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Thin Lens Equation

  1. Dec 1, 2006 #1
    1. The problem statement, all variables and given/known data

    3. A converging lens has a focal length of 10 cm. A screen is placed 30 cm from an object. Where should the lens be placed, in relation to the object, to produce a focused image?

    2. Relevant equations

    1/do + 1/di = 1f

    3. The attempt at a solution

    Do can not be greater than 2F, because then the image appears between F and 2F on the other side of the lens. From 2F to 2F is 40cm, the distance we want is 30cm.

    Do cannot be equal to 2F because then the image appears at 2F on the other side. 2F to 2F in this case is 40cm. So its not possible.

    Do cannot be between f (10cm) and 2f or the image will appear beyond 2f.
    F to 2F is 30cm, so the object cant be beyond F and the image cant be beyond 2F when there’s only 30cm between them.

    Do cannot be F because no image is formed when do is at F.

    Therefore, do must be less than F. However, this is not possible because the lens must be between the screen and the object, as specified by the question, and do being less than f results in the image appearing on the same side of the lens as the object.

    If I actually attempt it mathematically it turns into a sadistic guessing game.

    di + do = 30
    1/do + 1/di = 1/10

    How many combinations of 2 numbers sum 30? Am I expected to draw a ray diagram for all of them until I find a combo that works, or is there a formulaic way of approaching this that's going over my head?

    The problem being... do and di are defined in relation to the position of the lens. In this case, the position of the lens is the variable.
    Last edited: Dec 1, 2006
  2. jcsd
  3. Dec 1, 2006 #2
    You've got the right equations use the first one to eliminate di or do from the second one. Then work out di or do, and hence find the othervalue.
  4. Dec 1, 2006 #3
    Yes.. I guess the real problem is I've been trying to do that for 4 hours with no success which is why I posted here.
  5. Dec 1, 2006 #4
    Of course you could just think how far from a lens a focused image is :).
  6. Dec 1, 2006 #5
    Can you tell me how to do it mathematically? I need to know how to solve this type of problem.
  7. Dec 1, 2006 #6
    Please explain how to do this?
  8. Dec 1, 2006 #7
    Doesnt it depend on where the object is?
  9. Dec 1, 2006 #8
    I suspect that your getting confused with signs. You do know how to cross multiiply right?

    [tex] \frac{1}{d_i} - \frac{1}{30 - d_i} = \frac{1}{f} [/tex]
  10. Dec 1, 2006 #9
    I dunno.. That latex graphic thing didnt seem to work.

    There was a problem with the question though. Its been changed to di+do=60cm

    So could you teach me how to evaluate this?

    1/di + 1/(60-di) = 1/10
  11. Dec 1, 2006 #10


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