Astrophsics Option A

Option A chapter 1Telescopes

AQA Physics A A2 Chapter map Nelson Thornes Ltd 2009 1

Specification link-up A1.1 Lenses and optical telescopesA1.2 Non-optical telescopes

Chapter teacher guide

Chapter 1 provides opportunities for practical study, individual research and class discussions.

Even if the students have already encountered lenses and the lens formula (Topic 1.1) during their GCSE work, it is worth quickly revisiting the basic ideas here. You can demonstrate the action of converging and diverging lenses by using cylindrical lenses and ray streaks, or by using spherical lenses on an optical bench. Most optical bench experiments are more successful when there is at least a partial blackout. With a convex lens on the optical bench, you can use a small array of differently coloured LEDs (or just the LED display of a digital clock) as the light source. If the object is not symmetrical it is easier to explain the idea of image inversion. Students might need to be reminded how to check the focal length of a convex lens quickly by focusing the view outside a window on the opposite wall. Practice with the lens formula can be related to familiar uses, for instance a camera. Most digital camera lenses are marked with their focal length. Students might gain from a class discussion about the size and type of image produced in a mobile phone camera, where the lens typically has a focal length of about 5 mm.

A good way to introduce the refracting telescope (Topic 1.2) is as a combination of camera and magnifying glass. The objective lens forms a real image, which is then viewed using the eyepiece lens. If students attempt the experiment with tracing paper described in the chapter, it helps to reduce the room lighting if possible. If you do not have suitable demonstration lenses, a refracting telescope can be put together using lenses from an unwanted pair of prismatic binoculars.

The manufacturers of binoculars and telescopes often give useful data such as comparative brightness (twilight factor) and lens diameter for their products on their websites. If the idea is new to them, it may be worth asking the students to spend a few minutes with a calculator or a spreadsheet to investigate the range of small angles for which (radians) = tan is valid.

Students may not have seen real images formed by concave mirrors and a few simple experiments can be instructive here. As with the convex lens, students can roughly check the focal length of a concave mirror by focusing the image of the view outside a window. Much useful information about reflecting telescopes (Topic 1.3), including some excellent animations, is available from the Internet. One way to introduce the idea of spherical aberration is to ask students to study the cusp reflected inside a teacup. The effects of chromatic aberration can be simply demonstrated with a convex lens on the optical bench by alternately focusing the light from red and blue LEDs.

Once students are familiar with the idea of the resolving power of an optical telescope (Topic 1.4), they might usefully discuss the resolving power of a radio telescope (Topic 1.5). Further data about the radio wavelengths and diameters used can be found on the Internet. The idea of combining radio telescopes as an array to produce larger effective diameters, and hence better resolution, makes a good discussion topic. Small air-cooled CCD imagers are available for use by amateur astronomers; the manufacturers websites include useful information.

Common misconceptionsStudents sometimes forget to apply the sign convention when calculating with the lens formula. Errors can usually be avoided by taking extra care when working with virtual images.

Chapter map

Option A chapter 1Telescopes


Chapter map

Specification link-up A1.1: Lenses and optical telescopes What is a converging lens and what is its focal length? How does a converging lens form an image? How can we predict the position and magnification of an image formed by a converging lens?

1.1 Lenses

Specification link-up A1.1: Lenses and optical telescopes What is a refracting telescope?What do we mean by angular magnification?How does the angular magnification depend on the focal lengths of the two lenses?

1.2 The refracting telescope

Specification link-up A1.1: Lenses and optical telescopes What do we mean by angular separation?Why does a wide telescope resolve two stars that cannot be resolved by a narrower telescope? What is the Rayleigh criterion for resolving two point objects?

1.4 Resolving power

Specification link-up A1.1: Lenses and optical telescopes; A1.2: Non-optical telescopesWhat is a charge-coupled device (CCD) and why is it important in astronomy? How does a CCD work?What are non-optical telescopes used for?How do non-optical telescopes compare with each other and with optical telescopes?

1.5 Telescopes and technology

Specification link-up A1.1: Lenses and optical telescopes What is a Cassegrain reflecting telescope?What is meant by spherical aberration and chromatic aberration? What are the relative merits of a reflecting telescope and a refracting telescope?

1.3 Reflecting telescopes

AQA A2 Physics A Nelson Thornes 2009

Astrophysics

Chapter 1 Telescopes 1.1 Lenses Learning objectives: What is a converging lens and what is its focal

length? How does a converging lens form an image? How can we predict the position and

magnification of an image formed by a converging lens?

The converging lens Lenses are used in optical devices such as the camera and the telescope. A lens works by changing the direction of light at each of its two surfaces. Figure 1 shows the effect of a converging lens and of a diverging lens on a beam of parallel light rays.

Figure 1 Focal length A converging lens makes parallel rays converge to a focus. The point where parallel rays are

focused to is called the principal focus or the focal point of the lens. A diverging lens makes parallel rays diverge (i.e. spread out). The point where the rays

appear to come from is the principal focus or focal point of this type of lens.

In both cases, the distance from the lens to the principal focus is the focal length of the lens. In this option, we consider the converging lens only.

Nerojaan Chandran

Nerojaan Chandran

Nerojaan Chandran

Nerojaan Chandran

Nerojaan Chandran


Astrophysics

Note The plane on each side of the lens perpendicular to the principal axis containing the principal focus is called the focal plane. Investigating the converging lens The arrangement in Figure 2 can be used to investigate the image formed by a converging lens. Light rays illuminate the crosswires which form the object. These light rays are refracted by the lens such that the rays form an image of the crosswires.

Figure 2 Investigating images 1 With the object at different distances beyond the principal focus of the lens, the position

of the screen is adjusted until a clear image of the object is seen on the screen. The image is described as a real image because it is formed on the screen where the light rays meet. If the object is moved nearer the lens towards its principal focus, the screen must be moved further from the lens to see a clear image. The nearer the object is to the lens, the larger the image is.

2 With the object nearer to the lens than the principal focus, a magnified image is formed. The lens acts as a magnifying glass. But the image can only be seen when you look into the lens from the other side to the object. The image is called a virtual image because it is formed where the light rays appear to come from.

Ray diagrams The position and nature of the image formed by a lens depends on the focal length of the lens and the distance from the object to the lens.

If we know the focal length, f, and the object distance, u, we can find the position and nature of the image by drawing a ray diagram to scale in which:

the lens is assumed to be thin so it can represented by a single line at which refraction takes place

the straight line through the centre of the lens perpendicular to the lens is called the principal axis

the principal focus F is marked on the principal axis at the same distance from the lens on each side of the lens

the object is represented by an upright arrow as shown in Figure 3. Note that the horizontal scale of the diagram must be chosen to enable you to fit the object, the image and the lens on the diagram.

Formation of a real image by a converging lens To form a real image, the object must be beyond the principal focus F of the lens. The image is formed on the other side of the lens to the object.


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Figure 3 Formation of a real image by a converging lens To locate the tip of the image, three key construction rays from the tip of the object are drawn, through the lens. The tip of the image is formed where these three rays meet. The image is real and inverted.

1 Ray 1 is drawn parallel to the principal axis before the lens so it is refracted by the lens through F.

2 Ray 2 is drawn through the lens at its centre without change of direction. This is because the lens is thin and its surfaces are parallel to each other at the axis.

3 Ray 3 is drawn through F before the lens so it is refracted by the lens parallel to the axis. Figure 4(a) and 4(b) show ray diagrams for the object at 2F and between F and 2F respectively. The results for Figures 3 and 4 are described in the table below. Notice that the image is:

diminished in size when the object is beyond 2F as in Figure 3 the same size as the object when the object is at 2F as in Figure 4(a) magnified when the object is between F and 2F as in Figure 4(b).

Figure 4 Using ray diagrams to locate an image


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Formation of a virtual image by a converging lens The object must be between the lens and its principal focus, as shown in Figure 5. The image is formed on the same side of the lens as the object.

Figure 5 Virtual image by a converging lens Figure 5 shows that the image is virtual, upright and larger than the object. The image is on the same side of the lens as the object and can only be seen by looking at it through the lens. This is how a magnifying glass works. If the object is placed in the focal plane, light rays from any point on the object are refracted by the lens to form a parallel beam. A viewer looking at the object through the lens would therefore see a virtual image of the object at infinity. Object position Image position Nature

of image

Magnified or diminished

Upright or inverted

Application

beyond 2F between F and 2F real diminished inverted camera 2F 2F real same size inverted inverter between F and 2F beyond 2F real magnified inverted projector < F same side as object virtual magnified upright magnifying

lens

Table 1 Image formation by a converging lens Note

The linear magnification of the image = object theofheight image theofheight

It can be shown that this ratio is equal to distanceobject thedistance image the

The image is said to be magnified if the image height is greater than the object height and diminished if it is smaller.

