electromagnetismoptics
I'm working on an optics project that I'm using Fresnel Equation for Intensity. but I've got a question, is copper a non-magnetic ?
You know because in Fresnel Equation, for non-magnetic materials which their refractive index is a real number, the formula is more simple.
Thanks in advance
Metals refractive index is always complex number (and not only for metals). Imagine part shows the extinction coefficient $k$ - absorption in a material.
Real and imagine part isn't connected.
P.S. For engineering calculations real part sometimes is less than 1.
Theoretically even for Fresnel reflection in dielectric we must use full formula with complex part:
$R=\frac{(n-1)^2+k^2}{(n-1)^2+k^2}$
for dielectrics (glasses) in visual diapason $k<<1$ so we don't use it, but u got metal ;-)
Suppose we define the following $\zeta = \ln{\varepsilon}$ and $\xi = \ln{\mu}$, where $\varepsilon$ and $\mu$ are the permittivity and permeability, respectively. In a system with no sources (i.e., $\mathbf{j} = 0$ and $\rho_{c} = 0$), then we know that $\nabla \cdot \mathbf{D} = 0$, where $\mathbf{D} = \varepsilon \ \mathbf{E}$ and $\mathbf{B} = \mu \ \mathbf{H}$. After a little vector calculus we can show that: $$ \nabla \cdot \mathbf{E} = - \nabla \zeta \cdot \mathbf{E} \tag{0} $$ Using this and some manipulation of Faraday's law and Ampêre's law, we can show that the general differential equation in terms of electric fields only is given by: $$ \left( \mu \varepsilon \frac{ \partial^{2} }{ \partial t^{2} } - \nabla^{2} \right) \mathbf{E} = \left( \mathbf{E} \cdot \nabla \right) \nabla \zeta + \left( \nabla \zeta \cdot \nabla \right) \mathbf{E} + \nabla \left( \zeta + \xi \right) \times \left( \nabla \times \mathbf{E} \right) \tag{1} $$
We can get a tiny amount of reprieve from this by assuming that the permeability is that of free space, i.e., $\nabla \xi = 0$. If we further argue that the only direction in which gradients matter is along $\hat{x}$ and that the incident wave vector, $\mathbf{k}$, is parallel to this, then we can further simplify Equation 1 to: $$ \left( \mu_{o} \varepsilon \frac{ \partial^{2} }{ \partial t^{2} } - \frac{ \partial^{2} }{ \partial x^{2} } \right) \mathbf{E} = \left( \mathbf{E} \cdot \nabla \right) \zeta' \hat{x} + \left( \zeta' \frac{ \partial }{ \partial x } \right) \mathbf{E} + \zeta' \hat{x} \times \left( \nabla \times \mathbf{E} \right) \tag{2} $$ where $\zeta' = \tfrac{ \partial \zeta }{ \partial x }$.
After some more manipulation, we can break this up into components to show that: $$ \begin{align} \text{x : } \left( \mu_{o} \varepsilon \frac{ \partial^{2} }{ \partial t^{2} } - \frac{ \partial^{2} }{ \partial x^{2} } \right) E_{x} & = E_{x} \zeta'' + \zeta' \frac{ \partial E_{x} }{ \partial x } \tag{2a} \\ \text{y : } \left( \mu_{o} \varepsilon \frac{ \partial^{2} }{ \partial t^{2} } - \frac{ \partial^{2} }{ \partial x^{2} } \right) E_{y} & = 0 \tag{2b} \\ \text{z : } \left( \mu_{o} \varepsilon \frac{ \partial^{2} }{ \partial t^{2} } - \frac{ \partial^{2} }{ \partial x^{2} } \right) E_{z} & = 0 \tag{2c} \end{align} $$
In the limit where the incident wave is entirely transverse, then $E_{x} = 0$ and the x-component (Equation 2a) is entirely zero.
Next you assume that $\mathbf{E} = \mathbf{E}_{o}\left( x \right) e^{i \omega t}$, where $\omega$ is the frequency of the incident wave. Then there will be incident, reflected, and transmitted contributions to the total field at any given point (well the transmitted is always zero in the first medium, of course). Any incident and transmitted contributions with have $\mathbf{k} \cdot \mathbf{x} > 0$ while reflected waves will satisfy $\mathbf{k} \cdot \mathbf{x} < 0$. You define the ratio of the reflected-to-incident fields (well impedances would be more appropriate) to get the coefficient of reflection.
Simpler Approach
A much simpler approach is to know where to look for the answer to these types of questions. As I mentioned in the comments, there has been a ton of work on this very topic (i.e., spatially dependent index of refraction) done for the ionosphere. If we look at, for instance, Roettger [1980] we find a nice, convenient equation for the reflection coefficient, $R$, as a function of the index of refraction, given by:
$$
R = \int \ dx \ \frac{ 1 }{ 2 \ n\left( x \right) } \frac{ \partial n\left( x \right) }{ \partial x } \ e^{-i \ k \ x} \tag{3}
$$
There is no analytical expression for $R$ for your specific index of refraction. However, numerical integration is not difficult if one knows the values for $d$ and $\epsilon$. Note that if we do a Taylor expansion for small $\epsilon$, then the integrand (not including the exponential) is proportional to cosine, to first order in $\epsilon$ (cosine times sine if we go to second order).
References
Best Answer
Copper is a diamagnetic material. Individual copper atoms have one unpaired electron in the valence shell, and thus might be considered paramagnetic, but when many copper atoms are combined into the bulk metal, their valence electrons are sent into a cloud that forms metallic bonds among the copper atoms, and the metal is diamagnetic. All the answers given to this similar question explain the diamagnetic nature of copper in more detail: Why is copper diamagnetic?
Although a diamagnetic material opposes a magnetic field in which it is placed, most interact only very weakly with a magnetic field. The diamagnetic property of copper metal is so weak that it is considered non-magnetic.