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Homework Help: Archived Optics - magnification of a converging lens

  1. Dec 19, 2012 #1
    Optics -- magnification of a converging lens

    1. The problem statement, all variables and given/known data
    Dioptry of converging lens is $D=3$. What is magnification ##u##?

    2. Relevant equations



    3. The attempt at a solution
    ##\frac{1}{f}=D## - dioptry.
    [tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{l}[/tex]
    [tex]u=\frac{l}{p}[/tex]
    [tex]l=up[/tex]
    [tex]D=\frac{l+p}{lp}=\frac{p(u+1)}{up^2}[/tex]
    Putting ##d=0.25m##, I get
    [tex]0.25\cdot 3\cdot u=u+1[/tex]
    and ##u=-4##. Where I make a mistake?
     
  2. jcsd
  3. Oct 29, 2016 #2

    Ibix

    User Avatar
    Science Advisor

    A real image will be inverted (i.e., will have a negative magnification). So, using the sign convention implied by $$\frac 1f= \frac 1p+\frac 1l $$where both ##l## and ##p## are positive for a real image, the magnification ought to be ##u=-l/p##.

    Carrying that through gives you $$D=\frac { (u-1)}{up} $$and hence u=4 given D=3 and p=0.25, which I presume was the expected answer.
     
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