Physics › Astrophysics › The astronomical refracting telescope

The astronomical refracting telescope

An astronomical refracting telescope is two converging lenses sharing a focal plane, and in normal adjustment the final image lies at infinity. Angular magnification is a ratio of angles, the image's to the object's, and in that adjustment equals the objective's focal length divided by the eye lens's. The two focal lengths added give the tube length.

Year 13AQA 3.9.1.1

Telescopes and image formation, part 1 of 2. Part 2 is Reflecting telescopes.

Builds on Refraction and Snell's law.

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.

COMMON MISCONCEPTION

A telescope produces an enlarged image of a distant object, the way a magnifying glass enlarges print.

For an object at infinity the image in normal adjustment is also at infinity: nothing is enlarged. The telescope wins by angle, making the image subtend M = f_o/f_e times the angle the object does.

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, $f_{o} + f_{e}$ .

$A refracting telescope in normal adjustment: the objective and the eye lens share a focal plane, so parallel light comes in and steeper parallel light goes out$

Describe the drawing

Three parallel rays arrive at a large objective lens, cross at a real image point on the shared focal plane, then leave a smaller eye lens as a parallel beam tilted several times more steeply: the ray diagram for a refractor in normal adjustment, with the chief ray through the lens centre drawn in amber.

FIG. 1Normal adjustment: parallel rays from a distant point converge to a real image on the shared focal plane, then leave the eye lens as a parallel beam tilted more steeply. The amber ray, through the centre of the objective, passes straight through undeviated.

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:

M = \frac{f_{o}}{f_{e}}

IN THE AQA DATA BOOKLET

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 $h/f_{o}$ , the steeper angle going out is $h/f_{e}$ , and dividing one by the other leaves $f_{o}/f_{e}$ .

Angular magnification as geometry: the image height h subtends a small angle at the objective and a four times steeper one at the eye lens, so M equals fo over fe — FIG. 2The geometry of angular magnification. One image height h serves both lenses: it subtends the small angle α at the objective, one long focal length away, and the steeper angle β at the eye lens, one short focal length away. The drawn focal lengths sit in the ratio four to one, so β is four times α, which is M = $f_{o}/f_{e}$ read straight off the drawing.

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 = $f_{o}/f_{e}$ = 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 Moon the size of a pea 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 has moved to the next part of this lesson.

Reflector or refractor has moved to the next part of this lesson.

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 $f_{o} + f_{e}$ .
M is a ratio of angles, never of sizes, and it equals $f_{o}/f_{e}$ only in normal adjustment. Say so before you use it.

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 = $f_{o}/f_{e}$ = 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.

15 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the astronomical refracting telescope questions page.

5 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

CHECK YOUR PROGRESS

Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.

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.

Open the full revision checklist to track your progress across the whole unit.

The astronomical refracting telescope

The astronomical refractor

Question workbook

Question workbook

Answer book