• 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 = -$\frac{f_o}{f_e}$(1 + $f_e$/D).

Resolving power = $\frac{D}{1.22λ}$, 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).