Download Chapter 1 Telescopes 1.1 Lenses

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.
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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
2
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.
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
2
3
Ray 1 is drawn parallel to the principal axis before the lens so it is refracted by the lens
through F.
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.
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
beyond 2F
2F
between F and 2F
<F
between F and 2F
2F
beyond 2F
same side as object
Nature
of
image
real
real
real
virtual
Magnified or
diminished
Upright or
inverted
Application
diminished
same size
magnified
magnified
inverted
inverted
inverted
upright
camera
inverter
projector
magnifying
lens
Table 1 Image formation by a converging lens
Note
The linear magnification of the image =
height of the image
height of the object
It can be shown that this ratio is equal to
the image distance
the object distance
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!
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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
1 1 1
 
u v
f
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
Hence
1
1
1
1 1 1
 
 
gives
0.200 v 0.150
u v
f
1
1
1


= 6.67 − 5.00 = 1.67
v 0.150 0.200
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 =
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height of the image
image distance
=
height of the object
object distance
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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 viewer’s 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 viewer’s 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 1, the angle subtended by the telescope image to the eye would be 3. 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 = 
 The angle subtended by the distant object to the unaided eye = 
The angular magnification of the telescope in normal adjustment =
angle subtended by the final image at infinity to the viewer 

angle subtended by the distant object to the unaided eye

Figure 5 Angular magnification
h
h1
and tan  = 1 , where h1 is the
fo
fe
height of the real image and fo and fe are the focal lengths of the objective and eyepiece lenses
respectively.
From the inset diagram in Figure 5, it can be seen that tan  =
h1
fe fo
tan 
Combining these two equations to eliminate h1 gives


h
tan
fe
1
fo
Assuming angles  and  are always less than about 10, applying the small angle approximation
tan  =  in radians and tan  =  in radians gives
Therefore:
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 fo

 fe
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the angular magnification of a telescope in normal adjustment =
fo
fe
Notes
1 The height h1 of the real image = fo tan  = fo × ( in radians)
Remember 360 = 2 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 0.40. 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 =
f o 0.840

 7.0
f e 0.120
ii Distance between the lenses = fo + fe = 0.840 + 0.120 = 0.960 m
b i
angular magnification =

where  = 0.40

Therefore  =  × angular magnification = 7 = 2.8
ii h1 = fo tan  = 0.840  tan 0.40 = 5.9 × 10−3 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
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of the diameter. Therefore, in comparison with the unaided eye, a telescope with an objective
lens:
  60  2 
 of diameter 60 mm would collect 36 times      more light per second from a star
  10  


  120  2 
 of diameter 120 mm would collect 144 times   
  more light per second from a star.
  10  


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 viewer’s 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 0.15 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 4.8 to the
viewer. Calculate:
i the angle subtended by the tower to the viewer’s 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
2
3
reflected by the concave mirror, then
reflected by the convex mirror onto the small hole at the centre of the concave mirror into the
eyepiece, then
refracted by the eyepiece into a parallel beam which enters the viewer’s 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 topheavy.
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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Astrophysics
For light of wavelength  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
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  for the first
dark ring is approximately 6.3 × 10−6 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  for the resolution of two stars is given approximately by the
condition:


D
where  = the wavelength of light, and D = the diameter of the circular aperture.
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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 don’t forget to convert angles to radians if
their values are wanted in radians or given in degrees.
Remember: 2 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 × 10−5 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.
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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 × 10−6 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 × 10−5 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.
Earth–Moon distance = 3.8 × 108 m
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Astrophysics
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 lightsensitive 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
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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 Earth’s 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

20
, where  is the
wavelength of the radio waves.
The dish diameter, D, determines
 the collecting area of the dish (=
1
4
D2)
 the resolving power of the telescope (=

)
D
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
(= 14 of 0.2) 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 proton’s 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
Earth’s 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 Earth’s 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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Astrophysics
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
optical
ground or
satellite
radio
ground
Wavelength
range
350–650 nm
Resolution
(degrees)
−5
10 for HST
1 mm to 10 m
0.2 for
Lovell
infrared
Key advantages
Major disadvantages
gives very detailed
images,
can detect distant
galaxies
radio waves pass
through dust in
space and through
the atmosphere
can detect warm
objects that do not
emit light,
can detect dust
clouds in space
ground telescopes
suffer from
atmospheric refraction
ultraviolet
X and
gamma
0.2 for
INTEGRAL
large, supporting
structure needed for a
steerable dish.
mirror needs to be
cooled
must be above the
Earth’s atmosphere
e.g. on a satellite
must be above the
Earth’s 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.
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Astrophysics
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 s−1 × 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 Earth’s orbit in this time.
By measuring the angular shift of a star’s 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 Earth’s 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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Astrophysics
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 star’s position relative to stars much further away.
Figure 2 Parallax angle
The parallax angle  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 star’s line of sight
over six months.
 From the triangle consisting of the three lines between the Sun, the star and the Earth as
R
shown in Figure 2, tan  .
d
R
 Since  is always less than 10°, using the small angle approximation gives   , where 
d
R
is in radians. So, d  . Note that 360° = 2 radians.

1 degree
. For this
3600
reason, star distances are usually expressed for convenience in terms of a related non-SI unit
called the parsec (abbreviated as pc).
Parallax angles are generally measured in arc seconds where 1 arc second =
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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Astrophysics
Since 1 arc second =
equation d 
R

1 degree
= 4.85 × 10−6 radians and 1 AU = 1.496 × 1011 m, using the
3600
gives:

1.496  1011 m 
 = 3.26 light years
 1 parsec = 3.08 × 1016 m  
6

4
.
85

10
radians


 The distance, in parsecs, from a star to the Sun =
1

, where  is the parallax angle of the star
angle, in arc seconds
d (in parsecs) 
1
 (in arc seconds)
The smaller the parallax angle of a star, the further away the star is. For example:
  = 1.00 arc second, d = 1.00 pc
  = 0.50 arc seconds, d = 2.00 pc
  = 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 star’s 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 Earth’s surface is about 1400 W m−2. 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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Astrophysics
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
I
than from Y. Generalising this rule gives X  100
IY
m
5
, where m = mY − mX
Taking base 10 logs of this equation gives:
m


I 
log X   log100 5   log 100 0.2 m  0.2m log100  0.4m


 IY 




I
Multiplying both sides of the equation by 2.5 gives 2.5 log X
 IY

  m

Hence
I 
mY − mX = 2.5 log  X 
 IY 
The absolute magnitude, M, of a star is defined as the star’s 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:
d 
m − M = 5 log  
 10 
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  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
2
I
d
Earth, the ratio of their received intensities X would be   .
 10 
IY
Therefore, the difference between their apparent magnitudes,
mY − mX = 2.5 log
IX
IY
d
= 2.5 log  
 10 
2
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Astrophysics
d
= 5 log  
 10 
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
d
So, mY − MY = 5 log  
 10 
More generally, for any star at distance d, in parsecs, from the Earth:
d
m − M = 5 log  
 10 
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
 d  gives M = m −5 log  d 

 
 10 
 10 
Rearranging m − M = 5 log 
Hence M = 6.0 − 5 log
80
= 6.0 − 5 log 8 = 1.5 (= 1.48 to 3 significant figures)
10
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.
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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.
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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.
Wien’s law
The wavelength at peak intensity, P, is inversely proportional to the absolute temperature T of
the object, in accordance with the following equation known as Wien’s law:
maxT = 0.0029 m K
Therefore, if max 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 max T = 0.0029 m K gives T 
0.0029 m K
500  10 9 m
= 5800 K
Stefan’s 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 Stefan’s law
P = AT4
where  is the Stefan constant which has a value of 5.67 × 10−8 W m−2 K−4. 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
 = 5.67 × 10−8 W m−2 K−4
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 = AT4 gives A 
Hence A 
b i
P
T 4
6.0 10 28
5.67 10 8  34004
 7.9 10 21 m 2
For a sphere of radius R, its surface area A = 4R2
Rearranging this equation gives R 2 
A 7.9 10 21

 6.3  10 20 m 2
4π
4π
Hence R = 2.5 × 1010 m
ii Ratio of radius to Sun’s radius 
2.5 1010 m
7.0 108 m
 36
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 = AXTX4, where AX = surface area of X and TX = surface temperature of X
 power output of Y = AYTY4, where AY = surface area of Y and TY = surface temperature of Y
For equal power output, AXTX4 = AYTY4
4
Hence
AX TY

AY TX 4
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.
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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
O
B
A
F
G
K
M
Intrinsic Colour
Temperature (K)
Prominent Absorption Lines
blue
blue
blue-white
white
yellow-white
orange
red
25 000–50 000
11 000–25 000
7500–11 000
6000–7500
5000–6000
3500–5000
 2500–3500
He+, He, H
He, H
H (strongest), ionised metals
ionised metals
ionised & neutral metals
neutral metals
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.
Don’t forget temperature in Wien’s law and Stefan’s law is
always in kelvin.
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Summary questions
Wien’s law constant = 0.0029 m K,  = 5.67 × 10−8 W m−2 K−4
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 star’s 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 star’s 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 Sun’s 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 Hertzsprung–Russell 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 m−2. 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 s−1 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 sphere’s surface.
Since the surface area of a sphere of radius r is equal to 4r2, the power output of the Sun is
therefore 4.0 × 1026 J s−1 (= 1400 J s−1 m−2 × 4 × (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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4.0 × 1028 J s−1 (= 100 × the power output of the Sun). Substituting this value of power output and
the star’s surface temperature into Stefan’s 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.
Stefan’s 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
 = 5.67  10−8 W m−2 K−4
A K-class star has a power output of 4.0 × 1028 J s−1 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 = AT4 gives A 
Hence A 
4.0 10 28
5.67 10 8  40004
P
T 4
 2.8 10 21 m 2
ii For a sphere of radius R, its surface area A = 4R2
Rearranging this equation gives R 2 
A 2.8 10 21

