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Use the law of reflection and index of refraction to find apparent d

  1. Mar 9, 2013 #1
    1. The problem statement, all variables and given/known data

    See image

    2. Relevant equations

    n1sin(theta1) = n2sin(theta2)
    1/s + 1/s' = 1/f

    3. The attempt at a solution

    My question is for the second part. I think I've figured out why I got it wrong, but I'm not sure.

    So since the object distance (s) should be approximately 21 cm, then you apply the refraction theory to find

    s'=s*n1/n2
    s'=21*(1.33)
    s'=~28 cm

    Is this correct?
     

    Attached Files:

    Last edited: Mar 9, 2013
  2. jcsd
  3. Mar 9, 2013 #2

    Doc Al

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    Staff: Mentor

    No, not correct. How did you do the first part, since it's the same kind of problem? For the second part, where is the image of the fish in the mirror? That's your "object".

    Once you have the correct depth of the "object", do this. Imagine a small cone of light coming from that object extending up towards the surface. If that cone has an angle θ, what will be the angle of the cone after it leaves the water? Use that and a bit of trig to figure out the apparent depth of the object. (Assume the angles involved are small.)
     
  4. Mar 9, 2013 #3
    For the first part, I used

    (Na/Sa) + (Nb/Sb) = (Nb - Na) / R

    1.3/6 + 1/x = nb - na / R

    .217 + 1/x = nb - na / infinity

    1/x = .217

    x = 4.61 cm
     
  5. Mar 9, 2013 #4

    Doc Al

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    Staff: Mentor

    That's an excellent way to solve the problem. But don't round off until the end and you'll get the correct answer.

    The second part can be handled the same way, once you find the correct image location.
     
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