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Optics - Image formation (lenses)

  1. Jul 4, 2014 #1

    DataGG

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

    A lense with a focal length of ##-48.0cm## forms an image ##17.0cm## to the right of the lense. Where is the object positioned?

    2. Relevant equations

    $$\dfrac{1}{d_0}+\dfrac{1}{d_i}=\dfrac{1}{f}$$

    3. The attempt at a solution

    Well, since the focal length is negative, we know that the lense is divergent (concave). The problem statement says that the image is to the right of the lense, so the ##s_i=17cm (>0)##


    We can compute the position of the object by using the formula above:

    $$\dfrac{1}{d_0}+\dfrac{1}{17}=- \dfrac{1}{48}$$

    And doing the calculations, it comes down to:

    $$s_0=-12.55$$

    However, the solutions say the solution is ##s_0=26.3##. One can get that if we take ##s_i## to be negative, which is wrong, in my opinion. (?)

    So, why am I wrong? Or am I correct?
     
  2. jcsd
  3. Jul 4, 2014 #2

    Doc Al

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    Which way are you assuming the light is traveling?
     
  4. Jul 4, 2014 #3

    DataGG

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    From left to right.
     
  5. Jul 4, 2014 #4

    Doc Al

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    If you assumed right to left, the book answer would make some sense.

    What book are you using?
     
  6. Jul 4, 2014 #5

    DataGG

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    It's not a book answer. It's in a file full of exercises (made by the professor). He also gives the solutions.

    We're using Hecht though.
     
  7. Jul 4, 2014 #6

    Doc Al

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    Ah, he probably just messed up then.
     
  8. Jul 6, 2014 #7
    Since the image is formed to the right of the mirror and the lens in question is diverging ,then the object has to be on the right side of the lens .

    In the Cartesian coordinate sign convention , the direction of incident rays is considered positive ,so right to left is positive .Both focal length and image distance are negative.

    You should get ##s_0=-26.3## ,which means the object is to the right of the lens.
     
  9. Jul 6, 2014 #8

    Doc Al

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    Yes, but that assumes the light is going from right to left, as I had pointed out earlier. (Of course, nothing in the problem statement prohibits that.)
     
  10. Jul 6, 2014 #9
    Is it possible , that with concave lens , image is formed towards right while the object is to the left ?
     
  11. Jul 6, 2014 #10

    Doc Al

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    I don't see how.
     
  12. Jul 6, 2014 #11
    This is why I am saying that light travels from right to left .We are not making any assumptions. I don't know about other sign conventions ,but I think the problem is fine.

    But I think you meant in post#6 that there is something wrong with the problem.

    Sorry if I am not understanding you correctly .
     
  13. Jul 6, 2014 #12

    Ibix

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    For what it's worth, I agree with Doc Al and the OP. The prof must have switched something around when doing the problem.
     
  14. Jul 6, 2014 #13

    Doc Al

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    You are correct. There's nothing wrong with the problem as stated. (But the light obviously does not move from left to right, as is the usual convention in simple lens problems.) I should have been clearer earlier.

    We are in agreement. :smile:
     
  15. Jul 6, 2014 #14

    Doc Al

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    That's what I suspect also. But, as stated, the problem is valid, even if unintentionally. But who knows? Perhaps the prof is more clever than we think!
     
  16. Jul 6, 2014 #15

    Ibix

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    Possibly there's been a sneakiness fail. "To the right of the lens" would usually imply a positive di, which leads to DataGG's answer under the usual orientation conventions for lens diagrams. If the question specified that the image was virtual, however, then di is negative and out drops the book answer.

    Either the question assumes standard conventions and the answer is wrong, or the question is sneaky and incomplete. Either way, one to raise with the instructor.
     
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