$A$ card sheet divided into squares each of size $1 \, mm^{2}$ is being viewed at a distance of $9 \, cm$ through a magnifying glass (a converging lens of focal length $10 \, cm$) held close to the eye.
$(a)$ What is the magnification produced by the lens? How much is the area of each square in the virtual image?
$(b)$ What is the angular magnification (magnifying power) of the lens?
$(c)$ Is the magnification in $(a)$ equal to the magnifying power in $(b)$? Explain.

(A) Area of each square,$A = 1 \, mm^{2}$.
Object distance,$u = -9 \, cm$.
Focal length of a converging lens,$f = 10 \, cm$.
For image distance $v$,the lens formula is $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{10} = \frac{1}{v} - \frac{1}{-9} \implies \frac{1}{v} = \frac{1}{10} - \frac{1}{9} = \frac{9-10}{90} = -\frac{1}{90}$.
Therefore,$v = -90 \, cm$.
Linear magnification,$m = \frac{v}{u} = \frac{-90}{-9} = 10$.
Area of each square in the virtual image $= m^{2} \times A = 10^{2} \times 1 \, mm^{2} = 100 \, mm^{2} = 1 \, cm^{2}$.
$(b)$ Magnifying power of the lens is given by $M = \frac{d}{|u|}$,where $d = 25 \, cm$ (near point).
$M = \frac{25}{9} \approx 2.78$.
$(c)$ No,the magnification in $(a)$ is not equal to the magnifying power in $(b)$.
Linear magnification $m = \frac{v}{u}$ depends on the image distance $v$,while magnifying power $M = \frac{d}{|u|}$ is the ratio of the angle subtended by the image at the near point to the angle subtended by the object at the near point.
They are equal only when the image is formed at the near point $(v = -25 \, cm)$.