Worked example · Physics

Inverse-square law: intensity from a point source

A small source radiates 1,200 W equally in every direction. What is the intensity 3 m away, and how far away has it fallen to 2 W m⁻²? The square is not an arbitrary rule: it comes from the area of the sphere over which the same power is spread.

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Inverse proportion
≈ 7 minutes

Step 1 — on paper

Why doubling distance quarters the intensity

At distance r, radiation from an ideal point source is spread over a sphere of area 4πr². Intensity is power per unit area:

I = \frac{P}{4 π r^{2}}

I intensity in W m⁻² · P total radiated power in W · r distance from the point source in m

Double r and the sphere's area becomes 2² = 4 times larger. The same power is divided among four times the area, so the intensity is one quarter. Triple the distance and intensity becomes one ninth. It is 1/r², not 1/r.

Relative intensity from the same ideal point source.
Distance	Sphere area factor	Intensity factor
r	1	1
2r	4	1/4
3r	9	1/9
10r	100	1/100

Step 2 — on the calculator

Intensity 3 m from a 1,200 W source

Substitute P = 1200 W and r = 3 m. Keep the entire sphere area in the denominator; the brackets make that visible:

1200
÷
(
4
×
π
×
3
x²
)
=

1200÷(4×π×3²) 10.61032954 W m⁻² — 10.6 W m⁻² to three significant figures

Ans on iPhone showing 1200 divided by, open bracket, 4 times pi times 3 squared, close bracket, and the result 10.61032954. — The expression remains above the result, so the squared distance and the whole 4πr² denominator can be checked before trusting the digits.

At 1 m the intensity would be 95.493 W m⁻². At 3 m, one ninth of that is 10.610 W m⁻², providing a quick independent check.

Do not square the power. The square belongs to distance because the receiving sphere grows in two dimensions. The source still emits 1,200 joules each second; only the area over which those joules are spread changes.

Step 3 — turn the formula round

Where does the intensity fall to 2 W m⁻²?

Rearrange before substituting. Multiply by r², divide by I, then take the positive square root because distance is positive:

r = \sqrt{\frac{P}{4 π I}} = \sqrt{\frac{1200}{8 π}} = 6.909882989 m

The threshold is reached at about 6.91 m. Beyond that, the idealised intensity is below 2 W m⁻²; closer than that, it is above. Substitution closes the loop: 1200÷(4π×6.909882989²) ≈ 2.

Intensity from the 1,200 W ideal source.
Distance	Intensity	Compared with 1 m
1 m	95.493 W m⁻²	1
2 m	23.873 W m⁻²	1/4
3 m	10.610 W m⁻²	1/9
6 m	2.653 W m⁻²	1/36
6.910 m	2.000 W m⁻²	about 1/47.75
10 m	0.95493 W m⁻²	1/100

Step 4 — on the graph

Find the threshold as an intersection

Plot the intensity against positive distance and add the 2 W m⁻² threshold. The function rises without bound as x approaches zero, so start the window at 1 m rather than asking the graph to include the point-source singularity. Then keep the ceiling low: y from 0 to 12 still shows the 10.6 W m⁻² reading at 3 m, and leaves the 2 W m⁻² crossing clear of the axis. A window tall enough for the 95 W m⁻² at 1 m would squash that crossing into the bottom edge. Use x from 1 to 10, then choose Intersection.

f₁(x) = 1200/(4*pi*x^2) f₂(x) = 2

Ans Graph showing an inverse-square intensity curve falling with distance and crossing the horizontal 2 watts per square metre line near 6.91 metres. — **Intersection** — the curve meets 2 W m⁻² at about **6.91 m**, agreeing with the rearranged square-root formula.

The plot makes the sensitivity visible. Close to the source, a small move changes intensity sharply; farther away the same move matters much less. The curve never reaches zero, although at a large enough distance other sources and background effects will dominate the ideal model.

When the inverse-square model is valid

The source is small compared with the distance, so it can be treated as a point.
It radiates equally in every direction. A reflector, lens, loudspeaker enclosure or antenna pattern can concentrate power instead.
The medium does not absorb or scatter energy, and there are no reflections or obstacles.
The quoted 1,200 W is the power carried by the radiation of interest, not necessarily the device's electrical input power.
The receiver is small and faces the incoming radiation normally; orientation changes the intercepted power.
Near an extended source or in the electromagnetic near field, geometry can differ and the simple 1/r² law need not apply.

Next: photon energy from wavelength for another radiation calculation, or projectile motion for a model whose assumptions change the answer visibly.

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Worked on Ans — a scientific calculator you can own and Ans Graph. One checks the π-and-square arithmetic; the other shows the whole inverse-square relationship. Paid for once.