• 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:

• ?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; when the light is twice as far away, it has spread over four times the area, hence the intensity is four times smaller

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

Step 3:?Rearrange for distance?d, and calculate