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Optics - Fish in a bowl

  1. Sep 14, 2011 #1
    1. The problem statement, all variables and given/known data
    I asked for help on the concept of this problem in another forum section on this site. I went through the problem and I am getting an answer that KIND OF makes sense, but not really..

    The problem says..
    A small goldfish is viewed through a spherical glass fish-bowl 30 cm in diameter. Determine the apparent position and magnification of the fish’s eye when its actual position is a) at the center of the bowl and b) nearer to the observer, halfway from center to glass, along the line of sight. Assume that the glass is thin enough so that its effect on the refraction may be neglected.


    2. Relevant equations
    The Mirror Equation
    [itex]\frac{1}{s}+\frac{1}{s'}=\frac{-2}{R}[/itex]


    3. The attempt at a solution

    http://img217.imageshack.us/img217/2995/imgcqu.jpg [Broken]

    Yes I know its REALLY sloppy! I am getting an image that is closer to the observer, which makes sense, yet I am getting a magnification of a smaller fish which doesn't make sense. I followed the sign convention in my book. "s" is negative because "O" is to the right of "V". (So says my book.) Can any one help?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 15, 2011 #2

    PeterO

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    Homework Helper

    Case 1.
    If the fish's eye is at the centre of the bowl, all light leaving the eye will be travelling along a radius of the spherical bowl of water, and thus hit the surface at right angles and pass straight through will it not??
     
    Last edited by a moderator: May 5, 2017
  4. Sep 15, 2011 #3
    Hmm.. that seems to make sense. So is my answer of 15cm correct? Why is the magnification smaller though?
     
  5. Sep 16, 2011 #4
    Can anyone help me with this? What is incorrect about my logic?
     
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