A telescope looks at things that are far away and that you cannot bring closer. It has two jobs, and it helps to keep them apart. It gathers light, so faint things become visible, and it resolves detail, so close pairs can be told apart (see the Rayleigh criterion). Magnifying the result, the eyepiece's job, comes last. Try it in the Science Lab's Optics Bench: line up a long-focus lens and a short-focus lens and build your own.
Key idea
A refracting telescope with objective focal length $f_o$ and eyepiece focal length $f_e$, set $f_o + f_e$ apart, has angular magnification $$M = -\frac{f_o}{f_e}.$$ Its light-gathering power grows with the aperture's area, $D^2$, and its finest detail is $\theta_R = 1.22\lambda/D$. A bigger aperture shows fainter and finer things; a stronger eyepiece only makes what is there look bigger.
Why it works
Light from a distant star arrives as parallel rays at a small angle $\theta$ to the axis. The objective brings them to a focus in its focal plane, a height $h = f_o\theta$ off the axis. The eyepiece, placed one focal length beyond that image, sends the rays out parallel again, now at the angle $\theta' = h/f_e = \frac{f_o}{f_e}\theta$. So angles are multiplied by $\frac{f_o}{f_e}$, and the image is inverted, hence the minus sign. The light a telescope collects is all the light falling on its objective: twice the diameter catches four times the light.
Worked examples
Worked example: Magnification and length
A telescope's objective has $f_o = 100$ cm and its eyepiece $f_e = 2.5$ cm. Find the magnification and the length of the tube.
- $M = -\frac{100}{2.5} = -40$.
- Length: $f_o + f_e = 102.5$ cm.
Answer: $M = -40$ (inverted); about 1.0 m long
Worked example: Choosing an eyepiece
With an objective of focal length 1.20 m, which eyepiece focal length gives a magnification of 50?
- $f_e = \frac{f_o}{|M|} = \frac{120\ \text{cm}}{50}$.
Answer: $f_e = 2.4$ cm
Worked example: Light gathering
How much more light does a telescope 20 cm across collect than a dark-adapted eye with a 7.0 mm pupil?
- Collected light goes as area, so as $D^2$.
- $\left(\frac{200\ \text{mm}}{7.0\ \text{mm}}\right)^2 = 28.6^2 = 816$.
Answer: About 820 times as much light
Common mistake
Thinking the magnification is what lets a telescope show faint objects. Magnification spreads the same light over a bigger image, which makes an extended object like a galaxy look dimmer, not brighter. The light comes from the aperture: area $\propto D^2$. That is why astronomers rate telescopes by their diameter, not their magnification. Check: a 10 cm telescope at $50\times$ and at $200\times$ collects exactly the same light.
Common mistake
Writing the magnification as $f_e/f_o$. A long-focus objective and a short-focus eyepiece give a large magnification, so $f_o$ goes on top.