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Optics - distance between object and image

  1. Jul 21, 2009 #1
    hello,

    1. The problem statement, all variables and given/known data
    A lens has a focal length of 20 cm and a magnification of 4. How far apart are the object and the image?

    2. Relevant equations
    m=di(image)/do(object)
    1/di + 1/do = 1/f


    3. The attempt at a solution
    Well, I don't really know what to do, I feel like I don't have enough information. Your help will be very appreciated!
     
  2. jcsd
  3. Jul 21, 2009 #2

    Redbelly98

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    Re: Optics-Lens

    Since you have 2 equations in 2 unknowns ... solve 1 equation for 1 of the unknowns, then substitute that into the 2nd equation.
     
  4. Jul 21, 2009 #3
    Re: Optics-Lens

    Yes, I was thinking about that but how do I do it? Seeing as if i did m=di/do 4=di/do --but it leads me to a dead end

    or if i used 1/di+1/do=1/f i can only go as far as 1/di+1/do=1/20 then once again i get to a dead end!
     
  5. Jul 21, 2009 #4

    Redbelly98

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    Re: Optics-Lens

    Okay, so far so good. Now solve that equation for di:

    di = _____ ?​
     
  6. Jul 21, 2009 #5
    Re: Optics-Lens

    umm.. di=4(do) but since we dont have do we can't solve for di
     
  7. Jul 21, 2009 #6

    Redbelly98

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    Re: Optics-Lens

    You can now substitute 4do for di in the other equation ... that eliminates di, leaving do as the only unknown.
     
  8. Jul 21, 2009 #7
    Re: Optics-Lens

    so the equation is : 1/di +1/do =1/f

    1/4do + 1/do = 1/20

    Now I'm guessing we want to make do single... but i have to get rid of the 4 so since i'm dividing....

    1/do^2 = 1/20 - 1/4
     
  9. Jul 21, 2009 #8

    rl.bhat

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    Re: Optics-Lens

    Multiply do to both sides and solve for do.
     
  10. Jul 22, 2009 #9

    Redbelly98

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    Re: Optics-Lens

    Yes, exactly. It's just algebra at this point.
     
  11. Jul 25, 2009 #10
    Re: Optics-Lens

    sorry for the late reply but, thanks a lot for your help, I really appreciated it! :)
     
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