This mini-lesson covers the AQA 3.9 Astrophysics option: telescopes and resolving power, apparent and absolute magnitude, Wien’s law and Stefan’s law, the classification of stars and the HR diagram, supernovae, neutron stars and black holes, and the Doppler shift, Hubble’s law and the Big Bang.
Optional topic — examined in Paper 3 Section B.
Optional topic — AQA sections 3.9–3.13 are options: you study exactly ONE of the five (Astrophysics, Medical physics, Engineering physics, Turning points in physics, Electronics). It is examined in Paper 3 Section B. Check with your teacher which option your school teaches before revising this one.
Work through each screen, answer the questions as you go (many are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
3.9.1 Β· telescopes
Telescopes and resolving power
A refracting telescope uses two converging lenses in normal adjustment: the focal points coincide, so parallel light in gives parallel light out.
M = angular magnification = fo ÷ fefo = objective focal length · fe = eyepiece focal length · separation of lenses = fo + fe
Cassegrain reflector: a concave parabolic primary mirror plus a convex secondary. Mirrors avoid chromatic aberration entirely (no refraction, so no dispersion) and can be made huge, light and supported from behind.
Chromatic aberration (lenses only): different wavelengths are refracted differently, so coloured fringes appear. Spherical aberration: rays far from the axis focus at a different point — fixed by a parabolic mirror.
Resolving power (Rayleigh criterion): two objects are just resolved if θ ≈ λ ÷ D. A larger aperture D means better resolution and more collected light.
Collecting power ∝ D² — doubling the diameter collects four times the light, so fainter objects become visible.
Worked example
fo = 1200 mm, fe = 25 mm → M = 1200 ÷ 25 = ×48.
Calculate
Your turn — calculation 1
1A refracting telescope in normal adjustment has an objective of focal length 1200 mm and an eyepiece of focal length 25 mm. Calculate the angular magnification.
Γ
Hint: M = fo ÷ fe = 1200 ÷ 25.
Quick check
Think it through
?Why do all large modern research telescopes use mirrors rather than lenses?
3.9.2 Β· magnitudes
Apparent and absolute magnitude
Apparent magnitude m — how bright a star looks from Earth. A smaller (or more negative) m means brighter.
The scale is logarithmic: a difference of 5 magnitudes = a factor of exactly 100 in intensity. So one magnitude step = a factor of 1001/5 = 2.51.
Absolute magnitude M — the apparent magnitude the star would have at a standard distance of 10 parsecs. It is a true measure of the star’s output.
m − M = 5 log(d ÷ 10)d in parsecs · 1 pc = 3.08×1016 m = 3.26 light years
Worked example
A star has M = 1.0 and lies at d = 100 pc.
m = M + 5 log(100/10) = 1.0 + 5 log(10) = 1.0 + 5 = 6.0 — just at the limit of naked-eye visibility.
Calculate
Your turn — calculation 2
2A star has an absolute magnitude of M = 1.0 and is 100 parsecs away. Calculate its apparent magnitude m.
Hint: m = M + 5 log(d ÷ 10) = 1.0 + 5 log(10), and log(10) = 1.
Quick check
Think it through
?Star A has apparent magnitude 1 and star B has apparent magnitude 6. How much brighter does A appear than B?
3.9.2 Β· black-body radiation
Wien’s law, Stefan’s law and stellar classification
Stars radiate approximately as black bodies, so their spectra tell us their temperature and size.
λmaxT = 2.90 × 10−3 m KWien’s displacement law: hotter stars peak at shorter wavelengths — they look blue
P = σAT4Stefan’s law: σ = 5.67×10−8 W m−2 K−4 · A = 4πr² for a star · P is the star’s luminosity
Spectral classes O, B, A, F, G, K, M run from hottest (O, ~25 000 K+, blue) to coolest (M, ~3000 K, red). The Sun is a G star (~5800 K).
Hydrogen Balmer absorption lines are strongest in class A (~10 000 K): hot enough to excite many electrons to n = 2, but not so hot that hydrogen is fully ionised.
Worked example — the Sun
T = 5800 K → λmax = 2.90×10−3 ÷ 5800 = 5.0×10−7 m = 500 nm (green-yellow, the peak of human vision — not a coincidence).
Calculate
Your turn — calculation 3
3The Sun’s surface temperature is about 5800 K. Use Wien’s law to calculate the wavelength at which it emits most strongly, in nm. (Wien constant = 2.90×10−3 m K)
nm
Hint: λmax = 2.90×10−3 ÷ 5800 = 5.0×10−7 m. Convert m → nm.
Calculate
Your turn — calculation 4
4A star has radius 7.0 × 108 m and surface temperature 5800 K. Calculate its luminosity using Stefan’s law. Give the number in front of ×1026 W. (σ = 5.67×10−8 W m−2 K−4)
Γ 10Β²βΆ W
Hint: A = 4πr² = 6.16×1018 m². T4 = 58004 = 1.13×1015. P = σAT4.
3.9.2 Β· HR diagram
The HR diagram and stellar evolution
The Hertzsprung–Russell diagram plots absolute magnitude (up the axis = brighter) against temperature — with temperature decreasing to the right, and usually on a logarithmic scale.