When you draw a ray diagram, make sure you choose a suitably large scale that enables you to fit the object and the image on your diagram and use a ruler to make sure your lines are straight!


Astrophysics

The lens formula For an object on the principal axis of a thin lens of focal length f at distance u from the lens, the distance, v, from the image to the lens is given by

fvu111

Notes 1 Proof of the lens formula is not required for this specification. 2 When numerical values are substituted into the formula, the sign convention

real is positive; virtual is negative is used for the object and image distances. The focal length, f, for a converging lens is always assigned a positive value. A diverging lens is always assigned a negative value.

Worked example An object is placed on the principal axis of a convex lens of focal length 150 mm at a distance of 200 mm from the centre of the lens.

a Calculate the image distance. b State the properties of the image. Solution a f = +0.150 m, u = +0.200 mm

Using the lens formula fvu11 1 gives

150.011

200.01 v

Hence 200.01

150.011 v = 6.67 5.00 = 1.67

Therefore v = +0.600 m b The image is real (because v is positive), inverted and magnified (because v > u).

Summary questions 1 a i Copy and complete the ray diagram in Figure 6 to show how a converging lens in a camera forms

an image of an object.


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Figure 6 ii State whether the image in Figure 6 is real or virtual, magnified or diminished, upright or

inverted. b i Draw a ray diagram to show how a converging lens is used as a magnifying glass. ii State whether the image in your diagram is real or virtual, magnified or diminished, upright or

inverted. 2 An object is placed on the principal axis of a thin converging lens at a distance of 400 mm from the

centre of the lens. The lens has a focal length of 150 mm. a Draw a ray diagram to determine the distance from the image to the lens. b State whether the image is: i real or virtual ii upright or inverted. c Use the lens formula to check the accuracy of your ray diagram. 3 An object is placed on the principal axis of a thin converging lens at a distance of 100 mm from the

centre of the lens. The lens has a focal length of 150 mm. a Draw a ray diagram to determine the distance from the image to the lens. b State whether the image is: i real or virtual ii upright or inverted. c Use the lens formula to check the accuracy of your ray diagram. 4 An object of height 10 mm is placed on the principal axis of a converging lens of focal length 0.200 m. a Calculate the image distance and the height of the image for an object distance of: i 0.150 m ii 0.250 m. b In each case above, calculate the distance between the object and the image and state whether the

image in each case is real or virtual and upright or inverted.

The linear magnification of the image = object theofheight image theofheight =

distanceobject distance image


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1.2 The refracting telescope Learning objectives: What is a refracting telescope? What do we mean by angular magnification? How does the angular magnification depend on

the focal lengths of the two lenses?

The astronomical telescope consisting of two converging lenses To make a simple refracting telescope, two converging lenses of differing focal lengths are needed. The lens with the longer focal length is referred to as the objective because it faces the object. The viewer needs to look through the other lens, the eyepiece, as shown in Figure 1. Light from the object enters his or her eye after passing through the objective then through the eyepiece into the viewers eye. By adjusting the position of the inner tube in the outer tube, the distance between the two lenses is altered until the image of the distant object is seen in focus. If the telescope is used to view a distant terrestrial object, the viewer sees an enlarged, virtual and inverted image.

Figure 1 The simple refracting telescope To understand why the viewer sees a magnified virtual image, consider the effect of each lens on the light rays from the object that enter the telescope:

The objective lens focuses the light rays to form a real image of the object. This image is formed in the same plane as the principal focus of the objective lens which is where the light rays cross each other after passing through the objective lens. If a tracing paper screen is placed at this position, as shown in Figure 2, the real image formed by the objective can be seen directly on the paper without looking through the eyepiece.


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The eyepiece gives the viewer looking through the telescope a magnified view of this real image with or without the tracing paper present. If the tracing paper is removed, the viewer sees the same magnified view of the real image except much brighter. This magnified view is a virtual image because it is formed where the rays emerging from the eyepiece appear to have come from.

The virtual image is inverted compared with the distant object. This is because the real image formed by the objective is inverted and the final virtual image is therefore inverted compared with the distant object.

Figure 2 Investigating the simple refracting telescope The ray diagram in Figure 3 shows in detail how the viewer looking through the eyepiece sees the final virtual image. The diagram shows the telescope in normal adjustment which means the telescope is adjusted so the virtual image seen by the viewer is at infinity. In this situation, the principal focus of the eyepiece is at the same position as the principal focus of the objective. In other words, in normal adjustment:

the distance between the two lenses is the sum of their focal lengths This is because:

the real image of the distant object is formed in the focal plane of the objective (because the light rays from each point of the object are parallel to each other before entering the objective lens)

the eyepiece is adjusted so its focal plane coincides with the focal plane of the objective. As a result, the light rays that form each point of the real image leave the eyepiece parallel to one another. To the viewer looking into the eyepiece, these rays appear to come from a virtual image at infinity.

Figure 3 Ray diagram for a simple refracting telescope in normal adjustment


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Notes 1 The light rays from each point of the distant object:

are effectively parallel to each other by the time they reach the telescope leave the telescope as a parallel beam which therefore appears to the viewer to come from

a distant point. 2 The real image formed by the objective lens is inverted and diminished in size. The eyepiece

in effect acts as a magnifying glass with the real image being viewed by it. The viewer sees a magnified virtual image which is upright compared with the real image and therefore inverted compared with the distant object.

3 Notice that all the light rays from the object that pass through the eyepiece all pass through a circle referred to as the eye-ring. This is the best position for the viewers eye as the entire image can be seen by the eye at this position.

Angular magnification Application Investigating the simple refracting telescope Use two suitable converging lenses in holders to make a simple refracting telescope. Adjust the position of the eyepiece so an image of a distant object is seen in focus. The image of the object is inverted and it should be magnified.

Place a tracing paper screen between the lenses and locate the real image of the distant object formed by the objective lens. Observe the image directly and through the eyepiece to see that the eyepiece gives a magnified virtual image of the real image. The virtual image becomes brighter if the screen is removed.

View the distant object directly with one eye and through the telescope with the other eye, as in Figure 4. You should be able to estimate how large the image appears to be compared with the object viewed directly (i.e. without the aid of the telescope). This comparison is referred to as the angular magnification (or magnifying power) of the telescope.

Figure 4 A telescope test

Suppose a telescope in normal adjustment makes a distant object appear to be three times larger. Its angular magnification would therefore be 3. If the angle subtended by the distant object to the unaided eye is 1q, the angle subtended by the telescope image to the eye would be 3q. Figure 5 shows the idea. The diagram shows only one light ray from the top of the object entering the


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telescope at the objective lens and leaving in a direction as if it was from the tip of the virtual image seen by the viewer. The distant object and the image are meant to be at infinity so the angle subtended by the distant object to the unaided eye is effectively the same as the angle subtended by the object to the telescope.

The angle subtended by the final image at infinity to the viewer = E The angle subtended by the distant object to the unaided eye = D The angular magnification of the telescope in normal adjustment =

DE eye unaided the toobject distant the by subtended angle

viewer the to infinityat image final the by subtended angle

Figure 5 Angular magnification

From the inset diagram in Figure 5, it can be seen that tan D = o

1

fh and tan E =

e

1

fh , where h1 is the

height of the real image and fo and fe are the focal lengths of the objective and eyepiece lenses respectively.

Combining these two equations to eliminate h1 gives e

o

o1

e1

tantan

ff

fh

fh

DE

Assuming angles D and E are always less than about 10q, applying the small angle approximation

tan D = D in radians and tan E = E in radians gives e

o

ff D

E

Therefore:


Astrophysics

the angular magnification of a telescope in normal adjustment = e

off

Notes 1 The height h1 of the real image = fo tan D = fo (D in radians)

Remember 360q = 2S radians. 2 The objective is the lens with the longer focal length. If you use a telescope the wrong way

round, you will see a diminished image!

Always check your calculator is in the correct angle mode when carrying out calculations involving angles.

Worked example A refracting telescope consists of two converging lenses of focal lengths 0.840 m and 0.120 m.

a If the telescope is used in normal adjustment, calculate: i its angular magnification ii the distance between its lenses.

b The telescope in normal adjustment is used to observe the Moon when the angle subtended by the lunar disc is 0.40q. Calculate:

i the angle subtended by the image of the lunar disc ii the diameter of the real image of the lunar disc formed by the objective lens.

