A 2.0 cm tall object is placed 40 cm from a perging lens of focal length 15 cm. Find the position and size of the image.

Focal length, $f$ = $-$15 cm

Object distance from the lens, $u$ = $-$40 cm

Height of the object, $h$ = $+$2.0 cm

To find: Position of the image, $v$, and height of the object, $h'$.

Solution:

From the lens formula, we know that-

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

Substituting the given values, we get-

$\frac {1}{v}-\frac {1}{(-40)}=\frac {1}{(-15)}$

$\frac {1}{v}+\frac {1}{40}=-\frac {1}{15}$

$\frac {1}{v}=-\frac {1}{15}-\frac {1}{40}$

$\frac {1}{v}=\frac {-8-3}{120}$

$\frac {1}{v}=-\frac {11}{120}$

$v=-\frac {120}{11}$

$v=-10.90cm$

Thus, the image distance from the lens is 10.90 cm, and the negative sign $(-)$ implies that it is formed behind the lens (on the left side).

Now, 

From the magnification formula of the lens, we know that-

$m=\frac {v}{u}=\frac {h'}{h}$

Substituting the given values, we get-

$\frac {-10.90}{-40}=\frac {h'}{2}$

$\frac {10.90}{40}=\frac {h'}{2}$