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I am going through Electromagnetic waves. In footnote, the author was giving illustration on transparency of materials.

Was going through some references here as well as in Google. Many people attribute that photons get absorbed in material and that's why light does not pass through. Fine!

But I am looking for an answer from wave approach (not particle). For this to happen, the dielectric constant has to be $\infty$, and then the speed will go to zero inside the metal. But if I consider AM/FM radio waves, they pass through easily, and from that angle I am under impression that dielectric constant has to be low as not to disturb (decrease) the speed of incident wave too much!

Q. 1) I deduce, dielectric constant has to be frequency dependent. But is it that sensitive to incident frequency?

I have another angle to look into this phenomenon. Which is wave attenuation. This can happen, if the material introduces damping or so to say. For glass, the conductivity is zero (almost). In that case we can take the damping as $e^{-kz}$, where $k$ ~ $0$, and $z$ is the direction of wave vector - direction of propagation. In that case, also I am unable to differentiate the fact that light is allowed whereas radio waves are not as damping factor is almost equal to $1$ in both cases.

Q. 2) If attenuation also can't explain on why light is allowed whereas radio waves are not, then what I am missing, OR what is (are) wrong in the above steps.

Your support is kindly apreciated.

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    $\begingroup$ Some photons get absorbed depending on their energy. There is the frequency dependence. $\endgroup$ – AHusain Jul 13 '16 at 19:50
  • $\begingroup$ Could you kindly elaborate on this? $\endgroup$ – user3001408 Jul 13 '16 at 19:52
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    $\begingroup$ See "Why isn't everything transparent", and related questions: physics.stackexchange.com/q/247084 $\endgroup$ – Peter Diehr Jul 13 '16 at 20:14
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Dielectric materials have a loss tangent : at a given frequency, the response to EM radiation will have a phase lag relative to the incident wave. For pure dielectric materials, the phase angle is zero; when there are losses in the material, the loss angle will be non-zero.

For a given loss angle (usually given as the tangent of the angle) the amplitude of the wave can be given as

$$A_x = A_0 e^{jkx}$$

where $k$ is a complex number, $k = k_0(1+\alpha j)$. Now you can expand that, and you get

$$A_x = A_0 e^{jk_0 x}e^{-k_0\alpha x}$$

You can see there is now a decay term that depends on both the thickness of the sample, and the wave number ($k=\frac{2\pi}{\lambda}$). From which it follows that EM radiation with a longer wavelength (lower wavenunber) will be less absorbed in material of a given thickness, even when the loss tangent is the same.

But there's a second factor. A lot of the "lack of transparency" of materials like cardboard is due to scatter - basically because the cardboard is not homogeneous, there are many changes in refractive index; at every change there is a probability of light scattering, with the probability greater when the change in refractive index is greater (incidentally this is why wet paper is more translucent: although light has the same number of transitions, there are smaller changes in refractive index when air is replaced with water). BUT for long wavelengths, this scatter mechanism doesn't apply: Rayleigh scattering has a strong wavelength dependence (which is why the sky is blue).

So - long wavelengths don't scatter off small imperfections; and when the total path traversed comprises fewer wavelengths, there is less attenuation.

Both these factors play a role. And then there are wavelength-dependent absorption properties (depending on the exact molecular structures) - but we don't need to invoke those to answer your question. It is covered in some of the linked answers given in the comments.

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    $\begingroup$ "From which it follows that materials with a lower wavelength will be less absorbed in material of a given thickness", you meant EM radiation instead of the 1st instance of "materials" ? $\endgroup$ – Manu de Hanoi Sep 24 '18 at 11:47
  • $\begingroup$ @ManudeHanoi you are absolutely right. I have modified the wording. Thank you! $\endgroup$ – Floris Sep 24 '18 at 14:15

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