Astrophysics option: A-level Physics AQA revision notes

3.9.1

Telescopes

Telescopes 3.9.1.1–3.9.1.4

Definitions

Resolving power: a telescope's ability to distinguish two close objects as separate rather than as one blurred point — improves with larger aperture and shorter observing wavelength.

Key results

Angular magnification (refracting telescope, normal adjustment): $M = \frac{f _{o}}{f _{e}}$ .

Notes

A refracting telescope uses two converging lenses: an objective lens of long focal length $f_{o}$ gathers light and forms a real image, and an eyepiece lens of short focal length $f_{e}$ magnifies that image for the eye.
A reflecting telescope replaces the objective lens with a concave mirror, avoiding the chromatic aberration inherent to lenses (different wavelengths refracting by slightly different amounts).
A mirror also allows much larger apertures than a lens-based design can practically support, since it only needs to be supported and shaped accurately on one reflective surface rather than manufactured as flawless glass throughout its volume.
Better resolving power is the central practical reason astronomers build ever-larger telescopes — but a larger aperture also collects more light (light-gathering power increases with the square of the diameter), letting fainter, more distant objects be observed at all, a distinct advantage from resolving power even though both improve together as diameter increases.
Radio telescopes, observing at much longer wavelengths than visible light, need correspondingly enormous dishes to achieve comparable resolution — a direct consequence of resolving power improving with aperture and worsening with wavelength.
The atmosphere is transparent to visible light and radio waves, but strongly absorbs most infrared, ultraviolet and X-ray radiation, so infrared, ultraviolet and X-ray telescopes are generally most effective mounted on high-altitude sites or, better, carried above the atmosphere entirely on satellites or space telescopes.
Each part of the electromagnetic spectrum reveals different physical processes — radio and infrared trace cooler gas and dust, visible and ultraviolet trace hot stars, X-rays trace extremely energetic events (e.g. matter falling into black holes) — so a complete picture of an astronomical object usually needs observations from telescopes operating in several different wavebands, not just one.

Worked examples

Worked example 3.9.1 · 3 marks

A refracting telescope, used in normal adjustment, has an objective lens of focal length 1.20 m and an eyepiece lens of focal length 25 mm.

Calculate the angular magnification of the telescope.

Show worked solution

M = \frac{f _{o}}{f _{e}}

Converting the eyepiece focal length to metres,

f_{e} = 25 mm = 0.025 m

So:

M = \frac{1.20}{0.025} = 48

(no units, since magnification is a ratio of two lengths).

Mark scheme · 3 marks

Converts the eyepiece focal length to metres (25 mm = 0.025 m) 1 mark
Substitutes correctly into $M = \frac{f _{o}}{f _{e}}$ : $M = \frac{1.20}{0.025}$ 1 mark
Calculates the correct final value $M = 48$ , with no units 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.9.1 · 4 marks

Large modern research telescopes are almost always reflecting telescopes rather than refracting telescopes.

Explain two reasons why a reflecting design is preferred over a refracting design once a very large aperture is required.

Show worked solution

A refracting telescope's objective lens suffers from chromatic aberration, since different wavelengths of light refract by slightly different amounts as they pass through the glass, blurring the image.

A mirror avoids this problem because reflection does not depend on wavelength in the same way.

Separately, a mirror only needs one accurately shaped and supported reflective surface, whereas a lens must be manufactured as flawless glass throughout its entire volume — this makes very large apertures practically achievable for a mirror but not for a lens, and a larger aperture gives both better resolving power and greater light-gathering power.

Mark scheme · 4 marks

States that a refracting telescope's lens suffers chromatic aberration, since different wavelengths refract by different amounts 1 mark
States that a mirror avoids chromatic aberration 1 mark
States that a mirror needs only one accurately shaped reflective surface, whereas a lens needs flawless glass throughout its volume 1 mark
Concludes this makes very large apertures practical for mirrors but not lenses, giving better resolving power and light-gathering power 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.9.1 · 3 marks

Radio telescope dishes are typically tens or even hundreds of metres across, far larger than the mirrors of optical telescopes, yet radio telescopes often still achieve worse angular resolution than optical telescopes.

