Case 1
- Known intensity I₁
- 100
- Distance it was measured at, d₁
- 1
- Distance you want, d₂
- 2
25.0000
Open with these valuesInverse Square Law Calculator
25.0000
Result: 25.0000Intensity 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 2
- Known intensity I₁
- 1000
- Distance it was measured at, d₁
- 2
- Distance you want, d₂
- 10
40.0000
Open with these valuesCase 3
- Known intensity I₁
- 500
- Distance it was measured at, d₁
- 5
- Distance you want, d₂
- 1
12,500.0000
Open with these valuesHow it's calculated
I₂ = I₁ × (d₁ ÷ d₂)²
- StepEnter the intensity you know and the distance it was measured at.
- StepEnter the distance you want, in the same length unit.
- ResultDivide d₁ by d₂, square it, multiply by I₁ — that is the answer.
Reference table
| I₁, d₁, d₂ | What happens | Intensity at d₂ |
|---|---|---|
| 100, 1, 1 | Same spot, nothing changes | 100.0000 |
| 100, 1, 2 | Twice as far — a quarter | 25.0000 |
| 100, 1, 3 | Three times as far — a ninth | 11.1111 |
| 1000, 2, 10 | Five times as far — a twenty-fifth | 40.0000 |
| 500, 5, 1 | Five times closer — twenty-five times | 12500.0000 |
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
Information, not professional advice.