How does optical reflection work?

  1. How does reflection work in non-metals and metals?

    If you take, for example, a polished brass surface (or whatever really) and an incident photon, is the photon being absorbed and re-emitted by an atom at the surface of the reflective brass?

    If so, how is it that the re-emitted photon is always the same wavelength as the incident photon?

    The only plausible way I can think to picture what's going on is that the incoming EM radiation causes electrons to oscillate and therefore re-emit EM radiation of the same frequency. I don't understand how this vibration process could be 100% energy efficient though. Also, can electrons oscillate at any given frequency? Aren't these vibrations quantized?

    I become even more confused when thinking about non-metals that totally don't have free electron gases. (For instance how does reflection in water work like when I see my face in the pond?)

    Edit: I want to be clear about what I'm asking and why I can't find it (at least not easily) in the archives here or on the google.

    I have come across many answers that essentially point to the Huygens Fresnel Principle

    such as this.
    https://www.physicsforums.com/showthread.php?t=680129&highlight=reflection
     
    Last edited: Mar 29, 2013
  2. jcsd
  3. Conservation of energy - remember that the energy of a photon is inversely proportional to its wavelength. The incoming photon transfers its energy to the electron and the energy (photon) that is released out must be the same, therefore the wavelength is the same. This may be oversimplified, but I think it answers at least part of your question.
     
  4. Simon Bridge

    Simon Bridge 15,041
    Science Advisor
    Homework Helper
    Gold Member

    A metal, like brass, shares some electrons across the whole material ... so the "surface" is covered by a cloud of electrons. Reflection of visible light happens at different depths in this cloud.
    It isn't always - but it is mostly.
    It's due to quantum mechanics - what you get is an electron absorbs a photon, which gives it more energy for a bit. After that bit, it gets rid of the energy. The quickest way is to just dump one photon.
    You'll notice that brass has a yellow tint to it's reflections? Not all wavelengths are reflected with equal probability. There are quite a lot of things that can happen besides reflection.
    In a metal the energy levels of the conduction electrons are very close together.

    I've been a more detailed discussion of reflection:
    https://www.physicsforums.com/showthread.php?t=570003
    For refractive substances, reflections are more easily understood in terms of the refractive index.
    http://scienceline.ucsb.edu/getkey.php?key=2816

    Richard Feynman has a good set of descriptions of reflection on the quantum scale.
    Fundamentally, reflection is an emergent behavior - something that happens on average.
     
    Last edited: Mar 30, 2013
  5.  
  6. Thanks for the answers and the links guys!

    This clears things up a lot.
     
  7. Claude Bile

    Claude Bile 1,479
    Science Advisor

    The key here is to understand that photons do not undergo absorption and re-emission upon reflection. Atoms at a surface can be thought of as driven oscillators - the incoming wave(function) polarizes the atoms, which produce a response field - the reflected wave(function). The response field must be the same frequency to preserve continuity of the electric field. This applies whether the atoms are part of a conductor or dielectric.

    In fact, there is not real need to invoke QM, as reflection is readily explained via classical optics.

    If you insist on a QM description, think of the atoms at a surface as being in a virtual energy state during photon reflection. This is why there is no quantization; the states are virtual. This is quite different to absorption and re-emission from a real energy state, where the emitted photon is random in time, phase, frequency and direction.

    Claude.
     
    Last edited: Apr 12, 2013
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?

0
Draft saved Draft deleted