$$\mathbf{H}=\frac{1}{\mu\omega} \mathbf{k\times E}\;\;\;\;(incident)\;\;(2.49)$$

$$\mathbf{H^{'}}=\frac{1}{\mu\omega} \mathbf{k^{'}\times E^{'}}\;\;\;\;(reflected)\;\;(2.50)$$

$$\mathbf{H^{''}}=\frac{1}{\mu\omega} \mathbf{k^{''}\times E^{''}}\;\;\;\;(transmitted)\;\;(2.51)$$

It should be noted that the above equations apply either to the instantanous values of the fields or to the amplitudes, since the exponential factors $\exp i(\mathbf{k\cdot r} -\omega t)$, and so forth, are common to both the electric and associated magnetic fields.

I am trying to understand the paragraph written in the picture.

I understand that when the light incident at a boundary, there will be part of the light that gets transmitted and some of it gets reflected and We get these equations.

I don't understand the reason for the statement that "above equations apply either to the instantaneous values of the fields or to the amplitudes".

Can someone please help me to understand?


The instantaneous electric and magnetic fields are \begin{align} \mathbf{E}(\mathbf{r},t) &= \mathbf{E}_0 e^{i(\mathbf{k}\cdot\mathbf{r} - \omega t)}\\ \mathbf{H}(\mathbf{r},t) &= \mathbf{H}_0 e^{i(\mathbf{k}\cdot\mathbf{r} - \omega t)} \end{align} and $\mathbf{E}_0$, $\mathbf{H}_0$ are the amplitudes.

If we have \begin{align} \mathbf{H} = \frac{1}{\mu \omega}\mathbf{k}\times \mathbf{E}, \end{align} for the fields, then this means \begin{align} \mathbf{H}_0 e^{i(\mathbf{k}\cdot\mathbf{r} - \omega t)} = \frac{1}{\mu \omega}\mathbf{k}\times\mathbf{E}_0 e^{i(\mathbf{k}\cdot\mathbf{r} - \omega t)}. \end{align} Since the same (scalar, non-zero) factor $e^{i(\mathbf{k}\cdot\mathbf{r} - \omega t)}$ appears on both sides, we can cancel it to get \begin{align} \mathbf{H}_0 = \frac{1}{\mu \omega} \mathbf{k}\times \mathbf{E}_0 \end{align} So the relation $\mathbf{H} = \frac{1}{\mu \omega}\mathbf{k}\times \mathbf{E},$ applies to both the instantaneous fields $\mathbf{E}(\mathbf{r},t)$, $\mathbf{H}(\mathbf{r},t)$ and the amplitudes $\mathbf{E}_0$, $\mathbf{H}_0$.

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  • $\begingroup$ thank you @d_b. $\endgroup$ – Quantum_boy Nov 25 '19 at 1:25

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