Astronomy and Cosmology Revision Notes — 9702 Paper 4

Luminosity and radiant flux intensity F = L/(4πd²)

Syllabus requirement · §25.1

“

recall and use F = L / (4πd²)

”

Two very different numbers about the same star

Point a telescope at two stars and one plain fact is obvious immediately: one looks brighter than the other. But "looks brighter" is not, by itself, a statement about the stars at all — a feeble nearby star and a genuinely powerful star lying much further away can appear equally bright, purely by coincidence of distance. Untangling how much power a star actually emits from how bright it happens to look from here is the very first thing this whole topic has to do, because every other idea in this note — standard candles, stellar radii, even Hubble's law — depends on being able to separate the two.

Luminosity: the star's own number, unaffected by distance

The luminosity $L$ of a star is the total power it radiates, in all directions, measured in watts. It is a property of the star alone: two identical stars have identical luminosity, however far apart they are placed, because $L$ says nothing about any observer — it is simply how much energy the star's own nuclear furnace is releasing as electromagnetic radiation, every second.

Radiant flux intensity: the observer's number, which DOES depend on distance

The radiant flux intensity $F$ at a point is the power received per unit area at that point — measured in $\text{W m}^{-2}$ . Unlike $L$ , this is explicitly a statement about a particular observer at a particular distance $d$ from the star: move the same detector twice as far away, and $F$ falls, even though the star's own $L$ has not changed by a single watt. $F$ is what a telescope or a light-meter actually measures; $L$ is what a physicist actually wants to know.

Why F falls as 1/d² — spreading the same power over a growing sphere

A star radiates its power $L$ equally in every direction (this note treats every star as a point source radiating isotropically). Picture an imaginary sphere of radius $d$ centred on the star: by the time the radiation has travelled outward to reach that sphere, the entire luminosity $L$ must be passing through its surface — energy is neither created nor destroyed on the way, so none of it is "lost" between the star and the sphere.

The surface area of a sphere of radius $d$ is:

A = 4\pi d^2

this is exactly the same geometry already met for a wave spreading out from a point source in AS Waves, just written for a sphere instead of a general spreading wavefront.

Radiant flux intensity is power divided by the area it is spread over, so divide the star's total power by this sphere's surface area:

F = \frac{L}{4\pi d^2}

Nothing about this equation is a new fact about starlight specifically — it is the inverse-square law, forced on any source that radiates power equally in all directions through empty space. Move the observing sphere out to $2d$ and its area becomes $4\pi(2d)^2 = 16\pi d^2$ — four times larger — so the same total power $L$ is now spread four times more thinly: $F$ at $2d$ is a quarter of $F$ at $d$ , not a half.

F = \frac{L}{4\pi d^2}

Radiant flux intensity — L is the star's total power output (W), d is the distance from the star (m), F is the power received per unit area (W m⁻²).

The same total luminosity L, radiated equally in all directions, is spread over a sphere of surface area 4πd² at distance d — and over four times that area at distance 2d, so F there is only a quarter as much.

Invented demo — flux intensity at two distances

A star has luminosity $L = 4.0\times10^{26}\ \text{W}$ . Calculate the radiant flux intensity (a) at a distance of $2.0\times10^{18}\ \text{m}$ , and (b) at twice that distance, $4.0\times10^{18}\ \text{m}$ .

