Archived Optics - magnification of a converging lens

[matematikuvol]

Optics -- magnification of a converging lens

Homework Statement

Dioptry of converging lens is $D=3$. What is magnification $u$?

Homework Equations

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?

 

[Ibix]

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.