1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Optics - magnification of a converging lens
Loading...