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Double Lens System: Converging/Diverging

  1. Mar 28, 2015 #1
    1. The problem statement, all variables and given/known data

    Image: http://i.imgur.com/NMS7BUK.jpg?1

    a) Find the position of the final image of the 1.0-cm-tall object.
    b) Find the size of the final image of the 1.0-cm-tall object.

    h = 1 cm
    f1 = 5 cm
    do = 4 cm
    f2 = -8 cm
    L = 12 cm
    di = ?

    2. Relevant equations

    1/f = 1/do + 1/di


    3. The attempt at a solution

    a) di1 = 1/(1/f1 - 1/do) = 1/(1/5cm - 1/4cm) = -20 cm (virtual image through a converging lens?)
    di2 = 1/(1/f2 - 1/(l - do)) = 1/(1/-8cm - 1/(12-(-2))) = -6.4 cm

    This answer is not right. I would really appreciate any help with double lens systems. We're covering them now in class, and I'm very confused by optics!
     
    Last edited: Mar 28, 2015
  2. jcsd
  3. Mar 28, 2015 #2

    TSny

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    Check the above calculation.

    [EDIT: I think this might just be a typo. I also get -6.4 cm for the image distance of the second lens. Did the question state how the answer should be given? That is, are you supposed to give the answer relative to the second lens or relative to the first lens?]
     
    Last edited: Mar 28, 2015
  4. Mar 28, 2015 #3
    Oh! It's the distance from the object! So, I added 4 cm + (12 cm - 6.4 cm) which gave me the correct answer of 9.6 cm. (Also: Yep, that was a typo!)

    Thank you! :D
     
  5. Apr 3, 2015 #4
    I have this same problem, and I also get di1 as -20cm, what am I missing? I'm having trouble determining what "do2" I should use in the second calculation? How do I get there?
     
  6. Apr 3, 2015 #5

    TSny

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    -20 cm is correct for di1. The image of the first lens is treated as the object of the second lens.
     
  7. Apr 3, 2015 #6
    I get that far but when I do the calculation, the answer is wrong. 1/f - 1/di1 = 1/di2 --> 1/-8 - 1/-20 = 1/di2 --> -5/40 - (-2/40) = 1/di2 --> 40/-3 = di2 =.13.3333cm , which is incorrect according to previous replies in this forum.
     
  8. Apr 3, 2015 #7

    TSny

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    When you treat the image of the first lens as the object of the second lens, it doesn't mean that the image distance of the first lens equals the object distance for the second lens.

    Draw a diagram. Locate the image of the first lens in your diagram. If that image is treated as the object of the second lens, then what should you use for the object distance for the second lens?
     
  9. Apr 3, 2015 #8
    Again, this is the piece I do not understand.
     
  10. Apr 3, 2015 #9

    TSny

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    Can you describe in words what it means to say that the image distance of the first lens is -20 cm?
     
  11. Apr 3, 2015 #10
    If the lens equation gives a negative image distance, then the image is virtual on the same side of the lens as the object?
     
  12. Apr 3, 2015 #11

    TSny

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    Yes. So, where did you draw the image of the first lens?
     
  13. Apr 3, 2015 #12
    Would it be to the left of the first lens?
     
  14. Apr 3, 2015 #13

    TSny

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    Yes. How far to the left of the first lens?
     
  15. Apr 3, 2015 #14
    16cm left?
     
  16. Apr 3, 2015 #15

    TSny

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    No. How did you get that?
     
  17. Apr 3, 2015 #16
    20 - 4 since do was already 4cm away from the first lens? Clearly I do not understand, any real explanation you could give would help..
     
  18. Apr 3, 2015 #17

    TSny

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    You need to review the basic definition of "image distance of a lens (di)". It's defined to be the distance from the lens to the image (with images on the left defined as having negative image distances).

    So, when you find that di1 = -20 cm, where is the image of the first lens located in relation to the first lens?
     
  19. Apr 3, 2015 #18
    20cm left of the first lens
     
  20. Apr 3, 2015 #19

    TSny

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    Yes. Good. So, next you treat this image of the first lens as the object of the second lens. What should you use for do2?
     
  21. Apr 3, 2015 #20
    -20cm - 12cm(distance b/w lenses), -32cm?
     
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