Syllabus Edition

First teaching 2023

First exams 2025

|

Apparent Brightness & Luminosity (HL IB Physics)

Revision Note

Katie M

Author

Katie M

Last updated

Apparent Brightness & Luminosity

  • The apparent brightness b of a star is defined as:

The intensity of radiation received on Earth from a star 

  • Apparent brightness is measured in watts per metre squared (W m−2)
  • The apparent brightness of a star depends on two main factors:
    • How much light the star emits
    • How far away the star is (more distant stars are usually fainter than nearby stars)
  • How much light the star emits is given by the luminosity L of the star, which is defined as:

The total power output of radiation emitted by a star

  • Luminosity is measured in units of watts (W)

What is the difference between apparent brightness and luminosity?

Luminosity and Apparent Brightness, for IB Physics Revision Notes

The luminosity is the total power output of the star, whereas the apparent brightness is what is measured on Earth

  • Knowing the luminosity and apparent brightness of a star is useful because it allows us to determine how far away it is from the Earth
  • This is because
    • Luminosity tells us how bright the star is at its surface
    • Apparent brightness tells us how bright the star is as observed from the Earth
  • Therefore, by the time the radiation from the distant star reaches the Earth, it will have spread out over a very large area
    • This means the intensity of the radiation detected on Earth will only be a fraction of the value of the star's luminosity

Inverse Square Law of Radiation

  • A light source which is further away appears fainter because the light it emits is spread out over a greater area
  • The moment the light leaves the surface of the star, it begins to spread out uniformly through a spherical shell
    • The surface area of a sphere is equal to r2
  • The radius r of this sphere is equal to the distance d between the star and the Earth
    • By the time the radiation reaches the Earth, it has been spread over an area of d2
  • This is called following an inverse square law
    • This phrase can be used to refer to any quantity whose intensity reduces by a factor equal to the square of the distance to the observer (e.g. the intensity of ionising radiation follows an inverse square law)
  • The inverse square law of radiation can be written as:

b space equals space fraction numerator L over denominator 4 straight pi d squared end fraction

  • Where:
    • b = apparent brightness, or observed intensity on Earth (W m−2)
    • L = luminosity of the source (W)
    • d = distance between the star and the Earth (m)
  • This equation assumes:
    • The power from the star radiates uniformly through space
    • No radiation is absorbed between the star and the Earth
  • This equation tells us:
    • For a given star, the luminosity is constant
    • The intensity of the emitted light follows an inverse square law
    • For stars with the same luminosity, the star with the greater apparent brightness is closer to the Earth

Inverse square law of radiation

Inverse Square Law of Flux, for IB Physics Revision Notes

When the light is twice as far away, it has spread over four times the area, hence the intensity is four times smaller

Worked example

A star has a known luminosity of 9.7 × 1027 W. Observations of the star show that the apparent brightness of light received on Earth from the star is 114 nW m–2.

Determine the distance of the star from Earth.

Answer: 

Step 1: Write down the known quantities

  • Luminosity, L = 9.7 × 1027 W
  • Apparent brightness, b = 114 nW m–2 = 114 × 10–9 W m–2

Step 2: Write down the inverse square law of radiation and rearrange for distance d

b space equals space fraction numerator L over denominator 4 straight pi d squared end fraction

d space equals space square root of fraction numerator L over denominator 4 straight pi b end fraction end root

Step 3: Substitute in the values and calculate the distance d

d space equals space square root of fraction numerator 9.7 space cross times space 10 to the power of 27 over denominator 4 straight pi space cross times space open parentheses 114 cross times 10 to the power of negative 9 end exponent close parentheses end fraction end root

distance, d = 8.2 × 1016 m

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Katie M

Author: Katie M

Expertise: Physics

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.