Solution a i The objective is the lens with the longer focal length.

Angular magnification = 0.7120.0840.0

e

o ff

ii Distance between the lenses = fo + fe = 0.840 + 0.120 = 0.960 m

b i angular magnification = DE where D = 0.40q

Therefore E = D angular magnification = 7D = 2.8q ii h1 = fo tan D = 0.840 u tan 0.40q = 5.9 103 m

Image brightness A star is so far away that it is effectively a point object. When viewed through a telescope, a star appears brighter than when it is viewed by the unaided eye. This is because the telescope objective is wider than the pupil of the eye so more light from a star enters the eye when a telescope is used than when the eye is unaided.

The pupil of the eye in darkness has a diameter of about 10 mm. The light entering the eye pupil or the objective is proportional to the area in each case and the area is proportional to the square


Astrophysics

of the diameter. Therefore, in comparison with the unaided eye, a telescope with an objective lens:

of diameter 60 mm would collect 36 times

2

1060 more light per second from a star

of diameter 120 mm would collect 144 times

2

10120 more light per second from a star.

This is why many more stars are seen using a telescope than using the unaided eye. The greater the diameter of the objective of a telescope, the greater the number of stars that can be seen.

Planets and other astronomical objects in the solar system are magnified using a telescope (unlike stars which are point objects and are seen through telescopes as point images no matter how large the magnification of the telescope is). Yet the image of a planet viewed using a telescope is not significantly brighter than the planet when it is viewed directly. This is because, although more light per second enters the eye when a telescope is used, the virtual image is magnified so is spread over a larger part of the field of view. As a result, the amount of light per second per unit area of the virtual image is unchanged.

Warning! Never view the Sun using a telescope or directly. The intensity of sunlight entering the eye would damage the retina of the eye and cause blindness. How science works

Galileo on trial Although the telescope was first invented by the English astronomer Thomas Digges, it was not generally known about until after its rediscovery in 1609 by the Dutch lens-maker Hans Lippershey. When Galileo first heard about it, he rushed to make his first telescope so he could demonstrate it before anyone else to his patrons in Venice observing incoming ships would enable them to buy the ships cargoes before their competitors could! After being rewarded accordingly, Galileo went on to make more powerful telescopes and used them to observe the stars and the planets.

His discoveries of craters on the lunar surface and of the four inner moons of Jupiter (now referred to as the Galilean moons Io, Callisto, Ganymede and Europa) convinced him that the Copernican model of the solar system published by Copernicus more than seventy years earlier was correct the planets orbit the Sun and the Earth itself is a planet. After Galileo published his discoveries in 1610, his support for the Copernican model was challenged by members of the Inquisition and he had to rely on his friends in the Church to defend him. As a result of a further publication Dialogue on the Two Chief World Systems which he wrote in 1629, Galileo was tried by the Inquisition for heresy and forced to confess. He was sentenced to life imprisonment which his friends in the Church managed to reduce to confinement at his home in Tuscany. However, before he died in 1642, he wrote a textbook on his scientific theories and experiments in which he established the scientific method used by scientists worldwide ever since.


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Figure 6 Galileo

Summary questions 1 Draw a ray diagram of a telescope consisting of two converging lenses to show how an image is formed

of a distant object. Show clearly on your ray diagram the principal focus of the lenses, the position of the viewers eye and label the two lenses.

2 A telescope consists of two converging lenses of focal lengths 60 mm and 450 mm. It is used in normal adjustment to view a distant object that subtends an angle of 0.15q to the telescope.

a Explain what is meant by the term normal adjustment. b Calculate: i the angular magnification of the telescope ii the angle subtended by the virtual image seen by the viewer. 3 Explain the following observations made using a telescope. a A star too faint to see with the unaided eye is visible using the telescope. b The Galilean moons of Jupiter can be observed using a telescope but not by the unaided eye. 4 A telescope consisting of two converging lenses has an eyepiece of focal length 40 mm. When used in

normal adjustment, the angular magnification of the telescope is 16. a Calculate: i the focal length of the objective lens ii the separation of the two lenses. b The image of a tower of height 75 m viewed through the telescope subtends an angle of 4.8q to the

viewer. Calculate: i the angle subtended by the tower to the viewers unaided eye ii the distance from the tower to the viewer.


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1.3 Reflecting telescopes Learning objectives: What is a Cassegrain reflecting telescope? What is meant by spherical aberration and

chromatic aberration? What are the relative merits of a reflecting

telescope and a refracting telescope?

The Cassegrain reflecting telescope A concave mirror instead of a converging lens is used as the objective of a reflecting telescope. The concave reflecting mirror is referred to as the primary mirror because a secondary smaller mirror reflects light from the concave reflector into the eyepiece.

The shape of a concave mirror is such that parallel rays directed at it are reflected and focused to a point by the mirror. The principal axis of the mirror is the line normal to its reflecting surface through its centre. If rays are parallel to the principal axis of the concave mirror then the point where the reflected rays converge is called the principal focus F (i.e. the focal point) of the mirror. Figure 1 shows the idea. The distance from the principal focus to the centre of the mirror is the focal length, f, of the mirror. The light rays from a distant point object are effectively parallel when they reach the mirror. So a concave mirror will form a real image of a distant point object in the focal plane, the plane containing the principal focus.

Figure 1 The focal length of a concave mirror In a Cassegrain reflecting telescope, the secondary mirror is a convex mirror positioned near the focal point of the primary mirror between this point and the primary mirror itself. The purpose of the convex mirror is to focus the light onto or just behind a small hole at the centre of the concave reflector. The light passing through this small hole then passes through the eyepiece which is behind the concave mirror centre, as shown in Figure 2. The distance from the concave mirror to the point where it focuses parallel rays is increased by using a convex mirror instead of a plane mirror as the secondary mirror. This distance is the effective focal length of the objective.


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Figure 2 Ray diagram for a Cassegrain reflector When the telescope is directed at a distant object, a viewer looking into the eyepiece sees a virtual image of the distant object. The light from the distant object is:

1 reflected by the concave mirror, then 2 reflected by the convex mirror onto the small hole at the centre of the concave mirror into the

eyepiece, then 3 refracted by the eyepiece into a parallel beam which enters the viewers eye. Consequently, the viewer sees the virtual image at infinity.

Notes on the Cassegrain telescope 1 The effective focal length of the objective is increased by using a secondary convex mirror.

Therefore, the angular magnification (= focal length of objective focal length of eyepiece) is also increased.

2 In a typical Cassegrain reflector, the image of a distant object is usually brought into focus by adjusting the position of the secondary convex mirror along the principal axis.

3 The primary mirror should be parabolic in shape rather than spherical to minimise spherical aberration due to the primary mirror. This effect happens with a spherical reflecting surface because the outer rays of a beam parallel to the principal axis are brought to a focus nearer the mirror than the focal point, F, as shown in Figure 3(a). In comparison, the parabolic mirror in Figure 3(b) focuses all the light rays to F.

Figure 3 Spherical aberration


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Comparison of refractors and reflectors Reflecting telescopes in general have a key advantage over refracting telescopes because they can be much wider. This is because high-quality concave mirrors can be manufactured much wider than a convex lens can. The wider the objective is, the greater the amount of light they can collect from a star, enabling stars to be seen that would be too faint to see even with a refractor. As explained in Topic 1.2, the light collected by a telescope is proportional to the area of the objective. As the area is proportional to the square of its diameter, a reflector with an objective of diameter 200 mm can collect 25 times as much light as a refractor with an objective of diameter 40 mm.

Telescopes with wide objectives usually have a concave mirror as the objective rather than a convex lens. The high quality of a wide concave mirror compared with a wide convex lens is because:

image distortion due to spherical aberration is reduced if the mirror surface is parabolic unwanted colours in the image are reduced. Such unwanted colours are due to splitting of

white light into colours when it is refracted. The result is that the image formed by a lens of an object is tinged with colour, particularly noticeable near the edge of the lens. The effect is known as chromatic aberration. Figure 4 illustrates the effect. Notice the blue image is formed nearer the lens than the red image; this is because blue light is refracted more than red light.

Figure 4 Chromatic aberration Also, a wide lens would be much heavier than a wide mirror and would make the telescope top-heavy.

Further comparisons between refractors and reflectors are summarised below.

