The focal length of the eye-piece of an astronomical telescope is f1, while the focal length of the object is…

Analyze the question: The eye-piece has focal length f₁ and the objective has focal length f₂. We need the angular magnification of an astronomical telescope in normal adjustment, where the final image is at infinity.

Concept: Angular magnification is

M = (angle subtended at the eye by the final image) / (angle subtended at the eye by the object directly)

In normal adjustment:

• The objective forms a real image of the distant object at its focal point.
• The eye-piece is positioned so that this image lies at the eye-piece's focal point, so the final image is at infinity.
• The distance between the lenses is therefore f_objective + f_eyepiece.

Solution:

Let the object subtend angle α at the objective. The intermediate image has height h, so

α ≈ h / f_objective

The eye-piece views this image from its focal point, so the angle subtended by the final image is

β ≈ h / f_eyepiece

Therefore:

M = β/α = (h / f_eyepiece) ÷ (h / f_objective) = f_objective / f_eyepiece

Using the question's notation, the objective is f₂ and the eye-piece is f₁:

M = f₂ / f₁

This matches option A. The question's wording says "focal length of the object" but means the objective lens.

Options B and C do not match any standard telescope formula. Option D is the inverted ratio, which would be wrong because the objective must have the longer focal length for magnification greater than 1.

Tip: Remember "Objective over Eye-piece": M = f_o / f_e. The objective is a long-focus lens and the eye-piece is a short-focus lens, which makes M greater than 1. The tube length in normal adjustment is f_o + f_e.