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Optics, reflexion over water: why?

  1. Mar 11, 2010 #1

    fluidistic

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    Starting from Fermat's principle, one can reach Snell's law. In the derivation, one find that there can be a critical angle at which light doesn't pass anymore from a material to another. That is, if you consider 2 materials and the light starts in the material of greater refractive index.
    For example if we have air and water. Water has a greater refractive index than air, mathematically we have [tex]n_{\text{water}}>n_{\text{air}}[/tex].
    However there is no critical angle if we are in the air and throw a light beam to the water. It means that any light beams heading into the water will cross the water surface and eventually be under water' surface. Then why can we see a reflexion over water surface? It may be our own face, Sun or clouds, but there is a reflexion of light over water. How is that possible?
     
  2. jcsd
  3. Mar 11, 2010 #2
    If you have light going from air to water it´s true that any incidence angle is O.K. but it doesn´t mean the total amount of energy goes into the water. At the air-boundary interface a set of boundary relationships of the electric and magnetic fields must be satisfied. The resulting set of equations (Fresnel equations) say that some energy gets into the water and some energy is reflected. That´s why you can see a reflected image.
     
  4. Mar 11, 2010 #3

    fluidistic

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    Ok thanks for the reply.
    Does this imply that Fermat's principle is not a fundamental law of the Universe?
     
  5. Mar 13, 2010 #4
    Snell's law only tells you about refraction, it says nothing about reflection. You can also derive the law of reflection from Fermat's principle. In that sense, Fermat's principle remains a "fundamental law."
     
  6. Mar 13, 2010 #5

    fluidistic

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    In fact I've derived Snell's law and the law of reflexion via Fermat's principle. But I've understood that if light passes from a medium with a refractive index into another medium with a greater refractive index, then there is no critical angle. It means that the totality of a light beam should get refracted and no partly reflected and partly reflexion as I can see with the case of air-water.

    By fundamental law, I meant a law that always holds, no matter the situation.
     
  7. Mar 13, 2010 #6

    Born2bwire

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    Unfortunately, that is a misunderstanding on your part then. Snell's law and the law of specular reflection make no assertions on the amount of energy transmitted and reflected. They are merely a means of deriving the path of the light. The critical angle marks the point where the path of light cannot proceed through the interface, that is the only reason why we can make any assertions about the fact that no light is transmitted.
     
  8. Mar 13, 2010 #7

    Matterwave

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    Don't confuse total internal reflection with regular reflection. Light reflects off of mirrors, not because they are incident at angles greater than the critical angle.

    You can think of total internal reflection as ANOTHER mechanism by which light rays may reflect instead of passing through the medium.
     
  9. Mar 14, 2010 #8

    fluidistic

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    Ok thanks for the information. I'm new to optics and I think I'm going to learn more about how light behaves.
     
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