Refracting telescopes: use lenses only and do not contain secondary mirrors and supporting frames which would

otherwise block out some of the light from the object have a wider field of view than reflectors of the same length because their angular

magnification is less. Astronomical objects are therefore easier to locate using a refractor instead of a reflector of the same length.

Reflecting telescopes: are shorter and therefore easier to handle than refractors with the same angular magnification have greater angular magnification than refractors of the same length and therefore produce

greater magnification of distant objects such as the Moon and the planets.

Summary questions 1 Draw a ray diagram to show the passage of light from a distant point object through a Cassegrain

reflecting telescope. Show the position of the eye of the observer on your diagram and label the parts that make up the telescope and the effective focal point of the objective.

2 a State what is meant by chromatic aberration.


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b Explain why the objective of a refracting telescope produces chromatic aberration whereas that of a Cassegrain reflector does not.

3 State and explain one disadvantage and one advantage, other than reduced chromatic aberration, a Cassegrain telescope has in comparison with a simple refractor telescope.

4 A Cassegrain telescope has a primary mirror of diameter 80 mm. a Calculate the ratio of the light energy per second it collects to the light energy per second collected

by the eye when the eye pupil is 8 mm in diameter. b The telescope objective has an effective focal length of 2.8 m and its eyepiece has a focal length of

0.07 m. Calculate its angular magnification.


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1.4 Resolving power Learning objectives: What do we mean by angular separation? Why does a wide telescope resolve two stars

that cannot be resolved by a narrower telescope?

What is the Rayleigh criterion for resolving two point objects?

Diffraction The extent of the detail that can be seen in a telescope image depends on the width of the objective. Imagine viewing two stars near each other in the night sky. The angular separation of the two stars is the angle between the straight lines from the Earth to each star, as shown in Figure 1.

Figure 1 Angular separation Suppose the two stars are viewed through a telescope and their images can just be seen as separate images. In other words, the telescope just resolves the two stars. If the telescope is replaced by one with a narrower objective, the images of the two stars would overlap too much and the observer would not be able to see them as separate stars. This is because:

the objective lens or mirror is in an aperture (i.e. a gap) which light from the object must pass through and diffraction of light always occurs whenever light passes through an aperture

instead of focusing light from a star (or other point object) to a point image, diffraction of light passing through the objective causes the image to spread out slightly.

the narrower the objective, the greater the amount of diffraction that occurs when light passes through the narrower objective. So the greater the spread of the image.

Diffraction at a circular aperture Diffraction at a circular aperture can be observed on a screen when a narrow beam of light passes through a small circular aperture before reaching the screen. Figure 2 shows the diffraction pattern on the screen. The pattern consists of a central bright spot surrounded by alternate concentric dark and bright rings. The bright rings are much fainter than the central spot and their intensity decreases with distance from the centre.

The objective of a telescope is a circular aperture containing a convex lens or a concave mirror. Diffraction occurs when light from a star passes through the aperture. As the light is focused by the objective, the star image showing the same type of pattern as in Figure 2 is observed in the focal plane of the objective. An observer looking through the eyepiece would see a magnified view (i.e. a magnified virtual image) of the star image formed by the objective.


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For light of wavelength O passing through a circular aperture of diameter D, it can be shown that an approximate value of the angle of diffraction, in radians, of the first dark ring is given by D

O .

Link Topic 13.6 of AS Physics A looks at single slit diffraction.

Prove for yourself that, for an objective of diameter 80 mm , the angle of diffraction T for the first dark ring is approximately 6.3 106 radians (= 0.00036 degrees) for light of wavelength 500 nm. In comparison, the corresponding angle for an objective of diameter 20 mm would be four times larger (i.e. 0.0014(4) degrees).

Figure 2 Diffraction of a small circular aperture

Resolving two stars Two stars near each other in the night sky can be resolved (i.e. seen as separate stars) if the central diffraction spots of their images do not overlap significantly. This condition can be expressed numerically using the Rayleigh criterion which states that resolution of the images of two point objects is not possible if any part of the central spot of either image lies inside the first dark ring of the other image. As shown in Figure 3, this means that the angular separation of the two stars must be at least equal to the angle of diffraction of the first dark ring.

In other words, using the above approximation for the angle of diffraction of the first dark ring, the least angular separation T for the resolution of two stars is given approximately by the condition:

DOT |

where O = the wavelength of light, and D = the diameter of the circular aperture.


Astrophysics

Figure 3 Resolving two stars For example, a telescope with an 80 mm diameter objective will just be able to resolve two stars with an angular separation of 0.000 36 degrees, assuming an average value of 500 nm for the wavelength of light. Without the telescope, the human eye would not be able to resolve them as the typical eye pupil diameter is about 8 mm which is a tenth of the width of an 80 mm wide telescope. The unaided eye can resolve two stars only if their angular separation is at least 0.0036 degrees (i.e. ten times greater than that with an 80 mm wide telescope).

When you use the formula, make sure your calculator is in radian mode and dont forget to convert angles to radians if their values are wanted in radians or given in degrees.

Remember: 2S radians = 360 degrees.

Notes 1 Resolution or resolving power are both used sometimes to describe the quality of a

telescope in terms of the minimum angular separation it can achieve. For example, a telescope described as having a resolution or resolving power of 0.004 degrees can resolve two stars which have an angular separation of at least 0.004 degrees.

2 The Rayleigh criterion applies to the detail visible in extended images as well as to stars. For example, a telescope with a resolving power of 5 105 radians (= 0.003 degrees) is capable of seeing craters on the lunar surface which have an angular diameter of 0.003 degrees. As the Moon is about 380 000 km from Earth, such craters are about 20 km in diameter.

3 Refraction due to movement of air in the atmosphere causes the image of any star seen through a telescope to be smudged slightly. As a result, ground-based telescopes with objectives of diameter greater than about 100 mm do not achieve their theoretical resolution.


Astrophysics

The stunning clarity of images from the Hubble Space Telescope is because the telescope has an objective mirror of diameter 2.4 m and is above the atmosphere and therefore does not suffer from atmospheric refraction. Hence it achieves its theoretical resolution which is about 240 times greater than that of a 100 mm wide telescope.

How science works and application The Hubble Space Telescope

Figure 4 A HST image of a cluster of galaxies After it was first launched in 1990, HST images were found to be poor because of spherical aberration in its primary mirror due to a manufacturing fault. This was corrected in 1993 when a space shuttle mission was launched to enable astronauts to fit small secondary mirrors to compensate exactly for the fault and give amazing images that have dramatically increased our knowledge of space.

The Hubble Space Telescope detects images at wavelengths from 115 nm to about 1000 nm, thus giving infrared, visible and ultraviolet images.

Summary questions 1 a What is the name for the physical phenomenon that causes the image formed by a lens or mirror of a

point object to be spread out? b i Sketch the pattern of the image of a distant point object formed by a lens. ii Describe how the pattern would differ if a wider lens of the same focal length had been used? 2 State and explain what is meant by the Rayleigh criterion for resolving two point objects using a

telescope.

3 Two stars have an angular separation of 8.0 106 rad. a Assuming light from them has an average wavelength of 500 nm, calculate an approximate value for

the diameter of the objective of a telescope that can just resolve the two stars. b Discuss how the image of the two stars would differ if they were viewed with a telescope with an

objective of twice the diameter and the same angular magnification.

4 The Hubble Space Telescope has an objective of diameter 2.4 m. a Show that the theoretical resolution of the HST is 1.2 105 degrees. b Hence estimate the diameter of the smallest crater on the Moon that can be seen using the telescope.

Assume the wavelength of light is 500 nm. EarthMoon distance = 3.8 108 m


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1.5 Telescopes and technology Learning objectives: What is a charge-coupled device (CCD) and why

is it important in astronomy? How does a CCD work? What are non-optical telescopes used for? How do non-optical telescopes compare with

each other and with optical telescopes?

Charge-coupled devices Astronomers have always used photographic film to capture images ever since photography was first invented in the 19th century. However, the charge-coupled device invented in the late 20th century fitted to a telescope has dramatically extended the range of astronomical objects that can be seen as well as providing images of stunning quality.

Figure 1 Using a CCD (a) A CCD in a telescope (b) A CCD image of a spiral galaxy The CCD is an array of light-sensitive pixels which become charged when exposed to light. After being exposed to light for a pre-set time, the array is connected to an electronic circuit which transfers the charge collected by each pixel in sequence to an output electrode connected to a capacitor. The voltage of the output electrode is read out electronically then the capacitor is discharged before the next pulse of charge is received. In this way, the output electrode produces a stream of voltage pulses, each one of amplitude in proportion to the light energy received by an individual pixel. Figure 2 shows part of an array of pixels.


