Optical Instruments

Lessons

Senior 4 · Physics

Optical Instruments

4 lessons

Lesson 3 of 4

The Astronomical Telescope

25 min

Bringing the Moon closer

On a clear night far from city lights, the Moon covers only about half a degree of the sky, and Jupiter looks like a bright star. Point even a small telescope at them and the Moon's craters and Jupiter's four largest moons appear. A telescope cannot make the Moon itself bigger, and it does not bring it nearer; what it does is make the Moon subtend a larger angle at your eye, and collect far more light than your pupil can. This lesson explains how two lenses do both.

How the astronomical telescope works

An astronomical (refracting) telescope has two converging lenses (Figure 3.1). The objective has a long focal length $f_{o}$ and a large diameter; the eyepiece has a short focal length $f_{e}$ .

Light from a distant object arrives as parallel rays. The objective forms a real, inverted, diminished image $I_{1}$ in its focal plane.
The eyepiece is used as a magnifying glass to view $I_{1}$ .
In normal adjustment, $I_{1}$ lies at the focal point of the eyepiece as well, so the rays leave the eyepiece parallel and the final image is at infinity: the eye views it relaxed. The lenses are then $f_{o} + f_{e}$ apart.

Figure 3.1: The astronomical telescope in normal adjustment. The final image is at infinity and inverted.

Magnifying power in normal adjustment. The object (a star, the Moon) is so far away that the unaided eye sees it at the same angle $\alpha$ as the objective does. The ray through the centre of the objective is undeviated, so the image $I_{1}$ , of height $h$ , satisfies $\tan\alpha = \frac{h}{f_o}$ . The ray from the top of $I_{1}$ through the centre of the eyepiece leaves at the angle $\beta$ of the final image, with $\tan\beta = \frac{h}{f_e}$ . For small angles $\tan\theta \approx \theta$ , so

M = \frac{\beta}{\alpha} = \frac{h/f_e}{h/f_o} = \frac{f_o}{f_e}, \qquad \text{length of telescope } L = f_o + f_e

A long objective focal length and a short eyepiece give high magnification. Astronomers change the magnification by swapping eyepieces.

Final image at the near point. If the eyepiece is pushed in slightly so that the final image is at $D = 25\,\mathrm{cm}$ , the eyepiece magnifies more (it acts as a magnifier with $1 + \frac{D}{f_e}$ ), and

M = \frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right)

This is a small gain, and the eye must strain to focus at its near point, so normal adjustment is the usual setting.

Definition

A telescope is in normal adjustment when the final image is formed at infinity, so that a relaxed normal eye can view it; the focal points of the objective and eyepiece then coincide. The eye ring (exit pupil) is the image of the objective formed by the eyepiece: the small circle through which all the light leaving the telescope passes, and the best place for the observer's pupil. Example: a telescope with $f_o = 90\,\mathrm{cm}$ and $f_e = 3.0\,\mathrm{cm}$ in normal adjustment is 93 cm long and magnifies 30 times.

Worked example 1: looking at the Moon

A telescope has an objective of focal length 100 cm and an eyepiece of focal length 5.0 cm. The Moon subtends 0.52° at the Earth. Find (a) the magnifying power and length in normal adjustment, (b) the angle the Moon's final image subtends, (c) the diameter of the Moon's image $I_{1}$ in the focal plane of the objective.

\text{(a)}\;\; M = \frac{100}{5.0} = 20, \qquad L = 100 + 5.0 = 105\,\mathrm{cm}

\text{(b)}\;\; \beta = M\alpha = 20 \times 0.52^\circ = 10.4^\circ

\text{(c)}\;\; h = f_o\,\alpha = 100\,\mathrm{cm} \times (0.52 \times \tfrac{\pi}{180}) = 100 \times 0.00908 = 0.91\,\mathrm{cm}

The real image of the whole Moon is less than a centimetre across; the eyepiece then views that small image from close up.

Worked example 2: final image at the near point

The same telescope is adjusted so that the final image is 25 cm from the eye. Find the new magnifying power and the new separation of the lenses.

