# 5.40 Inverse Square Law of Flux

## 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

• 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

#### 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.

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  ### Get unlimited access

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