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Figure 2 Inside a CCD Each pixel has three small rectangular metal electrodes (labelled A, B and C in Figure 2) which are separated by a thin insulating layer of silicon dioxide from p-type silicon which is the light-sensitive material underneath. The electrodes are connected to three voltage supply rails.

The rectangular electrodes and the insulating layer are thin enough to allow light photons to pass through and each liberate an individual electron in the light-sensitive material underneath.

When collecting charge, the central electrode in each pixel (labelled B in Figure 2) is held at +10 V and the two outer electrodes at +2 V. This ensures the liberated electrons accumulate under the central electrode.

After the pixels have collected charge for a certain time, the charge of each pixel is shifted towards the output electrode via the adjacent pixels. This is achieved by altering the voltage level of each electrode in a sequence of three-step cycles, as shown in Figure 2.

The quantum efficiency of a pixel is the percentage of incident photons that liberate an electron. About 70% of the photons incident on a pixel each liberates an electron. Therefore, the quantum efficiency of a pixel is about 70%. In comparison, the grains of a photographic film have a quantum efficiency of about 4% as only about 4 in every 100 incident photons contributes to the darkening of each grain. So a CCD is much more efficient than a photographic film and hence it will detect much fainter astronomical images than a film.

Further advantages of a CCD Its use in recording changes of an image. It can record a sequence of fast-changing

astronomical images which can be seen by the eye but could not be recorded on a photographic film.

Its wavelength sensitivity from less than 100 nm to 1100 nm is wider than that of the human eye which is from about 350 nm to 650 nm. Hence it can be used with suitable filters to obtain infrared images.

The quantum efficiency of a CCD is the same at about 70% from about 400 nm to 800 nm, reducing to zero below 100 nm and at 1100 nm.

However, CCDs for use in astronomy need to have a larger number of pixels in a small area and are therefore expensive compared with CCDs in most electronic cameras. More significantly, CCDs used in astronomy are often cooled to very low temperatures using liquid nitrogen


Astrophysics

otherwise random emission of electrons causes a dark current which does not depend on the intensity of light.

Radio telescopes Single-dish radio telescopes each consist of a large parabolic dish with an aerial at the focal point of the dish. A steerable dish can be directed at any astronomical source of radio waves in the sky. The atmosphere transmits radio waves in the wavelength range from about 0.001 m to about 10 m. When the dish is directed at an astronomical source that emits radio waves in the above wavelength range, the waves reflect from the dish onto the aerial to produce a signal. The dish is turned by motors to enable it to scan sources and to compensate for the Earths rotation.

Figure 3 A single-dish radio telescope The amplitude of the signal is a measure of the intensity of the radio waves received by the dish. The signal from the aerial is amplified and supplied to a computer for analysis and recording. As the dish scans across the source, the signal is used to map the intensity of the radio waves across the source to give a radio image of the source.

The dish surface usually consists of a wire mesh which is lighter than metal sheets and just as

effective in terms of reflection, provided the mesh spacing is less than about 20O , where O is the

wavelength of the radio waves.

The dish diameter, D, determines the collecting area of the dish (= 4

1 SD2)

the resolving power of the telescope (= DO )

The Lovell radio telescope at Jodrell Bank in Cheshire has a 76 m steerable dish which gives a resolution of 0.2 degree for 21 cm wavelength radio waves. In comparison, the Arecibo radio telescope in Puerto Rica is a 300 m fixed concave dish set in a natural bowl. As it is four times wider than the Lovell telescope, it can therefore resolve radio images to about 0.05 degrees (= 4

1 of 0.2q) and detect radio source 16 times fainter (as it collects 16 times as much radio


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energy per second than the Lovell telescope does). However, the Arecibo telescope can only detect radio sources when they are close to its principal axis.

Uses of radio telescopes Locating and studying strong radio sources in the sky The Sun, Jupiter and the Milky Way are all strong sources of radio waves. Some galaxies are also relatively strong emitters of radio waves. Such galaxies are usually elliptical or spherical without spiral arms. Many radio galaxies are found near the centre of clusters of galaxies and their optical images often show evidence of violent events such as two galaxies merging or colliding or a galaxy exploding or emitting immensely powerful jets of matter.

Mapping the Milky Way galaxy Hydrogen atoms in dust clouds in space emit radio waves of wavelength 21 cm. These are emitted when the electron in a hydrogen atom flips over so its spin changes from being in the same direction as the protons spin to a lower energy level in the opposite direction. The Milky Way is a spiral galaxy with the Sun in an outer spiral arm. Dust clouds in the spiral arms prevent us from seeing stars and other radio sources, such as hot gas behind the dust clouds, as dust absorbs light. However, radio waves are not absorbed by dust so radio telescopes are used to map the Milky Way.

Link Electromagnetic waves were looked at in Topic 1.3 of AS Physics A.

Infrared telescopes Infrared telescopes have a large concave reflector which focuses infrared radiation onto an infrared detector at the focal point of the reflector. Objects in space such as planets that are not hot enough to emit light emit infrared radiation. In addition, dust clouds in space emit infrared radiation. Infrared telescopes can therefore provide images from objects in space that cannot be seen using optical telescopes.

Ground-based infrared telescopes A ground-based infrared telescope needs to be cooled to stop infrared radiation from its own surface swamping infrared radiation from space. Water vapour in the atmosphere absorbs infrared radiation, so an infrared telescope needs to be sited where the atmosphere is as dry as possible and as high as possible. The 3 m diameter infrared telescope on a mountain in Hawaii is located there because the atmosphere is very dry and the water vapour that is present has less effect than if the telescope was at a lower level.

Infrared telescopes on satellites An infrared telescope on a satellite in orbit above the Earth is not affected by water vapour. However, the telescope still needs to be cooled to a few degrees above absolute zero to be able to detect infrared radiation from weak infrared sources.

IRAS, the first infrared astronomical satellite, launched in 1978, discovered bands of dust in the solar system and dust around nearby stars. It carried a 60 cm wide infrared telescope fitted with a detector capable of detecting infrared wavelengths from 0.01 mm to about 1 mm.


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The Hubble Space Telescope with its objective at 2.4 m wide is capable of detecting infrared wavelengths from 700 nm to about 1000 nm (= 0.001 mm). It can form images of warm objects such as dying stars and planets in other solar systems that emit thermal radiation but not light.

Ultraviolet telescopes Ultraviolet (UV) telescopes must be carried on satellites because UV radiation is absorbed by the Earths atmosphere. As UV radiation is also absorbed by glass, a UV telescope uses mirrors to focus incoming UV radiation onto a UV detector. UV radiation is emitted by atoms at high temperatures, so UV telescopes are used to map hot gas clouds near stars and to study hot objects in space such as glowing comets, supernova and quasars. Comparing a UV image of an object with an optical or infrared image gives useful information about hot spots in the object.

The International Ultraviolet Explorer (IEU) launched in 1978 carried a 0.45 m wide Cassegrain telescope with a UV detector instead of an eyepiece in its focal plane.

The Hubble Space Telescope uses a CCD to detect images at wavelengths from 115 nm to about 1000 nm, giving ultraviolet images as well as visible and infrared images according to the filters used over the CCD.

The XMM-Newton space observatory, launched in 1999 and still in operation, carries a 30 cm wide modified Cassegrain reflector fitted with a detector with a wavelength range from 170 nm to 650 nm. So, it can give ultraviolet as well as optical images.

Figure 4 A combined UV and optical image of the galaxy M82 galaxy (UV in blue).

X-ray and gamma-ray telescopes They need to be carried on satellites as the Earths atmosphere absorbs X-rays and gamma rays. Discoveries using such telescopes include:

X-ray pulsars, stars that emit X-ray beams that sweep round the sky as they spin X-ray and gamma-ray bursters billions of light years away which emit bursts of gamma

rays.

X-ray telescopes work by reflecting X-rays off highly-polished metal plates at grazing incidence onto a suitable detector. Gamma ray telescopes work by detecting gamma photons as they pass through a detector containing layers of pixels, triggering a signal in each pixel it passes through it. The direction of each incident gamma photon can be determined from the signals. The


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International Gamma Ray Astrophysics Laboratory (INTEGRAL) launched in 2002 is being used to study supernova, gamma ray bursts and black holes. As gamma rays and X-rays are very short wavelength, diffraction is insignificant and image resolution is determined by the pixel separation.

Summary questions The table below is an incomplete comparison of different types of astronomical telescopes.

