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Reflection of light

  1. Jul 5, 2007 #1
    If reflection of light is an interaction with atoms how we get always angle i=angle r?
  2. jcsd
  3. Jul 5, 2007 #2


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    We don't always get that. For example, in a diffraction grating, the angle of incidence is not equal to the angle of reflection.
  4. Jul 5, 2007 #3
    Ok, but suppose the case we always get reflection (plane mirror, polished surface metal ... ! What happens at atomic level when light varies it´s angle i and angle r is automatically adjusted to the same value as angle i?
  5. Jul 5, 2007 #4


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    It's not an interaction with individual atoms. You have to consider the atoms of a solid or liquid as a collective system which interacts with photons differently than simply as the sum of independent interactions with individual atoms.

    To get an idea of what is involved, see post #4 of the Physics Forums FAQ, although it specifically addresses refraction rather than reflection.
  6. Jul 6, 2007 #5
    Think of the photon as a billiard ball hitting a bumper. The electrons in the atoms in the mirror scatter an electron just like the bumper scatters the ball. It doesn't know to go at the right angle, it is merely worked upon by the mirror - which recoils.
  7. Jul 7, 2007 #6
    Feynman's book "QED" decribes this nicely. Photons can scatter in a broad range of angles off every point on the mirror, but the paths add in phase. If you sum up the contributions to the electric field at a detector position, you find that the phases of most paths cancel, and the paths that don't cancel are those very near the point on the mirror where the incident angle equals the reflected angle. But the other parts matter too - as you make the mirror narrower and narrower (or as you look more and more carefully), you will start to see diffraction artifacts in the reflection. Theta(i) = Theta(r) is a ray-based simplification that captures enough detail for most purposes, but a correct treatment is much more complicated.

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