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4.6.5Electrons, Waves and Photons

Intensity of a Wave

5 min read 5 parts Lesson 46 of 62 Unit quiz

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Part 1 of 5Defining Intensity

4.6.5 Intensity of a Wave

Intensity tells you how concentrated a wave's energy is where it arrives.

Definition
  • Intensity is the radiant power passing through a surface per unit area perpendicular to the direction of travel.
Equation
  • I = P/A, with I in watts per square metre (W m⁻²), P in watts and A in square metres.
    What the symbols mean: I = intensity (W m−2) · P = power (W) · A = area the power falls on (m2)
Meaning
  • A higher intensity means more energy arrives each second on each square metre of the surface.
Examples
  • Sunlight, a laser beam and a loudspeaker all deliver energy over an area, so their intensities can be compared.

Part 2 of 5Intensity and Amplitude

Intensity and Amplitude
Proportionality
  • For a given wave, intensity is directly proportional to the square of the amplitude: I ∝ A².
Doubling
  • Doubling the amplitude makes the intensity four times greater.
Energy link
  • The energy carried by the wave each second grows as the oscillation becomes larger, so a larger amplitude means more power per area.
Application
  • Louder sounds and brighter light correspond to larger amplitudes of the oscillation.

Part 3 of 5Intensity and Distance

1
A point source sends energy outwards equally in all directions, so the power spreads over a growing sphere.
2
The surface area of a sphere of radius r is 4πr², so intensity is I = P / 4πr².
3
Doubling the distance spreads the same power over four times the area, so the intensity falls to a quarter.
4
This inverse square law only applies to a point source in a medium that does not absorb the energy.
5
A narrow parallel beam, such as a laser, loses intensity far more slowly with distance.

Part 4 of 5Worked Example

Question

  • A small lamp emits light with total radiant power 5.0 W equally in all directions.
  • Calculate the intensity at 2.0 m and at 4.0 m from the lamp.

  1. At 2.0 m: area = 4π × 2.0² = 50.3 m², so I = 5.0 / 50.3 = 0.099 W m⁻².
  2. At 4.0 m: area = 4π × 4.0² = 201 m², so I = 5.0 / 201 = 0.025 W m⁻².
  3. The distance doubled, and the intensity fell to one quarter: 0.099 / 4 ≈ 0.025 W m⁻².

Answer 0.099 W m⁻² at 2.0 m and 0.025 W m⁻² at 4.0 m.

Part 5 of 5Exam Checklist & Tips

4.6.5 – Key Points for the Exam

  • Intensity Radiant power per unit area, measured in W m⁻².
  • Inverse square law For a point source, intensity is inversely proportional to the square of the distance.
  • Amplitude
    • The maximum displacement of a wave;
      • intensity is proportional to its square.
  • Point source A source small enough that its energy spreads out uniformly in all directions.

Common Mistakes

  • Saying intensity is proportional to amplitude; it is proportional to amplitude squared.
  • Using the circle area πr² instead of the sphere area 4πr².
  • Treating intensity and power as the same quantity.
  • Applying the inverse square law to a parallel beam.
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