1. Given Data (from Ex 9.10)

• Focal length of Convex Lens (f₁): +30 cm
• Focal length of Concave Lens (f₂): -20 cm
• Separation (d): 8.0 cm
• Object Size (h₁): 1.5 cm
• Object Distance (u₁): -40 cm (from convex lens)

2. Part (a): Effective Focal Length

For a system of two lenses separated by a distance d, the effective focal length (F) is given by:

1/F = 1/f₁ + 1/f₂ - d/(f₁f₂)

Substituting the values:

• 1/F = 1/30 + 1/(-20) - 8 / (30 × -20)
• 1/F = 1/30 - 1/20 - 8 / (-600)
• 1/F = 1/30 - 1/20 + 8/600

Using a common denominator of 600:

• 1/F = (20 - 30 + 8) / 600
• 1/F = -2 / 600
• 1/F = -1 / 300
• F = -300 cm

Dependence on Direction: The value of the effective focal length (-300 cm) remains the same regardless of which side the light enters from. However, the position of the focal point relative to the lenses will change.

Usefulness: Since the effective focal length (-300 cm) is very large compared to the separation (8 cm), the system behaves somewhat like a single diverging lens with a long focal length. However, because the system is "thick," treating it as a single thin lens at a fixed point is not entirely accurate for precise optical design.

3. Part (b): Image Formation

Step A: Refraction at First Lens (Convex)

Using the lens formula: 1/v₁ - 1/u₁ = 1/f₁

• 1/v₁ = 1/30 + 1/(-40)
• 1/v₁ = (4 - 3) / 120 = 1/120
• v₁ = 120 cm

This image acts as a Virtual Object for the second lens.

Step B: Refraction at Second Lens (Concave)

The distance of this virtual object from the second lens is:

u₂ = v₁ - d = 120 - 8 = +112 cm

Using the lens formula: 1/v₂ - 1/u₂ = 1/f₂

• 1/v₂ = 1/(-20) + 1/112
• LCM of 20 and 112 is 560.
• 1/v₂ = (-28 + 5) / 560
• 1/v₂ = -23 / 560
• v₂ ≈ -24.35 cm

Step C: Total Magnification

Magnification by lens 1 (m₁) = v₁/u₁ = 120 / -40 = -3

Magnification by lens 2 (m₂) = v₂/u₂ = (-560/23) / 112 = -5/23

Total Magnification (M) = m₁ × m₂

M = (-3) × (-5/23) = 15/23 ≈ 0.652

Step D: Image Size

Size (h₂) = M × h₁

h₂ = 0.652 × 1.5

h₂ ≈ 0.98 cm