I was reading the following:

What happens to photons after they hit objects?

And this was stated, but wasn't really backed up with mathematics (in the answer):

A photon reflected from a white wall can certainly be reflected from another white wall. However, the reflection can’t occur forever, because no real wall can reflect 100% of the light shone on it.

And was wondering, what law of physics states that 100% of light cannot be reflected off a surface? Does anyone have a proof or calculation for this? Also, currently known to man, what is the most reflective surface we have-- at what magnitude and in what units is it's reflective capability?


2 Answers 2


Total reflection means that the dimensionless reflection coefficient $R$ is unity. The formula for reflectivity at perpendicular incidence from vacuum onto a metal or dielectric is $R=|\frac{1- n} {1+n}|^2$. $R$ tends to unity for large n, but obviously never quite gets there. Notably metals have large n values for infrared light. This is likely the motivation for the assertion that $R$ can never reach unity, that is, total reflection. However there is the phenomenon of total internal reflection. I don't know if that would could as reflection from a wall, though.

  • $\begingroup$ Just to add on. Total reflection can occur in real world materials where the angle of incidence is not perpendicular. One would then think we can fire a photon at the angle of total reflection and have it bounce about two mirrors so that this angle is always preserved. The Lagrangian formalism of Quantum Mechanics states that all paths for the photon are possible. Even paths in which it bounces into the wall at different angle are possible. The chance of this happening may be extremely small while the region of paths close to the classical one are, naturally, the most likely. $\endgroup$
    – Finesagan
    Commented Aug 24, 2018 at 2:09
  • $\begingroup$ @finesagan if it bounced to the next atom deeper into the wall-- could it keep transferring between those two atoms indefinitely? $\endgroup$ Commented Aug 24, 2018 at 4:21

First, we need to clarify a few things. When a photon interacts with an atom, three things can happen:

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

  2. inelastic scattering, the photon gives part of its energy to the atom and changes angle

  3. absorption, the photon gives all its energy to the atom and the absorbing electron moves to a higher energy level as per QM.

In the case of reflection it is elastic scattering, but this happens with mirrors, not white walls so much.

In your case, with white wall, it is all three things that happen to the photons. Now the ratio of the three with the white wall is:

  1. absorption-reemission, this is the dominant in the case of the white wall, and this causes most of the white color

  2. elastic scattering, this is called reflection, and this is not so dominant with the white wall, the ratio of photons elastically scattered off the white wall is less then the ratio of absorbtion-reemission

  3. inelastic scattering, the ratio of these photons with a white wall is the smallest. This mostly causes molecular vibrations, thermal energy given to the wall.

So we need to define that white color light is a combination of all wavelengths. There is no white wavelength photon.

The main reason the white wall is white is that:

  1. most of the photons of all visible wavelengths are absorbed, or inelastically scattered (transform into thermal energy of the molecular vibrations)

  2. most of the re-emitted photons are a combination that creates perception of white light.

So, you would think that all wavelength photons are just absorbed or inelastically scattered, and only same wavelength are re-emitted too.

But that is not always true. When an electron gets excited by a certain wavelength photon, it moves to a higher energy level as per QM. And then it relaxes, to a lower level and emits a photon. But what wavelength photon does it emit? The same wavelength as the absorbed one? Well, not always.

The excited photon's wavelength can be:

  1. the same as the absorbed one

  2. different, in this case it is called that the electron relaxes in more then one step. It emits more than one photon of different wavelengths.

Now, when a certain wavelength photon is absorbed on the white wall, it is usually re-emitted in multiple steps, a combination of all wavelengths.

You would think that white walls are always white, in all conditions, not true. They are only white, because we are used to be in Sunlight environment.

If you shine red light on a white wall, the wall will seem reddish. You can do it with other color lights.

Sunlight is a combination of all colors, most people think the Sun is yellow. No, the Sun is white light, a combination of all colors. That is why a white wall is white, because white light is shining on it from the Sun, and it absorbs and re-emits all combinations.

Now why do we see certain color lights, that is because our eyes have RGB receptors, and a certain combination of those seems like certain colors. Usually white is the combination that excites all receptors.

Now to your question, why can't a surface reflect 100% of light? Now to do that, since reflection is elastic scattering, and that is what a mirror does, you would need a perfect mirror. That would mean, no inelastic scattering, and no absorption. There is no perfect mirror, that would only elastically scatter. All mirrors will heat up after a while, because of inelastic scattering and absorption.

It is probabilities. The ratio on a mirror is highest for elastic scattering. Some of the photons will though be inelastically scattered, and absorbed.

The best dialectic mirror can reflect almost all of photons. Please see here: https://en.wikipedia.org/wiki/Perfect_mirror

  • $\begingroup$ Wow! Thats actually a really simple and elegant way of explaining it. So in summary different atomic structures recieve X amount of temporary energy from a photon, which can also potentially absorb some of the energy-- reducing the wavelength of the photon? This is why certain materials cause the photons to enter different wavelengths in color when the photon finally gets to us? Sort of how mass spectrometers work? Higher order atoms return less wavelength and appear to have less eneregy? $\endgroup$ Commented Aug 24, 2018 at 4:02

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