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Magnification of an Image formed by a converging lens

  1. Aug 30, 2012 #1
    1. The problem statement, all variables and given/known data
    A converging lens with a focal length of 14.2 cm is used to inspect a coin. The lens forms a virtual image 30.1 cm away. What is the magnification?


    2. Relevant equations
    (1/Si) + (1/So) = (1/f)
    M = (hi/ho) = (-Si/So)


    3. The attempt at a solution
    1/So = 1/f - 1/Si

    1/(1/So) = So = 26.88cm

    M = (-Si/So) = -30.1/26.88 = 1.12


    I'm pretty sure that I have the solution and answer correct, but I just wanted to double check. This homework is online and the site is telling me that I have the answer wrong and I just wanted to make sure that I wasn't missing anything before I bother my teacher about it. As of right now I think it's just an error in the code of the site, but I'm just double checking. Thanks.
     
  2. jcsd
  3. Aug 30, 2012 #2

    TSny

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    Have you accounted for the fact that it's a virtual image?
     
  4. Aug 30, 2012 #3

    PeterO

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    This object [the coin] is placed inside the focus [giving a virtual image] so some/all of those distances may be negative.
    I am never confident with that (1/Si) + (1/So) = (1/f) formula, so I always use a couple of alternates, and a ray diagram.
     
  5. Aug 30, 2012 #4

    TSny

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    Check your notes or your textbook. There are specific rules for deciding on the signs of the various quantities.
     
  6. Aug 30, 2012 #5
    Ah, right, I forgot about the signs. Looks like the summer off has made me a little rusty. I just accounted for the signs and got the right answer. Thanks for your help.
     
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