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We know that light is an electromagnetic wave, so it depends on the permittivity and permeability of a substance. The permittivity and permeability of free space(vacuum) are constants.

So, propagation of an electromagnetic wave through a medium depends on the permittivity and permeability of an object. But in a superconductor, all magnetic flux fields are repelled from the object(no magnetic fields can pass through it). So, can light or any other electromagnetic wave travel through a superconductor?

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  • $\begingroup$ Visible light can't even pass through metals, which are highly conductive but not superconductive. A superconductor would have even better ability to reject EM waves. $\endgroup$
    – The Photon
    Apr 23, 2021 at 5:59
  • $\begingroup$ My guess: for a field (magnetic/electric) to be removed from a superconductor the charge particles (cooper pairs) have to re-arrange to counter-balance. They are massive objects, have inertia. Therefore I expect that a very-high frequency radiation can penetrate (any) objects. At high frequencies the mass of charged particles inside the object prevents them ro react (displace themselves) quickly enough. $\endgroup$
    – F. Jatpil
    Apr 23, 2021 at 11:29
  • $\begingroup$ they say - NO, light can not travel through a superconductor>> physics.stackexchange.com/questions/360178/… $\endgroup$
    – pan91
    Apr 23, 2021 at 15:52

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The answer is more complicated than the comments. Here I'll ignore evanescent wave effects which allow light to go through metals and superconductors thin films. Talking about bulk materials only, it depends on which light you are talking about.

The electromagnetic spectrum goes from dc, to radiofrequency, to microwaves, to the far-infrared (or its new trendy name THz), to the infrared, visible, UV, deep-UV, soft x-rays, hard-x-rays, to gamma-rays (and I might be missing a few).

Gamma-rays go trough trough any matter we know with a transmission coefficient close to 100%. Even metals and superconductors. A hand-waving explanation is that the frequency of the oscillation is so high that no solid excitation can follow it.

Being more reasonable, let's talk about visible and UV frequencies. In conducting media a quantity called plasma frequency $\Omega_p$ becomes very important:

$$ \Omega_p = \frac{ne^2}{\varepsilon_0 m} $$

where $n$ is the free carrier charge density, $m$ its (effective mass), $e$ the electron charge, and $\varepsilon_0$ the vacuum permittivity.

In a very good metal, roughly described by the Drude model, light of frequencies up to $\Omega_p$ are almost 100% reflected. The charge density in metals, puts $\Omega_p$ in the UV. That is why the aluminum mirror in your bathroom reflects your image. Note that above $\Omega_p$ metals are quite transparent. Hard x-rays go through most metals.

The "almost" 100% comes from what is called the scattering rate (the inverse of the time between two collisions of an electron in a solid). If the scattering rate is zero (infinite time between collisions) then one would have a perfect 100% reflectivity up to $\Omega_p$.

Now let's talk about superconductors. The short answer is that they are the same as metals. No visible light goes through. But there is a twist.

Below the critical temperature, a superconductor becomes sort of a perfect metal with a zero scattering rate. And experimentally, we see that the "almost 100%" reflectivity becomes "really 100%" reflectivity. But there is another twist.

A superconductor is "sort-of-a-perfect-metal" but it is not one. It has a superconducting gap. The perfect conductivity comes from electrons forming pairs (Cooper pairs) and then condensing in a single macroscopic quantum state. Well, if you give a Cooper pair an energy equivalent to the superconducting gap, you break it. And light can give that energy to a Cooper pair.

So, the reflectivity of a superconductor is exactly 100% for all frequencies whose energy $h \nu$ is smaller than the gap $2\Delta$ ($\Delta$ is the energy gain for one electron so $2\Delta$ for a pair). $2\Delta$ is in the microwaves for the very low $T_c$ materials and move into the far-infrared in the highest-$T_c$'s (let's ignore the room temperature superconductors at high pressures). Above that energy, the reflectivity falls down to the "almost 100%" and stays that way up to $\Omega_p$. Then it falls down steeply and the material becomes transparent for those frequencies. And this is a common way to measure the superconducting gap.

In the above, there are a lot of simplifications (I did not consider any absorption by interband transitions, I considered a fully symmetric so-called $s$-wave gap, and what-not). But the overall description is this one.

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