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Homework Help: Help with light and lens problem

  1. Feb 5, 2006 #1
    Here is the problem:

    A converging lens with a focal length of 12.0 cm forms a virtual image 8.00 mm tall, 17.0 cm to the right of the lens.

    I am supposed to find the position of the object. I tried using the lens eq. 1/s + 1/s' = 1/f where s is the distance of the object from the lens, s' is the dist of the image from the lens, and f is the focal point. I got s to be 40.82 cm but I'm being told it's wrong. Is there something im missing here? Any tips and points would be great.

    Thanks!!
     
  2. jcsd
  3. Feb 5, 2006 #2

    George Jones

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    Careful with the signs!

    For a converging lens, where must an object be for a virtual image to be produced.

    Regards,
    George
     
  4. Feb 5, 2006 #3
    to the right of the lens correct? does this change the equation in some way? :confused:
     
  5. Feb 5, 2006 #4

    George Jones

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    And where with respect to the focal point?

    George
     
  6. Feb 5, 2006 #5
    Well my obvious guess would just be (s' - f) centimeters to the right of the focal point.. :confused:
     
  7. Feb 5, 2006 #6

    George Jones

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    In order for a converging lens to produce a virtual image, the object must be between the focal point and then lens. I have been trying to get you to see that your answer of 40.8 cm can't be correct. I am reasonably certain that you have made a sign mistake, but I used a different sign convention (and thus different formulae) last semster, and I can't remember, off the top of my head, the other sign convention, so I'm not quite sure where you're going wrong.

    When I use the convention with which I am familiar, I can:

    1) get an answer that is between the focal point and the lens;

    2) get 40.8 cm when I make a sign mistake.

    Regards,
    George
     
  8. Feb 5, 2006 #7

    nrqed

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    What sign did you use for f and for s'? That`s your mistake. f is positive for a converging lens. What is the sign for s' for a *virtual* image?

    Pat
     
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