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Geometric Optics: A Diverging and Converging Lense in Series

  1. Nov 3, 2009 #1
    A romantic, lighted candle is placed 39 cm in front of a diverging lens. The light passes through the diverging lens and on to a converging lens of focal length 10 cm that is 5 cm from the diverging lens. The final image is real, inverted, and 21 cm beyond the converging lens. Find the focal length of the diverging lens.

    -1/S + 1/S' = 1/f
    Here's my approach which led me to the wrong answer:
    I started by just looking a the converging lense. The image's location is 21 to the right of the lense, or at +21cm. The primary focal point of the converging lense should be to the left, or at -10cm. Here we solve for S using the equation I listed about.
    -1/S=1/f - 1/S'
    -1/S= 1/10 - 1/21
    So S= -19.09cm

    Then I take into account that the diverging and converging lense are 5 cm apart. So I am going to subtract 5 from the S solved about.
    So S' for the diverging lense= 14.090cm. Since I have to create a new coordinate system, this image is to the left of the diverging lense, so it is at -14.090cm.
    Then I use -1/S + 1/S' = 1/f again to solve for f.
    so -1/-39cm + 1/-14.090cm= 1/f
    I solved for f, and got f= -22.0598, which makes sense because the image should be closer than the focal point, but it's wrong...

    Any ideas?
    Last edited: Nov 3, 2009
  2. jcsd
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