Question

If the distance of an object from the first focus of an equiconvex lens is x, and the distance of its real image from second focus is 4x. If the focal length of the lens is nx, then find the value of n :

Answer 1

$u=-(x+f)$ and $v=+(4x+f)$
From $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{(4x+f)}+\frac{1}{(x+f)}=\frac{1}{f}$
On solving, $f=2x$