Type Location Wavelength range

Resolution (degrees)

Key advantages Major disadvantages

optical ground or satellite

350650 nm 105 for HST gives very detailed images, can detect distant galaxies

ground telescopes suffer from atmospheric refraction

radio ground 1 mm to 10 m 0.2 for Lovell

radio waves pass through dust in space and through the atmosphere

large, supporting structure needed for a steerable dish.

infrared can detect warm objects that do not emit light, can detect dust clouds in space

mirror needs to be cooled

ultraviolet must be above the Earths atmosphere e.g. on a satellite

X and gamma

0.2 for INTEGRAL

must be above the Earths atmosphere e.g. on a satellite

1 Copy this table and use the information in this topic to complete columns 2 and 3. 2 a Use the information in the previous pages to estimate the resolution in degrees of: i HST at a wavelength of 0.001 mm ii XMM-Newton at a wavelength of 170 nm b Use your estimates to complete column 4 of your table. 3 Complete column 5 by giving two key advantages of: a UV telescopes b X-ray and gamma ray telescopes. 4 The collecting power of a telescope is a measure of how much energy per second it collects. This

depends on the area of its objective as well as the power per unit area (intensity) of the incident radiation.

a For the same incident power per unit area, list the following telescopes in order of their collecting power:

Hubble Space Telescope (2.4 m in diameter) INTEGRAL (0.60 m diameter) IRAS (0.60 m diameter) Lovell telescope (76 m diameter) XMM-Newton (0.30 m diameter) b The Lovell radio telescope is linked to other radio telescopes in England so they act together as an

effective radio telescope of much greater width. Discuss without calculations how the resolving power and the collecting power of the linked system compare with that of the Lovell telescope on its own.

Option A chapter 2Surveying the stars


Specification link-up A1.3 Classification of stars

Chapter teacher guide

Chapter 2 provides an alternative starting point for the option and gives opportunities for individual research and class discussions.

An evenings star-gazing, or a trip to a planetarium, can introduce the students to the range of brightness, distance and sheer number of stars in the sky. The idea of parallax (Topic 2.1) can be demonstrated simply by placing a retort stand in the centre of the laboratory to represent the near star. Two metre rules can act as different sight lines from the back of the laboratory to show the apparent shift of the star against a fixed pattern of stars marked on the front wall. Students who are not confident with logarithmic scales can practise converting the apparent magnitude of some real stars into absolute magnitude using data available on the Internet.

When introducing Wiens law (Topic 2.2) and relating star colour to temperature, it helps to remind the students of the relationship between frequency and energy for electromagnetic waves. The higher the frequency, the closer the colour is to the high energy end of the visible spectrum and, naturally, higher frequencies relate to shorter wavelengths. Stefans law may not be so intuitive, but students should realise its importance for calculating the size of stars. If you have a hydrogen discharge tube it is worth letting students have another look at the Balmer lines.

The HertzsprungRussell diagram (Topic 2.3) shows how using the absolute magnitude of stars ensures a fair comparison. Students should be warned that they may encounter different forms of the diagram where the axes show other properties, for instance luminosity (y-axis) and colour or spectral class (x-axis). The study of star types and stellar evolution is an excellent topic for Internet research students could use their findings to give a presentation or to inform class discussion.

As with other aspects of stellar evolution, the supernova (Topic 2.4) makes a good topic for individual research. The idea of a star running out of fuel, exploding and leaving behind a neutron star or black hole is a fascinating subject. As a result, a large amount of data is available via the Internet but not all of it is totally accurate. As an exercise, you could ask the students to consider the reliability of the different sources of information they use.

Common misconceptionsMany students struggle with the HertzsprungRussell diagram. The star groupings are not constellations in the sky, nor is there a time axis showing stellar evolution. It may help to explain that the diagram basically illustrates the way stars can be grouped when you plot their colour against their brightness.

Chapter map

Option A chapter 2Surveying the stars


Chapter map

Specification link-up A1.3: Classification of starsHow is the distance to a nearby star measured? What do we mean by apparent and absolute magnitude?How can we calculate the absolute magnitude of a star?

2.1 Star magnitudes

Specification link-up A1.3: Classification of starsWhat does the colour of a star tell us about the star? How can we classify stars?What can we tell from the absorption spectrum of a star?

2.2 Classifying stars

Specification link-up A1.3: Classification of starsWhy is a supernova called a supernova? What is a neutron star? How is a black hole formed?

2.4 Supernovae, neutron stars and black holes

Specification link-up A1.3: Classification of starsWhat does the colour of a star tell us about the star? How do stars form? Why do we think the Sun will eventually become a white dwarf star?

2.3 The HertzsprungRussell diagram


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Chapter 2 Surveying the stars 2.1 Star magnitudes Learning objectives: How is the distance to a nearby star measured? What do we mean by apparent and absolute

magnitude? How can we calculate the absolute magnitude of

a star?

Astronomical distances One light year is the distance light travels through space in 1 year and equals 9.5 1015 m. Light from the Sun takes 500 s to reach the Earth, about 40 minutes to reach Jupiter, about 6 hours to reach Pluto and about 4 years to the nearest star, Proxima Centauri.

As there are 31.536 million seconds in one year, it follows that one light year = speed of light time in seconds for one year = 3.00 108 m s1 3.15 107 s = 9.45 1015 m.

The Sun and nearby stars are in a spiral arm of the Milky Way galaxy. The galaxy contains almost a million million stars. Light takes about 100 000 years to travel across the Milky Way galaxy.

Galaxies are assemblies of stars prevented from moving away from each other by their gravitational attraction. Galaxies are millions of light years apart, separated from one another by empty space.

The most distant galaxies are about ten thousand million light years away and were formed shortly after the Big Bang. The Universe is thought to be about 13 thousand million (i.e. 13 billion) years old. The most distant galaxies are near the edge of the observable Universe.

Measurement of the distance to a nearby star Astronomers can tell if a star is relatively near us because nearby stars shift in position against the background of more distant stars as the Earth moves round its orbit. This effect is referred to as parallax and it occurs because the line of sight to a nearby star changes direction over six months because we view the star from diametrically opposite positions of the Earths orbit in this time. By measuring the angular shift of a stars position over six months, relative to the fixed pattern of distant stars, the distance to the nearby star can be calculated as explained below.

The Earths orbit round the Sun is used as a baseline in the calculation, so accurate knowledge of the measurement of the mean distance from the centre of the Sun to the Earth is required. This distance is referred to as one astronomical unit (AU) and is equal to 1.496 1011 m.

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Figure 1 Star parallax To calculate the distance to a nearby star, consider Figure 2 which shows the six month angular shift of a nearby stars position relative to stars much further away.

Figure 2 Parallax angle The parallax angle T is defined as the angle subtended by the star to the line between the Sun and the Earth, as shown in Figure 2. This angle is half the angular shift of the stars line of sight over six months.

From the triangle consisting of the three lines between the Sun, the star and the Earth as

shown in Figure 2, dR Ttan .

Since T is always less than 10, using the small angle approximation gives dR T , where T

is in radians. So, TRd . Note that 360 = 2S radians.

Parallax angles are generally measured in arc seconds where 1 arc second = 3600degree 1 . For this

reason, star distances are usually expressed for convenience in terms of a related non-SI unit called the parsec (abbreviated as pc).

1 parsec is defined as the distance to a star which subtends an angle of 1 arc second to the line from the centre of the Earth to the centre of the Sun.

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Since 1 arc second = 3600degree 1 = 4.85 106 radians and 1 AU = 1.496 1011 m, using the

equation TRd gives:

1 parsec = 3.08 1016 m

uu radians1085.4

m10496.16

11

= 3.26 light years

The distance, in parsecs, from a star to the Sun = T1 , where T is the parallax angle of the star

angle, in arc seconds

1(in parsecs)(in arc seconds)

d T

The smaller the parallax angle of a star, the further away the star is. For example:

T = 1.00 arc second, d = 1.00 pc T = 0.50 arc seconds, d = 2.00 pc T = 0.01 arc seconds, d = 100 pc Notes 1 For telescopes sited on the ground, the parallax method for measuring distances works up to

about 100 pc. Beyond this distance, the parallax angles are too small to measure accurately because of atmospheric refraction. Telescopes on satellites are able to measure parallax angles much more accurately and thereby measure distances to stars beyond 100 pc.

