# Where do the conductivity and the frequency come in when solving for reflection coefficient of a plane wave hitting a conducting surface?

I have a plane wave hitting a surface of copper at normal incidence. I know that

$$R=\frac{I_R}{I_I}= \left ( \frac{E_{0,R}}{E_{0,I}} \right )^2=\left (\frac{n_1-n_2}{n_1+n_2} \right )^2$$

How does one reach a definition of the reflection coefficient that includes $\sigma$ and $\omega$? $\sigma$ is the conductivity of the material and $\omega$ is the frequency of the incident wave.

My attempt:

I used the equation:

$$E_{0,R}=\left |\frac{v_2-v_1}{v_2+v_1} \right |E_{0,I}$$

and plugged it into the equation for the reflection coefficient. Then I needed to decide what to use for $v_2$ and $v_1$. I know that $v=c/n$, where $n$ is the index of refraction.

• Make the anszats, take the dot products, assume continuity and you are done, if I remember my homework from 7 years ago right. – Emil Apr 26 '17 at 4:27
• I'm sure that's probably correct, but it's a bit vague for me (a beginner) to follow.... – loltospoon Apr 26 '17 at 4:30

For a given $\omega$, the useful quantity here is the generalized (complex) impedance $\eta$, defined by $$\eta=\sqrt{\frac{\mu}{\epsilon}}\left(1-i\frac{\sigma\omega}{\epsilon}\right)^{-1/2}\, .$$
The reflection coefficient and reflectivity are given by $$\Gamma=\frac{\eta_2-\eta_1}{\eta_2+\eta_1}\, ,\qquad\qquad R=\vert \Gamma\vert^2\, . \tag{1}$$ Usually $\mu,\epsilon$ and $\sigma$ can all have some $\omega$-dependence so these quantities are tabulated at specific frequencies. There are also microscopic models to evaluate some of these.

For lossless media, where $\sigma=0$, the impedance $\eta$ and index of refraction $n$ are related by $$\eta=\sqrt{\frac{\mu}{\epsilon}}=\sqrt{\frac{\mu_r}{\epsilon_r}}\,\mu_0\sqrt{\frac{1}{\mu_0\epsilon_0}}:=\frac{c\mu_0}{n}=\frac{\eta_0}{n}$$ Inserting this in (1) yields $$\Gamma\to \frac{n_1-n_2}{n_1+n_2}$$ as you found. One can also write $\epsilon(\omega)$ as done by @user157879 to get an effective $n(\omega)$ expression.

Note that, for a perfect conductor, $\eta\to 0$ so the reflection coefficient is $\Gamma=-1$ for a wave propagating from medium $1$ hitting a perfectly conducting medium $2$.

The frequency dependence comes into the index of refraction $n=n(\omega)$. The index of refraction is not only a function of incident light frequency, but is also complex. Allowing the index to be complex means you are now allowing absorption, as evanescent decaying waves effectively have imaginary rather than real wave vectors. The frequency dependence of the index is expected, you don't see through metals in the visible range, but you can do so in the x-ray range.

You may be wondering when you made any assumption regarding frequency. The assumption was made when you considered the reflectivity of the surface. By definition the reflectance of the surface is the fraction of the incoming light frequency that is reflected into the exact same frequency. You are ignoring the possibility that the frequency of light changes as it is reflected (I.e. Raman scattering). However, you have not specified any specific frequency, but rather that all the light waves stay at the same frequency through the reflectance process.

The relation to the frequency dependent optical conductivity is through the following definitions that come the Fourier transform of Maxwell's equations and the assumption of linear response (Ohm's law, dielectric function linear in E-field):

$$n(\omega)^2=\epsilon(\omega) \mu(\omega) \approx \epsilon(\omega)$$ $$\epsilon(\omega) = \epsilon_0 + \frac{i \sigma(\omega)}{\omega}$$