Question

Where should an object be placed from a converging lens of focal length $15cm$, so as to obtain, a real image of same size as that of the object?

Answer 1

For a converging lens
$f=15cm$

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$m=\frac{v}{u}=-1$ (real image)

$v=-u$

$\frac{1}{f}=\frac{1}{-u}-\frac{1}{u}=-\frac{2}{u}$

$\frac{1}{15}=-\frac{2}{u}$

$u=\frac{-15}{2}cm$
Object should be $7.5cm$ in front of the lens.