1. Given Data

• Object Distance (u): -9 cm
• Focal Length (f): +10 cm (Assumed standard value)
• Area of each square (A_o): 1 mm²
• Least Distance of Distinct Vision (D): 25 cm

2. Part (a): Linear Magnification & Area

Step A: Find Image Position (v)

Using the Lens Formula (1/v - 1/u = 1/f):

1/v = 1/f + 1/u

1/v = 1/10 + 1/(-9)

1/v = 1/10 - 1/9

1/v = (9 - 10) / 90

1/v = -1 / 90

v = -90 cm (Virtual image formed at 90 cm)

Step B: Linear Magnification (m)

m = v / u

m = -90 / -9

m = 10

Step C: Area of Square in Image

The Area Magnification is the square of the linear magnification (m²).

A_image = m² × A_object

A_image = (10)² × 1 mm²

A_image = 100 × 1

A_image = 100 mm²

3. Part (b): Angular Magnification

The Angular Magnification (Magnifying Power) is the ratio of the angle subtended by the image at the eye to the angle subtended by the object at the near point (D).

M = D / u

Substituting the values:

M = 25 / 9

M ≈ 2.8

4. Part (c): Are they equal?

No, the magnification in (a) is not equal to the magnifying power in (b).

Explanation:

• Linear Magnification (m = 10): This is the ratio of the image size to the object size (v/u). It depends on the image distance v. Since the image is formed far away (90 cm), the linear size is very large.
• Angular Magnification (M = 2.8): This compares how "large" the object looks through the lens versus looking at it with the naked eye at the near point (D/u).

These two quantities are only equal when the image is formed exactly at the Near Point (v = D = 25 cm). Since the image is at 90 cm, the linear magnification is significantly larger.