Explain why such large dishes are needed for radio astronomy.

Show worked solution

A telescope's resolving power improves as its aperture (diameter) increases, and worsens as the observing wavelength increases.

Radio waves have a very much longer wavelength than visible light.

To achieve a resolving power comparable to an optical telescope, a radio telescope must therefore have a correspondingly much larger aperture to compensate for its far longer observing wavelength.

Mark scheme · 3 marks

States that resolving power improves with increasing aperture and worsens with increasing wavelength 1 mark
States that radio waves have a much longer wavelength than visible light 1 mark
Concludes a much larger dish is needed for radio telescopes to achieve resolution comparable to an optical telescope 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

3.9.2

Classification of stars

Stars 3.9.2.1–3.9.2.6

Definitions

Apparent magnitude, $m$ : how bright a star appears from Earth, on a reversed logarithmic scale — a smaller (or more negative) number means a brighter apparent appearance.
Absolute magnitude, $M$ : the apparent magnitude a star would have if viewed from a standard distance of 10 parsecs, removing the effect of distance so stars' true (intrinsic) brightnesses can be compared directly.

Key results

Wien's law: $λ_{m a x} T = constant$ .
Stefan–Boltzmann law: $L = 4 π r^{2} σ T^{4}$ .
Relationship between apparent and absolute magnitude: $m - M = 5 lo g_{10} (\frac{d}{10})$ , with $d$ in parsecs — used to find a star's distance once both magnitudes are known.

Notes

A star's peak emission wavelength depends on its surface temperature via Wien's law — hotter stars peak at shorter (bluer) wavelengths, cooler stars at longer (redder) wavelengths.
This is why star colour is a genuine, direct indicator of surface temperature rather than an arbitrary label.
Luminosity depends extremely steeply on temperature (to the fourth power in the Stefan–Boltzmann law), which is why relatively modest temperature differences between stars correspond to enormous differences in luminosity.
Two stars can have identical absolute magnitude (the same true brightness) yet very different apparent magnitude, simply because they lie at different distances from Earth — apparent magnitude alone says nothing about a star's actual power output without also knowing its distance.
The Hertzsprung–Russell diagram plots luminosity (or absolute magnitude) against surface temperature (conventionally decreasing left to right) for large numbers of stars, revealing that most stars fall along a diagonal band called the main sequence, with giants and supergiants above it and white dwarfs below.
A star's position and movement across this diagram over its lifetime traces its evolutionary path: a main-sequence star fuses hydrogen into helium in its core for most of its life, with its eventual fate depending on its initial mass.
A star's radius can be found by combining Wien's law (giving temperature) with the Stefan–Boltzmann law (giving radius from luminosity and temperature) — a frequently examined two-step calculation linking both equations together.
A star of roughly solar mass or below, having exhausted its nuclear fuel, sheds its outer layers and collapses to a white dwarf — a dense, Earth-sized remnant supported against further collapse by electron degeneracy pressure, slowly cooling and fading over a very long time.
A much more massive star ends its life very differently: once fusion can no longer support the core against gravity, the core collapses catastrophically in seconds while the outer layers are blown off explosively as a supernova — briefly outshining an entire galaxy, and the source of most elements heavier than iron.
What remains after a supernova depends on the original star's mass: a core left below about 1.4 solar masses (the Chandrasekhar limit) collapses no further than a white dwarf; a somewhat more massive core collapses further still, with gravity overwhelming even electron degeneracy pressure, into an extremely dense neutron star supported by neutron degeneracy pressure; and a core above roughly 3 solar masses collapses completely, with no known force able to halt it, forming a black hole from which not even light can escape.

Worked examples

Worked example 3.9.2 · 6 marks

A star has luminosity:

L = 3.8 \times 1 0^{27} W

and its emission spectrum peaks at wavelength:

λ_{ma x} = 580 nm

Using Wien's law (:

b = 2.90 \times 1 0^{- 3} m K

) and the Stefan–Boltzmann law (:

σ = 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4}

), calculate the star's surface temperature and radius.

