# How does increasing conductivity increases radar cross section?

This is for the context of radar detection. I am trying to figure out exactly why increasing the conductivity of a metallic object increases its radar cross section. I've been directed to looking at reflection coefficients which I understand is the ratio of the reflected wave to the incident wave. I've even read a paper on reducing the conductivity and whilst I don't understand all the theoretical explanations through Maxwell's equations, it was said that higher conductivity means more loss and thus increases reflection at the interface but I would like to understand why. Lower conductivity suggests that the incident energy would turn more into heat but again, I can't really visualize what is going on and how it all relates.

I've also heard that increasing the conductivity increases the power scattered back from the target.

I feel like I'm missing some vital principles I need to understand and would be grateful if anyone could help me.

PS: I can't use the tag "conductivity". If anyone would like to help and add that to this question then that would be amazing.

The quantities you want are the Fresnel reflection and transmission coefficients, respectively denoted by $\Gamma$ and $\tau$. They are defined as the ratio of the reflected to the incident electric field, and the ratio of the transmitted to the incident electric field. For normal incidence \begin{align} E_{r}&=\left(\frac{\eta_{c2}-\eta_{c1}}{\eta_{c2}+\eta_{c1}}\right)E_{is}:=\Gamma E_{i}\, ,\\ E_{t}&=\left(\frac{2\eta_{c2}}{\eta_{c2}+\eta_{c1}}\right)E_{i}:= \tau E_{i}\, , \end{align} where $\eta_{ck}$ are the complex impedences of the media $1$ and $2$. These depend on the conductivity of the respective media. The impedence of vacuum is $\eta_0\approx 120\pi$ (expressed in $\Omega$, since impedence as the same units as resistance.)

The expression for $\eta_{c}$ is messy in general, but if we assume the medium is a good conductor this simplifies to \begin{align} \eta_c \approx \sqrt{\frac{\mu\omega}{2\sigma}}\left(1+j\right)\, , \tag{1} \end{align} Thus, for a wave incident in air on medium $2$ that is a good good conductor, $\eta_{c1}\approx 120\pi$ and $\eta_{c2}$ given by (1) we find the reflection coefficient $$\Gamma=\frac{-1+\frac{1}{120\pi}\sqrt{\frac{\mu\omega}{2\sigma}}\left(1+j\right)}{1+\frac{1}{120\pi}\sqrt{\frac{\mu\omega}{2\sigma}}\left(1+j\right)}\, .$$ In the limit of perfect conductor, $\sigma\to \infty$, $\Gamma\to -1$ and all the wave is reflected.

Details on the exact expression for $\eta_{c}$ can be found in any intermediate-level textbook on electromagnetic theory.

I saw engineer in the OP name so I will answer with that in mind.

Wood, fiberglass and glass are insulators and bounce back virtually no signal.

In the same way human bones are essentially metallic due to their nano level structure and opaque to gamma rays, metallic objects are also highly relfective of RF singnals.

Salt or water on the other hand are highly absorbant but that is a function of the scattering angle.

This comes down to at least two factors, angle of incidence and either dielectric constant or conductivity althought my area of expertise is the Qm OF SEMICONDUCTORS.