(a)

Hint: \(m=\frac{v}{u}\)
Step 1: Find the magnification of the card.
Given that:
Object distance, u = −9 cm
Focal length, f = 10 cm
Using the lens formula:

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}$
$\frac{1}{10}=\frac{1}{\mathrm{v}}+\frac{1}{9}$
$\frac{1}{\mathrm{v}}=-\frac{1}{90}$
$\therefore \mathrm{v}=- 90 \mathrm{cm}$
$\mathrm{Magnification}, \mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}$
$=\frac{-90}{-9}=10$

Step 2: Find the area of each square in the virtual image.
Area of each square, A = 1 mm²
Area of each square in the virtual image = (10)²A = 10² x 1 = 100 mm² = 1 cm²

(b)
Hint: \(\alpha=\frac{d}{u}\)

Step: Find the magnifying power of the lens.
The magnifying power of the lens =$\frac{\mathrm{d}}{\left|\mathrm{u}\right|}=\frac{25}{9}=2.8$

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

Step: Compare the magnification and magnifying power.
The magnification in (a) is not the same as the magnifying power in (b).

The two quantities will be equal when the image is formed at the near point (25 cm).