Show worked solution

Using Wien's law,

T = \frac{b}{λ _{ma x}} = \frac{2.90 \times 1 0 ^{- 3}}{580 \times 1 0 ^{- 9}} = 5000 K

Rearranging the Stefan–Boltzmann law for radius,

r = \frac{L}{4 πσ T ^{4}}

Since:

T^{4} = 500 0^{4} = 6.25 \times 1 0^{14} K^{4}

r = \frac{3.8 \times 1 0 ^{27}}{4 π \times 5.67 \times 1 0 ^{- 8} \times 6.25 \times 1 0 ^{14}} = 8.5 \times 1 0^{18} \approx 2.9 \times 1 0^{9} m

Mark scheme · 6 marks

Uses Wien's law rearranged for temperature: $T = \frac{b}{λ _{ma x}}$ 1 mark
Correct temperature value $T = 5000 K$ 1 mark
Selects the Stefan–Boltzmann law rearranged for radius: $r = \frac{L}{4 πσ T ^{4}}$ 1 mark
Correctly evaluates $T^{4} = 6.25 \times 1 0^{14} K^{4}$ 1 mark
Correct substitution of $L$ , $σ$ and $T^{4}$ into the rearranged equation 1 mark
Correct final radius $r \approx 2.9 \times 1 0^{9} m$ (2 s.f.) 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.9.2 · 4 marks

A star has an apparent magnitude of $m = + 7.2$ and an absolute magnitude of $M = + 1.4$ .

Calculate the distance to the star, in parsecs.

Show worked solution

Rearranging:

m - M = 5 lo g_{10} (\frac{d}{10})

7.2 - 1.4 = 5.8 = 5 lo g_{10} (\frac{d}{10})

so:

lo g_{10} (\frac{d}{10}) = 1.16

Taking inverse logs,

\frac{d}{10} = 1 0^{1.16} = 14.5

so $d \approx 145 pc$ .

Mark scheme · 4 marks

Substitutes correctly into $m - M = 5 lo g_{10} (\frac{d}{10})$ : $5.8 = 5 lo g_{10} (\frac{d}{10})$ 1 mark
Rearranges to isolate $lo g_{10} (\frac{d}{10}) = 1.16$ 1 mark
Takes the correct inverse log: $\frac{d}{10} = 1 0^{1.16} = 14.5$ 1 mark
Correct final distance $d \approx 145 pc$ (2–3 s.f.) 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.9.2 · 4 marks

A supernova leaves behind a stellar core of mass 2.1 solar masses.

State and explain, with reference to the Chandrasekhar limit, what kind of object this core will become.

Show worked solution

The core's mass, 2.1 solar masses, exceeds the Chandrasekhar limit of about 1.4 solar masses, so electron degeneracy pressure cannot support it as a white dwarf.

However, its mass remains below about 3 solar masses, so it does not collapse all the way to a black hole.

Instead, gravity overwhelms electron degeneracy pressure and the core collapses further, forming an extremely dense neutron star supported by neutron degeneracy pressure.

Mark scheme · 4 marks

States that the core mass (2.1 solar masses) exceeds the Chandrasekhar limit of about 1.4 solar masses 1 mark
States this means electron degeneracy pressure cannot support the core as a white dwarf 1 mark
States that the core mass remains below about 3 solar masses, so it does not collapse to a black hole 1 mark
Concludes the core collapses further to form a neutron star, supported by neutron degeneracy pressure 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

3.9.3

Cosmology

Redshift and Hubble's law 3.9.3.1–3.9.3.2

Key results

Redshift: $z = \frac{Δ λ}{λ} \approx \frac{v}{c}$ (for speeds much less than $c$ ).
Hubble's law: $v = H_{0} d$ .

