# How does one determine whether an object will make an EM wave refract in a qualitative way?

for example, i have a vague notion that the actual answer is that the permittivity and permisivity are different in each different material, so all waves refract at every boundary, but we only call it that if it makes it out with any real magnitude left which depends on the skin depth or something like that, but is there a simple off the cuff way to estimate based on the ratio of the wavelength and that of the object?

for example, will a radio wave refract through a tree?

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Do you want to know just whether the EMW will be refracted by the object (spoiler: yes) or estimate the refraction index? Is direct measurement allowed? –  Maksim Zholudev Jan 8 '12 at 16:19

From one engineer to (I suspect) another, I'll give you my practical perspective in this largely theoretical forum.

If you're talking about radio refraction by objects in the real world, the "back of the envelope" or intuitive way to think about it is in terms of conductivity or lossiness of the medium. The concept of skin depth you mention is also useful. In classical EM theory, an EM wave will refract through anything that's not a perfect conductor. However, in real everyday materials, there is the concept of skin depth, which is the spatial rate of exponential falloff of the fields that penetrate the object. It's like the fields go as $e^{-x/\delta}$ as you go $x$ units into a material with $\delta$ skin depth. This is because the wave is robbed of energy through ohmic losses, whereby the wave's energy is used to move around electron currents in the material, which jostle around creating heat energy. Since energy is conserved, the heat generated takes away energy from the EM wave.

The skin depth is related to the conductivity or lossiness of the material you're talking about. You can calculate it for materials if you know the material permittivity, permeability, conductivity, and the frequency of the EM wave. If you're used to decibels, a rule of thumb is that each skin-depth you penetrate into the material takes about 8.7 dB out of the wave (the exact number is $20 \log_{10}(e)$=8.685889638065...).

So, if you're concerned with effects, say, within 90 dB of dynamic range of the incident power levels, then about 10 skin depths is about the largest thing through which you have to worry about refraction. It boils down to picking a cutoff for "what's weak enough" to neglect, then figuring how many skin depths corresponds to that level of attenuation in your particular material and frequency. If your tree is larger than that, you can neglect the refractive propagation through the tree.

Incidentally, very lossy materials also reflect waves at their interface in addition to attenuating what does penetrate them. To be more rigorous, you'd have to include that effect as well. This reflection coefficient is again related to permittivity, permeability, and conductivity. And in fact, higher conductivity means less penetration (more reflection) to begin with, so the bottom line is that if the material is even slightly conductive and the frequency is VHF or higher, you don't usually consider refraction through the material as a propagation mechanism of concern. This doesn't mean that it doesn't physically occur, just that people working in radio don't usually consider this effect when thinking of radio propagation.

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If you're not looking for a physical explanation, rather a quick way to know what will happen, you need only look up the complex index if refraction of that material which accounts for all of those factors. Look at this site: http://refractiveindex.info. A material's index of refraction is different for varying wavelengths, so you have to take that into consideration. The real part, $n$, represents how much light is slowed (or sped up, if $n<1$) in the medium, which determines how much refraction occurs as well. The imaginary part, listed there as $k$, relates how much the radiation is absorbed in the medium. The skin depth is how far a wave penetrates before the magnitude of the wave falls off by a factor of $e$. A material with a high $k$ for a given wavelength doesn't just absorb the light, however, it's usually reflected. Silver, for example, has a $k$ of around 3 to 4 for visible light, which it's why used in mirrors. Absorption occurs more due to resonance than conductive properties, but I'm near the limit of my understanding there.