Critical angle and area

  1. Aug 24, 2011 #1
    1. The problem statement, all variables and given/known data
    calculate the critical angle for light at a water (n=1.33) and air (n=1) interface.
    if a fish is 2m below the surface of the water , calculate the area at the surface through which the fish can viw the world above the water.

    2. Relevant equations

    none given

    3. The attempt at a solution

    critical angle = sin theta c= n2/n1 = 48.75 deg

    area of surface=

    using basic trig find out all the sides and angles of triangle ..
    then area of tribgle is .5*2.28*2 = 2.28 m^2

    am i right in all this??
  2. jcsd
  3. Aug 24, 2011 #2


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    the Area on the SURFACE that the fish sees thru ...
    it looks like a circular disk of bright blue sky directly above the fish ... out to 48deg.
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