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Refraction problem, involving mirrors

  1. Sep 21, 2007 #1
    1. The problem statement, all variables and given/known data

    A tank whose bottom is a mirror is filled with water to a depth of 19.3 cm. A small fish floats motionless 7.50 cm under the surface of the water.

    What is the apparent depth of the reflection of the fish in the bottom of the tank when viewed at normal incidence?

    2. Relevant equations

    s = (n1/n2) * s'
    therefore
    s' = s/(n1/n2)

    3. The attempt at a solution

    I found the apparent depth of the fish to be 5.64 (viewed from the normal). Then i found the aparent depth of the mirror to be 14.5, and then used the property of mirrors (virtual image distance = real distance, s=s')

    (14.5-5.64)*2=17.72, incorrect
     
  2. jcsd
  3. Sep 21, 2007 #2
    nevermind, i got it

    i did [[(19.3*2) -7.5] /1.33]
     
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