 2.2 10 20 m 2
4π
4π
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 = AXTX4, where AX = surface area of X and TX = surface temperature
of X
 power output of the Sun, PS = ASTS4, where AS = surface area of the Sun and TS = its surface
temperature.
power output of X, PX
A T
A
Therefore,
 X X4  X
power output of the Sun, PS ASTS
AS
4
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T 
  X 
 TS 
4
Astrophysics
4
A
P T 
Rearranging this gives X  X   X  = (power output ratio)  (temperature ratio)4
AS PS  TS 
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 420½ 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 Hertzsprung–Russell 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 Hertzsprung–Russell (or HR) diagram.
Figure 2 The Hertzsprung–Russell 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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 Giant stars have absolute magnitudes in the range of about +2 to −2 so they emit more
power than the Sun and are 10 to 100 times larger. Red giants are cooler than the Sun.
 Supergiant stars have absolute magnitudes in the range from about −5 to −10 and are much
brighter and larger than giant stars. They have diameters up to 1000 times that of the Sun.
They are relatively rare compared with giant stars.
 White dwarf stars have absolute magnitudes between +15 and +10 and are hotter than the
Sun but they emit much less power. They are much smaller in diameter than the Sun.
Worked example
A red giant and a main sequence star have the same absolute magnitude of 0. Their surface temperatures
are 3000 K and 15 000 K respectively. Show that the radius of the red giant is 25 times larger than that
of the main sequence star.
Solution
Power output of a star, P = AT4, where A = its surface area and T = its surface temperature.
The two stars have the same power output as they have the same absolute magnitude.
Therefore, AT4 for the red giant = AT4 for the main sequence star.
Cancelling  on both sides of this equation and rearranging gives
ARG
AMS

4
T MS
4
T RG
4
 15000 
4

  5  625
 3000 
Since the surface area A = 4(radius)2, the radius of the red giant is therefore 25 times (= 625 ½) the
radius of the main sequence star.
Stellar evolution
The Sun is a middle-aged star about 4600 million years old. It produces energy as a result of
nuclear fusion in its core converting hydrogen into helium. The core temperature must be of the
order of millions of kelvin to maintain fusion. The fusion reactions release energy which
maintains the core temperature. Radiation from the core heats the outer layers of the Sun causing
light to be emitted from its surface (photosphere).
Main sequence stars like the Sun are in a state of internal equilibrium in the sense that
gravitational attraction acting inwards is balanced by radiation pressure due to the outflow of
gases which expand and cool. The star will move from its position on the main sequence and
become a red giant star.
All stars evolve through a sequence of stages from their formation to the main sequence stage and
beyond.
Formation
A star is formed as dust and gas clouds in space contract under their own gravitational attraction
becoming denser and denser to form a protostar (a star in the making).
 In the collapse, gravitational potential energy is transformed into thermal energy as the atoms
and molecules in the clouds gain kinetic energy so the interior of the collapsing matter
becomes hotter and hotter.
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 If sufficient matter accretes to form the protostar, the temperature at the core of the protostar
becomes high enough for nuclear fusion to occur. If there is insufficient matter, the star does
not become hot enough for nuclear fusion to occur and it gradually cools once it has stopped
contracting.
 Energy released as a result of nuclear fusion of hydrogen to form helium increases the core
temperature so fusion reactions continue to occur as long as there are sufficient light nuclei.
As a result of continuing fusion reactions, the outer layers of the protostar become hot and a
light-emitting layer (the photosphere) is formed and the protostar becomes a star.
Main sequence
The newly-formed star reaches internal equilibrium as the inward gravitational attraction is
balanced by the outward radiation pressure. The star therefore becomes stable with constant
luminosity.
 Its absolute magnitude depends on its mass; the more mass it has, the greater its luminosity so
it joins the main sequence at a position according to its mass.
 The star remains at this position for most of its lifetime, emitting light as a result of
‘hydrogen burning’ in its core.
Application
Cepheid variables
Most stars have constant luminosity. In other words, their power output is constant and their
brightness does not vary. Some stars do vary in their luminosity because they pulsate. Cepheid
variables pulsate with a period of the order of days that depends on their average luminosity. By
measuring the period of all ‘nearby’ Cepheid variables at known distances (and therefore known
absolute magnitudes), the absolute magnitude of and hence distance to any other Cepheid variable
(e.g. in a distant galaxy) can be determined by measuring its period. Prove for yourself that the
Cepheid variable represented in Figure 3 is about 280 parsecs from the Sun. The graph in (b)
shows the relationship for metal-rich Cepheids (group 1) and metal-poor Cepheids (group 2). The
absolute magnitude for the period of the Cepheid variable represented in (a) is shown by the red
dashed line.
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Figure 3 Using Cepheid variables
Red giants
Once most of the hydrogen in the core of the star has been converted to helium, the core collapses
on itself and the outer layers of the star expand and cool as a result. The star swells out and moves
away from its position on the main sequence to become a giant or a supergiant star.
 The temperature of the helium core increases as it collapses and causes surrounding hydrogen
to form a ‘hydrogen-burning’ shell which heats the core further.
 When the core temperature reaches about 108 K, helium nuclei in the core undergo fusion
reactions in which heavier nuclei are formed, principally beryllium, carbon and oxygen. The
luminosity of the star increases and the wavelength at peak intensity increases because it
becomes cooler.
 The red giant stage lasts about a fifth of the duration of the main sequence stage. The
evolution of a star after the red giant stage follows one of two paths according to its mass.
Below a mass of about 8 solar masses, a red giant star sooner or later becomes a white dwarf.
A star of higher mass swells out even further to become a supergiant which explodes
catastrophically as a supernova.
White dwarfs
When nuclear fusion in the core of a giant star ceases, the star cools and its core contracts,
causing the outer layers of the star to be thrown off.
 The outer layers are thrown off as shells of hot gas and dust which form so-called planetary
nebulae around the star. This happens through several mechanisms including ionisation in the
star’s outer layers as the layers cool causing the layers to trap radiation energy which
suddenly breaks out.
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 If the mass of the red giant star is between 4 and 8 solar masses, the core becomes hot enough
to cause energy release, through further nuclear fusion, to form nuclei as heavy as iron in
successive shells. The process stops when the fuel (i.e. the light nuclei) has all been used up.
After throwing off its outer layers, the star is now little more than its core which at this stage is
white hot due to release of gravitational energy. If its mass is less than 1.4 solar masses, the
contraction of the core stops as the electrons in the core can no longer be forced any closer. The
star is now stable and has become a white dwarf which will gradually cool as it radiates its
thermal energy into space and eventually becomes invisible. In the next topic, we will see that if
its mass at this stage is greater than 1.4 solar masses, it does not form a white dwarf. Instead, it
explodes catastrophically as a supernova.
Application
The future of the Sun
From what is known about the stars, astronomers have predicted the evolutionary path of the Sun.
In about 5000 million years time, the Sun will become a red giant and swell out as far as the Earth.
Its increased luminosity will evaporate Mars and blaze away the gaseous atmospheres of the
planets beyond Mars. After the red giant stage which will last about 1 to 2 billion years, the Sun
will throw off most of its mass into space and evolve into a white dwarf not much wider than the
Earth and about ten times dimmer than at present. Over the next few billion years, it will become
fainter and fainter and gradually fade away.
Figure 4 Evolution of the Sun
Summary questions
1 a Sketch a Hertzsprung–Russell diagram to show the full range of main sequence, giant and
supergiant stars and white dwarfs. Show the relevant scales on each axis.
b Show on your diagram the present position of the Sun and its evolutionary path after it leaves the
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main sequence.
2 a Describe the formation of a star from gas and dust clouds.
b A protostar first becomes visible as a very dim cool star then moves onto a fixed position on the
main sequence.
i Indicate on your HR diagram the position where the protostar first becomes visible.
ii What physical property of a newly formed star determines its position on the main sequence?
3 When a certain red giant star evolves into a white dwarf, it becomes very hot without loss of brightness,
then it becomes fainter and it cools before stabilising as a white dwarf.
a Indicate on your HR diagram the evolutionary path of this star after the red giant stage.
b State the defining characteristics of a white dwarf star and list two other properties it possesses.
4 Three stars X, Y and Z have surface temperatures of 4000 K, 8000 K and 20 000 K respectively and
absolute magnitude −2, +4 and +10 respectively.
a List the stars in order of increasing power output.
b State the evolutionary stage of each star, giving your reason for each statement.
c Calculate the ratio of the diameter of X and of Y relative to Z.
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Astrophysics
2.4 Supernovae, neutron stars and black holes
Learning objectives:



