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What should be a very simple optics question is giving me problems (two lenses)

  1. Sep 29, 2011 #1
    I'm studying for the GREs, and I ran across this problem:

    Object located at x=0
    Lens 1 located at x=40cm
    Lens 2 located at x=70cm (30cm from lens 1)

    f1 = 20cm
    f2 = 10cm

    Both lenses are converging, thin lenses. So I have to find the final image location.

    The understanding I have of multiple lens problems is that you find the image from the first lens and treat that as the object for the second lens. I'm guessing this approach is wrong because it's leading to nonsensical answers:

    [itex]\frac{1}{f1}[/itex] = [itex]\frac{1}{o}[/itex] + [itex]\frac{1}{i}[/itex]

    [itex]\frac{1}{20}[/itex] = [itex]\frac{1}{40}[/itex] + [itex]\frac{1}{i}[/itex]

    [itex]\frac{2}{40}[/itex] - [itex]\frac{1}{40}[/itex] = [itex]\frac{1}{i}[/itex]

    40 = i = o' (this is 10 cm to the right of the second lens; on the focal point which means that I'm going to get the second image at x=-infinity)

    [itex]\frac{1}{f2}[/itex] = [itex]\frac{1}{o'}[/itex] + [itex]\frac{1}{i'}[/itex]

    [itex]\frac{1}{10}[/itex] = [itex]\frac{1}{40-30}[/itex] + [itex]\frac{1}{i'}[/itex]

    [itex]\frac{1}{10}[/itex] - [itex]\frac{1}{10}[/itex] = [itex]\frac{1}{i'}[/itex]

    i' = infinity

    The answer is actually i' = 5 cm to the right of the second lens. I have no idea how to get to this though. Where'd I go amiss?
    Last edited: Sep 29, 2011
  2. jcsd
  3. Sep 29, 2011 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Since the image is 10 cm to the right of the second lens, it must be treated as a virtual object: Use an object distance = -10 cm (not + 10 cm, which would represent a real object 10 cm to the left of the lens).
  4. Sep 29, 2011 #3
    That fixes the side (thank you), but won't that still result in +infinity as the location of the final image?

    Edit: Nevermind, I actually redid it and realized that I jumped the gun. That worked, thanks!
    Last edited: Sep 29, 2011
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