Notes

Observing that almost every distant galaxy shows redshift, and that more distant galaxies show greater redshift, is the direct observational basis for an expanding universe.
The same law and expanding-space picture applies from any galaxy's point of view — every observer sees every other galaxy receding, with speed proportional to distance.
Hubble's law does not imply that our galaxy occupies any special central position in the universe.
The cosmic microwave background — a near-uniform, extremely faint microwave glow observed in every direction in the sky, with a spectrum matching a black body at about 2.7 K — is strong independent evidence for a hot, dense early universe that has since expanded and cooled.
The CMBR is one of the central pieces of evidence supporting Big Bang cosmology, alongside Hubble's law and the observed abundance of light elements.
$1/ H_{0}$ gives a rough estimate of the universe's age, since it is the time it would have taken all galaxies to reach their current separations at their current recession speeds, assuming a constant expansion rate — a genuine (if simplified) calculation, not just a unit conversion.

Quasars 3.9.3.3

Definitions

Quasar (quasi-stellar object): an extremely luminous, compact, and highly redshifted object powered by a supermassive black hole actively accreting matter at the centre of a distant galaxy.

Notes

A quasar can outshine every star in its host galaxy combined, despite the emitting region being no larger than the Solar System — this combination of extreme luminosity and small physical size is what first made quasars so puzzling when discovered.
The enormous luminosity is powered by gravitational energy released as matter spirals into an accretion disc around a supermassive black hole, heating the disc to very high temperatures before it crosses the event horizon.
Quasars show very large redshifts, placing them at huge distances and therefore seen as they were very early in the universe's history — observing a quasar is effectively observing a galaxy in an extremely active, early phase of its evolution.
Because quasars are so luminous, they remain detectable across cosmological distances at which an ordinary galaxy's stars would be far too faint to observe individually, making them valuable distance markers and probes of the early universe.
Quasar redshifts provided some of the earliest strong evidence for cosmological redshift at very large recession speeds, extending the observational basis for an expanding universe well beyond the more nearby galaxies used to establish Hubble's law directly.

Exoplanets 3.9.3.4

Notes

Exoplanets are generally far too faint and too close to their host star, in angular terms, to be imaged directly with current technology.
Almost all detections rely on indirect methods that infer a planet's presence from its effect on the star it orbits.
The transit method detects a tiny, periodic dip in a star's observed brightness as a planet passes directly between the star and the observer, with the size of the dip related to the ratio of the planet's cross-sectional area to the star's.
The radial velocity method detects the small, periodic Doppler shift in a star's spectrum caused by its own slight wobble around the star-planet system's common centre of mass, as the orbiting planet's gravity tugs the star back and forth.
Both methods are strongly biased towards detecting large planets in close, short-period orbits — a large planet produces a deeper transit dip and a larger stellar wobble, and a short orbital period means multiple transits or wobble cycles can be confirmed within a realistic observing campaign.
This detection bias is an important caveat when interpreting the overall population statistics of exoplanets discovered so far, since it does not necessarily reflect the true underlying distribution of planet sizes and orbital distances.

Worked examples

Worked example 3.9.3 · 6 marks

Quantity	Wavelength / nm
Rest wavelength (laboratory), λ	656.3
Observed wavelength (galaxy), λ+Δλ	662.0

The table shows the rest (laboratory) wavelength and the observed wavelength of the same hydrogen absorption line in the spectrum of a distant galaxy.

Calculate

(a) the redshift $z$ ,

(b) the galaxy's recession velocity, and

H_{0} = 70 km s^{- 1} Mpc^{- 1}

Show worked solution

The wavelength shift is:

Δ λ = 662.0 - 656.3 = 5.7 nm

The redshift is:

z = \frac{Δ λ}{λ} = \frac{5.7}{656.3} = 8.69 \times 1 0^{- 3}

Since $z \approx \frac{v}{c}$ , the recession velocity is:

v = zc = 8.69 \times 1 0^{- 3} \times 3.00 \times 1 0^{8} = 2.61 \times 1 0^{6} m s^{- 1}

(2610 km s⁻¹).

