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As I know, when photons hit the surface of a mirror, atoms on the surface of the mirror are excited and when they back to some lower states, they emit photons. I have two questions about this process. First, why atoms back to a lower state while photons are hitting with the surface of the mirror continuously? The second question is that why emitted photons are emitted in a way that their angles with the mirror are the same as the angle of the photons before hitting the mirror? In another word, why the image in the mirror is not disordered?

Qmechanic
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user201949
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1 Answers1

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Very good questions. The comments are right that there is no consensus on this site about what happens when photons hit a mirror, so I will give you the version that seem the most logical and will answer your questions.

You figured out by yourself, that when you try to describe how photons are creating a mirror image with absorption, re-emission, it will not give you the correct image.

Why? Because there are three things that can happen when a photon hits a mirror and interacts with the atoms:

  1. elastic scattering, the photon keeps its energy and phase and changes angle

  2. inelastic scattering, the photon will give part of its energy to the atom, and will change angle

  3. absorption, the photon will give all its energy to the atom, and the valence electron will move to a higher energy level as per QM

Now in the case of a mirror, it cannot be absorption, because of what you say:

  1. the photons would change angle, decoherently

The emission can also change angle decoherently, so the original photon's angle will not be the opposite of the re-emitted angle, and all the photons will be re-emitted in different angles.

  1. the photons would not keep their energies, and phases

The problem with absorption is that when the atom re-emits the photon, it can re-emit in multiple steps. So the absorbed photon's energy will be divided into more then one photon and the re-emitted photons will have energies that will sum up the original photon. So the re-emitting electron will come down as per QM to a lower energy level in more then one step. This way a mirror image cannot be built.

so a mirror image will not be built with absorption-re-emission.

The only way to keep the photon's energy and phase is elastic scattering, that is Rayleigh scattering, when the photon's wavelengths' are big compared to the size of atoms.

With mirrors, some photons will be elastically scattered, some will be inelstically scattered, and some will be absorbed. In the case of a mirror, most of them will be elastically scattered, and that will build up the image.

Now mirror images are not only built by metals. If you look at glass, some can give you a really good mirror image. So why is that?

It is because it is elastic scattering, and it is almost the same thing in the case of mirror as refraction.

What is the difference in the case of glass between reflection (mirror image in glass) and refraction (image traveling in glass)?

The only difference is the angle. Reflected photons build up an image that is reflected by an angle almost opposite to the original angle. Refracted photons build up an image that is refracted by an angle that is almost the same as the original one.

It is just the angle, and that with refraction, photons travel inside the new medium, glass. With reflection, photos travel back in air. Both cases it is elastic scattering. It is called specular reflection.

Árpád Szendrei
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