Show full working

1
(a) Substitute $L$ and $d=2.0\times10^{18}\ \text{m}$ into $F=L/(4\pi d^2)$ :
$F = \frac{4.0\times10^{26}}{4\pi\times(2.0\times10^{18})^2}$
The distance is squared FIRST, inside the denominator, before anything else is evaluated — this is the step most easily rushed.
2
Evaluate the denominator, then divide:
$4\pi\times(2.0\times10^{18})^2 = 4\pi\times4.0\times10^{36} = 5.03\times10^{37}$ $F = \frac{4.0\times10^{26}}{5.03\times10^{37}} = 7.95\times10^{-12}\ \text{W m}^{-2}$
Splitting the denominator into its own line, before dividing, keeps the huge power-of-ten arithmetic checkable in two smaller pieces rather than one long calculation.
3
(b) Repeat with $d=4.0\times10^{18}\ \text{m}$ (double the previous distance):
$F = \frac{4.0\times10^{26}}{4\pi\times(4.0\times10^{18})^2} = \frac{4.0\times10^{26}}{2.01\times10^{38}} = 1.99\times10^{-12}\ \text{W m}^{-2}$
Same equation, same L — only d has changed, so this is purely a test of the 1/d² scaling.
4
Compare the two results:
$\frac{F(\text{at }2d)}{F(\text{at }d)} = \frac{1.99\times10^{-12}}{7.95\times10^{-12}} \approx \frac14$
Doubling the distance drops F to a QUARTER, confirming the inverse-SQUARE law directly from the numbers, not just from the algebra.

Answer

F(d) ≈ 8.0×10⁻12 W m⁻²; F(2d) ≈ 2.0×10⁻12 W m⁻² — a quarter of the first value.

Whenever a question changes only the distance, the new flux can be found by scaling the old one by (old d / new d)² — it is often faster than recomputing F = L/(4πd²) from scratch, and a good check on a full recalculation.

A straight-line trick, for later

Because $F=(L/4\pi)\times d^{-2}$ has exactly the shape $y=mx$ (with $y=F$ , $x=d^{-2}$ , and gradient $m=L/4\pi$ ), plotting $F$ against $d^{-2}$ for several measurements of the same star gives a straight line through the origin, whose gradient is $L/4\pi$ — so $L=4\pi\times\text{gradient}$ . This is a far more reliable way to find $L$ from real data than trusting a single $(F,d)$ pair, and it is exactly the technique behind the real exam graph met in §05, where it is combined with Wien's law to find a star's radius.

Plotting F against d⁻² turns the inverse-square law into a straight line through the origin, with gradient L/4π.

A real flux-intensity calculation

9702/43 M/J 2023 Q10(b)2 marks

A star of luminosity $3.8\times10^{31}\ \text{W}$ is a distance of $1.8\times10^{24}\ \text{m}$ from the Earth.

Calculate the radiant flux intensity at the Earth of the radiation emitted by the star.

Show full working

1
Write down $F=L/(4\pi d^2)$ and identify each quantity:
$L = 3.8\times10^{31}\ \text{W}, \qquad d = 1.8\times10^{24}\ \text{m}$
Naming exactly what the question has given, in the equation's own symbols, before any arithmetic — this is what stops a candidate substituting L where d belongs.
2
Square the distance:
$d^2 = (1.8\times10^{24})^2 = 3.24\times10^{48}\ \text{m}^2$
Squaring first, as its own explicit step, is what the invented demo above practised — the single most error-prone move in this whole equation.
3
Substitute into $F=L/(4\pi d^2)$ :
$F = \frac{3.8\times10^{31}}{4\pi\times3.24\times10^{48}}$
Everything is now a plain number substituted into the formula, with nothing left to rearrange.
4
Evaluate the denominator, then divide:
$4\pi\times3.24\times10^{48} = 4.07\times10^{49}$ $F = \frac{3.8\times10^{31}}{4.07\times10^{49}} = 9.3\times10^{-19}\ \text{W m}^{-2}$
The denominator evaluated on its own line, exactly as before, keeps the final division a single manageable step.

Answer

F = 9.3×10⁻19 W m⁻².

A star's flux intensity at Earth is always an almost absurdly small number in W m⁻² — that is expected, and not a sign of an arithmetic error, given just how enormous 4πd² becomes at interstellar distances.