Why is a supernova called a supernova?
What is a neutron star?
How is a black hole formed?
The death of a high-mass star
As discussed in the previous topic, when nuclear fusion ceases in the core of a red giant star, the
outer layers of the star are thrown off and, if mass of the core and remaining matter is less than
1.4 solar masses, the star stabilises as a white dwarf. The repulsive force between the electrons in
the core pushing outwards counterbalances the gravitational force pulling the core inwards.
Nuclear fusion ceases when there are no longer any nuclei in the core that release energy when
fused. This happens when iron nuclei are formed by fusion as they are more stable than any other
nuclei so cannot fuse to become even more stable.
If the core mass exceeds 1.4 solar masses, the electrons in the iron core can no longer prevent
further collapse as they are forced to react with protons to form neutrons. The equation for this
reaction is:
p + e −  n + e
The sudden collapse of the core makes the core more and more dense until the neutrons can no
longer be forced any closer. The core density is then about the same as the density of atomic
nuclei, about 1017 kg m−3. The core suddenly becomes rigid and the collapsing matter surrounding
the core hits it and rebounds as a shock wave propelling the surrounding matter outwards into
space in a cataclysmic explosion. The exploding star releases so much energy that it can outshine
its host galaxy. The event is referred to as a supernova as it is much brighter than a nova or
‘new’ star in the same galaxy.
Figure 1 The Crab Nebula.
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Figure 1 shows a supernova remnant of a star that exploded in AD 1054, about 2000 parsecs
away. It is now about 2–3 parsecs in width.
How science works
Novae and supernovae
A nova is a star that suddenly becomes brighter, often having been too dim to be visible so it
appears as a ‘new’ star. This can happen if a white dwarf draws matter from an invisible
companion star and suddenly overheats and expels the excess matter as a result. Supernovae are,
as described above, exploding stars that scatter much of their matter into space. Although
astronomers discover many supernovae in distant galaxies every year using large telescopes and
other detectors, supernovae visible to the unaided eye are rare. The last such supernova seen in the
Milky Way occurred in 1604. A supernova observed in 1987 in a nearby galaxy became visible to
the unaided eye and has been studied extensively since.
A supernova is typically a thousand million times more luminous than the Sun. Its absolute
magnitude is therefore between −15 and −20. In comparison, the absolute magnitude of the Sun is
+4.8. This increase of luminosity occurs within about 24 hours. Measurements of their
subsequent luminosity show a gradual decrease on a time scale of the order of years. Thus the
tell-tale sign of a supernova is a sudden and very large increase in luminosity of the star
corresponding to a change of about 20 magnitudes in its absolute magnitude.
A supernova explosion throws the matter surrounding the core into space at high speeds.
Elements heavier than iron are formed by nuclear fusion in a supernova explosion. Such fusion
reactions occur as the shock wave travels through the layers of matter surrounding the neutronfilled core. The supernova explosion scatters the matter surrounding the core into space. Thus the
supernova remnants in space contain all the naturally occurring elements. Note that helium is
formed from hydrogen in fusion reactions in main sequence stars. Other elements as heavy as iron
are formed progressively in fusion reactions in red giant stars. As explained earlier, elements
heavier than iron cannot be formed in main sequence and red giant stars. Their existence in the
Earth tells us that the Solar System formed from the remnants of a supernova.
A supernova explosion also causes an intense outflow of neutrinos and gamma photons.
Neutrinos from supernova 1987A were detected three hours before light was detected from it.
The light seen from the explosion was produced when the shock wave hit the outer layers of the
star. In contrast, the neutrinos produced by nuclear fusion as the shock wave made its way
through the interior travelled much faster than the shock wave, reaching the surface hours before
the shock wave.
More about supernovae
Supernovae are classified into several types according to their line absorption spectra.
 Type I supernovae have no strong hydrogen lines present and are further subdivided into
three groups.
 Type Ia supernovae show a strong absorption line due to silicon. They rapidly reach peak
luminosity of about 109 times the Sun’s luminosity then decrease smoothly and gradually.
They are thought to occur when a white dwarf star in a binary system attracts matter from
a companion giant star causing fusion reactions to restart in which carbon nuclei form
silicon nuclei. The fusion process becomes unstoppable as further matter is drawn from
the giant star and the white dwarf explodes.
 Type Ib supernovae show a strong absorption line due to helium; these are thought to
occur when a supergiant star without hydrogen in its outer layers collapses. After reaching
peak luminosity, their light output decreases steadily and gradually.
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 Type Ic supernovae lack the strong lines present in types Ia and Ib; these are thought to
occur when a supergiant without hydrogen or helium in its outer layers collapses. After
reaching peak luminosity, their light output also decreases steadily and gradually.
 Type II supernovae have strong hydrogen lines; these are thought to occur when a supergiant
which has retained the hydrogen or helium in outer layers collapses. Their peak luminosity is
not as high as type Ia supernovae and their light output decreases gradually but unsteadily.
Table 1 summarises the characteristics of the different types of supernovae.
Type
Ia
Ib
Ic
II
Spectrum
no hydrogen lines; strong silicon
line
no hydrogen lines; strong
helium line
no hydrogen; no helium lines
strong hydrogen and helium
lines
Light output
decreases steadily
Origin
white dwarf attracts matter and explodes
decreases steadily
supergiant collapses then explodes
decreases steadily
decreases unsteadily
supergiant collapses then explodes
supergiant collapses then explodes
Table 2 Types of supernova
Type Ia supernovae reach a known peak luminosity and are characterised by the presence of a
strong silicon absorption line, so they are used to find the distance to their host galaxy. A
supernova can temporarily outshine its host galaxy, so the detection of a type Ia supernova in a
galaxy at unknown distance enables the distance to the galaxy to be found. This method of
measuring distances to distant galaxies has led to the prediction of a new form of energy referred
to as dark energy.
Neutron stars and black holes
A neutron star is the core of a supernova after all the surrounding matter has been thrown off
into space. A neutron star is extremely small in size compared with a star such as the Sun. If its
mass was the same as that of the Sun:
 its diameter would be about 30 km
 its surface gravity would be over two thousand million times stronger than at the surface of
the Sun.
The first evidence for neutron stars came with the discovery in 1967 of pulsating radio stars or
pulsars. The radio pulses are at frequencies of up to about 30 Hz. A typical pulsar is less than
100 km in diameter and has a mass of about two solar masses. From their observations including
the discovery of extremely strong magnetic fields in pulsars, astronomers deduced that pulsars are
rapidly rotating neutrons stars that produce beams of radio waves. These sweep round the sky as
the neutron star rotates like the light beam from a lighthouse.
Application
What causes the radio beams from a pulsar?
Each time the beam sweeps over the Earth we receive a pulse of radio waves. The radio beams are
thought to be generated by charged particles spiralling in the intense magnetic field above the
magnetic poles of the neutron star. The magnetic axis and the rotation axis are different, so the
radio beams sweep round as the star spins about its rotation axis.
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Figure 2 Radio waves from a pulsar
A black hole is an object so dense that not even light can escape from it. A supernova core
contains neutrons only but if its mass is greater than about three solar masses, the neutrons are
unable to withstand the immense forces pushing them together. The core collapses on itself and
becomes so dense that not even light can escape from it. The object is then a black hole. It can’t
emit any photons and it absorbs any photons that are incident on it.
The event horizon of a black hole is a sphere surrounding the black hole from which nothing can
ever emerge. The radius of this sphere is called the Schwarzschild radius, RS, of the black hole.
Einstein’s general theory of relativity gives the following equation for the Schwarzschild radius
of a black hole of mass M
RS 
2GM
c2
where G is the universal constant of gravitation and c is the speed of light in free space.
What happens inside a black hole can not be observed. A black hole attracts and traps any
surrounding matter, increasing its mass as a result. Matter falling towards a black hole radiates
energy until it falls within the event horizon. Inside the black hole, matter is drawn with everincreasing density towards a singularity at its centre, a point where the laws of physics as we
know them may not apply.
The key characteristic of a black hole is its mass. It may also be charged and it may or may not be
rotating. Matter that falls into a black hole contributes its mass, its charge if any and its rotational
motion if any to the black hole. Any other property carried by infalling matter is lost. For
example, the properties of a black hole are unaffected by the chemical elements in the matter
dragged into the black hole. This information about the infalling matter is lost in the black hole.
Evidence for black holes
Evidence for black holes formed from collapsed neutron stars was found in 1971 using the first
satellite-mounted X-ray telescope. The satellite pinpointed an X-ray source, labelled Cygnus X-1,
in the same location as a supergiant star 2500 parsecs away. The intensity of the X-rays varied
irregularly on a time scale of the order of 0.01 seconds, indicating a source diameter of the order
of 3000 km (= speed of light × 0.01 s) which is smaller than the Earth. When the position of the
supergiant was found to vary slightly, it was realised the supergiant and the X-ray source must be
orbiting each other as a binary system. The mass of the X-ray source was estimated at about
7 solar masses or a quarter of the mass of the supergiant. Its mass is above the upper limit of
3 solar masses for a neutron star, so astronomers think that Cygnus X-1 is a black hole which
attracts matter from the supergiant. As the matter falls towards the black hole, it becomes so hot
that it emits X-rays.
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Figure 3 Evidence for a black hole
Further similar evidence for black holes has been found from several other X-ray sources. These
findings indicate that black holes may form in binary systems where one of the stars explodes as a
supernova, leaving a core of mass greater than about three solar masses that collapsed to become
a black hole. The other star may not have reached or gone beyond the giant stage as in the above
examples.
A further possibility is that a white dwarf or a neutron star might have pulled matter off a binary
companion star and turned into a black hole when its mass exceeded three solar masses or both
stars in a binary system might have become neutron stars and merged to become a black hole.
Supermassive black holes
Supermassive black holes of almost unimaginable mass are thought to exist at the centre of many
galaxies. At the centre of a galaxy, stars are much closer together than they are at the edges of the
galaxy. A supermassive black hole at the centre could pull millions of millions of stars in. Such
black holes can therefore gain enormous quantities of matter and are referred to as supermassive
black holes. Strong evidence now exists that there are supermassive black holes at the centre of
many galaxies.
 The Andromeda galaxy, M31: detailed observations of the central region of the Andromeda
galaxy, the nearest large galaxy to the Milky Way, show that stars near the galactic centre are
orbiting the centre at speeds of the order of 100 km s−1 at distances of no more than about
5 parsecs from the centre. These stars must therefore be orbiting a central object of diameter
less than 5 parsecs. Applying satellite theory to this object gives a central mass of about
10 million solar masses which is thought to be a supermassive black hole.
 The Milky Way Galaxy: images using infrared radiation and radio waves from the centre of
the Milky Way indicate stars there that are orbiting the galactic centre at speeds of more than
1500 km s−1 orbiting at distances of about 2 parsecs from the galactic centre. This information
indicates a supermassive black hole of mass equal to about 2.6 million solar masses. Strong
evidence has also been found of other local galaxies that have a supermassive black hole at
the centre.
 Distant galaxies have also yielded evidence of a supermassive black hole at each centre. The
Sombrero galaxy, M104, has fast-moving stars in orbits close to its centre, indicating a
supermassive black hole of mass equal to 1000 million solar masses.
Topic 3.3 of these notes will return to the subject of supermassive black holes.
Summary questions
G = 6.67 × 10−11 N m2 kg−2, c = 3.0 × 108 m s−1
1 a What change in a giant star causes its core to collapse?
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b Why does infalling matter rebound when the core of a giant star collapses?
2 a A neutron star is made of neutrons. State two other characteristics of a neutron star.
b Explain why a neutron star has a mass which is:
i more than 1.4 solar masses
ii less than about 3 solar masses.
3 a What astronomical observation indicates that a supernova has occurred?
b What astronomical observation indicates that a particular supernova is due to an explosion of a
white dwarf star rather than the collapse of a red giant star?
4 a i What is a black hole and what are its physical properties?
ii Where should astronomers look to locate a supermassive black hole?
b For a black hole of the same mass as the Sun, which is 2.0 × 1030 kg, calculate:
i its Schwarzschild radius
ii the mean density inside its event horizon.
c By carrying out appropriate calculations, compare the density of a supermassive black hole of mass
10 million solar masses with your answer to b ii.
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Chapter 3 Cosmology
3.1 The Doppler effect
Learning objectives:



