An erect image 2.0 cm high is formed 12 cm from a lens, the object being 0.5 cm high. Find the focal length of the lens.

Image distance, $v$ = $-$12 cm          (Image is erect)

Height of the image, $h'$ = 2 cm         (Image is erect)

Height of the object, $h$ = 0.5 cm

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

Solution:

From the magnification formula, we know that-

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

Substituting the given values in the formula we get-

$\frac {-12}{u}=\frac {2}{0.5}$

$2u=0.5\times {(-12)}$               (by cross multiplication)

$u=\frac {-6}{2}$

$u=-3cm$

Thus, the object $u$ is placed at a distance of 3 cm from the convex lens.

Now,

From the lens formula, we know that-

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

Substituting the value of $f$ and $v$ in the formula we get-

$\frac {1}{(-12)}-\frac {1}{(-3)}=\frac {1}{f}$

$-\frac {1}{12}+\frac {1}{3}=\frac {1}{f}$

$\frac {1}{f}=\frac {1}{3}-\frac {1}{12}$

$\frac {1}{f}=\frac {4-1}{12}$

$\frac {1}{f}=\frac {3}{12}$

$\frac {1}{f}=\frac {1}{4}$

$f=4cm$

Thus, the focal length $f$ of the convex lens is 4 cm.