Inverse Square Law of Flux (Edexcel International A Level (IAL) Physics): Revision Note

Exam code: YPH11

Katie M

Author

Katie M

Last updated

Inverse Square Law of Flux

  • The moment the light leaves the surface of the star, it begins to spread out uniformly through a spherical shell

    • Light sources which are further away appear fainter because the emitted light has been spread over a greater area

  • The surface area of a sphere is equal to 4πr2

    • The radius r of this sphere is equal to the distance d between the star and the Earth

    • Therefore the radiation received at Earth has been spread over an area of 4πd2

  • The inverse square law of flux can therefore be calculated using:

F equals fraction numerator L over denominator 4 pi d squared end fraction

  • Where:

    • F = radiant flux intensity, 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 radiant flux follows an inverse square law

    • The greater the radiant flux (larger F) measured, the closer the star is to the Earth (smaller d)

Inverse Square Law, downloadable AS & A Level Physics revision notes

Inverse square law; 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 luminosity that is known to be 4.8 × 1029 W. A scientist observing this star finds that the radiant flux intensity of light received on Earth from the star is 2.6 nW m–2. Determine the distance of the star from Earth.

Answer:

Step 1: Write down the known quantities

Luminosity, L = 4.8 × 1029 W

Radiant flux intensity, F = 2.6 nW m–2 = 2.6 × 10–9 W m–2

 Step 2: Write down the inverse square law of flux

F equals fraction numerator L over denominator 4 pi d squared end fraction

Step 3: Rearrange for distance d, and calculate

d equals square root of fraction numerator L over denominator 4 pi F end fraction end root equals square root of fraction numerator 4.8 space cross times space 10 to the power of 29 over denominator 4 pi space cross times space left parenthesis 2.6 space cross times space 10 to the power of negative 9 end exponent right parenthesis end fraction end root space equals space 3.8 space cross times space 10 to the power of 18 space m

For teachers

Ready to test your students on this topic?

  • Create exam-aligned tests in minutes
  • Differentiate easily with tiered difficulty
  • Trusted for all assessment types
Explore Test Builder
Test Builder in a diagram showing questions being picked from different difficulties and topics, and being downloaded as a shareable format.
Katie M

Author: Katie M

Expertise: Physics Content Creator

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.