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Combinination of two lenses, converging and diverging

  1. Mar 17, 2007 #1
    1. The problem statement, all variables and given/known data
    A converging lens and a diverging lens, separated by a distance of 30 cm, are used in combination. The converging lens has a focal length of 15 cm. The diverging lens is of unknown focal length. An object is placed 19 cm in front of the converging lens; the final image is virtual and is formed 12 cm before the diverging lens. What is the focal length of the diverging lens?


    2. Relevant equations
    I know for sure I'm suppose to use the equations:
    1/f1 = 1/p1 + 1/q1
    1/f2 = 1/p2 + 1/q2
    p2 = s - q1

    3. The attempt at a solution
    Here's what I tried to do. I understand that I'm suppose to do one lens at a time:
    1/q1 = 1/f1 - 1/p1 = 1/15cm - 1/19cm = 71.25cm
    p2 = s - q1 = 30cm - 71.25cm = -41.25cm
    f2 = 1/q2 + 1/p2 = 16.923cm

    I tried changing signs of each number part, but I'm still not getting the right answer.:frown: I tried reading class notes and textbook and I still can't figure out what I did wrong, or if there's something I'm missing

    There's another similar problem to this I'm having difficulty with (which adds height), but I think I'll understand that too once I get this.

    Any help would be greatly appreciated. :)
     
  2. jcsd
  3. Mar 17, 2007 #2

    hage567

    User Avatar
    Homework Helper

    "p2 = s - q1 = 30cm - 71.25cm = -41.25cm"

    I'm not sure I understand what you are trying to do here.

    I think you've found q1 properly. I would now use the location of the image from the first lens, find out where that is in relation to the second lens. This will give you some information that you can put into your first equation again and this time solve for f2. Try that approach. Be careful with your signs.
     
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