A lens produces an erect image, of size$12\mathit{ mm}$, when an object of size $6\mathit{ mm}$ is placed $12\mathit{ cm}$ from its optical centre. Calculate the power of this lens.

A lens produces an erect image, of size$12\mathit{ mm}$, when an object of size $6\mathit{ mm}$ is placed $12\mathit{ cm}$ from its optical centre. 

The power of the lens will be $5\mathit{ D}$.
The power of a lens is the inverse of focal length. 

Let us calculate the focal length of the lens. 
given$\mathit{ u }= 12\mathit{ cm}$
          $v= ?$
$\mathrm{magnification}=\frac{\mathrm{v}}{\mathrm{u}}=\frac{\mathrm{image} \mathrm{size} }{\mathrm{object} \mathrm{size}}$ 

$\Rightarrow \frac{v}{12}=\frac{12}{6}$
$\Rightarrow v= 24\mathit{ cm}$
Using the lens formula, we can find out the focal length of the lens.
      $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ 

(Where f = focal length, v = image distance, u = object distance)
$\Rightarrow \frac{1}{f}=\frac{1}{24}+\frac{1}{12}$

$\therefore \frac{1}{f}=-\frac{1}{24}$ $f=-24\mathit{ cm}$ 

Now, the power of the lens will be, 
$P=\frac{1}{f}$ 

$\Rightarrow P=\frac{1}{f}\left(m\right)$ 

$\Rightarrow P=\frac{1}{0.24}$

 $\mathrm{P}=\mathrm{ }4.16 \mathrm{D}$