Problem: 1
An object is placed at a distance of 10 cm from a convex lens of focal length 12 cm .Find the position and nature of the image.
Solution:
Given, u$=$-10cm
f$=$12cm
v$=$?
we know
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\frac{1}{12}=\frac{1}{v}-\frac{(-1)}{10}$
$\frac{1}{v}=\frac{5-6}{60}$
$v=-60cm$
Hence, image formed behind the mirror at a distance 60 cm and the image is virtual and magnified beyond 2F.

Problem:2
A 4 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 24 cm. The distance of the object from the lens is 16 cm. Find the position, size and nature of the image formed, using the lens formula.
Solution:
Given  -    $f=+24cm,u=-16cm$ ,As $|u|(16cm)<|f|(24cm)$ ,it means object lies between $F$ and $C$ , in this position of object the rays from object cannot meet at any point on the other side of lens ,which is also clear from the position of image found below ,
From lens equation  $v=\frac{uf}{u+f}=\frac{-16\times 24}{-16+24}=-48cm$ ,-ive sign shows that image will be formed on the same side of lens , where the object is placed .
Now linear magnification     $m=I/O=v/u=-48/-16=+3$ ,or                                        $I=3\times 4=12cm$       (given  $O=4cm$) ,since $m=+ive$ and $m>1$ ,therefore image will be virtual ,erect and magnified .