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Lens Help Needed

  1. Feb 22, 2009 #1
    Lens Help Needed!!

    A converging lens with a focal length of 50cm and a diverging lens with a focal length of -52 cm are 227cm apart. A 3.9-cm-tall object is 70cm in front of the converging lens.


    Calculate the distance between image and diverging lens.
    Express your answer using two significant figures.
    d = 26cm
    correct



    Part B
    Calculate the image height.
    Express your answer using two significant figures.
    h = 1.3 wrong
    one attempt remaing.


    Work:
    for final image distance:

    1/f1 = 1/p1 + 1/q1
    1/50 - 1/70 = 1/q1
    q1 = 175 cm

    227-175=52cm

    1/f2 = 1/p2 + 1/q2

    -1/52 - 1/52 = 1/q2

    q2 = -26 -----> 26 cm final image distance


    for image height:


    M1 = -q1/p1 = 2.5

    l-2.5l = h1(image)/h1(ogject)

    9.75cm = h1(image)

    M2 = -q2/p2 = .5

    .5 = h2(image)/h1(image)

    4.9 = h2(image)


    is this right? I have only one attempt left

    Please help....
     
  2. jcsd
  3. Feb 22, 2009 #2
    Re: Lens Help Needed!!

    any help??? please
     
  4. Feb 23, 2009 #3

    Redbelly98

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    Re: Lens Help Needed!!

    4.9 cm is the correct magnitude.

    However, note that M1 should be negative (the positive lens forms an inverted image). I don't know if you are expected you to bother with +/- signs to indicate upright/inverted images.
     
  5. Feb 23, 2009 #4
    Re: Lens Help Needed!!

    is it suppose to be positve b/c M =absolute[h(image)/h(object)]

    or is it neg.?
     
  6. Feb 24, 2009 #5

    Redbelly98

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    Re: Lens Help Needed!!

    M = -q/p

    If q and p are both positive or both negative, M is negative.

    If q and p are of opposite sign, M is positive.

    In this case, I get that the final image height is negative.
     
  7. Feb 25, 2009 #6
    Re: Lens Help Needed!!

    the answer was positive

    Thanks for all Help guys!!!
     
  8. Feb 25, 2009 #7

    Redbelly98

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    Re: Lens Help Needed!!

    You're welcome!

    Sounds like they were interested in the absolute value.
     
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