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Inverse Square Law Calculator

Your numbers

Change any value to update the result.

Intensity at d₂

25.0000

Intensity at d₂: 25.0000
Examples
How the result moves

Your current valid inputs, with one value varied across the range.

Distance you want, d₂ → Intensity at d₂
Distance you want, d₂: 2; Intensity at d₂: 25

How the result moves

Held fixed: Known intensity I₁ 100.0000, Distance it was measured at, d₁ 1.000.

Distance you want, d₂Intensity at d₂
1.50044.4444
1.75032.6531
2.000Your value25.0000
2.25019.7531
2.50016.0000

Intensity falls with the square of the distance: I₂ = I₁ × (d₁ ÷ d₂)². A reading of 100 at 1 m is 25 at 2 m and 11.11 at 3 m. Moving closer works the same way in reverse — half the distance gives four times the intensity. Both distances must use one unit.

Worked examples

Case 1

Known intensity I₁
100
Distance it was measured at, d₁
1
Distance you want, d₂
2

25.0000

Open with these values

Case 2

Known intensity I₁
1000
Distance it was measured at, d₁
2
Distance you want, d₂
10

40.0000

Open with these values

Case 3

Known intensity I₁
500
Distance it was measured at, d₁
5
Distance you want, d₂
1

12,500.0000

Open with these values

How it's calculated

I₂ = I₁ × (d₁ ÷ d₂)²

  1. StepEnter the intensity you know and the distance it was measured at.Known intensity I₁=100·Distance it was measured at, d₁=1
  2. StepEnter the distance you want, in the same length unit.Distance you want, d₂=2
  3. ResultDivide d₁ by d₂, square it, multiply by I₁ — that is the answer.25.0000

Reference table

I₁, d₁, d₂What happensIntensity at d₂
100, 1, 1Same spot, nothing changes100.0000
100, 1, 2Twice as far — a quarter25.0000
100, 1, 3Three times as far — a ninth11.1111
1000, 2, 10Five times as far — a twenty-fifth40.0000
500, 5, 1Five times closer — twenty-five times12500.0000

Questions

What is the inverse square law?

The intensity of a point source — light, sound, gravity or radiation — falls with the square of the distance. The same energy spreads over a sphere whose area grows as distance squared, so the intensity per unit area drops as 1/d². Twice as far gives a quarter of the intensity.

How do I calculate the intensity at a new distance?

Multiply the known intensity by the square of the distance ratio: I₂ = I₁ × (d₁/d₂)². For 100 units at 1 m, the value at 2 m is 100 × 0.25 = 25.

What happens when you double the distance?

It drops to one quarter, because the ratio is 1/2 and its square is 1/4. Tripling gives one ninth and quadrupling one sixteenth. Halving the distance quadruples the intensity instead.

What units does this use?

The intensity can be any linear unit — lux, watts per square metre, a dose rate, raw counts — and the result comes back in that same unit. The two distances must share one unit, because they appear only as the ratio, which cancels it. Decibels do not work: that scale is logarithmic.

When does the inverse square law not apply?

It assumes an ideal point source radiating freely into empty space. It breaks down for focused beams such as spotlights, lasers and directional antennas, and very close to a source that is large compared with the distance. Reflection and absorption — a reverberant room, fog, shielding — change the fall-off too.

Sources and last check

  1. en.wikipedia.org

Information, not professional advice.