Lessons â€ē 3 Waves and Optics â€ē 3.1 Wave PhenomenaPrevious lessonNext lesson

3.1.4 Inverse Square Law

By the end of this lesson you can:

  • Explain why wave intensity falls with the square of distance
  • Calculate intensity at different distances
  • Apply the law to sound and light sources
Find your course: NSW Year 11: Light and soundNSW Year 12QLD Unit 2: WavesVICAPIB
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Learn the idea: a short explanation, then a check.

I am sure you have noticed that the apparent brightness of a light source changes depending on your distance from it. For example, a car headlight can appear dimmer than a candle if the candle is right next to your eye while the car is far away.

A fundamental principle of waves is that while the total energy emitted by a source remains constant, its intensity depends on distance:

  • Power (P): A light source emits energy (E) over time, giving its power output in Watts (W = J/s). Power is an intrinsic property of the source and is independent of viewing distance.
  • Intensity (I): Intensity is defined as power per unit area (I = P / A), measured in W/m2. Unlike total power, intensity depends directly on distance.

As light radiates outward in a spherical shell, the surface area expands (A = 4Ī€r2). Consequently, as distance increases by a factor of r, the observed intensity decreases by a factor of r2—a relationship known as the Inverse Square Law. For instance, doubling your distance reduces the observed intensity to one-quarter (1/4) of its original value (I ∝ 1/r2).

Watch the following video which explains this in greater detail.
📝Practise
Worked examples first, then problems to try, then worksheets.

We are now ready to try a sample problem
Below is a sample problem with a video that explain how to solve it. It is suggested you try the problem beforehand, as this actually aids understanding, even if you are unsure if you are correct.

Picture

Check Your Understanding: Inverse Square Law

1. What does the inverse square law say about the intensity of light from a point source?
2. A lamp is moved to twice the distance from a screen. What happens to the intensity of the light on the screen?
3. What is the main reason the intensity of light from a point source falls as you move further away?

Problems to Try

1. The intensity of light 2.0 m from a small lamp is 8.0 W/m2. What is the intensity at a distance of 4.0 m?

Given: I1 = 8.0 W/m2, r1 = 2.0 m, r2 = 4.0 m

Formula: I1r12 = I2r22 → I2 = I1 (r1 / r2)2

Calculation: I2 = 8.0 × (2.0 / 4.0)2 = 8.0 × 0.25 = 2.0 W/m2

Answer: 2.0 W/m2
2. A point source produces an intensity of 100 W/m2 at a distance of 1.0 m. At what distance will the intensity be 4.0 W/m2?

Given: I1 = 100 W/m2, r1 = 1.0 m, I2 = 4.0 W/m2

Formula: I1r12 = I2r22 → r2 = r1 √(I1 / I2)

Calculation: r2 = 1.0 × √(100 / 4.0) = 1.0 × √25 = 5.0 m

Answer: 5.0 m
3. A point source radiates 100 W of light equally in all directions. What is the intensity of the light at a distance of 2.0 m from the source?

Given: P = 100 W, r = 2.0 m

Formula: The light spreads over a sphere of area 4Ī€r2, so I = P / (4Ī€r2)

Calculation: I = 100 / (4Ī€ × 2.02) = 100 / 50.27 ≈ 1.99 W/m2

Answer: 2.0 W/m2
Worksheet
🕹ī¸Explore
Test your predictions in a simulation.
🍎For teachers
Ideas for using this lesson in class, and where to find the answers.

The above experiment can be done as part of online work at home, but can also be done in class. You can access a worksheet to go with this lab.