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NotesPhysicsTopic 5.5Luminosity, apparent brightness and distance
Back to Physics Topics
5.5.33 min read

Luminosity, apparent brightness and distance

IB Physics • Unit 5

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Contents

  • Luminosity vs apparent brightness
  • The inverse-square law: b = L / (4π d²)
  • Measuring distance by parallax
  • Exam-style question
The big idea: Two stars can look equally bright in the sky yet pour out wildly different amounts of power — one might just be much closer.

So we keep two separate ideas:

- Luminosity (L) = the total power the star radiates (its own property, in watts). - Apparent brightness (b) = the power we receive per square metre here on Earth (it depends on distance).

Luminosity L

  • The total power the star pours out, in all directions
  • A property of the star itself — it does not depend on how far away you are
  • Measured in watts (W), just like a light-bulb's power rating

Apparent brightness b

  • The power you actually receive, spread over each square metre
  • Depends on how far away you are — the same star looks dimmer farther off
  • Measured in watts per square metre (W m⁻²)
New words, plainly: Luminosity (L) = how powerful the star is in total — like a bulb's wattage. It never changes with where you stand.

Apparent brightness (b) = how bright it looks to us — the power landing on each m² of our detector. Move farther away and b drops, even though L is unchanged.

Inverse-square law = brightness falls off as 1 ÷ (distance)².
As the light travels outward…What happens to the brightness
The star sends out a fixed total power L every secondL never changes — it's the star's own property
That power spreads over a sphere of area 4π d²The sphere gets bigger as d grows
Each square metre therefore catches a smaller shareThe apparent brightness b falls
Double the distance → 4× the area → ¼ the brightnessThis is the inverse-square law
Why distance changes b but not L: The star always emits the same total power L.

But by the time that light reaches us it has spread over a huge sphere of area 4π d². The farther away we are, the bigger that sphere, so each square metre gets a smaller slice — and b is smaller.

The light isn't destroyed; it's just spread thinner.

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Because the star's power L spreads evenly over a sphere of area 4π d², the brightness we measure is simply L divided by that area. That single formula links luminosity, brightness and distance.

Apparent brightness — the inverse-square law (given in the data booklet). b is the power received per m², L is the star's luminosity, d is the distance.
apparent brightness — power received per unit area at Earth (W m⁻²)
luminosity — total power the star radiates in all directions (W)
distance from the star to the observer (m)
surface area of the sphere the light has spread over by distance d (m²)

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The shortcut you'll use most: Because b depends on 1/d², distance changes brightness fast:

- 2× farther → ¼ as bright - 3× farther → 1/9 as bright - 10× farther → 1/100 as bright

Watch your powers of ten — luminosities and distances are huge numbers in standard form.
IB-style questionCalculate[2 marks]

A star has a luminosity of L = 2.5 × 10²⁸ W. It is at a distance of d = 4.0 × 10¹⁸ m from Earth. Calculate the apparent brightness of the star as seen from Earth.

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How we know how far a star is: As Earth orbits the Sun, a nearby star seems to shift slightly back and forth against the distant background stars over a year. That shift is called parallax.

The closer the star, the bigger the shift. Measure the shift angle and you get the distance — no need to travel there.
Parallax — the ideaWhat it tells you
Earth moves to the other side of its orbit over six monthsWe view the star from two ends of a baseline
A nearby star appears to shift against the distant backgroundThe shift angle is bigger for closer stars
The parallax angle p = half that total yearly shift, in arc-secondsSmaller p → the star is farther away
Distance in parsecs = 1 ÷ p1 arc-second of parallax = 1 parsec away, by definition
Stellar parallax distance (given in the data booklet). d is the distance in parsecs; p is the parallax angle in arc-seconds. Smaller angle → farther star.
distance to the star, in parsecs (pc)
parallax angle — half the star's apparent shift over a year, in arc-seconds (″)
Units matter here: This formula only works when p is in arc-seconds and you read d straight out in parsecs.

1 parsec (pc) ≈ 3.26 light-years ≈ 3.1 × 10¹⁶ m.

1 arc-second (″) = 1/3600 of a degree — far too small to see by eye, which is why parallax needs telescopes.
IB-style questionCalculate[2 marks]

A star has a measured parallax angle of p = 0.025 arc-seconds. Calculate its distance from Earth, in parsecs.

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How this is tested — usually a quick Paper 1A calculation, occasionally inside a longer Paper 2 star question:

Paper 1A

  • Calculate: a parallax angle → distance with d = 1/p, or a brightness with b = L/(4π d²).
  • Determine: two stars look equally bright but one is N× more luminous — how much farther is it? (Equal b ⇒ d ∝ √L.)

Paper 2

  • Brightness / luminosity / distance feeds into a bigger black-body question (linking to L = σAT⁴ in 5.5.4).
The classic trap: Mixing up luminosity (total power out — fixed) with apparent brightness (received per m² — depends on distance). Read which one the question gives you.
IB-style questionDetermine[2 marks]

Two stars, A and B, appear equally bright from Earth (the same apparent brightness b). Star B is 9 times more luminous than star A. Determine how many times farther away star B is than star A.

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Test yourself on Luminosity, apparent brightness and distance. Write your answer and get instant AI feedback — just like a real IB examiner.

A star has a measured parallax angle of 0.040 arc-seconds.

the distance to the star, in parsecs.
[2 marks]

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