A faint star in the night sky isn’t necessarily a weak one — it might be a powerhouse that just happens to be extremely far away. Telling these two possibilities apart is exactly what brightness and luminosity are for.
📘 What you need to know
Apparent brightness, b, is the intensity of radiation received on Earth from a star, measured in W m⁻²
Luminosity, L, is the total power output radiated by a star, measured in W
Apparent brightness depends on both a star’s luminosity and its distance from Earth
Radiation follows an inverse square law: b = L⁄(4πd²)
For two stars of equal luminosity, the one with the greater apparent brightness is the closer one
Apparent Brightness vs Luminosity
Luminosity describes how bright a star truly is at its surface — a fixed property of the star itself. Apparent brightness describes how bright that star appears from Earth, which depends just as much on distance as it does on luminosity. By the time light from a distant star reaches us, it has spread out over an enormous area, so only a tiny fraction of its total luminosity actually reaches any given square metre of a telescope’s detector.
A star’s luminosity radiates in every direction, but Earth only intercepts a small fraction of it — that fraction is the apparent brightness
The Inverse Square Law of Radiation
As light leaves a star, it spreads out uniformly over an ever-expanding spherical shell. The surface area of that shell is 4πr², so by the time the light has travelled a distance d, it’s been spread across an area of 4πd². This is why apparent brightness falls off as an inverse square law — a quantity whose intensity reduces in proportion to the square of the distance from its source.
Inverse square law of radiationb = L⁄(4πd²)
Where b is apparent brightness in W m⁻², L is luminosity in W, and d is the distance between the star and Earth in metres. This assumes the star radiates uniformly in all directions and that no radiation is absorbed on its way to Earth.
🧭 Recipe: Applying the Inverse Square Law
Identify what’s being asked for — apparent brightness, luminosity, or distance
Write down b = L ÷ (4πd²) and rearrange for the unknown quantity if needed
Check units carefully — apparent brightness is often given in nW m⁻² or similar small units
Substitute and calculate, keeping track of powers of ten
Quick recap: b = L/(4πd²). Doubling the distance to a star cuts its apparent brightness to a quarter. Two stars with equal luminosity — the brighter one is nearer.
WE 1
A star has a known luminosity of 4.5 × 10²⁶ W. A telescope measures its apparent brightness on Earth as 82 nW m⁻². Determine the distance from Earth to the star.
Step 1 — Rearrange for distanced = √[L ÷ (4πb)]Step 2 — Substitute the known valuesd = √[(4.5 × 10²⁶) ÷ (4π × 82 × 10⁻⁹)]≈ 2.09 × 10¹⁶ m
WE 2
A star located 3.0 × 10¹⁷ m from Earth has an apparent brightness of 5.4 × 10⁻⁹ W m⁻². Calculate the star’s luminosity.
Step 1 — Rearrange for luminosityL = b × 4πd²Step 2 — Substitute the known valuesL = (5.4 × 10⁻⁹) × 4π × (3.0 × 10¹⁷)²≈ 6.11 × 10²⁷ W
💡 Top tips
Don’t forget the factor of 4π when rearranging the inverse square law — a very common slip
Apparent brightness values are often given in very small units, like nW m⁻² — convert to W m⁻² carefully before substituting
A star can be faint either because it has low luminosity, or because it’s simply very far away — the inverse square law is how you tell these apart
Doubling distance divides apparent brightness by four, not by two — it’s a square law
⚠ Common mistakes
Forgetting to square the distance when substituting into b = L⁄(4πd²)
Mixing up apparent brightness (a per-area quantity) with luminosity (a total power)
Misplacing powers of ten when converting between nW, μW and W
Assuming a dimmer-looking star must have lower luminosity, without accounting for distance
Up next: Wien’s Displacement Law — where we look at how a star’s colour reveals its surface temperature.
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