The big idea: Stand right by a campfire and its warmth is fierce; step back and the same fire barely reaches you — how much radiation power lands on each square metre is its intensity.
Intensity and distance
- The Sun pours out power in all directions.
- The further away you are the more thinly it spreads, so the intensity you receive is smaller.
- Unit: W m⁻² (watts per square metre).
Intensity triangle: P = power (W), I = intensity (W m⁻²), A = area (m²). Cover the one you want — two side by side → multiply; one above the other → divide.
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Spreading thins it out: The Sun's power spreads over the surface of an ever-growing sphere as it travels outward.
Same total power, bigger area → smaller intensity. That is why it is hotter on Mercury than on Earth.
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Intensity = the radiation power divided by the area it is spread over. It is a simple quotient, so use a formula triangle.
- intensity (W m⁻²)
- radiation power passing through the area (W)
- area the power spreads over (m²)
Spreading over a sphere: A source sending power out equally in all directions spreads it over the surface of a sphere.
A sphere of radius d has area A = 4πd², so the intensity a distance d away is:
- intensity at the distance d (W m⁻²)
- total power radiated by the source (W)
- distance from the source (m)
The solar constant: The solar constant is the intensity of the Sun's radiation arriving at Earth's distance, measured just above the atmosphere.
Its value is S = 1.36 × 10³ W m⁻² — given in the data booklet.
Luminosity and apparent brightness: For a star the total power radiated is its luminosity L (in watts, or in multiples of the Sun's luminosity L⊙). The intensity it produces at a distance d is its apparent brightness b:
b = L ÷ 4πd² (given in the data booklet). The same star looks four times dimmer from twice as far away.
A small lamp radiates 60 W of light equally in all directions. Find the intensity 2.0 m away.
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Power from a solar panel: Power falling on the panel = intensity × area. Electrical power out = that × efficiency.
900 W m⁻² on 1.5 m² at 20 % efficiency: 900 × 1.5 × 0.20 = 270 W. Turn kW m⁻² into W m⁻² first.
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How this is tested — the solar constant and intensity turn up in the climate / energy-balance questions:
Paper 1A
- A solar panel's power output from intensity × area × efficiency.
- Or how intensity changes when you double the distance.
Paper 2
- A 'state' mark for what the solar constant means.
- Then a 'show that' — work the Sun's total power back with P = I × 4πd².
The classic trap: Inverse-square: double the distance → intensity drops to a quarter, not a half (the d is squared).
Power, intensity and a sphere: To get the total power a source radiates, multiply the intensity at distance d by the whole sphere area 4πd²:
P = I × 4πd².
For solar panels, the useful output is the incident power on the panel × its efficiency.
The words to know
| Word | What it means |
|---|---|
| Intensity | Power per unit area perpendicular to the rays (W m⁻²) |
| Luminosity L | Total power radiated by a star (W) |
| Apparent brightness b | Intensity received from a star, b = L/4πd² |
| Solar constant S | Intensity of the Sun's radiation at the Earth's distance, above the atmosphere (about 1.36 × 10³ W m⁻²) |
The solar constant is S = 1.36 × 10³ W m⁻². The Earth orbits the Sun at a distance d = 1.5 × 10¹¹ m. Show that the total power radiated by the Sun is about 4 × 10²⁶ W.
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