Common mistakes

Treating "the star looks twice as bright" as meaning its luminosity is twice as large.
Apparent brightness (radiant flux intensity F) depends on BOTH luminosity and distance — a star can look bright because it truly is powerful, or simply because it happens to be close.
This is the entire reason luminosity and radiant flux intensity have to be two separate quantities with two separate symbols — conflating them is the single most common conceptual slip in this sub-topic.
Halving F when d doubles, e.g. writing F(2d) = ½F(d).
F obeys an inverse-SQUARE law: doubling d makes F a QUARTER, not a half — always square d before dividing.
The demo above exists specifically to make this scaling concrete with real numbers, since the algebra alone is easy to mis-simplify under pressure.
Substituting d in kilometres, light-years, or parsecs directly into F = L/(4πd²) without converting to metres.
Convert d to metres FIRST, as its own explicit step, exactly like any other SI substitution in this note.
Astronomical distances are often quoted in non-SI units in textbooks and popular science, but the data-sheet form of this equation is strictly in SI units throughout.

Your turn

A routine invented calculation, a real one-mark recall question, and a real rearrangement for distance.

12 marks
A star has luminosity $9.0\times10^{27}\ \text{W}$ . Calculate the radiant flux intensity of the star's radiation at a distance of $3.0\times10^{17}\ \text{m}$ .
Stuck? Show hint
Square the distance first, as its own step, before dividing.
Show solution
1. 1
  Square the distance:
  $d^2 = (3.0\times10^{17})^2 = 9.0\times10^{34}\ \text{m}^2$
  As always, squaring is kept as its own visible step.
2. 2
  Substitute into $F=L/(4\pi d^2)$ and evaluate the denominator:
  $4\pi\times9.0\times10^{34} = 1.13\times10^{36}$
  The denominator evaluated separately, before the final division.
3. 3
  Divide:
  $F = \frac{9.0\times10^{27}}{1.13\times10^{36}} = 8.0\times10^{-9}\ \text{W m}^{-2}$
  A direct final division, giving the flux intensity in the required unit.
Answer
F = 8.0×10⁻9 W m⁻².
29702/42 F/M 2025 Q10(a)(i)1 mark
State what is meant by the luminosity of a star.
Stuck? Show hint
This is a definition, not a calculation — think about what makes luminosity different from radiant flux intensity.
Show solution
1. 1
  State the definition: the luminosity of a star is the total power of radiation it emits.
  The mark scheme awards the mark for 'total power (of radiation) emitted (by the star)' — the word 'total' matters, since it distinguishes luminosity from the power received in any one direction or over any one area.
Answer
Luminosity is the total power (of radiation) emitted by the star.
39702/41 O/N 2024 Q10(b)(i)2 marks
Stars in a distant galaxy emit radiation with a total luminosity of $1.90\times10^{36}\ \text{W}$ . Radiation from the galaxy is observed on the Earth with a radiant flux intensity of $8.42\times10^{-16}\ \text{W m}^{-2}$ .

Determine the distance of the galaxy from the Earth.
Stuck? Show hint
Rearrange F = L/(4πd²) to make d the subject BEFORE substituting any numbers.
Show solution
1. 1
  Rearrange $F=L/(4\pi d^2)$ to make $d^2$ the subject:
  $d^2 = \frac{L}{4\pi F}$
  Naming the rearrangement as its own step, before any numbers are touched — multiplying both sides by d² and dividing by F.
2. 2
  Substitute the given values:
  $d^2 = \frac{1.90\times10^{36}}{4\pi\times8.42\times10^{-16}}$
  Both L and F substituted directly from the question, with nothing simplified yet.
3. 3
  Evaluate the denominator, then divide:
  $4\pi\times8.42\times10^{-16} = 1.058\times10^{-14}$ $d^2 = \frac{1.90\times10^{36}}{1.058\times10^{-14}} = 1.80\times10^{50}\ \text{m}^2$
  The denominator evaluated on its own line, exactly as in the worked examples above.
4. 4
  Take the square root as the final, separate step:
  $d = \sqrt{1.80\times10^{50}} = 1.34\times10^{25}\ \text{m}$
  The square root is only taken once d² itself has been fully evaluated as a single number — never applied to L and F separately.
Answer
d = 1.34×10²⁵ m.

