Light (visible part of electromagnetic spectrum) is easily blocked by most materials. Radio waves are not but also X-rays are not. What is so special about this small part of the spectrum since both shorter and longer waves seem to behave differently?

  • 2
    $\begingroup$ A thin metal foil will block most radio waves just fine. You have to go to really low frequencies (on the order of 10kHz and less) to get trough more than a few mm of copper. You would need 50keV x-ray energy at the other end of the spectrum to do the same with high energy radiation. A thin sheet of copper can therefor block some 15 or 16 orders of electromagnetic frequencies. That's pretty good in my books. $\endgroup$
    – CuriousOne
    Dec 25, 2015 at 4:19

3 Answers 3


Blocking means there are interactions with the light that manage to absorb it completely.

Most materials are opaque to visible light due to the underlying quantum nature of matter. To emit light or absorb light there should be energy levels that will "trap" those frequencies

em spectrum

electromagnetic spectrum

Yes, x-rays and higher energy photons pass through thin layers of matter because they mostly scatter. Only very low frequency radio wave will go through anything. Most other frequencies will resonate with some electronic energy bands and particularly metals can trap them.

So it is not a "small part of the spectrum" that is blocked. What is true is that radio waves go through non metalic materials because there are no electronic energy levels to absorb them or reflect them. And the same is true for X rays and higher frequencies.

  • $\begingroup$ naively, do metallic grids absorb radio waves ? $\endgroup$
    – user46925
    Dec 25, 2015 at 10:50
  • $\begingroup$ @igael It depends on the grid dimensions and the wavelength of the radio wave. After all a metal antenna picks up the radio signal for the radio, so yes the electronic band structure of metal has energy bands that can absorb the radio frequency photons. Alternative picture classically, a dipole electric field can be built up on the antenna by the incoming em wave $\endgroup$
    – anna v
    Dec 25, 2015 at 19:19

Many frequencies of waves are blocked easily - for example, see this graph of atmospheric electromagnetic absorption --> https://en.wikipedia.org/wiki/File:Atmospheric_electromagnetic_opacity.svg

However, you also asked what is special about visible light - well, it's special to us because we can see it! We have evolved retinas that respond to this part of the spectrum, reacting to and interpreting the received light as visual information.

It is plausible this happened because the solar spectrum peaks in the region of visible light, making it the part of the spectrum which has the potential to provide the most information.

Is there something special in the visible part of electromagnetic spectrum?


Whilst your general statement about radio waves is true, X-rays do not in general pass through all materials better than visible light (eg consider the Earth's atmosphere or water).

I'll offer you a different take on the radio wave transparency issue.

The interaction of light with matter can be dealt with in terms of a macroscopic absorption coefficient that is related to microscopic radiation processes as $$\kappa = \frac{h\nu_{ji}}{4\pi}(n_i B_{ij} - n_j B_{ji}),$$ where $n_i$ and $n_j$ are populations in levels $i$ and $j$ and $B_{ij}$ and $B_{ji}$ are the Einstein absorption and stimulated emission coefficients for transitions between these levels.

It is important to note that stimulated emission is equivalent to negative absorption.

Now the Einstein coefficients are related to each other (see https://en.m.wikipedia.org/wiki/Einstein_coefficients), such that $$ g_i B_{ij} = g_j B_{ji},$$ where $g_i$ and $g_j$ are the statistical weights of the respective energy levels.

For a medium in thermal equilibrium, the populations of the levels are related by $$\frac{n_j}{n_i} = \frac{g_j}{g_i}\exp(-h \nu_{ji}/kT),$$ where $h\nu_{ji}$ is the energy difference between the levels. Thus the absorption coefficient could then be written $$\kappa =\frac{h\nu_{ji}}{4\pi}n_i B_{ij} \left(1 - \exp(-h\nu_{ji}/kT)\right)$$

Now radio wave transitions with wavelengths longer than say 1 m, have corresponding energies of $<1.2 \mu eV$. So even if temperatures are similar to the cosmic microwave background (and not much gets cooler than this), the term in brackets is very close to zero. Thus even where there are appropriate transitions at radio wave energies, any true absorption is almost exactly cancelled out by stimulated emission, resulting in no net absorption $\kappa \simeq 0$.

Note that this is only a partial explanation that deals with the lack of atomic or molecular absorption at discrete energies. However similar arguments pertain to continuum absorption processes.

A somewhat different argument can be employed for materials with free electrons and high conductivity. Here the classical skin depth is proportional to $\nu^{-1/2}$. So low frequencies penetrate further.


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