A lens forms a real image 3 cm high of an object 1 cm high. If the separation of object and image is 15 cm, find the focal length of the lens.

Height of object, $h$ = 1 cm

Height of image, $h'$ = $-$3 cm                                          (negative because image is real)

Distance between object and image, $(-u+v)$ = 15 cm

Therefore, Image distance, $v=15+u$

To find: Focal length of the lens $f$.

Solution:

Let the object = $-u$          $(\because the\ distance\ of\ the\ object\ is\ always\ taken\ negatively)$

Focal length = $+f$             $(\because the\ focal\ length\ is\ taken\ positive\ for\ a\ convex\ lens)$

From the magnification $(m)$ of the lens we know that-

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

Substituting the given values, we get-

$\frac {15+u}{u}=\frac {-3}{1}$

$-3\times {u}=15+u$                   (by cross multiplication)

$-3u=15+u$   

$3u+u=-15cm$

$4u=-15cm$

$u=-\frac {15}{4}$

$u=-3.75cm$

Thus, the object distance $u$ is 3.75 cm from the lens.

Putting the value of $u$ in the equation $v=15+u$ we get-

$v=15+(-3.75)$

$v=15-3.75$

$v=11.25cm$

Thus, the image distance $v$ is 11.25 cm from the lens.

Now, 

From the lens formula we know that-

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

Substituting the given values in the formula we get-

$\frac {1}{11.25}-\frac{1}{(-3.75)}=\frac{1}{f}$

$\frac {1}{11.25}+\frac{1}{3.75}=\frac{1}{f}$

$\frac {100}{1125}+\frac{100}{375}=\frac{1}{f}$

$\frac{1}{f}=\frac {100+300}{1125}$

$\frac{1}{f}=\frac {400}{1125}$

$\frac{1}{f}=\frac {400}{1125}$

$f=\frac {1125}{400}$

$f=+2.81cm$

Thus, the focal length $f$ of the lens is 2.81 cm.