2 1 parsec = 3.09 1016 m = 3.26 light years = 206 265 AU

Star magnitudes The brightness of a star in the night sky depends on the intensity of the stars light at the Earth which is the light energy per second per unit surface area received from the star at normal incidence on a surface. The intensity of sunlight at the Earths surface is about 1400 W m2. In comparison, the intensity of light from the faintest star that can be seen with the unaided eye is more than a million million times less. With the Hubble Space Telescope the intensity is more than 10 000 million million times less.

Astronomers in ancient times first classified stars in six magnitudes of brightness, a first magnitude star being one of the brightest in the sky and a sixth magnitude star being just visible on a clear night. The scale was established on a scientific basis in the 19th century by defining a difference of five magnitudes as a hundredfold change in the intensity of light received from the star. In addition, the terms apparent magnitude and absolute magnitude are used to distinguish between light received from a star and light emitted by the star respectively. The term absolute magnitude is important because it enables a comparison between stars in terms of how much light they emit.

On the scientific scale, stars such as Sirius which give received intensities greater than 100 times that of the faintest stars are brighter than first magnitude stars and therefore have zero or negative apparent magnitudes.

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Apparent magnitude The apparent magnitude, m, of a star in the night sky is a measure of its brightness which depends on the intensity of the light received from the star.

Consider two stars X and Y of apparent magnitudes mX and mY which give received intensities IX and IY. Every difference of 5 magnitudes corresponds to 100 times more light intensity from X

than from Y. Generalising this rule gives 5Y

X 100m

II ' , where 'm = mY mX

Taking base 10 logs of this equation gives:

mmII mm ' '

'' 4.0100log2.0100log100loglog 2.05

Y

X

Multiplying both sides of the equation by 2.5 gives mII '

Y

Xlog5.2

Hence

mY mX =

Y

Xlog5.2 II

The absolute magnitude, M, of a star is defined as the stars apparent magnitude, m, if it was at a distance of 10 parsecs from Earth.

It can be shown that for any star at distance d, in parsecs, from the Earth:

m M =

10log5d

To prove this equation, recall that the intensity I of the light received from a star depends on its distance d from Earth in accordance with the inverse square law (I v 1/d2). In using the inverse square law here, we assume the radiation from the star spreads out evenly in all directions and no radiation is absorbed in space.

Link The inverse square law for gamma radiation was looked it in Topic 9.3 of A2 Physics A.

Comparing a star X at a distance of 10 pc from Earth with an identical star Y at distance d from

Earth, the ratio of their received intensities Y

XII would be

2

10

d .

Therefore, the difference between their apparent magnitudes,

mY mX = 2.5 logY

XII

= 2.5 log2

10

d

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= 5 log

10d

Since the stars are identical, the absolute magnitude of X, MX = absolute magnitude of Y, MY. Also, because X is at 10 pc, its apparent magnitude mX = MX

So, mY MY = 5 log

10d

More generally, for any star at distance d, in parsecs, from the Earth:

m M = 5 log

10d

Proof of this formula is not required in this specification.

When you use this equation, make sure you use base 10 logs not base e.

Worked example A star of apparent magnitude m = 6.0 is at a distance of 80 pc from the Earth. Calculate its absolute magnitude M. Solution Rearranging m M = 5 log

10d gives M = m 5 log

10d

Hence M = 6.0 5 log 1080 = 6.0 5 log 8 = 1.5 (= 1.48 to 3 significant figures)

Summary questions 1 parsec = 206 000 AU

1 With the aid of a diagram, explain why a nearby star shifts its position over six months against the background of more distant stars.

2 a State what is meant by the absolute magnitude of a star. b A star has an apparent magnitude of +9.8 and its angular shift due to parallax over six months is

0.45 arc seconds. i Show that its distance from Earth is 4.4 pc. ii Calculate its absolute magnitude. 3 a Show that a star with an apparent magnitude i m = 3.0 at 100 pc has an absolute magnitude of 2.0 ii m = 1.4 at 2.7 pc has an absolute magnitude of +1.4 b Calculate the apparent magnitude of a star of absolute magnitude M = +3.5 which is 30 pc from

Earth.

4 The apparent magnitude of the Sun is 26.8.


Astrophysics

a Show that its absolute magnitude is +4.8. b Calculate the apparent magnitude of the Sun as seen from the planet Jupiter at a distance of 5.2 AU

from the Sun.


Astrophysics

2.2 Classifying stars Learning objectives: What does the colour of a star tell us about the

star? How can we classify stars? What can we tell from the absorption spectrum of

a star?

Starlight Stars differ in colour as well as brightness. Viewed through a telescope, stars that appear to be white to the unaided eye appear in their true colours. This is because a telescope collects much more light than the unaided eye thus activating the colour-sensitive cells in the retina. CCDs with filters and colour-sensitive photographic film show that stars vary in colour from red to orange and yellow to white to bluish-white.

Like any glowing object, a star emits thermal radiation which includes visible light and infrared radiation. For example, if the current through a torch bulb is increased from zero to its working value, the filament glows dull red then red then orange-yellow as the current increases and the filament becomes hotter. The spectrum of the light emitted shows that there is a continuous spread of colours which change their relative intensities as the temperature is increased. This example shows that:

the thermal radiation from a hot object at constant temperature consists of a continuous range of wavelengths

the distribution of intensity with wavelength changes as the temperature of the hot object is increased.

Figure 1 shows how the intensity distribution of such radiation varies with wavelength for different temperatures.

Figure 1 Black body radiation curves

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The curves are referred to as black body radiation curves, a black body being defined as a body that is a perfect absorber of radiation (absorbs 100% of radiation incident on it at all wavelengths) and therefore emits a continuous spectrum of wavelengths. Remember from GCSE that a matt black surface is the best absorber and emitter of infrared radiation. In addition to a star as an example of a black body, a small hole in the door of a furnace is a further example: any thermal radiation that enters the hole from outside would be completely absorbed by the inside walls. We can assume a star is a black body because any radiation incident on it would be absorbed and none would be reflected or transmitted by the star. In addition, the spectrum of thermal radiation from a star is a continuous spectrum with an intensity distribution that matches the shape of a black body radiation curve.

The laws of thermal radiation Black body radiation curves are obtained by measuring the intensity of the thermal radiation from a black body at different constant temperatures. Each curve has a peak which is higher and at shorter wavelength than the curves at lower temperatures. The following two laws of thermal radiation were obtained by analysing the black body radiation curves.

Wiens law The wavelength at peak intensity, OP, is inversely proportional to the absolute temperature T of the object, in accordance with the following equation known as Wiens law:

OmaxT = 0.0029 m K Therefore, if Omax for a given star is measured from its spectrum, the above equation can be used to calculate the absolute temperature T of the light-emitting outer layer, the photosphere, of the star. The photosphere is sometimes referred to as the surface of a star.

Notice that the unit symbol m K stands for metre kelvin not milli kelvin!

Worked example The peak intensity of thermal radiation from the Sun is at a wavelength of 500 nm.

Calculate the surface temperature of the Sun.

Solution Rearranging Omax T = 0.0029 m K gives

m10500Km0029.0

9u T = 5800 K

Stefans law The total energy per second, P, emitted by a black body at absolute temperature T is proportional to its surface area A and to T4, in accordance with the following equation known as Stefans law

P = VAT4 where V is the Stefan constant which has a value of 5.67 108 W m2 K4. In effect, P is the power output of the star and is sometimes referred to as the luminosity of the star. Therefore, if the absolute temperature T of a star and its power output P are known, the surface area A and the radius R of the star can be calculated.

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Worked example V = 5.67 108 W m2 K4 A star has a power output of 6.0 1028 W and a surface temperature of 3400 K.

a Show that it surface area is 7.9 1021 m2 b Calculate:

i its radius ii the ratio of its radius to the radius of the Sun.

radius of Sun = 7.0 108 m

Solution a Rearranging P = VAT4 gives 4T

PAV

Hence 221

48

28m109.7

34001067.5100.6 u uu

u A

b i For a sphere of radius R, its surface area A = 4SR2

Rearranging this equation gives 22021

2 m103.64109.7

4u u AR

Hence R = 2.5 1010 m

ii Ratio of radius to Suns radius 36m100.7m105.2

8

10

uu

Note Two stars that have the same absolute magnitude have the same power output. For two such stars X and Y:

power output of X = VAXTX4, where AX = surface area of X and TX = surface temperature of X power output of Y = VAYTY4, where AY = surface area of Y and TY = surface temperature of Y For equal power output, VAXTX4 = VAYTY4

Hence 4X

4Y

Y

X

TT

AA

Therefore, if their surface temperatures are equal, they must have the same radius. If their surface temperatures are unequal, the cooler star must have a bigger radius than the hotter star.

