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Lens magnification problem

  1. Oct 25, 2006 #1
    Here is the problem I am working on:

    Two lenses that are L = 38 cm apart are used to form an image, as shown in Figure 26-52. Lens 1 is diverging and has a focal length f1 = -7 cm; lens 2 is converging and has a focal length f2 = 16 cm.
    [​IMG]

    [a] Determine the distance from lens 1 to the final image. (Include the sign of each answer.)
    What is the magnification of this image?

    I got [a] correct with an answer of 62.3003 cm, using di1 = (1/f1 - 1/do1)^-1. For part I did the following:

    m = (m1)(m2) - (-do/do1)(-di2/do2) = (di1)(di2)/(do1)(0.38m-di1) = -0.188, which is wrong. Anyone able to help me out? Thanks.
     
  2. jcsd
  3. Oct 25, 2006 #2

    Doc Al

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    Not clear what you did wrong. What did you get for di1, do2, di2? And thus m1 and m2?
     
  4. Oct 25, 2006 #3
    di1 = -8.8421 cm
    do2 = (38cm - di1)
    di2 = 24.3003 cm
    m1 = -do/do1
    m2 = -di2/do2

    I used my values for di1 and di2 for part [a] and got the answer right, so I'm sure at least those 2 are correct. I am pretty sure the rest are as well, but let me know if it is an error in any of these values that is giving me a wrong answer.
     
  5. Oct 26, 2006 #4

    Doc Al

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    I get a different value for di1 (and thus di2 and part [a]), so one of us is making an error.
     
  6. Oct 26, 2006 #5

    OlderDan

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    I'm curious about what happened to this thread a couple of days ago. I was in the middle of responding and found that my preview was showing up in the middle of some totally unrelated thread. I restarted my browser to see if that would clear the problem and the whole thing was gone. I notified the webmaster that something had gone wrong, but I'd like to know if you deleted your first post of this problem, or if the system just lost it somehow. If you deleted it, I want to pass that infomation along to the webmaster.

    And of course Doc Al is right. Your di1 is not correct; that error propegates through your calculation. Also, although your final expression for the magnification is correct there are some (probably just typographical) errors in the intermediate steps. As a check on your yourself it would be good to keep in mind that the virtual image of a diverging lens is between the lens and the focal point. Your di1 is not.
     
  7. Oct 26, 2006 #6

    Doc Al

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    I was wondering the same thing myself. My guess is that when we moved to the new server, some recent data was vaporized. Several threads that I responded to just disappeared; in some other threads, my posts were gone but the thread remained. I hope it's stable now.
     
  8. Oct 26, 2006 #7
    Yeah, I think when the servers were moving, the thread was deleted, so I made a new one, because I couldn't find it anymore.

    If my di1 is not correct, how did I get a correct answer for part [a]? What did you get for di1 and di2? And are my other values that I am plugging in correct?
     
  9. Oct 26, 2006 #8

    Doc Al

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    Don't ask me how you got a correct answer for part [a]--I don't think you did! :smile: (It's not off by much.)

    Rather than debate it, just recalculate di1. It'll take you less than a minute.
     
  10. Oct 26, 2006 #9
    Using di1 = (1/f1 - 1/do1)^-1 = (1/-14cm - 1/24cm)^-1, I still get the same answer of -8.8421 cm.
     
  11. Oct 26, 2006 #10

    Doc Al

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    Where did that 14 cm come from?
     
  12. Oct 26, 2006 #11
    Double the 7cm, still keeping the negative sign. Is that incorrect? Am I supposed to just use -7cm instead of -14cm?
     
  13. Oct 26, 2006 #12

    Doc Al

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    Why in the world would you double it? The focal length is given as 7 cm, so use -7 cm (negative since it's a diverging lens).
     
  14. Oct 26, 2006 #13

    OlderDan

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    Stick with Doc Al on this, but FYI I don't think your first answer was correct. It was only off by about a centimeter, but it was off.
     
  15. Oct 26, 2006 #14
    Oh, my mistake lol. So the real answer for [a] is 63.3365 cm?, and di1 = -8.4194 cm and di2 = 25.3365 cm?

    So my answer for should be -0.2028?
     
  16. Oct 26, 2006 #15

    Doc Al

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    Please correct that value of di1, once and for all. (Yes, part [a] looks OK.)
     
  17. Oct 26, 2006 #16
    LOL okay. And is my part correct as well? Whoops, I meant, 0.1244 as my answer for .
     
    Last edited: Oct 26, 2006
  18. Oct 26, 2006 #17

    Doc Al

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    How did you arrive at that answer?


    Or that one?
     
  19. Oct 26, 2006 #18
    -0.1244, final answer?
     
  20. Oct 26, 2006 #19

    Doc Al

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    How did you get that answer? List all your values, and your final calculation.
     
  21. Oct 26, 2006 #20
    m = (m1)(m2) - (-do/do1)(-di2/do2) = (di1)(di2)/(do1)(0.38m-di1) = -0.1312 (sorry, this is my final answer lol)

    di1 = -0.054 m
    di2 = 0.253 m
    do1 = 0.24
    do2 = 0.38 m + 0.054 m
     
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