M = \frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right) = 20 \times \left(1 + \frac{5.0}{25}\right) = 24

For the eyepiece, $v_e = -25\,\mathrm{cm}$ : $\frac{1}{u_e} = \frac{1}{5.0} + \frac{1}{25}$ , so $u_e = 4.17\,\mathrm{cm}$ . The separation is $100 + 4.17 = 104.2\,\mathrm{cm}$ : the eyepiece moves in by 0.8 cm.

The eye ring and light-gathering

All the light entering the objective must leave through the eyepiece. The eyepiece forms a real image of the objective rim, the eye ring, just beyond itself. Treating the objective as an object at distance $u = f_{o} + f_{e}$ from the eyepiece,

\frac{1}{v} = \frac{1}{f_e} - \frac{1}{f_o + f_e} = \frac{f_o}{f_e(f_o + f_e)} \;\Rightarrow\; v = \frac{f_e(f_o + f_e)}{f_o}, \qquad \frac{\text{eye-ring diameter}}{\text{objective diameter}} = \frac{v}{u} = \frac{f_e}{f_o} = \frac{1}{M}

This gives a neat way to measure $M$ : point the telescope at a bright sky, catch the eye ring on a piece of tracing paper behind the eyepiece, and divide the objective's diameter by the eye ring's diameter. If the eye ring is about the size of the pupil, all the collected light enters the eye.

The amount of light collected is proportional to the area of the objective, so a telescope with an objective of diameter $d_{o}$ gathers $\left(\frac{d_o}{d_{\text{pupil}}}\right)^2$ times as much light as the eye. Faint stars and galaxies become visible. A larger objective also gives finer detail (better resolution), because diffraction blurring decreases as the aperture increases. That is why astronomers want large objectives, and why the biggest telescopes use mirrors, which can be made far larger than lenses (Unit 1, lesson 4).

Worked example 3: the eye ring

The objective of the telescope in Worked example 1 is 5.0 cm in diameter. Find the position and diameter of the eye ring, and how many times more light the telescope collects than a dark-adapted pupil 7.0 mm across.

v = \frac{f_e(f_o + f_e)}{f_o} = \frac{5.0 \times 105}{100} = 5.25\,\mathrm{cm}\ \text{beyond the eyepiece}, \qquad d = \frac{5.0\,\mathrm{cm}}{20} = 0.25\,\mathrm{cm} = 2.5\,\mathrm{mm}

\text{light-gathering ratio} = \left(\frac{50\,\mathrm{mm}}{7.0\,\mathrm{mm}}\right)^2 = 51

Other telescopes

Type	Objective	Eyepiece	Final image	Length in normal adjustment
astronomical (Kepler)	converging lens	converging lens	inverted	$f_{o} + f_{e}$
terrestrial	converging lens	converging lens, with an erecting lens between	upright	$f_{o} + f_{e} + 4 f$ (erecting lens of focal length $f$ used at 2F)
Galilean	converging lens	diverging lens	upright	$f_{o} - ∣ f_{e} ∣$ (short)
reflecting (Newtonian)	concave (parabolic) mirror	converging lens, with a small plane mirror to turn the light aside	inverted	about $f_{o}$

An inverted image does not matter when you look at stars, so astronomers use the simplest, shortest design. Binoculars are two terrestrial telescopes in which the erecting lens is replaced by the totally reflecting prisms of Unit 1, lesson 2; they turn the image upright and fold the long light path into a short case.

Percival Lowell at the eyepiece of the 24-inch (61 cm) refracting telescope of the Lowell Observatory, Arizona, observing Venus in daylight in 1914. The objective lens is at the far end of the long tube; the observer sits at the short-focus eyepiece at the lower end. Photo: unknown author, public domain, via Wikimedia Commons.

Warning

Safety. Never point a telescope or binoculars at the Sun or look at the Sun through them. The objective concentrates sunlight and can blind you instantly and permanently. Solar observing needs a certified solar filter over the objective, or projection of the image onto a card, under supervision.

Warning

Common mistakes. (1) Writing $M = \frac{f_e}{f_o}$ : the long objective focal length goes on top. (2) Thinking the telescope makes the object bigger: it increases the angle subtended. (3) Giving the length as $f_{o} - f_{e}$ for an astronomical telescope: that is for the Galilean type. (4) Forgetting that a larger objective improves brightness and resolution, not the magnification.

