A lens produces an erect image, of size 12 mm, when an object of size 6 mm is placed 12 cm from its optical centre. Calculate the focal length of the lens.

We know that, 
$\mathrm{m}=\frac{{\mathrm{h}}_{\mathrm{i}}}{{\mathrm{h}}_{\mathrm{o}}}\mathrm{ }=\frac{\mathrm{v}}{\mathrm{u}},\mathrm{ }$ 

Where, 

• h_i = height of the image (12 mm), 
• h_o = height of the object (6 mm),
•  v = image distance,
•  u = object distance (-12 cm = -120 mm).

Substituting values in the above formula, we get;
$\mathrm{m}=\frac{12}{6}\mathrm{ }=\frac{\mathrm{v}}{-120 \mathrm{mm}}$   

(According to the Cartesian sign convention the object distance u is the left-hand side of the lens that is to the negative X axis. Therefore u  is negative here.)
 $2\mathrm{ }=\frac{\mathrm{v}}{-120}$

=> v = - 240 mm
Using the lens formula, we can find out the focal length of the lens,
$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}$

$\frac{1}{\mathrm{f}}=-\frac{1}{240}+\frac{1}{120}$

$\frac{1}{\mathrm{f}}=-\frac{1}{240}+\frac{2}{240}$

$\frac{1}{\mathrm{f}}=\frac{-1+2}{240}$

$\mathrm{f} = 240 \mathrm{mm}$
Thus, f = 24 cm.