A metal (read mirror) behaves rather differently depending whether the frequency of the incoming light is above or below the plasmon frequency.
if the frequency is below the plasmon frequency, the electrons move in order to screen the electric field. This yields an approximate boundary condition $\mathbf E|_\text{boundary}=0$ and thus to reflection.
if the frequency is above the plasmon frequency, then the electrons cannot move (because of their inertia) and the wave essentially just goes through (just like in an ordinary dielectric).
As the plasmon frequency for typical metals is around $10^{15}\,$Hz, visible light is reflected and $X$-ray is transmitted.
Very reasonable question. I will try to answer it in an intuitive way.
If you have a scattering medium, photons are reflected in random directions; but when you have a refractive medium, something else happens. The photon is not absorbed and re-emitted: instead, the photon interacts with the electrons in the medium, and since these electrons are somewhat bound to the atoms, a displacement (due to the E field of the photon) results in a restoring force. We can measure the degree of displacement or polarization, and express it in terms of the dielectric constant of the material. If there is a lot of polarization, the dielectric constant is high. And the refractive index can be shown mathematically to scale with the square root of the dielectric constant.
What happens then is that the bound electrons move "behind in phase" with the photon, which results in a phase shift of the EM wave (the original phase of the photon which lost a bit of its amplitude, and a lagging phase of the electron). The photon didn't get destroyed - it got delayed, but maintained its direction.
Now when you have an interface between two materials of different refractive index, and light is incident at an angle to that interface, then a greater phase difference builds up between adjacent beams, which is why light is refracted; but if all the light is traveling through the same medium of constant refractive index, all beams will refract by the same amount and remain parallel.
Which is why you can see through a window.
But if you use ground glass, the surface is no longer flat but has been modified to change the direction - and this results in the image behind the glass becoming fuzzy. The same principle exists in the glass used in many showers (original image from Victoria Elizabeth Barnes's posts on bathroom remodeling:
You can clearly see how the bottom half of the window blurs the image of the trees outside. In the article they call this "pebbled" glass.
Best Answer
This question is really about how the refractive index of glass (and other materials) varies with frequency.
Here the term refractive index being used in its broadest sense it that it has a real part which is a measure of the speed of light through the material and an imaginary part which is called the attenuation coefficient.
For light the frequency dependence of refractive index can be seen with the dispersion of white light and absorption being frequency dependent with the absorption of ultra violet by glass.
I do not think that there is a material which is transparent to electromagnetic radiation at all frequencies and so it would seem that you cannot have a "one material fits all frequencies" lens.
So that is the answer to your question but for an explanation there are two very good answers related to this question in this forum but for a fuller explanation of the topic you could refer to Feynman I-31 and Feynman I-32.