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Optics(Converging lens and Diverging lens)

  1. Dec 20, 2011 #1
    1. The problem statement, all variables and given/known data
    http://www.imageurlhost.com/images/r0its2lyef7v3s0ppms2.png


    2. Relevant equations
    Mirror Equation
    (1/q) + (1/p) = (1/f) = (2/r)
    Magnification
    M = (q/p)

    for two lenses
    M1 & M2
    Total magnification = M1*M2


    3. The attempt at a solution
    we have p1 = 40cm and f1 = 30cm
    then the image distance of lens 1 will be q1 = 120

    then we have diverging lens 2
    with focal length f2 = 20cm

    then im stuck i dont know what to do with the object distance of the diverging lens
    i also dont know what to do with the fact that the object is 2cm high
     
    Last edited: Dec 20, 2011
  2. jcsd
  3. Dec 20, 2011 #2
    The image of the converging lens acts as a VIRTUAL object for the diverging lens.
     
  4. Dec 20, 2011 #3
    is it between the diverging and converging lens?

    this is how i do to find the image distance of the divergng lens
    between C lens and D lens (110cm)
    but there is an image distance q1 from the C lens (120cm)
    the object distance from the D lens p2 might be (110cm - 120cm)


    (1/q2) + (1/(110-120)) = (1/20)
    q2 = 6.67
    then i can find the magnification of the Divirging lens
    M2 = q2/p2
    total
    M = M1M2

    whats the use of that 2cm thing above the object?
     
  5. Dec 20, 2011 #4
    If the formula (1/u) + (1/v) = 1/f is to be used then the focal length of the diverging lens is taken as -20.
     
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