Tip

Telescope: "objective on top" ( $M = \frac{f_o}{f_e}$ ), and the eye ring is $M$ times smaller than the objective. Microscope: both lenses short, and the magnifications multiply.

The astronomical telescope (QuantumBoffin). Follow the ray diagram in normal adjustment and the derivation of the angular magnification from the two angles at the objective and the eyepiece.

✦

Practice

Draw a ray diagram of an astronomical telescope in normal adjustment and mark the angles $\alpha$ and $\beta$ .
A telescope has $f_o = 120\,\mathrm{cm}$ and $f_e = 4.0\,\mathrm{cm}$ . Find its magnifying power and length in normal adjustment.
A telescope in normal adjustment is 84 cm long and has a magnifying power of 20. Find the focal lengths of its lenses.
The telescope of question 2 has an objective 6.0 cm in diameter. Find the diameter of the eye ring and its distance from the eyepiece.
Jupiter subtends $2.2 \times 10^{-4}$ rad when viewed from the Earth. What angle does it subtend through the telescope of question 2?
A Galilean telescope has an objective of focal length 50 cm and a diverging eyepiece of focal length 5.0 cm. Find its magnifying power and length in normal adjustment.
An amateur astronomer must choose between telescope A (objective 60 mm, $f_o = 900\,\mathrm{mm}$ ) and telescope B (objective 150 mm, $f_o = 750\,\mathrm{mm}$ ) for viewing faint galaxies. Both come with a 10 mm eyepiece. Evaluate which is better, with reasons.

Show answersHide answers

See Figure 3.1: parallel rays at $\alpha$ meet at $I_{1}$ in the common focal plane and leave the eyepiece parallel at $\beta$ .
$M = \frac{120}{4.0} = 30$ ; $L = 124\,\mathrm{cm}$ .
$f_{o} + f_{e} = 84$ and $f_{o} = 20 f_{e}$ , so $21 f_{e} = 84$ : $f_e = 4.0\,\mathrm{cm}$ , $f_o = 80\,\mathrm{cm}$ .
$d = \frac{6.0}{30} = 0.20\,\mathrm{cm} = 2.0\,\mathrm{mm}$ ; $v = \frac{4.0 \times 124}{120} = 4.1\,\mathrm{cm}$ beyond the eyepiece.
$\beta = 30 \times 2.2 \times 10^{-4} = 6.6 \times 10^{-3}\,\mathrm{rad}$ (about 0.38°, nearly the Moon's size to the naked eye).
$M = \frac{50}{5.0} = 10$ ; $L = 50 - 5.0 = 45\,\mathrm{cm}$ ; the image is upright.
A: $M = 90$ ; B: $M = 75$ . For faint galaxies, light-gathering matters more than magnification: B collects $\left(\frac{150}{60}\right)^2 = 6.25$ times as much light and also resolves finer detail. B is the better choice; its slightly lower magnification is an advantage for extended faint objects.

Quick check

Key points

The objective (long $f_{o}$ ) forms a real image in its focal plane; the eyepiece (short $f_{e}$ ) magnifies it.
Normal adjustment: final image at infinity, $M = \frac{f_o}{f_e}$ , length $f_{o} + f_{e}$ ; with the image at the near point, $M = \frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right)$ .
The eye ring is the image of the objective, $M$ times smaller; light-gathering $\propto d_o^2$ .
Terrestrial and Galilean telescopes give upright images; reflecting telescopes use a concave mirror.

Sources and credits

26.5: Telescopes, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Photo: Percival Lowell observing Venus from the Lowell Observatory in 1914.jpg, unknown author, Wikimedia Commons, public domain.
Video: The Astronomical Telescope, QuantumBoffin, embedded from YouTube.

The derivations, the diagram, the table, the worked examples and the practice exercises are written by Ganzaa.

1Bringing the Moon closer

2How the astronomical telescope works

3The eye ring and light-gathering

4Other telescopes

Bringing the Moon closer

How the astronomical telescope works

The eye ring and light-gathering

Other telescopes