$(a)$ $A$ 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. Where should the lens be held in order to view the squares distinctly with the maximum possible magnifying power?
$(b)$ What is the magnification in this case?
$(c)$ Is the magnification equal to the magnifying power in this case? Explain.

(N/A) The maximum possible magnifying power is obtained when the image is formed at the near point $(d = 25 \; cm)$.
Image distance,$v = -25 \; cm$.
Focal length,$f = 9 \; cm$.
Using the lens formula,$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
$\frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{1}{-25} - \frac{1}{9} = \frac{-9 - 25}{225} = -\frac{34}{225}$.
$u = -\frac{225}{34} \approx -6.62 \; cm$.
Hence,the lens should be held $6.62 \; cm$ away from the card sheet.
$(b)$ Magnification $m = \left| \frac{v}{u} \right| = \frac{25}{225/34} = \frac{25 \times 34}{225} = \frac{34}{9} \approx 3.78$.
$(c)$ Magnifying power $M = \frac{d}{u} = \frac{25}{225/34} = 3.78$.
Yes,since the image is formed at the near point,the magnifying power is equal to the magnitude of the linear magnification.