• Tags: telescope, astronomical, resolving-power
• Difficulty: Moderate

An astronomical telescope uses a large-aperture objective (large f_o) and a small eyepiece (small f_e).

In normal adjustment (image at infinity): M = -f_o/f_e. Tube length = f_o + f_e. The negative sign indicates an inverted image.

When final image is at near point: M = -(f_o/f_e)(1 + f_e/D).

Resolving power = D/(1.22λ), where D is the aperture of the objective. Larger aperture gives better resolving power (ability to distinguish close objects).

Light-gathering power is proportional to $D^{2}$ (area of objective lens).