1. Given Data

• Focal length of Objective (f_o): 140 cm = 1.4 m
• Focal length of Eyepiece (f_e): 5.0 cm
• Height of Tower (h_obj): 100 m
• Distance of Tower (u): 3 km = 3000 m

2. Part (a): Separation between Lenses

For a telescope in normal adjustment, the separation (L) is the sum of the focal lengths.

L = f_o + f_e

L = 140 + 5

L = 145 cm

3. Part (b): Height of Image Formed by Objective

The objective lens forms a real, inverted image of the distant tower at its focal plane. We can calculate the height of this image (h₁) using the angle subtended.

Step A: Angle subtended by the tower (α)

α = Height of Tower / Distance

α = 100 / 3000 = 1/30 radians

Step B: Height of Image (h₁)

The angle subtended by the image at the objective is also α.

α = h₁ / f_o

h₁ = α × f_o

h₁ = (1/30) × 140 cm

h₁ ≈ 4.67 cm

4. Part (c): Height of Final Image

The image formed by the objective (h₁) acts as the object for the eyepiece. Since the final image is formed at the near point (25 cm), the eyepiece acts as a simple magnifier.

Step A: Magnification of Eyepiece (m_e)

m_e = 1 + D/f_e

m_e = 1 + 25/5

m_e = 1 + 5 = 6

Step B: Final Image Height (h_final)

h_final = m_e × h₁

h_final = 6 × 4.67

h_final ≈ 28 cm