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Intensity

Key idea: Define intensity as power per unit area, use I ∝ A², and apply the inverse-square law for point sources (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Use wave intensity, amplitude and inverse-square relationships with their assumptions.

1. Definitions (Must Know)

A. Intensity, I (W m⁻²)

Intensity, I, is the power transferred (radiated) per unit area, where the area is perpendicular to the direction of wave travel: I = P/A

B. Point source (inverse-square model)

If a source radiates power P equally in all directions (isotropic point source), then at distance r the wavefront is a sphere of area 4π r², so: I = P/(4π r²)

C. Intensity–amplitude relationship

For a progressive wave in the same medium (and no losses): I ∝ A²

2. Key Ideas (What Earns Marks)

  • Always state the model/assumptions:
    • inverse-square law needs “point source” + “no energy loss” + “spreads uniformly”.
  • Use ratio form to avoid algebra mistakes: I₁/I₂ = (r₂/r₁)²
  • If amplitude changes but the medium is the same: I₁/I₂ = (A₁/A₂)²
  • For EM waves, intensity relates to brightness; for sound, intensity relates to loudness (qualitatively).
Units check

I = P/A has units W m⁻² because P is in watts and area is in m².

3. Detailed Explanations

A. Why intensity follows an inverse-square law (point source)

If the same power P spreads over a larger spherical area as you move away:

Aₛₚₕₑᵣₑ = 4π r² ⇒ I = P/(4π r²) ⇒ I ∝ 1/r²

Inverse-square law (normalised)

Normalised intensity from a point source decreases as 1 over r squared.

Scroll across the graph to read all labels.

Normalised intensity from a point source decreases as 1 over r squared.Normalised intensity from a point source decreases as 1 over r squared.
For a point source with no absorption, doubling distance makes intensity one-quarter: I ∝ 1/r².
Open full-size graph
View figure data
Values for Inverse-square law (normalised)
Distance ratio, r/r₀ (unitless)I/I₀ = 1/(r/r₀)²
11
20.25
30.111
40.0625
50.04
60.0278

B. Why intensity depends on amplitude squared

Intensity measures the rate of energy transfer.

For many waves, doubling the amplitude makes the oscillation energy much larger, and the result is: I ∝ A²

You do not usually need to derive this relationship; you use it as a proportionality.

4. Common Mistakes

  • Using 4π r instead of 4π r² for spherical spreading.
  • Using the inverse-square law when the source is not a point source (or when energy is absorbed).
  • Forgetting that doubling distance makes intensity one-quarter (not half).
  • Mixing up amplitude ratios and intensity ratios.

5. Exam Tips

  • If the question gives two distances, use the ratio form: I₁/I₂ = (r₂/r₁)²
  • If the question asks for “how many times brighter/louder”, it’s usually an intensity ratio.
  • State your assumption: “treat the source as a point source radiating uniformly”.

6. Worked Examples

Modelled example 1

Intensity at a distance from a point source

Core

Problem

A small lamp radiates 60 W uniformly in all directions. Find the intensity at 2.0 m.
Study the worked solution

  1. Select the spreading model

    Method

    Use I = P/(4π r²).

    Reason

    An isotropic point source spreads the same power over a spherical wavefront.

    Working

    Aₛₚₕₑᵣₑ = 4π r²

  2. Substitute

    Method

    Divide the power by the area of the sphere at 2.0 m.

    Reason

    Intensity is power per area perpendicular to propagation.

    Working

    I = 60/4π(2.0)²

  3. Evaluate

    Method

    The intensity is approximately 1.19 W m⁻².

    Reason

    The source power is distributed over about 50.3 m².

    Working

    I ≈ 1.19 W m⁻²

Guided practice 2

Comparing intensities at two distances

About 4 min

Problem

A point-source wave has intensity I₁ at 3.0 m. Find I₂ at 12.0 m in terms of I₁.

Try this before viewing the solution

Hints

Hint 1: use the ratio model
I₂/I₁ = (r₁/r₂)² for unchanged isotropic power.
View solution step by step

  1. Find the distance factor

    Method

    The new distance is four times the original distance.

    Reason

    12.0/3.0 = 4.

    Working

    r₂ = 4r₁

  2. Apply inverse-square scaling

    Method

    The intensity becomes one sixteenth of its original value.

    Reason

    Intensity varies with the reciprocal square of distance.

    Working

    I₂/I₁ = (1/4)² = 1/16

Common misconception 3

Amplitude change and intensity

Find and correct the mistake

Learner claim

In the same medium, amplitude falls to 0.30A₁. A learner says intensity therefore falls to 0.30I₁. Diagnose the claim.

Try this before viewing the solution

View solution step by step

  1. Use the correct proportionality

    Method

    I ∝ A² in the same medium.

    Reason

    Wave energy transfer scales with the square of oscillation amplitude.

    Working

    I₂/I₁ = (A₂/A₁)²

  2. Square the amplitude factor

    Method

    The intensity becomes 0.090I₁.

    Reason

    (0.30)² = 0.090.

    Working

    I₂ = 0.090I₁

Examiner practice 4

Distance change for a required intensity drop

3 marks

Examination question

For a point source, intensity is I₁ at r₁. At what distance is it I₁/9? [3 marks]

Try this before viewing the solution

View solution step by step

  1. Set the inverse-square ratio

    1 mark

    Method

    Relate the intensity fraction to the squared inverse distance ratio.

    Reason

    The point-source power is assumed unchanged and uniformly spread.

    Working

    1/9 = (r₁/r₂)²

  2. Take the physical square root

    1 mark

    Method

    r₁/r₂ = 1/3.

    Reason

    Distances are positive magnitudes.

    Working

    square root of (1/9) = 1/3

  3. State the distance

    1 mark

    Method

    r₂ = 3r₁.

    Reason

    Tripling distance makes intensity one ninth.

    Working

    r₂ = 3r₁

Challenge 5

Using intensity change to infer amplitude change

Minimal support

Independent transfer

A point source produces I₁ at 2.0 m and I₂ at 8.0 m. Find A₂/A₁ in the same medium with no absorption.

Try this before viewing the solution

Hints

Hint 1: link two squared relationships
Use I ∝ 1/r² and I ∝ A² in sequence.
View solution step by step

  1. Find the intensity ratio

    Method

    I₂/I₁ = 1/16.

    Reason

    Distance increases by four and point-source intensity follows inverse-square spreading.

    Working

    I₂/I₁ = (2.0/8.0)² = 1/16

  2. Convert intensity to amplitude

    Method

    A₂/A₁ = 1/4.

    Reason

    Intensity ratio is the square of amplitude ratio in the same medium.

    Working

    A₂/A₁ = square root of (1/16) = 1/4

7. Mind Stretchers

Mind stretcher 1: Estimating power from intensity dataExtension

At a distance r from a point source, the measured intensity is I. Find the radiated power P in terms of I and r.

Show Answer

Rearrange I = P/(4π r²): P = 4π r² I

Mind stretcher 2: When does inverse-square fail?Extension

Give two reasons why measured intensity might not follow an exact inverse-square law in a real situation.

Show Answer

Examples include:

  • the source is not a point source / does not radiate uniformly (directional source)
  • energy is absorbed or scattered by the medium (so power is not conserved with distance)
  • reflections/interference in the environment change the measured intensity at a point

Mind stretcher 3: Optional (Enrichment)Extension

A. Decibels (sound level)

Sound level is often reported in decibels (dB) using a logarithmic scale. You generally do not need the decibel formula unless a question explicitly provides it.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027