Why does the wavelength of waves from a
moving source depend on the speed of the
source?
What is a Doppler shift?
How can we measure the velocity of the two
stars in a binary system?
Doppler shifts
The wavelengths of the light waves from a star moving towards the Earth are shorter than they
would be if the star was stationary. If the star had been moving away from the Earth, the
wavelengths of the light waves from it would be longer than if the star was stationary. This effect
applies to all waves and is known as the Doppler effect. It is the reason why the pitch of a siren
on an approaching emergency vehicle rises as the vehicle approaches then falls sharply as it
passes by. The pitch is higher as the source approaches and lower as it retreats because the source
moves a certain distance each time it emits each cycle of waves.
Consider a source of waves of frequency f moving at speed v. Figure 1 shows wave fronts
1
representing successive wave peaks emitted by the source at time intervals t = .
f
Figure 1 The Doppler effect
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The distance between successive wave peaks is the wavelength of the waves. In time t, each
c
wave peak shown travels a distance ct (= ) before the next wave peak is emitted and the
f
v
source travels a distance vt (= ).
f
 Waves emitted in the opposite direction to the motion of the source (‘behind’ the source) are
‘spaced out’. An observer in the path of these waves would therefore detect waves of longer
wavelength and therefore lower frequency. For light, this shift to longer wavelengths is
referred to as a red shift as it causes the lines of a line spectrum to shift towards the red end
of the visible spectrum.
 Waves emitted in the same direction as the motion of the source are ‘bunched together’ ahead
of the source. An observer in the path of these waves would therefore detect waves of shorter
wavelength and therefore higher frequency. For light, this shift to longer wavelengths is
referred to as a blue shift as it causes the lines of a line spectrum to shift towards the blue end
of the visible spectrum.
It can be shown that for a source moving at speed v relative to an observer,
towards the observer:
 the change of frequency f =
v
f
c
 the change of wavelength  = 
v

c
away from the observer:
v
f
c
v
 the change of wavelength  = 
c
 the change of frequency f = 
f

(or
).
f

Mathematically, red shifts and blue shifts are fractional changes in frequency or wavelength.
The Doppler shift, z, in frequency (or wavelength) is the fractional change
Table 1 summarises the fractional change, z, in frequency and wavelength.
Doppler shift, z
in frequency
in wavelength
Source moves
towards observer
f
f


+
v
c

v
c

v
c
+
v
c
Table 1 Summary of Doppler shifts
The frequency f is the frequency of the light emitted by the
source which is the same as the frequency emitted by an
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Source moves
away from observer
Astrophysics
identical source in the laboratory. The change of frequency
f is the difference between this frequency and the
observed frequency (the frequency of the light from the
source as measured by an observer).
Notes
1 The formulas above and in Table 1 can only be applied to electromagnetic waves at source
speeds much less than the speed of light c.
2 A star or galaxy may be moving through space with perpendicular velocity components
parallel and at right angles to the line from the Earth to the star. The first component is the
star’s radial speed and the second component its tangential speed. Throughout this topic,
speed v refers to its radial speed (i.e. the component of the star’s velocity parallel to the line
between the star and the Earth).
Astronomical velocities
The line spectrum of light from a star or galaxy is shifted to longer wavelengths if the star or
galaxy is moving away from us and to shorter wavelengths if it is moving towards us. By
measuring the shift in wavelength of a line of the star’s line spectrum, the speed of the star or
galaxy relative to Earth can be found. If the star is part of a binary system, its orbital speed can
also be found as explained later.
In practice, the line spectrum of the star or galaxy is compared with the pattern of the prominent
lines in the spectrum according to the star’s spectral class. The change of wavelength of one or
more prominent lines of known wavelength  in the spectrum is then measured and the Doppler

shift, z (=
) is then calculated.

For an individual star or galaxy, the speed v of the star or galaxy relative to a line between Earth
v
and the star is then calculated from z = .
c
The star or galaxy is moving:
 towards the Earth if the wavelength is shortened due to the star or galaxy’s relative motion
 away from the Earth if the wavelength is lengthened due to the star or galaxy’s relative
motion.
For binary stars in orbit about each other in the same plane as the line from the Earth to the stars,
the wavelength of each spectral line of each star changes periodically between:
 a minimum value of  −  when the star is moving towards the Earth
 a maximum value of  +  when the star is moving away from the Earth.
Worked example
A spectral line of a star is found to be displaced from its laboratory value of 434 nm by +0.087 nm. State
whether the star is moving towards or away from the Earth and calculate its speed relative to the Earth.
c = 3.0 × 108 m s−1
Solution
The star is moving away from the Earth because the wavelength of its light is increased.
Rearranging  
v
c 3.0 108  0.087 10 9
gives v 

 6.0 10 4 m s 1
9
c

434 10
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If the two stars cannot be resolved, they are referred to as a spectroscopic binary. Each spectral
line splits into two after the stars cross the line of sight then merge into a single line as the two
stars move towards the line of sight. Figure 2 shows the idea.
Figure 2 A spectroscopic binary
Note
If the stars are of different masses, they will move with the same period but at different speeds
and orbital radii. The change of wavelength will be greater for the faster star (less massive star)
than for the other star.
Worked example
c = 3.0 × 108 m s−1
A spectral line of a certain spectroscopic binary merges once every 1.5 years and splits to a maximum
displacement of 0.042 nm and 0.024 nm from their laboratory wavelength of 486 nm. Calculate:
a the orbital speed of each star
b the radius of orbit of the larger orbit.
Solution
a For the slower star,  = 0.024 nm
Rearranging  
v
c 3.0 108  0.024 10 9
gives v 

 1.5 10 4 m s 1
c

486 10 9
For the faster star,  = 0.042 nm
Rearranging  
v
c 3.0 108  0.042 10 9
gives v 

 2.6 10 4 m s 1
9
c

486 10
b The orbital speed of the faster star, v = 2r/T where r is its radius of orbit and T is the time period.
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Therefore, the radius of its orbit r 


vT
2.6 10 4 m s 1  1.5  365.25  24  60  60 s 

2π
2π
Hence r = 2.0 × 1011 m
Summary questions
c = 3.0 × 108 m s−1
1 Explain why the wavelengths of the light waves from a star moving away from the Earth are longer than
they would be if the star was stationary relative to the Earth.
2 A spectral line of a star is found to be displaced from its laboratory value of 656 nm by −0.035 nm. State
whether the star is moving towards or away from the Earth and calculate its speed relative to the Earth.
3 The spectral lines of a star in a binary system vary in wavelength.
a Explain why this variation is:
i periodic
ii over a narrow well-defined range of wavelengths.
b i State what measurements can be made by observing the variation in wavelength of a spectral line
from such a star.
ii Explain how the measurements can be used to find the radius of orbit of the star.
4 A spectral line of a certain star in a binary system changes from its laboratory wavelength of 618 nm by
±0.082 nm with a time period of 2.5 years.
Calculate:
a the orbital speed of the star
b its radius of orbit.
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Astrophysics
3.2 Hubble’s law and beyond
Learning objectives:




What do we mean by the term ‘red shift’?
Why do we think the Universe is expanding?
What evidence led to the acceptance of the Big
Bang theory
What is dark energy?
Galaxies
The Andromeda galaxy is the nearest large galaxy to the Milky Way. Andromeda can just be seen
by the unaided eye on a clear night. By taking photographs of Andromeda using a large telescope,
Edwin Hubble was able to identify Cepheid variable stars in Andromeda. These stars vary in
brightness with a period of the order of days and are named after the first one to be discovered,
-Cephei, the fourth brightest star in the constellation Cepheus. Their significance is that the
period depends on the absolute magnitude. Hubble measured the periods of the Cepheid variables
in Andromeda that he had identified. He then used data obtained on Cepheid variables of known
absolute magnitudes to find the absolute magnitude and hence the distance to each Cepheid
variable in Andromeda. He found that Andromeda is about 900 kiloparsec away, far beyond the
Milky Way galaxy which was known to be about 50 kiloparsec in diameter. His result settled the
issue of whether or not Andromeda is inside or outside the Milky Way galaxy.
Astronomers realised that many spiral nebula they had observed like Andromeda must also be
galaxies. The Universe consists of galaxies, each containing millions of millions of stars,
separated by vast empty spaces. Hubble and other astronomers studied the light spectra of many
galaxies and were able to identify prominent spectral lines as in the spectra of individual stars but
‘red-shifted’ to longer wavelengths. Hubble studied galaxies which were close enough to be
resolved into individual stars. For each galaxy, he measured:
 its red shift and then calculated its speed of recession (the speed at which it was moving
away)
 its distance from Earth by observing the period of individual Cepheid variables in the galaxy.
His results showed that galaxies are receding from us, each moving at speed v which is directly
proportional to the distance, d. This discovery, referred to as Hubble’s law, is usually expressed
as the following equation:
v = Hd
where H, the constant of proportionality, is referred to as the Hubble constant.
For distances in megaparsec (Mpc) and velocities in km s−1, the accepted value of H is
65 km s−1 Mpc−1.
In other words, the speed of recession of a galaxy at a distance of:
 1 Mpc is 65 km s−1 Mpc−1
 10 Mpc is 650 km s−1 Mpc−1
 100 Mpc is 6500 km s−1 Mpc−1
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Astrophysics
Figure 1 shows that the pattern of typical measurements of the speed of recession v and distance d
plotted on a graph is a straight line through the origin. (This example is from actual data
published in 2002. Notice the error bars on the distance estimates.) The slope of the graph is
equal to the Hubble constant H.
Figure 1 Speed of recession against distance for galaxies
Note
The galaxies local to the Milky Way galaxy such as Andromeda do not fit Hubble’s law because
their gravitational interactions have affected their direction of motion. Andromeda is known to be
on course to collide with the Milky Way galaxy billions of years in the future.
Worked example
c = 3.0 × 108 m s−1, H = 65 km s−1 Mpc−1
The wavelength of a spectral line in the spectrum of light from a distant galaxy was measured at
398.6 nm. The same line measured in the laboratory has a wavelength of 393.3 nm. Calculate:
a the speed of recession of the galaxy
b the distance to the galaxy.
Solution
a  = 398.6 − 393.3 = 5.3 nm
Rearranging  
v
c 3.0 108  5.3 10 9
gives v 