Using Hubble's law,

d = \frac{v}{H _{0}} = \frac{2610}{70} \approx 37 Mpc

Mark scheme · 6 marks

Calculates the wavelength shift $Δ λ = 662.0 - 656.3 = 5.7 nm$ 1 mark
Calculates the redshift $z = \frac{Δ λ}{λ} = 8.69 \times 1 0^{- 3}$ 1 mark
Uses $z \approx \frac{v}{c}$ to relate redshift to recession velocity 1 mark
Correct velocity value $v = 2.61 \times 1 0^{6} m s^{- 1}$ (2610 km s⁻¹) 1 mark
Substitutes into Hubble's law rearranged for distance: $d = \frac{v}{H _{0}}$ 1 mark
Correct final distance $d \approx 37 Mpc$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.9.3 · 5 marks

Planet	Fractional brightness dip
X	0.012%
Y	0.084%

Two candidate exoplanets, X and Y, are detected orbiting the same star using the transit method.

The table shows the fractional dip in the star's observed brightness during each transit.

(a) State which planet has the larger radius, explaining your reasoning.

(b) Describe how the radial-velocity method could provide independent confirmation of these planets, and state one limitation shared by both detection methods.

Show worked solution

The size of a transit's brightness dip is related to the ratio of the planet's cross-sectional area to the star's, so a larger dip indicates a larger planet (for the same star).

Since planet Y's dip (0.084%) is much greater than planet X's (0.012%), planet Y has the larger radius.

The radial-velocity method would detect the small periodic Doppler shift in the star's own spectrum, caused by the star wobbling around the star–planet system's common centre of mass as the orbiting planet's gravity tugs on it; a Doppler shift with the same period as each planet's transits would independently confirm its existence.

A shared limitation is that both methods are strongly biased towards detecting large planets in close, short-period orbits, so the population of exoplanets found this way does not necessarily reflect the true underlying distribution of planet sizes and orbital distances.

Mark scheme · 5 marks

States that transit dip size relates to the ratio of the planet's to the star's cross-sectional area, so a larger dip means a larger planet 1 mark
Correctly identifies planet Y as having the larger radius, since it produces the deeper transit dip 1 mark
Describes the radial-velocity method: periodic Doppler shift in the star's spectrum from its wobble about the common centre of mass with the planet 1 mark
States that a matching period between the transit and radial-velocity signals would give independent confirmation 1 mark
States a shared limitation: both methods are biased towards large planets in close, short-period orbits, so detections do not reflect the true underlying population 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.9.3 · 5 marks

Quasars were, for a long time, one of the most puzzling classes of astronomical object.

Explain what quasars are, why their early discovery was so puzzling, and why their high redshifts make them valuable for studying the early universe.

Show worked solution

A quasar is an extremely luminous, compact object powered by a supermassive black hole actively accreting matter at the centre of a distant galaxy; its huge luminosity comes from gravitational energy released as infalling matter heats up in an accretion disc before crossing the event horizon.

Their discovery was puzzling because a quasar can outshine every star in its host galaxy combined, despite the emitting region being no larger than the Solar System — an extreme luminosity from a remarkably small physical size.

Quasars show very large redshifts, placing them at huge distances, so they are observed as they were very early in the universe's history; because they remain so luminous, they stay detectable across distances at which an ordinary galaxy's individual stars would be far too faint to observe, making them valuable probes of galaxies in an early, highly active phase of evolution.

Mark scheme · 5 marks

States a quasar is an extremely luminous, compact object powered by a supermassive black hole accreting matter at a distant galaxy's centre 1 mark
States the luminosity comes from gravitational energy released as infalling matter heats an accretion disc before crossing the event horizon 1 mark
States that early discovery was puzzling because the huge luminosity (outshining a whole galaxy) came from a region no larger than the Solar System 1 mark
States that a quasar's large redshift places it at a huge distance, so it is observed as it was very early in the universe's history 1 mark
States that because quasars stay detectable where ordinary galaxy stars would be too faint, they act as valuable probes of the early universe 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.