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Magnification due to a Diverging Lens

  1. Apr 1, 2010 #1
    1. The problem statement, all variables and given/known data

    The focal length of the diverging lens is 74 cm. If an object with height h = 0.6 cm is located at d = 37 cm, what is the height of the image?

    2. Relevant equations

    Thin-Lens Equation:

    (1/do)+(1/di)=1/f

    Magnification Equation:

    m=-di/do

    3. The attempt at a solution

    I know that first I need to find the distance of the image, di. To do this I took: 1/[(1/74cm)-(1/37cm)]. From this I got -74cm.

    Next, I plugged this value into the Magnification equation: -(-74cm)/(37cm). From this I recieved a value of 2. Therefore the image should be 1.2 cm tall. When I entered this answer it was incorrect. I also know this is incorrect because images from a convex lens must be upright, reduced in size, and virtual.

    Can anyone help me out on where I am going wrong here?
     
  2. jcsd
  3. Apr 1, 2010 #2

    Doc Al

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    Staff: Mentor

    This isn't right. What's the sign of f for a diverging lens?
     
  4. Apr 1, 2010 #3
    Is it negative since it is to the left of the lens?
     
  5. Apr 1, 2010 #4

    Doc Al

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    Staff: Mentor

    Yes.
     
  6. Apr 1, 2010 #5
    Alright, so I redid the problem with a negative focal length and got a height of .4cm.
     
  7. Apr 1, 2010 #6

    Doc Al

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    Staff: Mentor

    What did you get for do? (Show how you got it.)
     
  8. Apr 1, 2010 #7
    So, I redid my equation: 1/[(1/-74cm)-(1/37cm)]. This gave me: -24.667. Then I redid my magnification equation: -(-24.667)/(37). This gave me .667 for my magnification. Then when multiplied by the height i got .4 cm.
     
  9. Apr 1, 2010 #8

    Doc Al

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    Staff: Mentor

    Actually, that's correct. (For some reason, I thought you meant it was wrong.)
     
  10. Apr 1, 2010 #9
    yay! thank you very much!
     
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