 4.0 10 6 m s 1
c

393 10 9
−1
b Converting v to km s−1 gives v = 4.0 × 103 km s
Rearranging v = Hd gives d 
v
4.0 103 km s 1

 62 Mpc
H 65 km s 1 Mpc1
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The Big Bang theory
Hubble’s law tells us that the distant galaxies are receding from us. The conclusion we must draw
from this discovery is that the galaxies are all moving away from each other and the Universe
must therefore be expanding. At first, some astronomers thought this expansion is because the
Universe was created in a massive ‘primordial’ explosion and has been expanding ever since.
This theory was referred to by its opponents as the Big Bang theory.
With no evidence for a primordial explosion other than an explanation of Hubble’s law, many
astronomers supported an alternative theory that the Universe is unchanging, the same now as it
ever was. This theory, known as the Steady State theory, explained the expansion of the Universe
by supposing matter entering the Universe at ‘white holes’ pushes the galaxies apart as it enters.
The Big Bang theory was accepted in 1965 when radio astronomers discovered microwave
radiation from all directions in space. Steady state theory could not explain the existence of this
microwave radiation but the Big Bang theory could.
Estimating the age of the Universe
The speed of light in free space, c, is 300 000 km s−1. No material object can travel as fast as light.
Therefore, even though the speed, v, of a galaxy increases with its distance d, no galaxy can
travel as fast as light.
The Hubble constant tells us that the speed of a galaxy increases by 65 km s−1 for every extra
million parsecs of distance or 3.26 million light years. Therefore, a galaxy travelling almost at the
300 000
speed of light would be almost at a distance of
× 3.26 million light years.
65
To reach this distance, light would need to have travelled for 15 000 million years. Thus the
Universe cannot be older than 15 000 million years.
Note
In mathematical terms, the speed of a galaxy v < c
Therefore, using the equation for Hubble’s law gives Hd < c or d <
c
H
c
represents the maximum expansion of the Universe and light could not have
H
travelled further than this distance since the Universe began.
The distance
The age of the Universe, T, is therefore given by equating the distance travelled by light in time T
c
(= cT) to the expansion distance
.
H
Hence cT =
T
c
gives
H
1
H
Substituting H = 65 km s−1 Mpc−1 = 2.1 × 10−18 s−1 (as 1 Mpc = 3.1 × 1022 m) therefore gives
1
1
= 4.7 × 1017 s = 15 000 million years.
T 
H 2.110 18 s 1
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Evidence for the Big Bang theory
The spectrum of microwave radiation
The spectrum of microwave radiation from space matched the theoretical spectrum of thermal
radiation from an object at a temperature of 2.7 K. Because the radiation was detected from all
directions in space with little variation in intensity, it was realised it must be universal or ‘cosmic’
in origin.
This background cosmic microwave radiation is explained readily by the Big Bang theory as
radiation that was created in the Big Bang has been travelling through the Universe ever since the
Universe became transparent. As the Universe expanded after the Big Bang, its mean temperature
has decreased and is now about 2.7 K. The expansion of the Universe has gradually increased the
background cosmic microwave radiation to its present range of wavelengths.
Relative abundance of hydrogen and helium
Stars and galaxies contain about three times as much hydrogen by mass as helium. In comparison,
other elements are present in negligible proportion. This 3 : 1 ratio of hydrogen to helium by mass
means that for every helium nucleus (of mass 4 u approximately) there are 12 hydrogen nuclei (of
mass 12 u in total). Thus there are 14 protons for every 2 neutrons (proton : neutron ratio of 7 : 1)
is ratio is because the rest energy of a neutron is slightly greater than that of the proton. As a
result, when the Universe cooled sufficiently to allow quarks in threes to form baryons, protons
formed from the quarks more readily than neutrons. Precise calculations using the exact
difference in the rest energies of the neutron and the proton yield a 7 : 1 ratio of protons to
neutrons.
Figure 2 Formation of hydrogen and helium nuclei
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Astrophysics
Dark energy
Astronomers in 1998 studying type Ia supernova were astounded when they discovered very
distant supernovae much further away than expected. To reach such distances, they must have
been accelerating. The astronomers concluded that the expansion of the Universe is accelerating
and has been for about the past 5000 million years. Before this discovery, most astronomers
expected that the Universe was decelerating as very distant objects would be slowed down by the
force of gravity from other galaxies. Many more observations since then have confirmed the
Universe is accelerating. Scientists think that no known force could cause an acceleration of the
expansion of the Universe and that a hitherto-unknown type of force must be releasing hidden
energy referred to as dark energy.
Evidence for accelerated expansion of the Universe is based on differing distance measurements
to type Ia supernova by two different methods:
1
2
The red shift method: measurement of the red shift of each of these distant type Ia supernova
and use of Hubble’s law gives the distance to each one.
The luminosity method: Type Ia supernova at peak intensity are known to be 109 times more
luminous that the Sun, corresponding to an absolute magnitude of about −18. The distance to
such a supernova can be calculated from its absolute magnitude M and its apparent magnitude
d
m using the formula m − M = 5 log   .
 10 
The two methods give results that are different and indicate that the distant type Ia supernova are
dimmer and therefore further away than their red shift indicates.
The nature of dark energy is unclear. It is thought to be a form of background energy present
throughout space and time. It is more prominent than gravity at very large distances because
gravity becomes weaker and weaker with increased distance whereas the force associated with
dark energy is thought to be constant. Current theories suggest it makes up about 70% of the total
energy of the Universe. The search for further evidence of dark energy will continue with
observations using larger telescopes and more sensitive microwave detectors on satellites.
Summary questions
c = 3.0 × 108 m s−1, H = 65 km s−1 Mpc−1
1 a State Hubble’s law.
b Explain why Hubble’s law leads to the conclusion that the Universe is expanding.
2 The wavelength of a spectral line in the spectrum of light from a distant galaxy was measured at
597.2 nm. The same line measured in the laboratory has a wavelength of 589.6 nm. Calculate:
a the speed of recession of the galaxy
b the distance to the galaxy.
3 State two pieces of experimental evidence other than Hubble’s law that led to the acceptance of the Big
Bang theory of the Universe.
4 A certain type Ia supernova has an apparent magnitude of +24.
a Calculate the distance to the supernova. (Assume the absolute magnitude of any type Ia supernova
is −18.)
b Outline why measurements on type Ia supernovae have led to the conclusion that the expansion of
the Universe is accelerating.
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Astrophysics
3.3 Quasars
Learning objectives:



How were quasars discovered?
What are the characteristic properties of a
quasar?
Why are there no nearby quasars?
The first quasar
The first quasar was announced two years after a previously discovered astronomical radio
source, 3C 273, was identified as a dim star in 1962 using an optical telescope. The star presented
a puzzle because its radio emissions were stronger than expected from an ordinary star and its
visible spectrum contained strong lines that could not be explained. Astronomers in California
realised the strong lines were due to a very large red shift of 0.15, corresponding to a light source
with speed of recession of 0.15 c (i.e. 15% of the speed of light) at a distance of over 2000 million
light years away.
Based on this distance, calculations showed that 3C 273 is 1000 times more luminous that the
Milky Way galaxy yet variations in its brightness indicated it is much smaller than the Milky
Way galaxy. Its variations on a time scale of the order of years or less tell us that its diameter
cannot be much more than a few light years.
Astronomers concluded that 3C 273 is more like a star than a galaxy in terms of its size yet its
light output is on a galactic scale or even greater. The object was referred to as a quasi-stellar
object or quasar. Many more quasars have been discovered moving away at speeds up to 0.85 c
or more at distances between 5000 and 10 000 light years away. The absence of quasars closer
than about 5000 million light years indicates a ‘quasar age’ that commenced 2000 to 3000 million
years after the Big Bang and lasted about 5000 million years.
Notes
1 To calculate the red shift of a quasar, the change of wavelength of one of its spectral lines of
  
known wavelength  is measured and then used to calculate the red shift z  
.
  
