An object 50 cm tall is placed on the principal axis of a convex lens. Its 20 cm tall image is formed on the screen placed at a distance of 10 cm from the lens. Calculate the focal length of the lens.

Given:

Height of object, $h$ = 50 cm

Height of image, $h'$ = $-$20 cm                      (real and inverted)

Distance of image from the lens, $v$ = 10 cm

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

Solution:

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

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

Substituting the given values, we get-

$\frac {10}{u}=\frac {-20}{50}$

$-20\times {u}=50\times {10}$                         (by cross multiplication)

$u=-\frac {500}{20}$

$u=-25cm$

Thus, the object distance $u$ is 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-

$\frac {1}{10}-\frac{1}{(-25)}=\frac{1}{f}$

$\frac {1}{10}+\frac{1}{25}=\frac{1}{f}$

$\frac{1}{f}=\frac {5+2}{50}$

$\frac{1}{f}=\frac {7}{50}$

$f=\frac {50}{7}$

$f=+7.14cm$

Thus, the focal length $f$ of the lens is 7.14 cm.