# Why are electromagnetic waves attenuated by thick barriers?

Atoms have a specific absorption spectrum that determines what frequencies of EM radiation they are most likely to absorb. This is why visible light cannot pass through a wall, but radio waves can. However, if you make the wall thick enough, the radio waves will eventually be attenuated. Since the radio frequencies are not in the absorption spectrum of the atoms in the wall, why does increasing the thickness of the wall attenuate the radio signal?

Each atom has a well-defined absorption spectrum $$\alpha(\lambda)$$. The most prominent features of these spectra are the absorption resonances at $$\{\lambda_1,\lambda_2, \ldots\}$$. However, this does not imply that absorption becomes zero between two resonances. Instead, the absorption for a wavelength between two resonances is merely "much smaller" than at the resonances. Hence, by increasing the width of the wall the exponential intensity law (Lambert's law) $$I(z) = I_0 e^{-\alpha(\lambda)\cdot z}$$ kicks in and the absorption becomes non negligible.

Mathematically there exists a very simply argument which shows that the absorption is not allowed to be zero within any finite wavelength interval: If the absorption would be zero within the interval $$[\lambda_0, \lambda_0+\epsilon]$$ for $$\epsilon>0$$ it must be zero everywhere.

Physically, the atomic transition can be modelled as an externally driven harmonic oscillator with eigenfrequency $$\omega_0$$, and damping constant $$\gamma$$. This leads to the complex part of the refractive index $$n$$ to be $$\textrm{Im}\{n\} =:\kappa = \frac{\lambda \alpha}{4\pi} \propto %= \frac{N q^2}{2\epsilon_0 m} \frac{\gamma/\omega_0}{(\omega^2 - \omega_0^2)^2 + (\gamma/2 )^2}$$ From this we see that the absorption never becomes truly zero.

• Does that mean an atom can absorb photons with energies less than its smallest energy gap? How does that wok? Jun 20, 2020 at 17:26
• @Matthew a single atom can have a narrow resonance. But when you combine many atoms together into a macroscopic object, those resonances spread out (because the levels they're based on are spread due to the Pauli exclusion principle) so that the macroscopic object can have strong absorption over a wide range of frequencies. Jun 20, 2020 at 18:33
• @ThePhoton To understand your last comment I’ll add that radio waves are radiation where zillions of polarized photons acting on the wall periodically. Radio waves are a special case of EM radiation. Using zillions photons from a thermal source with the same wavelength as in a radio wave does not the same with the wall. Simply no resonance happens and the photons get easier dissipated to heat. Jun 21, 2020 at 5:26
• @HolgerFiedler: I was not aware of that. Furthermore, I don't know how one could produce a EM in the radio band from a thermal source. Could you please provide a reference which compares the two scenarios. Jun 21, 2020 at 8:56
• @Semoi Take an electric bulb (with no filament but a straight wire) and power it with a radio generator. You’ll get a radio wave, where the photons are polarized and are emitted periodically. What I never can agree methodically is the point, that a periodically emission is set equal to its wavelength. The zillion of photons, emitted from the skin of the wire are of very different wavelengths, sometimes in the range of X-rays. That is why it is dangerous to be in friont of a radar. physics.stackexchange.com/questions/90646/… Jun 21, 2020 at 9:10

To answer Matthew's second question: If the substance of a wall is slightly electrically conductive, the impingement of radio waves upon it will induce currents to flow in it, which are dissipated by any electrical resistance that the wall material possesses. This mechanism does not invoke discrete energy level transitions occurring within the electronic orbital structure of the atoms comprising the wall.