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Homework Help: Critical Angle Geometric Optics

  1. Oct 20, 2008 #1
    There is a still-water lake and air interface. Light travels from the water to the air so that the incident angle is also a critical angle, making it so that the light runs along the surface of the water. Considering that this ray is reversible (air to water), a fish looking up at the surface at a critical angle would technically be able to see the shoreline. What kind of confused me is that this would mean that the light would travel along the surface of the water until it got to a certain point (the point the fish is looking at) and refract towards the fish...considering that the light is traveling along the surface, why wouldn't the light refract at any other point?
  2. jcsd
  3. Oct 20, 2008 #2


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    Hi Gear300! :smile:

    In practice, there would be waves, so light at a 0ยบ angle wouldn't go very far.

    If the water was perfectly "flat", then it would be the surface of a 4,000 mile radius sphere, so it would gradually get free, and in reverse the light would enter the sphere where it was tangent to it.

    If the water was perfectly "flat" and non-spherical, then (I think :rolleyes:) the light would stay "in" the surface as a surface wave, and in reverse the problem would be how to inject the light "into" the surface layer.

    (btw, there's an old Scientific American article on rainbows which says that rigorous quantum theory calculations of the angular distance of the arc show that the light is not internally reflected in raindrops as commonly believed, but becomes a wave going round the surface layer of the raindrop :wink:)
  4. Oct 20, 2008 #3
    Good information this is...Thanks
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