Physics › Astrophysics › Telescopes and image formation
Telescopes and image formation
A refracting telescope is two converging lenses sharing a focal point, and the ratio of their focal lengths gives the angular magnification. Reflecting telescopes do the same job with a mirror, which avoids chromatic aberration and folds a long focal length into a short tube.
Pick your board and the few notes written for the other boards quietly fold away, here and in the practice players. Nothing is deleted: every folded piece reopens on a tap.
Builds on Refraction and total internal reflection.
IN THIS TOPIC
- Draw the ray diagram for a refractor in normal adjustment, and say what normal adjustment means.
- Use angular magnification both as a ratio of angles and as one focal length over the other.
- Draw the Cassegrain arrangement and explain what each mirror contributes.
- Weigh reflectors against refractors, including chromatic and spherical aberration.
COMMON MISCONCEPTION
A good telescope is one with the biggest magnification.
The astronomical refractor
An astronomical refractor is two converging lenses on a shared axis. The objective, the large lens at the front, catches light from a distant object. That light arrives as near-parallel rays, so the objective brings it to a real image in its focal plane, one focal length behind the glass. The second, smaller lens, the eye lens, is then used as a magnifying glass to inspect that little image.
In normal adjustment the eye lens is positioned so that its focal plane coincides with the objective's. The real image then sits exactly one focal length in front of the eye lens, so the light leaves it parallel again and the final image forms at infinity. Your eye stays relaxed, focused on the far distance, for hours at the eyepiece. Add the two focal lengths and you have the tube length, .
Because the object and the final image both live at infinity, the telescope cannot make anything bigger in the ordinary sense. What it magnifies is angle. The angular magnification is defined as the angle the final image subtends at your eye, divided by the angle the object subtends at the unaided eye. In normal adjustment that ratio collapses to the focal lengths:
so a long-focus objective and a short-focus eye lens give the biggest angular gain. Look at the geometry in the figure and you can see where that comes from. Both angles share the same image height h in the focal plane. The small angle coming in is roughly , the steeper angle going out is , and dividing one by the other leaves .
WORKED EXAMPLE
What 48 times actually means
A refractor has an objective of focal length 1.2 m and an eye lens of focal length 25 mm. The Moon subtends about 9.0 × 10−3 rad to the naked eye. Find the magnification and the angle the Moon's image subtends through the telescope.
M = = 1.2 / 0.025 = 48.
Through the eyepiece the Moon subtends 48 × 9.0 × 10−3 = 0.43 rad. That is roughly 25°, so a coin at arm's length becomes a dinner plate filling much of your view.
Notice both focal lengths went in as metres. The ratio has no units, because it is one angle divided by another.
The Cassegrain reflector
Serious telescopes use mirrors, and the classic layout is the Cassegrain. A large concave primary mirror collects the light. Its surface is ground to a parabola, because a parabolic mirror brings every ray parallel to its axis to one sharp focus, however far from the axis the ray strikes. Before the light reaches that focus, a small convex secondary mirror intercepts it and reflects it back down the tube, through a central hole in the primary, to a focus just behind the main mirror where the eyepiece or camera sits.
The fold is the point. Reflecting the beam back on itself packs a long effective focal length into a short, stiff, steerable tube, and the eyepiece ends up in the most convenient place possible, behind the telescope where an observer or an instrument can sit. The angular magnification works exactly as before, with the effective focal length of the mirror pair playing the part of .
GUIDED PRACTICE
The folded focal length
A Cassegrain has an effective focal length of 2.0 m folded into a tube about half a metre long, and takes a 20 mm eyepiece. Find the magnification, and state the advantage the fold has bought.
Show the working
M = = 2.0 / 0.020 = 100.
A refractor with the same magnification and the same eyepiece would need a tube over two metres long. The fold delivers that focal length in a quarter of the tube, so the mount can be smaller and stiffer.
