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Thin Lens Enlargement

  1. Nov 13, 2008 #1
    If you want to produce an image of a bulb that is enlarged by a factor of 2, how far from the wall should the lens be placed. Focal length is 40cm.

    I know you use 1/d_o + 1/d_i =1/f and m= -d_i/d_o but how do you find d_o? -2/d_o +1/d_o =1/f ? but it comes out to 40cm and not 60cm for d_o. If someone could help me with that bit I can solve it from there. Thanks!!!
     
  2. jcsd
  3. Nov 13, 2008 #2

    Redbelly98

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    Actually it comes out to -40 cm.

    Here is a hint: what should m be, if the image is inverted and enlarged by a factor of 2?
     
  4. Nov 13, 2008 #3
    it should be -2?
     
  5. Nov 13, 2008 #4

    Redbelly98

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    Yes. Try that out.
     
  6. Nov 13, 2008 #5
    So m d_o = -d_i (-2)(-40)= -d_i so then it comes out to be -80cm but in the back of the book it says it's 1.2m.
     
  7. Nov 13, 2008 #6
    Why (-2)(-40)? do is not 40. The focal distance is 40.
    Use both equations and put m=-2 as Redbelly98 told you.
     
  8. Nov 13, 2008 #7

    Redbelly98

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    No. d_0 is not -40 cm. f is +40 cm.

    -2 d_o = -d_i

    and as you know, from the thin lens equation:

    1/d_o + 1/d_i = 1/(40cm)

    Take it from there.
     
  9. Nov 13, 2008 #8
    I understand that you use both equations, but the part I'm getting hung up on is how you solve for d_o with 1/do + 1/di = 1/F from my understanding it goes to -2/do + 1/do = 1/40 and the answer from that comes out to be -40 =do I then put that into the mdo=di equation and get 80. I'm assuming I'm solving for d_o wrong?
     
  10. Nov 13, 2008 #9

    Redbelly98

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    No. Use d_i = 2 d_o, as I said in post #7.
     
  11. Nov 13, 2008 #10
    so then it would be 1/do + 1/2do = 1/40?
     
  12. Nov 14, 2008 #11

    Redbelly98

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