(a)

Hint: Use the lens formula.

Step 1: Find the position of the image such that magnification is maximum.
For the maximum possible magnification, the image should be formed at the near point.

Step 2: Find the object's distance.
So, image distance, v = −d = −25 cm
The focal length, f = 10 cm
Let the object distance = u
According to the lens formula,

$$$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}$
$\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{f}}$
$=\frac{1}{-25}-\frac{1}{10}=\frac{-2-5}{50}=-\frac{7}{50}$
$\therefore \mathrm{u}=-\frac{50}{7}=-7.14 \mathrm{cm}$

Hence, the lens should be kept 7.14 cm away in order to view the squares distinctly.

(b)
Hint: \(m=\frac{v}{u}\)

Step: Find the magnification.
Magnification = $\left|\frac{\mathrm{v}}{\mathrm{u}}\right|=\frac{25}{{\displaystyle \frac{50}{7}}}=3.5$

(c)
Hint: Magnifying power = \(\frac{d}{u}\)

Step: Find the magnifying power.
Magnifying power=$\frac{\mathrm{d}}{\mathrm{u}}=\frac{25}{{\displaystyle \frac{50}{7}}}=3.5$

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