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Archived Optics - magnification of a converging lens

Optics -- magnification of a converging lens

1. Homework Statement
Dioptry of converging lens is $D=3$. What is magnification ##u##?

2. Homework 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?
 

Ibix

Science Advisor
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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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