2 To calculate the speed of recession v, the equation v = zc may be used only if v << c.
Otherwise, a relativistic equation relating v and z must be used. Knowledge of this relativistic
equation is not required in this option specification. Quasars generally have red shifts
between 1 and 5 corresponding to speeds from 0.6 c to about 0.95 c which are not
insignificant compared with the speed of light, c.
Quasar properties
Quasars are among the oldest and most distant objects in the Universe. A quasar is characterised
by:
 its very powerful light output, much greater than the light output of a star
 its relatively small size, not much larger than a star
 a large red shift indicating its distance is between 5000 and 10 000 light years away.
Many quasars are not like 3C 273 in that they do not produce strong radio emissions.
AQA A2 Physics A © Nelson Thornes 2009
Astrophysics
What are quasars? Detailed optical and radio images of quasars indicate fast-moving clouds of
gases and jets of matter being ejected. Quasars are found in or near galaxies which are often
distorted, sometimes with lobes either side. Such ‘active’ galaxies are thought to have a
supermassive black hole at their centres. As discussed in Topic 2.4, such a black hole could have
a mass of more than 1000 million solar masses. With many stars near it, matter would be pulled
in and would become very hot due to compression as it nears the event horizon. Overheating
would result in clouds of hot glowing gas being thrown back into space. A spinning supermassive
black hole would emit jets of hot matter in opposite directions along its axis of rotation.
Many astronomers think that a quasar is a supermassive black hole at the centre of a galaxy.
When we observe a quasar, we are looking back in time at a supermassive black hole in action.
The action ceases when there are no nearby stars for the black hole to ‘consume’. Fortunately, the
Milky Way galaxy and Andromeda and other galaxies close to us are relatively inactive because
each galaxy no longer has many stars left near the supermassive black hole at its centre.
Summary questions
c = 3.0 × 108 m s−1, H = 65 km s−1 Mpc−1
1 State three characteristics of a quasar.
2 Light from a certain quasar was found to contain a spectral line of wavelength 540 nm that had been
red-shifted from a normal wavelength of 486 nm.
a Show that the red shift of this quasar is 0.11.
b Calculate the speed of recession of this quasar, assuming its speed is much less than the speed of
light. Ignore relativistic effects.
3 a What features of the light from a quasar indicates a quasar is much more luminous than a star?
b What feature of the light from a quasar indicates a quasar is much smaller in size than a galaxy?
4 Outline why astronomers think certain galaxies have a supermassive black hole at their centres.
AQA A2 Physics A © Nelson Thornes 2009
Astrophysics
Answers
1.1
1 a i The top ray should refract at the lens and pass through F; the bottom ray should refract at
the lens and then become parallel to the principal axis; the image should be formed at
1.4(3)f on the right-hand side of the lens.
ii real, inverted, diminished
b ii virtual, magnified, upright
2 a and c v = + 0.240 m
b i real
ii inverted
3 a and c v = −0.300 m
b i virtual
ii upright
4 a i v = −0.600 m, image height = 40 mm
b i 0.450 mm, virtual and upright
ii v = +1.000 m, image height = 40 mm
ii 1.250 m, real and inverted
1.2
2 b i 7.5
ii 1.13 (1.125 to 4 s.f.)
4 a i 640 mm
ii 680 mm
1.3
4 a 100
b 40
1.4
3 a 0.06(3) m
4 b 80 m
1.5
2 a i 2 × 10−5 degree
ii 3 × 10−6 degree
4 a 4, 1, 2/3, 5
2.1
2 b ii +11.63
3 b +5.9
4 b −23.2
2.2
2 b 4700 K
4 b 2.0 × 109 m
2.3
AQA A2 Physics A © Nelson Thornes 2009
b i 0.3
ii 1.4(3) × 104 m
Astrophysics
4 a Z, Y, X
b X giant; Y main sequence; Z white dwarf
c X/Z = 6300, Y/Z = 100
2.4
4 b i 3.0 km
ii 1.8 × 1019 kg m−3
c 1.8 × 105
3.1
2 1.6 × 104 m s−1
4 a 4.0 × 104 m s−1
b 5.0 × 1011 m
3.2
2 a 3.9 × 106 m s−1
b 59 Mpc
4 a 2500 Mpc
Image acknowledgements
Section 1.4 figure 4: NASA, ESA, and the Hubble Heritage Team (STScI/AURA)
Section 2.4 figure 1: NASA, ESA, J. Hester and A. Loll (Arizona State University)
Section 2.5 figure 1: ESO; figure 3: CSIRO; figure 4: ESA
Section 3.2 figure 1: PNAS
AQA A2 Physics A © Nelson Thornes 2009
Astrophysics
Additional examination-style questions
1 (a)A lens used by Galileo has a range of focal lengths from 0.98 m to 0.92 m, depending on the
wavelength of the light passing through the lens.
(i)Calculate the power of the lens for red light.
(ii)Name the defect in the image which arises because a lens has different focal lengths for
different wavelengths of light.
(3 marks)
(b)The telescope with which Galileo discovered Io, one of the satellites of Jupiter, had an angular
magnification of 30. Calculate the maximum angular separation of the images of Io and Jupiter
when viewed through this telescope.
radius of the orbit of Io around Jupiter= 4.2 × 105 km
distance of Jupiter from the Earth
= 6.0 × 108 km.
(2 marks)
(c)A lens of focal length 0.95 m is used as the objective of an astronomical telescope. In normal
adjustment, the telescope has an angular magnification of 30. Calculate the distance between the
(2 marks)
objective and eyepiece lenses.
AQA, 2007
2 (a)Draw a ray diagram to show how a converging lens forms a diminished image of a real object.
Label the principal foci, the object and the image on your diagram.
(2 marks)
(b)A converging lens of power 12.5 D is used to produce an image of a real object placed 0.35 m
from the lens.
(i) Calculate the image distance.
(ii) State three properties of the image.
(4 marks)
AQA 2008
3 (a)Draw a ray diagram to show the path of two rays, initially parallel to the axis, through a
Cassegrain telescope, as far as the eyepiece.
(3 marks)
(b)The Bradford Robotic Telescope in Tenerife is a Cassegrain arrangement with an objective of
diameter 356 mm.
(i) Calculate the resolving power of this telescope when used with light of wavelength 570 nm.
(ii)The images are collected using a CCD. What feature of the structure of a CCD can affect the
resolution of the inal image obtained?
(iii)The quantum efficiency of a CCD is typically greater than 70%. What is meant by quantum
efficiency?
(3 marks)
AQA 2009
4 Stars of spectral classes A and B have strong hydrogen Balmer absorption lines in their spectra.
(a) Describe how Balmer absorption lines are produced.
You may be awarded marks for the quality of written communication in your answer. (4 marks)
(b) (i) Why do the spectra of stars in classes F and G not have strong Balmer absorption lines?
(ii) What is the prominent feature in the spectra of stars in classes F and G?
(2 marks)
AQA 2007
AQA Physics A A2 Level © Nelson Thornes Ltd 2009
1
Astrophysics
Additional examination-style questions
5 The data in the table gives some of the properties of the star Mu Cephei.
apparent magnitude
4.23
absolute magnitude
–6.81
surface temperature
3500 K
(a) (i) Calculate the wavelength of the peak in the black body radiation curve for Mu Cephei.
(ii)Sketch the black body radiation curve for Mu Cephei on the axes below. Label the
wavelength axis with a suitable scale.
relative
intensity
0
0
wavelength
(3 marks)
(b) Calculate the distance to Mu Cephei in light years.
(c)Mu Cephei is possibly the largest star yet discovered. Its radius is 1.2 × 109 km, which is about
the orbital radius of Saturn. Show that the power output of Mu Cephei is approximately 400 000
times that of the Sun.
surface temperature of the Sun= 5800 K
radius of the Sun
= 6.9 × 105 km
(2 marks)
(2 marks)
AQA 2007
Δf
6Tonantzintla 202 is a quasar with a red shift, ___
​     ​, of 0.366. When it was discovered in 1957 it as
f
wrongly assumed to be a white dwarf.
(a) Explain what is meant by
(i) a white dwarf,
(ii) a quasar.
AQA Physics A A2 Level © Nelson Thornes Ltd 2009
(5 marks)
2
Astrophysics
Additional examination-style questions
(b) Ignoring relativistic effects, calculate, for Tonantzintla 202
(i) its recessional speed, relative to the Earth,
(ii) its distance from the Earth.
(3 marks)
AQA 2007
7 (a) Explain what is meant by the terms Rayleigh criterion and Airy disc.
You may be awarded marks for the quality of written communication in your answer. (3 marks)
(b)The Very Large Telescope (VLT) facility in the Atacama desert in Chile is a combination of four
Cassegrain telescopes each of diameter 8.2 m. It is used to detect electromagnetic radiation of
wavelengths in the range 200 nm to 20 µm.
(i)Show that the combination has a similar light-collecting power to that of a single 16 m
diameter telescope.
(ii)The VLT is capable of an angular resolution similar to that of a 100 m diameter telescope.
Calculate the maximum angular resolution of the VLT.
(iii)The Atacama desert is possibly the driest place on Earth. What part of the electromagnetic
spectrum is significantly absorbed by water vapour?
(6 marks)
AQA 2008
8 (a)Sketch a Hertzsprung-Russell (HR) diagram on the axes below. Label the position of the main
sequence, dwarf and giant stars. Complete the spectral class axis by labelling the spectral classes.
absolute
magnitude
�10
�5
0
5
10
15
0
spectral class
(3 marks)
(b)Beta Hydri is a star with the same black body temperature as the Sun, but is approximately 3.5
times brighter.
(i) Label with the letter X the position of Beta Hydri on the HR diagram.
(ii) State and explain which star is larger, the Sun or Beta Hydri.
AQA Physics A A2 Level © Nelson Thornes Ltd 2009
(5 marks)
AQA 2008
3
Astrophysics
Additional examination-style questions
9I Zw 1 is an active galaxy, containing a supermassive black hole which produces a quasar as it
consumes its host galaxy.
(a) Explain what is meant by
(i) a quasar,
(ii) a black hole.
(3 marks)
(b) Analysis of radio waves from galaxy I Zw 1, suggest it is 800 million light years from Earth.
(i) Calculate the recessional speed of the galaxy.
(ii)The source of the radio waves is carbon monoxide molecules in the gas clouds of the galaxy.
When measured from a lab-based source, the waves have a frequency of 108 GHz. What is
the frequency of the waves detected from the galaxy?
(4 marks)
(c)The black hole at the centre of I Zw 1 could have a mass 100 million times greater than the Sun.
Calculate the radius of the event horizon of a black hole of this mass.
(2 marks)
AQA 2008
10The properties of some of the stars in Ursa Major are given in the table.
distance/
light year
spectral
class
Dubhe
1.8
124
K
Merak
2.4
79
A
Megrez
3.3
81
A
Mizar
2.1
78
A
Alkaid
1.9
101
B
(ii) Which star is the hottest? Explain your answer.
(2 marks)
(b) (i) Define absolute magnitude.
apparent
magnitude
(a) (i) Which of these stars appears dimmest? Explain your answer.
name
(ii) Which star has the brightest absolute magnitude? Explain your answer.
(2 marks)
(c) (i) Define the parsec.
(ii) Calculate the distance to Alkaid in parsecs.
(iii)Calculate the absolute magnitude of Alkaid.
(5 marks)
AQA 2009
AQA Physics A A2 Level © Nelson Thornes Ltd 2009
4
Astrophysics
Additional examination-style questions
11Eta Orionis is an eclipsing binary system. Analysis of the light from one of the stars shows that a
particular spectral line varies in wavelength as shown in Figure 1.
656.36
656.34
wavelength/nm
656.32
656.30
656.28
656.26
656.24
656.22
656.20
4
6
8
time/days
10
12
14
(a) (i) Show that the star has an orbital velocity of approximately 30 km s–1.
2
Figure 1
0
(ii) Calculate the diameter of the orbit of the star.
(4 marks)
(b)The graph of apparent magnitude against time (light curve) for this binary system is shown in
Figure 2.
(i) Label the time axis with a suitable scale.
apparent
magnitude
3.25
time/day
3.30
3.35
3.40
3.45
3.50
3.55
3.60
3.65
Figure 2
(ii)Explain, in terms of the movement of the two stars, how this light curve is produced.
(4 marks)
AQA Physics A A2 Level © Nelson Thornes Ltd 2009
5
Astrophysics
Answers to examination-style questions
Answers
Marks Examiner’s tips
1 (a) (i) For red light, f = 0.98 m
1 ​  = ____
P = ​ __
​  1   ​ = 1.02 dioptres
f 0.98
1
1
Although this is a fairly straightforward
calculation, it contains two aspects which
caused many students to fail to get full
marks. Choosing the right focal length
requires an understanding that ‘blue
bends best’, that is, that the shorter focal
length is for blue (or violet) light, and
therefore the longer (0.98m) is for red
light. The formula for power is relatively
easy, but, for full marks, the correct unit
for power had to be given.
1
Chromatic aberration occurs because a
lens can act like a prism, splitting white
light into its colours. The hint here is the
word ‘chromatic’ – related to colour.
1
This is effectively a two step calculation.
You need to work out the angular
separation of Io and Jupiter without the
telescope – this is found from dividing
their distance apart by how far away they
are.
(ii) Chromatic aberration
θ′
__
(b) Use of M = ​   ​ 
θ
30 × 4.2 × 105
____________
θ′ = ​ 
  
