Question

At what distance should an object be placed in front of a convex lens of focal length $0.4m$ so that the image is thrice the size ofthe object?

Answer 1

Given $f=0.4m,h_i=3h_0,u=?$
We know that, $m=\frac{h_i}{h_o}=\frac{-v}{u}$
$\frac{3h_o}{h_o}=\frac{-v}{u}$ or $v=-3u$
Also from lens formula, we have
$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$
$\frac{1}{f}=\frac{1}{u}+\frac{1}{-3u}=\frac{1}{u}-\frac{1}{3u}=\frac{3-1}{3u}=\frac{2}{3u}$
$\frac{1}{f}=\frac{2}{3u}$ or $3u=2f$
or $u=\frac{2\times f}{3}=\frac{2\times 0.4}{3}$
i.e., object distance, $u=0.267m.$
The object must be placed at $0.267m$ from the lens.