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Location final image diveraging and converging lens

  1. Aug 10, 2008 #1
    1. The problem statement, all variables and given/known data
    A 1.5 cm tall object is placed 0.50 m to the left of a diverging lens with a focal length of 0.20 m. A converging lens with a focal length of 0.17 m is located 0.08 m to the right of the diverging lens.
    a)What is the location of the final image with respect to the object? answer = final image is lcoated 1.3 m to the right of the object
    b)What is the height and orientation with respect to the orginal object of the final image?
    answer = 1.4 cm, inverted

    2. Relevant equations
    1/q+1/p =1/f
    M=hi/h =-q/p

    3. The attempt at a solution
    Ok, well i need to figure out part a before I can do part b...so
    I did 1/p1+1/q2=1/f
    1/0.5+1/q=-1/0.2 (negative focal length for diverging)
    solve for q = -0.14285
    I'm a little confused, when I draw my picture to help me visualize where the image is...i dont know whether or not to add the 0.08 or subtract it ...
    I thought I would add it (negative q means the image is in front of the lens (virtual))
    so i did...
    1/0.22285 +1/q2=1/0.17
    1/q2 = 1.39 but if i inverse it then it become 0.71 and that is not the answer....hm

    I can't seem to figure this out, please help!!! thanks in advance!
  2. jcsd
  3. Aug 10, 2008 #2


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    Homework Helper

    Hi crazyog,

    I think everything you've done so far looks fine, but you still have one more step to go. You have found the image distance for the second lens, which tells you where the final image is relative to the second lens. But the question asks you to find out how far the final image is from the object.
  4. Aug 10, 2008 #3
    Oh ok I think I get it, so since it is the distance from the object to the image I want to add 0.71683 + 0.08+0.5 = 1.3 and the image is to the right

    ok now for part b...
    I know M= -q/p so would I do -0.7168/.22285 = -3.216
    and we know the orginal height M=hi/h
    but the answer is not 1.4...what am i missing?
  5. Aug 10, 2008 #4
    Oh I just realized when I find the magnification due to the first lens, and then the second lens, the product of the two gave me the final magnification and I used this with the orginal height to find it to be 1.4 ....
    why does the product give me the final magnification? and inverted?
  6. Aug 10, 2008 #5
    Ok I just realized again haha...my first M is postive and 2nd negative

    hi=-1.4 so inverted
    ok I get the answer!! but I dont understand why it is the product why not add together?
  7. Aug 10, 2008 #6


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    The product gives the overall magnification, because of the way the system of lens works. For example, just to use easy numbers, let's say your first lens had a magnification of -2, and the second lens had a magnification of -3. So the image of the first lens is twice as large as its object and inverted, and the image of the second lens is three times larger than its object and inverted. But the image of the first lens is the object for the second lens, so overall the final image is 6 times larger than the (original) object; also, since its inverted twice, the final image is upright.

    or in equation form:

    m_{\rm total}&=m_1 m_2\nonumber\\
    m_{\rm total} &=(-2) (-3) = +6\nonumber
  8. Aug 10, 2008 #7
    Oh ok that makes a lot more sense! Thank you so much! You have been a great help :) !!
  9. Aug 10, 2008 #8


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    Homework Helper

    Sure, glad to help!
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