 ​ 
6 × 108
= 0.021 rad
1
When you use the telescope the distance
gets magnified – by a factor of 30 in this
case. This is how the answer is obtained.
Notice that the angular separation still
needs a unit – the radian (or rad). This is
the standard unit of angle used in A level
Physics.
fo
(c) Use of M = ​ __ ​
fe
0.95
​  30 ​ = 0.032
fe = ____
length = fo + fe = 0.98 m
AQA Physics A A2 Level © Nelson Thornes Ltd 2009
1
1
‘In normal adjustment’ just means ‘with
the image at infinity’. This is the standard
set up for the specification. The ray
diagram for the telescope needs to be
learned. From that you get the idea that the
separation of the lenses is the sum of the
focal lengths ( fo + fe), and the
fo
magnification is their ratio, __
​   ​. Putting
fe
these two ideas together gives the answer.
1
Astrophysics
Answers to examination-style questions
Answers
Marks Examiner’s tips
2 (a)Ray parallel to principal axis through
labelled principal focus
Ray through centre of lens to form
diminished image
F
object
1
1
__
(b) (i) ​ 1 ​  = 12.5
f
1   ​ + __
Use of __
​ 1 ​  = __
​ 1 ​gives 12.5 = ____
​ 0.35
​ 1 ​+ __
​ 1 ​
v
f u v
__
1
​   ​= 9.64
v
v = 0.10(4) m
(ii) diminished, inverted, real
AQA Physics A A2 Level © Nelson Thornes Ltd 2009
When drawing ray diagrams it is
important to use a ruler and to make your
labels clear.
It is also helpful to include the direction
of the rays. There are three possible rays
which could be drawn – only two are
needed however. Most commonly the
ones drawn are: a ray entering the lens
parallel to the principal axis passes
through the principal focus; a ray
travelling through the centre of the lens
travels in a straight line. The image is
found where these two cross, and is
drawn from that point to the principal
axis.
image
F
If the labels were missing, only 1 mark
would have been awarded.
1
1
1
1
This is a two stage question. The focal
length of the lens is found from the
power. This is then used in the lens
formula. A problem encountered by using
the lens formula is associated with the
‘sign’ of any distances. In this
specification there is only one rule to
follow – the distance is negative to a
virtual image. In this example the image
is real. This can be seen from the ray
diagram, and from the fact that the
answer is positive. Many candidates lost
a mark because they failed to invert their
__
final answer – that is, to change ​ 1 ​(9.64)
v
into v (0.10 m)
There are three properties of an image –
and each one has two alternatives. These
are: real/virtual; magnified/diminished;
upright/inverted. The question is not
asking for properties like position or
colour.
2
Astrophysics
Answers to examination-style questions
Answers
Marks Examiner’s tips
3 (a) concave primary mirror
convex secondary
two correct rays
1
1
1
eye piece
small convex mirror
light from
distant object
concave
mirror
λ
__
(b) (i) Use of θ = ​   ​= gives
d
570 × 10–9
_________
θ = ​  0.356 ​ 
 
= 1.6 × 10−6 rad
(ii) The size of the pixels
(iii)The ratio of the number of photons
falling on a device that produce a
signal to the total number of photons
falling on the device
AQA Physics A A2 Level © Nelson Thornes Ltd 2009
1
1
1
There are two optical telescope ray
diagrams which need to be learned – the
refracting telescope, and this one. The
curvature of the mirrors causes the
biggest problem for many students. The
objective is concave and the secondary is
convex. The secondary should not be
draw as a plane mirror – and it is
definitely wrong if drawn concave. The
problem with the objective is more
subtle. It should look like one continuous
mirror, with a gap in the middle. It is
quite commonly drawn as two separate
concave mirrors.
Problems with resolving power are often
related to the use of powers –
remembering that nm means 10−9 metres,
and the unit. It is quite common to see
‘watt’ as the unit here, which is an
obvious misunderstanding, rather than
radians.
In the past, the CCD has often been tested
in a whole question. This aspect of the
operation of a CCD requires an
understanding of both the structure of the
CCD and what is meant by resolution.
Each pixel of the CCD adds up the
photons hitting it (it accumulates charge)
over its whole area. Therefore the smaller
the pixel, the greater the resolution (that
is, the more detail which can be seen).
There are several definitions which need
to be learned in this specification. This is
just one of them. The most common error
is caused when students miss out the last
four words – i.e. they do not make it clear
that it is the fraction of those photons
falling on the device.
3
Astrophysics
Answers to examination-style questions
Answers
Marks Examiner’s tips
4 (a) • Hydrogen (in atmosphere of star) has
max 4 This is one of the few occasions where
electrons in n = 2 state.
there is a significant overlap between the
• Light of particular frequencies (from star
core content and the option. The need for
passing through atmosphere) is absorbed
the n = 2 state is so that the spectrum is
. . . corresponding to energy differences
visible. There are several parts to the
between orbits (E = hf).
hydrogen spectrum but only the Balmer
• When electrons return to lower energy
series needs to be understood for this
states, energy released in all directions so
option.
reduced intensity in original direction
It is also important to see that this is an
(or lower frequencies emitted as
absorption spectrum – that is, the whole
electrons can return to lower states in
spectrum is seen with gaps in it,
steps) . . . producing gaps in spectra.
corresponding to the Balmer series. It is
quite common to answer in terms of the
emission spectrum and so miss several
marks.
The reference to ‘passing through the
atmosphere’ can be confusing. The answer
is actually referring to the atmosphere of
the star. There are five marking points here
– you only need four for all four marks to
be awarded. It is worth writing out the full
answer however.
(b) (i)Temperature too low for hydrogen to
have electrons in n = 2 state
(ii) (Ionized) metal absorption lines
AQA Physics A A2 Level © Nelson Thornes Ltd 2009
1
1
In order for hydrogen atoms to have
electrons in the n = 2 state (i.e. not in the
ground state) the hydrogen needs to be
‘energised’. This means the atmosphere
of the star needs to be hot. F and G are
much cooler stars than A or B.
The specification contains a lot of detail
about spectral classes. This should be
learned.
4
Astrophysics
Answers to examination-style questions
Answers
Marks Examiner’s tips
0.0029
5 (a) (i) λmax = ______
​  3500 ​ 
1
= 8.3 × 10 m
–7
(ii)Graph with correct shape wavelength
axis labelled
with peak near 8 × 10–7 m.
1
1
relative
intensity
0
0
1000
2000
The value of the temperature is now used
to help label the temperature scale. The
calculated value in (i) shows the position
of the peak in the curve, rather than the
largest value of wavelength. The shape of
the curve is similar no matter what
temperature is chosen – the line should be
steeper on the left hand side.
3000
4000
wavelength/nm
d
___
(b) Use of m – M = 5 log ​ 10  ​ 
d
4.23 – (–6.81) = 5 log ​ ___
10  ​ 
d = 1614 pc = 5.26 × 103 lyr.
(c) Use of P = σAT4 and A = πr2
Pm _______
σAmTm4
​ ___
 ​
 
=
​ 
Ps σAsTs4 ​ 
r 2T 4
= ______
​  m2 m4 ​ 
rs Ts
= 401 000
or:
P = σAT4 and A = πr2
Ps= σAsTs4
= 5.67 × 10–8 × 2π(6.9 × 108)2 (5800)4
= 1.8 × 1026 (W)
Pm= σAmTm4
= 5.67 × 10–8 × 2π(1.2 × 1012)2 (3500)4
= 7.7 × 1031 (W)
P
7.7 × 1031
___
​ Pm ​=
  ________
​ 
 ​ 
1.8 × 1026
s
= 430 000
AQA Physics A A2 Level © Nelson Thornes Ltd 2009
This equation is unusual in that the
constant is given in the specification
rather than in the data list. Many students
and teachers fail to recognise that the ‘m’
in the unit of the constant (0.0029 m K)
refers to ‘metres’ rather than ‘milli’. This
means that they include an unnecessary
factor of 0.001 in their answer.
1
1
1
1
1
This calculation is not straightforward
and should be done in steps – with each
step written down. This means that if a
small mistake is made in any single step
the error can be identified and carried
forward so that the other marks can still
be obtained. The log referred to in the
equation is ‘base 10’. It is important to
know the difference between this and the
‘loge‘ or ‘ln’ button on a calculator. This
problem requires the use of the antilog –
sometimes shown as 10x. The value
obtained in the equation is in pc, and
must be converted to light year.
The relationship between the power
output of a star and its surface area has
often been tested in this option in the
past. It is important to make sure that the
correct constant is chosen from the list on
the data sheet. Notice there are two routes
to the answer – choose whichever one you
find more straightforward. The difference
in the two answers is due to rounding
errors. Both parts (b) an (c) of this
question are quite difficult and should be
studied carefully if you are thinking of
obtaining a high grade on this paper.
1
5
Astrophysics
Answers to examination-style questions
Answers
6 (a) (i) White dwarf:
• relatively hot (therefore white)
• but relatively dim (therefore small)
star.
(ii)
Quasar:
• very large power output
• large red shift therefore very distant
• relatively small for power output.
Marks Examiner’s tips
1
1
1
1
1
There are several astronomical objects
which are on the specification. The
properties asked for are related to the
objects themselves rather than how they
got to be there. In this case a white dwarf
is hot and dim – both properties which
can be remotely measured. Answers
related to their formation – such as the
core of a much larger star – would not
gain credit. Similarly restating the
properties in the question – for example,
‘small’ – would not gain the mark.
The quasar properties are a little more
controversial. Stating that it has a large
red shift and is very distant gains only
one mark as the first statement is a
measurable property and the second is an
inference from that property.
v
__
(b) (i) 0.366 = ​   ​
c
v = 0.366 × 3 × 108
= 1.1 × 108 m s–1.
(ii) Use of v = Hd and H = 65 km s–1 Mpc–1
1.1 × 105
v ________
d = ​ __
 