Nerojaan Chandran

Nerojaan Chandran

Nerojaan Chandran

Nerojaan Chandran

Nerojaan Chandran


Astrophysics

Stellar spectral classes The spectrum of light from a star is used to classify it as shown in Table 1. When the scheme was first introduced, stars were classified on an alphabetical scale A, B, C etc according to colour. The scale was re-ordered later according to surface temperature when the surface temperatures were first measured. As shown in Table 1, the main spectral classes in order of decreasing temperature are O, B, A, F, G, K and M. Spectral Class

Intrinsic Colour Temperature (K) Prominent Absorption Lines

O blue 25 00050 000 He+, He, H B blue 11 00025 000 He, H A blue-white 750011 000 H (strongest), ionised metals F white 60007500 ionised metals G yellow-white 50006000 ionised & neutral metals K orange 35005000 neutral metals M red | 25003500 neutral atoms, TiO Table 1 Characteristics of the main spectral classes

Figure 2 Star classification The spectrum of light from a star contains absorption lines due to a corona or atmosphere of hot gases surrounding the star above its photosphere. The photosphere emits a continuous spectrum of light as explained earlier. Atoms, ions and molecules in these hot gases absorb light photons of certain wavelengths. The light that passes through these hot gases is therefore deficient in these wavelengths and its spectrum therefore contains absorption lines.

The wavelengths of the absorption lines are characteristic of the elements in the corona of hot gases surrounding a star. By comparing the wavelengths of these absorption lines with the known absorption spectra for different elements, the elements present in the star can be identified. The last column in Table 1 shows how the elements present in a star differ according to the spectral class of the star.


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Figure 3 Excitation in the hydrogen atom Since the absorption lines vary according to temperature, they can therefore be used in addition to temperature to determine the spectral class of the star. Note that the hydrogen absorption lines correspond to excitation of hydrogen atoms from the n = 2 state to higher energy levels. These lines, referred to as the Balmer lines, are only visible in the spectra of O, B and A class stars as other stars are not hot enough for excitation of hydrogen atoms due to collisions to the n = 2 state. In other words, hydrogen atoms in the n = 2 state exist in hot stars (i.e. O, B and A class stars); such atoms can absorb visible photons at certain wavelengths hence producing absorption lines in the continuous spectrum of light from the photosphere.

Figure 4 The origin of the Balmer lines Note that hydrogen atoms in the n = 1 state (the ground state) do not absorb visible photons because visible photons do not have sufficient energy to cause excitation from n = 1.

Dont forget temperature in Wiens law and Stefans law is always in kelvin.


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Summary questions Wiens law constant = 0.0029 m K, V = 5.67 108 W m2 K4 1 With the aid of a diagram, explain what is meant by a black body spectrum and describe how such a

spectrum from a star is used to determine the temperature of the stars light-emitting surface.

2 a State the main spectral classes of a star and the approximate temperature range of each class. b The spectrum of light from a star has its peak intensity at a wavelength of 620 nm. Calculate the

temperature of the stars light-emitting surface. 3 A star has a surface temperature which is twice that of the Sun and a diameter that is four times as large

as the Suns diameter. Show that it emits approximately 250 times as much energy per second as the Sun.

4 Two stars X and Y are in the same spectral class. Star X emits 100 times more power that star Y. a State and explain which star, X or Y, has the bigger diameter. b X has a power output of 6.0 1026 W and a surface temperature of 5400 K. Show that its surface

area is 1.2 1019 m2 and calculate its diameter.


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2.3 The HertzsprungRussell diagram Learning objectives: What does the colour of a star tell us about the

star? How do stars form? Why do we think the Sun will eventually become

a white dwarf star?

The power of the Sun The intensity of solar radiation at the Earth is about 1400 W m2. This means that a solar panel (area = 1 m2) facing the Sun directly will receive 1400 J of solar energy per second. In practice, absorption due to the atmosphere occurs and there is also some reflection. So how much radiation energy does the Sun emit each second? The mean distance from the Earth to the Sun is 1 AU which is 1.5 1011 m.

Figure 1 Solar radiation Imagine the Sun at the centre of a sphere of radius 1.5 1011 m. Each square metre of surface of this sphere will receive 1400 J of solar energy per second.

The total amount of solar energy per second received by the sphere surface must be 1400 J s1 per square metre the surface area of the sphere. This must be equal to the amount of solar energy per second emitted by the Sun (its luminosity or power output) as no solar radiation is absorbed between the Sun and the spheres surface.

Since the surface area of a sphere of radius r is equal to 4Sr2, the power output of the Sun is therefore 4.0 1026 J s1 (= 1400 J s1 m2 4S (1.5 1011 m)2).

Dwarfs and giants Topic 2.2 looked at how the spectrum of a star can be used to find the surface temperature of the star and its spectral class. It also looked at how the output power of a star can be calculated if the surface temperature and diameter are known. However, star diameters except for the Sun cannot be measured directly and are determined by comparing the absolute magnitude of the star with that of the Sun which is 4.8.

For example, a G class star which has an absolute magnitude of 0.2 is five magnitudes more powerful than the Sun and is therefore 100 times more powerful. Therefore, its power output is

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Astrophysics

4.0 1028 J s1 (= 100 the power output of the Sun). Substituting this value of power output and the stars surface temperature into Stefans law therefore enables its surface area and diameter to be calculated.

A dwarf star is a star that is much smaller in diameter than the Sun. A giant star is a star that is much larger in diameter than the Sun. Stefans law gives the power output across the entire spectrum, not just across the visible spectrum. Absolute and apparent magnitudes relate to the visible spectrum. For a star that emits a significant fraction of its radiation in the non-visible spectrum, magnitude values that take account of non-visible radiation would need to be used. Such modifications are not part of the specification.

Worked example V = 5.67 u 108 W m2 K4 A K-class star has a power output of 4.0 1028 J s1 and a surface temperature of 4000 K.

a Calculate: i its surface area ii its diameter.

b The diameter of the Sun is 1.4 109 m. State whether it is a giant star or a dwarf star or neither. Solution a i Rearranging P = VAT4 gives 4T

PAV

Hence 221

48

28m108.2

40001067.5100.4 u uu

u A

ii For a sphere of radius R, its surface area A = 4SR2

Rearranging this equation gives 22021

2 m102.24108.2

4u u AR

Hence R = 1.5 1010 m so its diameter = 2R = 3.0 1010 m b The star is 21 times the diameter of the Sun and so it is a giant star.

A ready-reckoner To compare a star X with the Sun,

power output of X, PX = VAXTX4, where AX = surface area of X and TX = surface temperature of X

power output of the Sun, PS = VASTS4, where AS = surface area of the Sun and TS = its surface temperature.

Therefore, 4

S

X

S

X4

SS

4XX

S

X

Sun, theofoutput power X, ofoutput power

u T

TAA

TATA

PP

VV


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Rearranging this gives 4

S

X

S

X

S

X

y T

TPP

AA = (power output ratio) y (temperature ratio)4

For example, if the power ratio is 100 and the temperature ratio is 0.7, using the above expression gives 420 (to 2 significant figures) for the area ratio. So the diameter ratio is 420 as the area ratio is equal to the diameter ratio squared. So, the diameter of X is 20 times the diameter of the Sun.

Note X is 100 times more powerful than the Sun so its absolute magnitude = MS 5 where MS is the absolute magnitude of the Sun. Each magnitude difference of 1 corresponds to a power ratio of 1001/5 which is equal to 2.5.

In general, for two stars:

with the same surface temperature and unequal absolute magnitudes, the one with the greater power output has the larger surface area and hence diameter

with the same absolute magnitude and unequal surface temperatures, the hotter star has a smaller surface area and hence a smaller diameter.

The HertzsprungRussell diagram Stars of known absolute magnitude and known surface temperature can be plotted on a chart in which the absolute magnitude is plotted on the y-axis and temperature on the x-axis as shown in Figure 2. This was first undertaken independently by Enjar Hertzsprung in Denmark and Henry Russell in America. The chart is known as a HertzsprungRussell (or HR) diagram.

Figure 2 The HertzsprungRussell diagram The main features of the HR diagram are as follows:

The main sequence, a heavily-populated diagonal belt of stars ranging from cool low-power stars of absolute magnitude +15 to very hot high-power stars of absolute magnitude about 5. The greater the mass of a star, the higher up the main sequence it lies. Star masses on the main sequence vary from about 0.1 to 30 or more times the mass of the Sun.

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Astrophsics Option A

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