Practise luminosity and radiant flux intensity questionsReal past-paper questions · Luminosity and radiant flux intensity F = L / (4 pi d^2)

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Can you do all of these?

F = L/(4πd²): F (radiant flux intensity) is power PER UNIT AREA received at distance d; L (luminosity) is the star's TOTAL power output and does not depend on d at all
Doubling d makes F a QUARTER, not a half — F obeys an inverse-SQUARE law, not a simple inverse one
A standard candle is an object whose LUMINOSITY is known independently (from its type, not from measuring it at THIS distance) — that is the only thing that lets a measured F be inverted into a distance
d = √(L/4πF): take the square root as its own final step, after L/(4πF) has been evaluated as a single number — do not try to take the root of L and F separately
Wien's law λmaxT = constant is an INVERSE proportionality: hotter star ⇒ SMALLER λmax (peak shifts towards blue/short wavelengths), not larger
λmax is the wavelength of PEAK emission rate, not the shortest or longest wavelength the star emits at all — a star's spectrum is continuous and stretches either side of λmax
The numerical value of Wien's constant is not something to memorise confidently — most real questions use the RATIO method (comparing two stars, or a star against the Sun) so the constant cancels; only use a numeric value if the question actually gives one
Stefan-Boltzmann's 4πr² is the star's SURFACE area, treating it as a sphere — this is exactly the same 4πd² geometry as §01's inverse-square law, just applied to the star's own surface instead of an imaginary sphere at the observer's distance
σ (the Stefan constant, 5.67×10⁻8 W m⁻²K⁻4) IS given on the data sheet; the equation L = 4πr²σT⁴ itself is NOT — it must be recalled
Estimating a stellar radius is a THREE-source calculation: L comes from §01 (F and d), T comes from §03 (λmax), and only once BOTH are known separately can Stefan-Boltzmann be rearranged for r — never try to find r from F and λmax directly without finding L and T as their own explicit steps first
Redshift z = Δλ/λ always compares an OBSERVED wavelength against the same spectral line's known LABORATORY (rest) wavelength — identifying which lab line a shifted line corresponds to is itself part of the method
An INCREASE in observed wavelength (redshift, longer λ) means the source is RECEDING; a decrease (blueshift, shorter λ) would mean approaching — but every galaxy this note deals with is receding
v ≈ zc is only valid for recession speeds much less than c — never apply it uncritically to a huge redshift without checking this condition holds
Hubble's law v = H0d needs v and d found by TWO GENUINELY DIFFERENT methods for the SAME galaxy — v from its redshift (§06), d from a standard candle (§02) — never from the same measurement twice
On a v-against-d graph for several galaxies, the line is straight and passes through the ORIGIN; its GRADIENT is H0 — reading H0 off a single (v, d) pair is far less reliable than a gradient across several galaxies
The Big Bang argument: galaxies recede from us in EVERY direction ⇒ the expansion is isotropic (the same in all directions, so it is space itself expanding, not us at a special centre) ⇒ running time backwards, everything must once have been much closer together, hot and dense
H0 is not known precisely, so 1/H0 (a rough estimate of the universe's age) carries the SAME uncertainty — never quote an age from Hubble's law as if it were exact

Now do the questions

112 real Paper 4 parts from 2021-2025, sorted by difficulty, with mark schemes

Medical Physics

Astronomy and Cosmology

Luminosity and radiant flux intensity F = L/(4πd²)

Two very different numbers about the same star

Luminosity: the star's own number, unaffected by distance

Radiant flux intensity: the observer's number, which DOES depend on distance

Why F falls as 1/d² — spreading the same power over a growing sphere

A straight-line trick, for later

Standard candles and distances to galaxies

Wien's displacement law

Can you do all of these?

Luminosity and radiant flux intensity F = L/(4πd²)

Two very different numbers about the same star

Luminosity: the star's own number, unaffected by distance

Radiant flux intensity: the observer's number, which DOES depend on distance

Why F falls as 1/d² — spreading the same power over a growing sphere

A straight-line trick, for later

The rest of this note

Can you do all of these?