Reflector or refractor
Lenses carry two built-in flaws, and both are examinable by name. Chromatic aberration comes first. Glass refracts blue light more strongly than red, so a single lens gives each colour its own focal length and no one sharp focus exists. Every bright star wears a faint coloured fringe.
Then spherical aberration. A lens or mirror ground to a spherical surface focuses rays through its edge slightly short of rays through its centre, smearing the focus along the axis. Mirrors escape both problems more cheaply than lenses do. Reflection is independent of wavelength, so a mirror has no chromatic aberration at all, and grinding the primary to a parabola removes the spherical error for light arriving parallel to the axis.
| Refractor | Cassegrain reflector | |
|---|---|---|
| chromatic aberration | present, so each colour has its own focus | absent, because mirrors treat all colours alike |
| spherical aberration | present unless expensively corrected | removed by the parabolic primary |
| size limit | a lens can only be held by its rim, and a large one sags under its own weight | a mirror is supported across its whole back, so it can be built enormous |
| light lost | some absorbed crossing the glass | little lost at a coated surface |
| upkeep | sealed tube, little maintenance | mirror coatings need occasional renewal |
This is where the big-magnification boast collapses. Magnification only stretches the light you have already caught, and stretching a dim, blurred image gives a bigger dim, blurred image. What you actually pay for is how much light the telescope gathers and how fine the detail it can resolve. Both belong to the objective's diameter, the subject of the next lesson.
INDEPENDENT PRACTICE
Designing backwards
A refractor in normal adjustment magnifies 50 times and its tube is 1.02 m long. Find both focal lengths.
Show the working
Two facts, two unknowns. = 1.02 and = 50.
Substituting gives 51 = 1.02, so = 0.020 m and = 1.0 m.
The sanity check is built in. Your two answers must add back to the tube length, and 1.00 + 0.02 does.
ASSESSMENT FOCUS
- Three features carry the ray diagram marks. Rays arrive parallel, they cross at a real image on the shared focal plane, and they leave the eye lens parallel at a steeper angle. Label both focal lengths while you are there.
- Define normal adjustment in one sentence. The focal planes coincide, so the final image is at infinity and a relaxed eye views it. A second mark often hides in the tube length .
- M is a ratio of angles, never of sizes, and it equals only in normal adjustment. Say so before you use it.
- Aberration answers want the mechanism named, then the cure. Chromatic, blue refracted more, so the colours focus apart, cured by using a mirror. Spherical, edge rays focus short, cured by a parabolic surface. A merits question is a comparison, so answer in pairs, refractor then reflector. One-sided answers score half.
CHECK YOURSELF
A telescope has an objective of focal length 900 mm and an eye lens of focal length 15 mm. Jupiter subtends 2.4 × 10−4 rad to the naked eye. Find the magnification and the angle Jupiter subtends through the telescope, and state what normal adjustment means.
Show a hint
The ratio of focal lengths first, then remember what a telescope multiplies.
Show the answer
M = = 900 / 15 = 60. Millimetres are fine as long as both lengths use them.
Through the eyepiece Jupiter subtends 60 × 2.4 × 10−4 = 1.4 × 10−2 rad, about 1.6 times the angle the Moon covers unaided.
In normal adjustment the two focal planes coincide, so the final image sits at infinity and a relaxed eye views it.
The objective makes a real image; the eye lens turns it into steeper parallel light.
In normal adjustment M is the objective's focal length over the eye lens's, and the tube is the two added.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their mark schemes on the telescopes and image formation questions page.
CHECK YOUR PROGRESS
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- Draw the ray diagram for a refractor in normal adjustment, and say what normal adjustment means.
- Use angular magnification both as a ratio of angles and as one focal length over the other.
- Draw the Cassegrain arrangement and explain what each mirror contributes.
- Weigh reflectors against refractors, including chromatic and spherical aberration.
Open the full revision checklist to track your progress across the whole unit.