Main sequence — the broad diagonal band from hot/bright (top left) to cool/dim (bottom right). These stars are fusing hydrogen into helium in their cores. The Sun is one.
Red giants — top right: cool surfaces but very luminous, because they are enormous (A is huge in P = σAT4).
White dwarfs — bottom left: very hot but very dim, because they are tiny. Supported by electron degeneracy pressure.
Life of a Sun-like star: main sequence → core hydrogen runs out → core contracts, shell burning expands the envelope → red giant → outer layers puff off as a planetary nebula → the exposed core remains as a white dwarf, cooling forever.
Life of a massive star (> ~8 solar masses): main sequence → red supergiant → iron core forms and collapses → type II supernova → a neutron star, or a black hole if the remnant core is heavy enough.
Sort it
Main sequence, red giant or white dwarf?
Tap a statement, then tap the star type it describes.
βοΈ Main sequence
π΄ Red giant
βͺ White dwarf
3.9.2 Β· supernovae & black holes
Supernovae, neutron stars and black holes
Type Ia supernova: a white dwarf in a binary accretes matter until it passes the Chandrasekhar limit (~1.4 solar masses) and detonates. Crucially, every type Ia has almost the same peak absolute magnitude (about −19.3), which makes them standard candles for measuring cosmological distances.
Type II supernova: the core collapse of a massive star. The light curve is different from type Ia and is identified by the presence of hydrogen lines in the spectrum.
Neutron star: the remnant core, so dense that protons and electrons have combined into neutrons. Typically ~20 km across with the mass of the Sun. Rapidly rotating ones are seen as pulsars.
Black hole: so dense that the escape velocity exceeds c. Its event horizon lies at the Schwarzschild radius:
Rs ≈ 2GM ÷ c²set the escape velocity √(2GM/R) equal to c and rearrange
Supermassive black holes at galactic centres are detected from the orbits of nearby stars: measure the orbital radius and period, and Newtonian gravity gives the enclosed mass — millions of solar masses in a tiny volume.
Quick check
Think it through
?Why are type Ia supernovae so useful in cosmology?
3.9.3 Β· cosmology
Doppler shift, Hubble’s law and the Big Bang
z = Δλ ÷ λ = v ÷ cvalid for v « c · redshift = moving away · blueshift = approaching
v = H0dHubble’s law: recession speed is proportional to distance · H0 ≈ 65 km s−1 Mpc−1
Nearly all galaxies are redshifted, and the further away they are the faster they recede — the universe is expanding.
Run the expansion backwards and everything was once together: the Big Bang. The age of the universe ≈ 1/H0 (with H0 in SI units, s−1).
Other evidence: the cosmic microwave background at 2.7 K (a perfect black-body remnant of the hot early universe) and the observed abundance of helium.
Doppler shift also reveals binary stars (spectral lines shifting periodically) and exoplanets (the star wobbles about the common centre of mass).
Worked example
A line at 500.0 nm in the lab is seen at 500.5 nm from a galaxy: Δλ = 0.5 nm.
v = c × Δλ/λ = 3.00×108 × (0.5 ÷ 500) = 3.00×108 × 1.0×10−3 = 3.0×105 m s−1 = 300 km s−1, receding.
Calculate
Your turn — calculation 5
5A spectral line with a laboratory wavelength of 500.0 nm is observed from a distant galaxy at 500.5 nm. Calculate the galaxy’s recession speed in km s−1. (c = 3.00×108 m s−1)
km sβ»ΒΉ
Hint: v = c × Δλ ÷ λ = 3.00×108 × (0.5 ÷ 500.0) m s−1. Convert to km s−1.
Calculate
Your turn — calculation 6
6A galaxy is 200 Mpc away. Using H0 = 65 km s−1 Mpc−1, calculate its recession speed in km s−1.
km sβ»ΒΉ
Hint: v = H0d = 65 × 200.
Match it
Match each relationship to its law
Tap a card on the left, then its partner on the right.
Statement
Answer
Quick check
Think it through
?The event horizon of a black hole lies at the Schwarzschild radius Rs = 2GM/c². What does this radius represent?
Recap
The big ideas to know
Telescopes: M = fo/fe · reflectors avoid chromatic aberration · resolving power θ ≈ λ/D · collecting power ∝ D²
Magnitudes: smaller m = brighter · 5 magnitudes = ×100 · m − M = 5 log(d/10), d in parsecs
Black bodies: λmaxT = 2.90×10−3 m K · P = σAT4 · classes OBAFGKM, hottest to coolest
HR diagram: main sequence · red giants (cool, bright, huge) · white dwarfs (hot, dim, tiny)
Death of stars: type Ia = standard candle (M ≈ −19.3) · type II = core collapse · neutron star / black hole (Rs = 2GM/c²)
Cosmology: z = Δλ/λ = v/c · v = H0d · age ≈ 1/H0 · CMB at 2.7 K
You have covered the AQA 3.9 Astrophysics option — remember, you sit only ONE of the five options. Press Finish to see your score.
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