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This well-observed phenomenon has, besides several others, always been a fascination to me. We are well aware of several theories, experiments, and practical applications of this well-known phenomenon, but is it established what goes on at the grassroots of reflection?

My high school teacher once told me that whenever light is incident on any reflecting surface, its electrons absorb the energy of the photon and release back the same energy. If this is actually true, then I have plenty of questions to continue, but if it is not, then what is reflection, and how and why does it take place?

Thomas Lee Abshier ND
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Rijul Gupta
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2 Answers2

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The question What IS reflection? is really a duplicate of yours, but I suspect the answer may be a bit brief for you.

A light wave, like any electromagnetic wave, is a combination of an oscillating electric and magnetic field. These fields exert an oscillating force on the electrons in any material the wave hits, and those electrons start oscillating in response. However an oscillating electron emits electromagnetic radiation, and this radiation interferes with the incident light.

To calculate what happens at the interface, you have to take into account the incident light, the light reradiated away from the solid back into the vacuum, (i.e., the reflected light) and the light reradiated into the body of the solid, (i.e., the transmitted light). When you do this, you find light is reflected at the angle of incidence, and the light is transmitted at an angle given by Snell's law. The calculation is a bit messy, but if you are interested, lots of examples are yours for the cost of a quick Google.

Response to comment:

The oscillations of the electrons are driven by the incoming EM field, so the oscillation frequency is the same as the frequency of the incoming light. The phase need not be (which is the origin of the refractive index change).

In the real world, the reflection is not independent of colour (i.e., the frequency of the light), and this phenomenon is known as dispersion. It is also not independent of light intensity, which is known as non-linearity, though non-linearity is usually an extremely small effect.

The electrons do not react instantly when the light strikes. The fastest frequency at which they can react is the plasma frequency.

Response to response to comment:

Energy can be lost due to interactions with the lattice, and indeed this is the norm, so the sum of the reflected and transmitted waves is generally less than the incident wave. The lost energy ends up as heat i.e., lattice vibrations.

However the fact energy is lost does not change the frequency of oscillation of the electrons, because that is determined by the incident light. That means the frequency of the reflected and transmitted light is the same as the incident light i.e., the same colour.

If the incident light contains different frequencies, like white light, then the result can be a shift in the perceived colour. For example, if you shine white light onto gold, the reflected light is yellowish. However the process of reflection is not changing the frequency of the light, but rather it is changing the relative intensities of frequencies in the reflected light by absorbing some frequencies more strongly than others.

Thomas Lee Abshier ND
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John Rennie
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  • How come the reflected light remains constant even in terms of color and intensity (assuming negligible refraction) are oscillations always equal to the frequncy of light absorbed ? Do the electrons stop oscillating as soon as the light is stopped from being incident on reflector ? – Rijul Gupta Oct 22 '13 at 17:21
  • @rijulgupta: I've edited my answer to respond to your comment. – John Rennie Oct 22 '13 at 17:33
  • I accept that the electrons would start oscillating with same frequency, but if they were emitting energy at same frequency wouldn't that mean that they are sending back all the energy. Wouldn't some electron get excited to newer energy level or retain some energy in any way, not at all ? sounds a bit wrong. – Rijul Gupta Oct 22 '13 at 17:47
  • Some energy is send back as reflected light and some is sent forward as transmitted light. If you add up the reflected and transmitted energy it must add up to be equal to the energy of the incident wave. The electrons will only be excited to a higher energy level if the wavelength of the incident light exactly matches the excitation energy. At this energy the refractive index can behave strangely though I confess I forget the details and would have to look them up. From memory the refractive index goes to zero. – John Rennie Oct 22 '13 at 17:52
  • Cant simply the material be left over with some energy, like energized maybe. I remeber there was some experiment in which momentum was imparted by firing electrons at a substance, if this substance would made reflector it would reflect the light but would itself also gain KE not sending back entire energy, so definitely something must be different. – Rijul Gupta Oct 22 '13 at 17:58
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    Electrons can interact with the lattice surround them and dissipate energy that way. In that case the dissipated energy will end up as heat, i.e. lattice vibrations, and the sum of the reflected and transmitted light will be less that the incident light. The obvious example of this is the black body (http://en.wikipedia.org/wiki/Black_body), where all the light is absorbed so no light is reflected and none is transmitted. – John Rennie Oct 22 '13 at 18:01
  • Then why would we still get same characteristic reflected light even when we can atleast hypothesise the change or loss in energy ? – Rijul Gupta Oct 22 '13 at 18:04
  • @rijulgupta: I've edited my answer again to respond to your last point. – John Rennie Oct 22 '13 at 18:11
  • Why would frequency of reflected transmitted and incident light beam remain same ? We said that electrons start oscillating with the frequency of incident light, then also lose energy, this must have some impact on their oscillations and after this when they are emitting the energy why would it remain same ? – Rijul Gupta Oct 22 '13 at 18:16
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    @rijulgupta: No, energy loss reduces the amplitude of the oscillation but does not change its frequency. – John Rennie Oct 22 '13 at 18:19
  • Then that must bring some change in intensity right ? Even for monochromatic light – Rijul Gupta Oct 22 '13 at 18:23
  • @rijulgupta: Yes, it does change the intensity, but it does not change the frequency. – John Rennie Oct 22 '13 at 18:24
  • Cool ! Does everyone know this already or we found something new ? – Rijul Gupta Oct 22 '13 at 18:31
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    @rijulgupta: sadly I suspect many of us knew this, but then many of us are old hands at this game :-) – John Rennie Oct 22 '13 at 18:37
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The essence of your question is the fundamental principle of reflection of highschool level.

Reflection takes place at an interface with different index of reflection.

At atom level your picture of an electron absorbing energy of light is right. Think of it as a pendulum: the harmonic oscillator. Without attenuation the frequency of oscillator stays the same. So the reflected color of the mirror interface is the same.

In fact the mathematics about reflection starts with the movement of an electronic cloud, like the harmonic oscillator. Some frequencies are less attenuated and the element has high reflectivty. At some frequency the energy is enough to excite a new external state and the photon is less likely to be re-emitted. To be sound, an attentuation, dependant of the internal structure of the element has to be introduced. This acts like any dissipative force on the movement of the oscillator.

However with attenuation, like wavelength dependant absorption, gold mirrors differ from common aluminium mirrors in you bathroom. An explanation about the reflection of different metals should be read to dig deeper into the topic. For short: aluminium reflects the whole visible spectrum while gold lacks reflection in the blue regime.

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Stefan Bischof
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  • No attenuation, none at all ? How is that happening ? – Rijul Gupta Oct 22 '13 at 17:55
  • What possibilites of energy loss does an electron have? This question is pretty advanced stuff for highschool level. No Attenuation in this picture means it does not loose energy through lattice vibrations or radiation. – Stefan Bischof Oct 22 '13 at 17:59
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    And can you say that it would not loose energy through lattice vibrations or radiations even through the phenomenon in which momentum is imparted to the material when photons are fired upon it thus giving it KE ? – Rijul Gupta Oct 22 '13 at 18:01
  • @rijulgupta I provided more information about attenuation. – Stefan Bischof May 07 '14 at 07:05