(a) At what distance should the lens be held from the figure in

(a) The maximum possible magnification is obtained when the image is formed at the near point (d = 25 cm).

Image distance, v = −d = −25 cm

Focal length, f = 10 cm

Object distance = u

According to the lens formula, we have:

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{u}=\frac{1}{v}-\frac{1}{f}$

$=\frac{1}{-25}-\frac{1}{10}=\frac{-2-5}{50}=-\frac{7}{50}$

$\therefore u=-\frac{50}{7}=-7.14 \mathrm{~cm}$

Hence, to view the squares distinctly, the lens should be kept 7.14 cm away from them.

(b) Magnification $=\left|\frac{v}{u}\right|=\frac{25}{\frac{50}{7}}=3.5$

(c) Magnifying power $=\frac{d}{u}=\frac{25}{\frac{50}{7}}=3.5$

Since the image is formed at the near point (25 cm), the magnifying power is equal to the magnitude of magnification.