My book (Nanoscale Energy Transport & Conversion by Gang Chen, pg. 173) considers the following scenario of a TE wave reflecting at an interface:

enter image description here

It says the incident $\bf{E}$ field is: $$\textbf{E}_{\parallel i}exp\left[-i\omega\left(t-\frac{n_1xcos\theta_i+n_1zcos\theta_i}{c_0}\right)\right]$$

and that we can obtain

$$ H_i=\frac{n_1}{\mu c_0}E_{\parallel i}$$

from a Maxwell equation. How is this result obtained?

I assume we use

$$ \nabla \times \textbf{E} =-\frac{\partial \textbf{B}}{\partial t}$$


$$ \nabla \times \textbf{E} = \left( \frac{\partial E_x}{\partial z} - \frac{\partial E_z}{\partial x} \right)\hat{\textbf{y}}$$

and the the components $E_x$ and $E_z$ are unknown.


1 Answer 1


Those components are not unknown—in fact in linear media $\mathbf E_\text i$ is going to be perpendicular to $\mathbf k_\text i$ as shown in your figure.

It looks like the author has done something very strange with the actual components of their $\mathbf k$-vector; I would state that given the diagram you have, if the $+x$-direction is away from the interface and the $+z$-direction is otherwise in the direction of $\mathbf k_\text i$ as you have shown, then the right-handed $+y$-direction is into the plane of the figure, we have $\mathbf k_\text i = (-\cos\theta_\text i, 0, +\sin\theta_\text i),$ and then in somewhat more detail we have $$\mathbf E = E_0 ~\begin{bmatrix}\sin\theta_\text i\\ 0\\ \cos\theta_\text i\end{bmatrix}~\exp\left[-i\omega\left(t - \frac{-n_1 x \cos\theta_\text i + n_1 z \sin\theta_\text i}{c_0}\right)\right],$$notice the minus sign appearing by $n_1 x$.

Then indeed we can compute $\nabla \times\mathbf E$ to find $$ \nabla\times\mathbf E = -i\omega \frac{n_1}{c_0} E_0\begin{bmatrix}0\\-\cos^2\theta_\text i -\sin^2\theta_\text i\\0\end{bmatrix}~\exp\left[-i\omega\left(t - \frac{-n_1 x \cos\theta_\text i + n_1 z \sin\theta_\text i}{c_0}\right)\right], $$ and of course the quantity inside the column vector there simplifies to just $(0,-1,0).$ The time integral effectively divides by $-i\omega$ leading to $$ \begin{align}\mathbf B &= -\int dt~\nabla\times \mathbf{E}\\ &=-\frac{-i\omega}{-i\omega}\frac{E_0 n_1}{c_0} ~\begin{bmatrix}0\\-1\\0\end{bmatrix}~\exp\left[-i\omega\left(t - \frac{-n_1 x \cos\theta_\text i + n_1 z \sin\theta_\text i}{c_0}\right)\right]\\ &=\frac{E_0 n_1}{c_0} ~\begin{bmatrix}0\\1\\0\end{bmatrix}~\exp\left[-i\omega\left(t - \frac{-n_1 x \cos\theta_\text i + n_1 z \sin\theta_\text i}{c_0}\right)\right], \end{align}$$and then in linear media we have $\mathbf H = \mathbf B/\mu,$ hence $$\|\mathbf H\| = \frac{n_1}{\mu c_0}\|\mathbf E\|.$$Note that we haven't used the fact that there exists an interface yet—this is a completely general relation holding for all plane waves in linear media, not just ones that happen to be in a certain plane and happen to be heading towards an interface.

  • $\begingroup$ Your B field is polarized in the negative y-direction (into the screen). Shouldn't it be pointing in the positive y-direction (out of the screen)? $\endgroup$ Commented Nov 4, 2019 at 1:07
  • $\begingroup$ I'm saying this based on the figure - the B field is positive out of the screen, in the positive y-direction. Your B field seems to violate right hand rule? $\endgroup$ Commented Nov 4, 2019 at 1:34
  • $\begingroup$ @Drew You are correct that the sign was wrong; I have corrected it. $\endgroup$
    – CR Drost
    Commented Nov 4, 2019 at 19:56

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