Visible light (~500 THz) as well as gamma rays (~100 EHz) are electromagnetic radiation but we can reflect visible light using a glass mirror but not gamma rays. Why is that?

  • $\begingroup$ The frequency range for visible light is a lot smaller, only about 430 to 750 THz. This is about a factor of two in frequency or wavelength, not three orders of magnitude. The range you give would reach from far infrared to extreme ultraviolet. Still light, but most of it not visible. $\endgroup$
    – Dubu
    Jun 8, 2018 at 8:42
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    $\begingroup$ the same reason you can't reflect a truck with a trampoline $\endgroup$
    – Matt
    Jun 8, 2018 at 8:55
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    $\begingroup$ @Matt Except that gamma rays don't obliterate the mirror, where the truck, well, let's just say your mileage may vary. $\endgroup$
    – rubenvb
    Jun 8, 2018 at 11:29
  • $\begingroup$ Because a mirror is made for reflecting light. $\endgroup$ Jun 9, 2018 at 10:43

3 Answers 3


Look at the electromagnetic spectrum:

Visible frequencies have wavelengths of microns, $10^{-6}$ meters.

Gamma rays have a wavelength of $10^{-12}$ meters, picometers.

In physics, there are two mainframes, the classical frame, which includes Maxwell's electrodynamics, Newton's mechanics, and derivative theories, and the quantum mechanical frame which becomes necessary for small distances and high energies, where gammas (photons), electrons, atoms, nucleons, lattices belong.

The classical electromagnetic wave emerges from zillions of superposed photons. Maxwell's equations describe very well the behavior of light beams when scattering or reflecting or generally interacting for macroscopic distances and small energies. Reflection, classically, needs a very flat surface so that the phases of the reflected waves are retained. Depending on the material the classical beams may be absorbed, decohered in reflecting from many point sources, or reflected coherently if the scattering is elastic (mirrors elastically and coherently scatter incoming light).

Gamma rays though force us to go to the micro level, because of the very small wavelength that describes them as a light beam.

One has to look at the details of the surface, and whether a classical smooth surface for classical reflections can be modelled for gammas, and the answer is, no it cannot.

The spacing between atoms in most ordered solids is on the order of a few ångströms (a few tenths of a nanometer).

For micron wavelengths (optical light) the fields built up by atoms with angstrom distances in the lattice appear smooth and can be classically modelled.

Gamma rays considered as a classical light beam, with their picometer wavelengths see mostly empty space between the atoms of the solid.

An alternative analysis, still within the quantum frame, would be considering the photons which make up light, and the Heisenberg uncertainty $ΔpΔx$ in the location of the photon. For the small wavelengths of gamma rays, the photons see mostly empty space.

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    $\begingroup$ So a mirror made in the heart of a neutron star would reflect gamma rays? $\endgroup$
    – iceman
    Jun 8, 2018 at 2:30
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    $\begingroup$ @iceman A neutron star is made up of compacted neutrons, and no solid lattices can be imagined there. $\endgroup$
    – anna v
    Jun 8, 2018 at 4:12
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    $\begingroup$ @jamesqf Any gammas in the mess will be scattered but imo it is not possible to get a "surface" that would give coherent enough scattering to be called a mirror, even gammas hitting the surface of a neutron star. I may not have enough physics imagination. $\endgroup$
    – anna v
    Jun 8, 2018 at 4:48
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    $\begingroup$ A quick example of something similar: That grid mesh on the door of your microwave is sufficient to make the door opaque to microwave radiation, but it lets through the much-smaller-wavelength visible light just fine. So you are kept safe as you stare drooling at your microwave burrito heating up. $\endgroup$ Jun 8, 2018 at 16:41
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    $\begingroup$ This answer seems very different than my2cts's. What does the "distance between atoms" matter if it's the electron field that is doing the reflecting? $\endgroup$
    – Neil G
    Jun 9, 2018 at 4:08

The reason why is based in something called the plasma frequency of the metal of a mirror. A metal, as you may know, is composed of a series of atom (ion, effectively) cores - nuclei, together with some, but not all, of their bound electrons - which contribute the remaining outermost electrons of their unbound forms to a communally shared "electron sea" - kind of like a giant, distributed omnidirectional covalent bond that extends all throughout the whole metal crystal (here we're just considering a single crystal for simplicity). The electrons are quantumed out all over the full extent of the crystal and effectively form a sort of "gas" throughout and permeating the metal.

When an electromagnetic wave approaches that gas, the free charges within it - the electrons - start oscillating, and as they do so, they set up another wave going outward at the same time as the first is going in. This begins as soon as the first wave begins to impinge.

However, if the wave oscillation is fast enough, the electrons can't keep up due to their mass, and thus they are unable to form the reflected wave. The frequency at which this occurs is called the metal's plasma frequency (and is inversely proportional to the square root of the mass, so that a high mass particle would have a lower plasma frequency). The name comes from the fact that the metal can be thought of in a sense as a kind of "solid plasma" - ions with free electrons, the difference with what most people think of as a "plasma" being here the ions are not free to move about of their own accord.

From here:


the plasma frequency for copper is about 2.0 PHz, which would correspond to a wavelength of about 150 nm, in the ultraviolet range. If you subject copper to EM waves at a frequency much higher than this, they will pass right on through as the electrons effectively ignore them.

The idea mentioned here about photons "fitting between" atoms or "through" their electron fuzz is not quite right. The transparency shows up well before you get to wavelengths smaller than the inter-atomic distance - e.g. their 150 nm is on the order of roughly a thousand times an inter-atomic distance for a metal. It is true that if you were to make it that small it technically "sees" the structure of the atoms in that they are now larger than the blur radius (that is, the size below which an object will appear blurry and thus indistinct to the rays due to diffraction), but the actual transparency comes well before that point due to this effect.


Reflection is caused by electrons reacting to the electromagnetic field by oscillating at the same frequency. When they do this they emit radiation of the same frequency as the incoming light and this is observed as reflection. This works well if the EM frequency is near the eigen frequencies of the electrons. When the frequency is very high the electrons are simply too massive and the forces retaining them not strong enough - think of a mass on a spring - to follow the electric field. So gamma rays can pass through matter.


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