=E_{0t}\qquad \qquad$ Eq.(a)

$\frac{B_1}{\mu_1} = \frac{B_2}{\mu_2}\qquad \qquad$ (parallel $\vec B$ fields are discontinuous)

$\frac{E_i}{\mu_0 v_1} + \frac{E_{0r}}{\mu_0 v_1} = \frac{E_{0t}}{\mu_0 v_1} \qquad$ (B fields represented as electric field by dividing through by speed)

$E_{0i}-E_{0r} = \frac{v1}{v2} E_{0t} \qquad$
(cancel $\mu_0$ and multiply by $v_1$)

$E_{0i}-E_{0r} = \frac{n_2}{n_1} E_{0t} \qquad$ Eq. (b) (can swap $v$ for refractive index)

$2E_{0i}=\left(1+\frac{n_2}{n_1} E_{0t}\right)\qquad$ Eq.(c) (I can the get to here by adding equation a and b together)

I know I'm really close but I can't quite get to the reflection or transmission coefficient, how can i finish this off?

• Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. – ZeroTheHero Apr 29 '18 at 1:35
• ok done that, still stuck :( – tgmcroc Apr 29 '18 at 15:43
• Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. – ACuriousMind Apr 29 '18 at 15:43
• For starters you can rearrange Eq.(c) to $\frac{2 n_1}{n_1+n_2}E_{0i}=E_{0t}$ by multiplying both sides by $n_1$. The transmission coefficients are usually expressed in terms of $E_{0i}$ so I suspect your bottom right is slightly incorrect. – ZeroTheHero Apr 29 '18 at 16:30
• Your equation involving the B-field is incorrect. – Rob Jeffries Apr 29 '18 at 17:22