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v0.1.1065
NotesPhysics HLTopic 2.2Solar radiation, intensity and the solar constant
Back to Physics HL Topics
2.2.12 min read

Solar radiation, intensity and the solar constant

IB Physics • Unit 2

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Contents

  • What intensity is
  • Working out the intensity
  • Exam-style question
The big idea: Intensity is how much radiation power lands on each square metre.

The Sun pours out power in all directions. The further away you are, the more thinly that power is spread — so the intensity you receive is smaller.

Unit: W m⁻² (watts per square metre).

[Diagram: phys-formula-triangle] - Available in full study mode

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.

Intensity = the radiation power divided by the area it is spread over. It is a simple quotient, so use a formula triangle.

Intensity = power ÷ area. Given in the data booklet.
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:
Power spread over a sphere of radius d (so A = 4πd²). This is the inverse-square law: I ∝ 1/d².
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.

Worked example — intensity from a lamp

A small lamp radiates 60 W of light equally in all directions. Find the intensity 2.0 m away.

Solution

  1. Use the given formula for power spread over a sphere:
  2. Put in the numbers (P = 60, d = 2.0):
  3. Work it out — keep the unit:

Final answer

I = 1.2 W m⁻² (about 1.19 W m⁻²).

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How this is tested: The solar constant and intensity turn up in the climate/energy-balance questions.

- Paper 1A: quick MCQs — e.g. a solar panel's power output from incident 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 from the solar constant using P = I × 4πd².

Inverse-square trap: 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.

IB-style question — total power radiated by the Sun

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.

Solution

  1. The intensity at Earth's distance is spread over a sphere of radius d, so:
  2. Put in the numbers (I = S = 1.36 × 10³, d = 1.5 × 10¹¹):
  3. Work it out — keep the unit:

Final answer

P ≈ 3.8 × 10²⁶ W ≈ 4 × 10²⁶ W. The solar constant is just the Sun's total power spread over a sphere the size of Earth's orbit.

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A small heat lamp radiates 120 W of infrared power equally in all directions.

the intensity of the radiation 4.0 m from the lamp.
[2 marks]

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