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

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

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

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?

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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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