= 1.7 × 103 Mpc
H  ​= ​  65 ​ 
7 (a)Rayleigh criterion: two sources can just be
resolved.
If the 1st minimum of the diffraction
pattern of one source coincides with the
centre of the diffraction pattern of the
other.
The Airy disc is the bright central
maximum of the diffraction pattern of a
point source.
AQA Physics A A2 Level © Nelson Thornes Ltd 2009
1
1
1
1
1
1
The calculation of the recessional speed
from the red shift is very straightforward.
Careless errors can often be spotted if the
answer is faster than the speed of light.
The use of Hubble’s Law requires the
velocity to be in km s–1 and the distance
in Mpc. The value for Hubble’s constant
is provided on the data sheet. There is no
need to convert the final answer into light
year
This can be answered using a diagram.
Sometimes it is much easier to draw what
you mean rather than rely on just words.
The important point here is that the two
sources are just resolved for the Rayleigh
criteria and that it is the central maximum
of one that coincides with the first
minimum of the other. Few students
knew what was meant by the Airy disc.
6
Astrophysics
Answers to examination-style questions
Answers
Marks Examiner’s tips
(b) (i)(Light collecting power is proportional
to area.)
• Area of four telescopes of diameter
d 2
​ 2 ​   ​ = 211 m2
8.2m = 4π ​ __
• Single telescope
___ of this area has a
diameter = 2​ __
​ A
π ​ ​ = 16.4 m
or:
Light collecting power is proportional
to area, area is proportion to diameter2
therefore:
• diameter2 of four telescopes =
diameter2 of single telescope
4 × (8.2)2 = d 2
• d = 16.4 m
(  )
√
1
1
1
1
λ
200 × 10–9
​   ​gives θ = _________
​  100 ​ 
 
(ii)Use of θ = __
d
–9
= 2.0 × 10 rad
1
(iii)Infrared
1
8 (a)
absolute
magnitude
�10
As it is the maximum angular resolution
(i.e. the ‘best’), it is the smallest angle
that is wanted. Maximum resolution
occurs when close together objects can be
resolved. As the wavelength appears at
the top of the equation, it is the smallest
wavelength that is needed. Furthermore,
you need to be familiar with the prefixes
(nano, micro) and the unit of the final
answer (radians).
The HR diagram is a core aspect of this
specification. It should be understood in
terms of temperature and spectra. You
should be able to interpret the properties
of stars in different parts of the diagram
in some detail.
giants
�5
0
main sequence
5
Knowing that the collecting power is
proportional to area is the significant
point in this question. The answer then
becomes a simple issue of calculating the
areas of the two situations and showing
that they are similar. The answers quoted
are more rigorous in that they calculate
the diameter of the single telescope
objective.
10
dwarfs
15
O
B
A
F
G
K
M
spectral class
Main sequence correct
Dwarfs and giants
OBAFGKM
AQA Physics A A2 Level © Nelson Thornes Ltd 2009
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1
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Astrophysics
Answers to examination-style questions
Answers
(b) (i)Above G on spectral class, between 5
and 0 on absolute magnitude.
(ii) P = σAT 4
• Beta Hydri has a larger P (brighter),
same T, therefore it must have a
greater A.
• Beta Hydri is larger.
9 (a) (i)
or:
(argued from the HR diagram):
• At same spectral class (temperature),
larger stars are higher up the HR
diagram.
• Beta Hydri is above the Sun,
therefore it is larger.
Marks Examiner’s tips
1
1
1
For this question, you need to know the
position of the Sun on the HR diagram.
The fact that Beta Hydri has the same
temperature puts it in the same spectral
class – that is, G. The ‘3.5 times brighter’
means that its absolute magnitude is
between 1 and 2 less than that of the Sun.
This is because a difference of 1 on the
magnitude scale is a factor of 2.5 in
brightness – and the scale is inverted
(smaller = brighter). The allowed values
in the mark scheme are actually quite
generous but many students over-estimate
the difference and lose the mark.
Being able to use Stefan’s Law
quantitatively in this way requires a clear
understanding of what the equation tells
you.
1
1
Quasar: two from:
• very powerful/bright
• radio source
• large red shift/very distant
• relatively small
max 3 The properties of quasars are
controversial, and it is worth
understanding the ones accepted within
the specification. Notice that large red
shift and distant are seen as only one
answer. The black hole property is the
(ii) Black hole:
only one acceptable. References to
• escape velocity greater than speed of
singularities etc are usually ignored as the
light
significant property is related to the
escape velocity.
800 × 106
________
 
(b) (i) d = ​  3.26 ​ 
= 2.454 × 108 pc = 245 Mpc
use of v = Hd and H = 65 km s–1 Mpc–1
v = 65 × 245 = 1.59 × 104 km s–1
= 1.59 × 107 m s–1
AQA Physics A A2 Level © Nelson Thornes Ltd 2009
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1
There are several opportunities to go
wrong in this calculation. The important
thing to do is take it one step at a time.
The calculation of distance in Mpc
requires knowledge of what the M stands
for. The use of the correct value of H
(given on the data sheet) leading to an
answer in km s–1 can also cause problems.
8
Astrophysics
Answers to examination-style questions
Answers
Marks Examiner’s tips
∆f v
​   ​gives
(ii) use of ​ __ ​ = __
f c
108 × 109 × 1.59 × 107
 ​
  
  
∆f = ___________________
​ 
3 × 108
= 5.742 × 109 Hz
1
measured frequency = f – ∆f
= 108 × 109 – 5.74 × 109
= 1.02 × 1011 Hz
1
2GM
 ​ 
 
(c) Use of R = ​ _____
c2
–11
2(6.67 × 10 )(108 × 2 × 1030)
= ​ _________________________
   
 ​
  
(3 × 108)2
= 2.96 × 1011 m.
This question contains a unit prefix that
you need to learn. Giga stands for 109.
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1
There are two common errors which
occur when the calculation of the radius
of the event horizon is asked for. The
mass of the black hole is often
miscalculated: either the mass of the Sun
is missed off altogether, or the mass of
the Earth is substituted; the other is that
students forget to square the speed of
light.
10(a) (i)Megrez, highest value of apparent
magnitude.
1
This question you to know that the
brightness is related to the apparent
magnitude, and that the scale is inverted
(smaller number = brighter star).
(ii) Alkaid, as it is in spectral class B
1
This question requires knowledge that
spectral class is related to temperature
and that the order (from hottest to
coolest) is OBAFGKM.
(b) (i)Inherent brightness or brightness seen
from 10 pc
1
There are a few definitions that need to
be learned and this is one of them.
References to just ‘magnitude’ fail to get
the mark. Similarly luminosity and
brightness are not the same thing.
1
Something which looks brightest and is
also furthest away, must actually be the
brightest. It is time consuming but full
credit was given to students who
calculated all of the absolute magnitudes
and showed that Dubhe was the brightest.
1
The most common mistake here is to
simply state the parsec in light year or
metres. The definition of the parsec is
‘the distance at which 1 AU subtends an
angle of 1 second of arc’. This is another
example of where a diagram would make
things much easier to describe.
(ii)Dubhe: it appears the brightest from
Earth, and it is the furthest away
(c) (i)The distance from which the Earth and
Sun would appear to be separated from
one another by 1 second of arc
AQA Physics A A2 Level © Nelson Thornes Ltd 2009
9
Astrophysics
Answers to examination-style questions
Answers
Marks Examiner’s tips
101
(ii) ​ ____
3.26  ​ = 31 (parsecs)
d
(iii)Use of m – M = 5 log ___
​ 10  ​ to give
(  )
31
​ 10 ​   ​
M = 1.9 – 5 log ​ ___
= –0.56
11(a)
Neutron star: any two from:
• extremely dense
• small (typically 10 km diameter)
• dim
• (spinning) radio source
Black hole:
• object which has an escape velocity
greater than speed of light
2GM
_____
 
(b) Use of Rs = ​  2 ​ 
c
2(6.67 × 10–11)(40 × 2 × 1030)
= ​ ________________________
   
 ​
  
(3 × 108)2
= 1.19 × 105 m.
1
Changing the distance into parsecs is
necessary for the final part. To be given it
to do separately here makes the final part
much easier. This step may not always be
given to the student to do like this.
1
1
1
Common errors here include getting the
m and the M the wrong way round. The
use of the incorrect base for the
logarithms was less common, but missing
out the minus sign in the final answer did
lose a mark.
max 3 There are many aspects of neutron stars
which are related to the theories about
how they are formed or how they
produce the phenomena which are
detected (e.g. the pulsed radio waves).
The properties quoted here are the ones
currently accepted within this
specification. Similarly, black holes are
exciting phenomena which can stimulate
a lot of interesting research. The
fundamental property, however, is related
to the escape velocity.
1
1
This calculation of the Schwarzschild
radius is relatively straightforward but is
still prone to the same errors as seen
before. i.e failing to use the correct mass
– either missing out the mass of the Sun
altogether, or using the mass of the Earth
and forgetting to square the speed of
light.
Nelson Thornes is responsible for the solution(s) given and they may not constitute the only possible solution(s).
AQA Physics A A2 Level © Nelson Thornes Ltd 2009
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