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Lens Problem, Help from Anyone! Thank you, I appreciate it very much!

  1. Jan 16, 2008 #1
    [SOLVED] Lens Problem, Help from Anyone! Thank you, I appreciate it very much!

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

    An object is placed 30 mm in front of a lens. An image of the object is located 90 mm behind the lens.
    a) What is the focal lenght of the lens?
    b) On an axis (the x-axis) draw the lens at position x=0. draw at least two rays and locate the image to show the situation described above.

    2. Relevant equations

    1/f = 1/Do + 1/Di

    3. The attempt at a solution

    if i use -90mm then f= -45.
     
    Last edited: Jan 16, 2008
  2. jcsd
  3. Jan 16, 2008 #2

    dlgoff

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    "... i am not sure of is whether or not i should say -90mm because it says behind the lens."
    Distances in the gaussian form to the thin lens equation you gave are real positive values.

    "...but since i got 22.5 as the focal, my drawing isint coming out right."
    Remember, the focal length is the distance in front or behind where rays from a point source will converge. In your problem, they are wanting you to show how the arrow will look at 90mm, if I understand correctly.
     
  4. Jan 16, 2008 #3
    but don't you draw the rays thorugh the focal point?
     
  5. Jan 16, 2008 #4

    dlgoff

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  6. Jan 16, 2008 #5
    actually i think that picture just confused me more. but thanks for tyring to help.
     
  7. Jan 16, 2008 #6
    wait so would the resulting image be real or virtual?
     
  8. Jan 16, 2008 #7

    dlgoff

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    Sorry the link caused confusion but the pic is exactly what you have here.

    Edit: removed quote about virtual images.
     
    Last edited: Jan 16, 2008
  9. Jan 16, 2008 #8
    what do u mean by "Edit: removed quote about virtual images."
     
  10. Jan 16, 2008 #9

    dlgoff

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    This is what I was wanting to quote in the previous post:

    "...if an object is placed at a distance S1 along the axis in front of a positive lens of focal length f, a screen placed at a distance S2 behind the lens will have an image of the object projected onto it, as long as S1 > f. This is the principle behind photography. The image in this case is known as a real image."
     
  11. Jan 16, 2008 #10
    oooooooooooo okay i understand you now, thank you,
    but is my answer to part a) right?
     
  12. Jan 16, 2008 #11

    dlgoff

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    Yep. Glad I could help.
     
  13. Jan 16, 2008 #12
    so 22.5 is right?
    and thanks
     
  14. Jan 16, 2008 #13

    dlgoff

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    Yes. 22.5mm. And since this is image distance is grater than this focal length, the image is real.
     
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