$A$ card sheet divided into squares each of size $1 \; mm^{2}$ is being viewed through a magnifying glass (a converging lens of focal length $9 \; cm$) held close to the eye. What should be the distance between the object and the magnifying glass if the virtual image of each square in the figure is to have an area of $6.25 \; mm^{2}$? Would you be able to see the squares distinctly with your eyes very close to the magnifier?

(A) Area of the virtual image of each square,$A = 6.25 \; mm^{2}$.
Area of each square,$A_{0} = 1 \; mm^{2}$.
Linear magnification $m$ is given by $m = \sqrt{A/A_{0}} = \sqrt{6.25/1} = 2.5$.
Since $m = v/u$,we have $v = 2.5u$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ with $f = 9 \; cm$:
$\frac{1}{9} = \frac{1}{2.5u} - \frac{1}{u} = \frac{1}{u} (0.4 - 1) = \frac{-0.6}{u}$.
$u = -9 \times 0.6 = -5.4 \; cm$.
The object should be placed at a distance of $5.4 \; cm$ from the magnifying glass.
The image distance is $v = 2.5 \times (-5.4) = -13.5 \; cm$.
The virtual image is formed at $13.5 \; cm$ from the lens. Since this distance is less than the near point of a normal eye $(25 \; cm)$,the squares cannot be seen distinctly.