# Archived Optics - magnification of a converging lens

#### matematikuvol

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.
$$\frac{1}{f}=\frac{1}{p}+\frac{1}{l}$$
$$u=\frac{l}{p}$$
$$l=up$$
$$D=\frac{l+p}{lp}=\frac{p(u+1)}{up^2}$$
Putting $d=0.25m$, I get
$$0.25\cdot 3\cdot u=u+1$$
and $u=-4$. Where I make a mistake?

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#### Ibix

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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