Why intensity falls with the square of distance
A point source radiates energy uniformly in all directions, creating a spherical wavefront. The total power emitted (P) is constant and spread over the surface of that sphere. The surface area of a sphere grows as the square of the radius: A = 4πr². Therefore, the intensity — power per unit area — at any distance r is I = P / (4πr²). Doubling the distance quadruples the area, so intensity drops to one quarter.
This gives the inverse-square law directly: I ∝ 1/r². Taking the ratio at two distances d₁ and d₂ eliminates the source power P and yields the dimensionless form used by this calculator: I₂/I₁ = (d₁/d₂)². The formula works for any intensive quantity that radiates isotropically from a point: radiation dose rate (mSv/h, mrem/h), light illuminance (lux), sound intensity (W/m²), gravitational field strength, or electrostatic field intensity.
The inverse formula solves for the required distance: d₂ = d₁ × √(I₁/I₂). This answers the question: how far must I be from this source to reduce the dose rate to my target value?
When the inverse-square law applies (and when it doesn't)
The law applies strictly when three conditions hold: (1) the source is a point — or can be approximated as one because the observation distance is much larger than the source dimensions; (2) emission is isotropic — the source radiates equally in all directions; (3) there is no significant attenuation by the medium — absorption and scattering are negligible between d₁ and d₂.
The law does NOT apply to: (a) collimated beams — a laser or a tightly focused X-ray beam does not spread spherically, so intensity can remain nearly constant with distance; (b) extended sources — a large gamma-emitting tank, a long pipe, or a surface source; the law underestimates dose near the source and overestimates it at great distance; (c) media with significant attenuation — water, concrete, lead, or dense air paths where scattering and absorption reduce intensity faster than the geometric 1/r² term.
For radiation, air attenuation is generally negligible for high-energy gamma (>1 MeV) at distances under ~10 m, making the law a reasonable first approximation in open air for such sources. For low-energy photons (X-rays, low-energy gamma) or distances over tens of metres, air attenuation becomes significant.
Practical use in radiation protection — and its limits
In radiation protection, the inverse-square law provides a quick estimate for the dose rate at a new distance when a measurement is already known at a reference distance. This is valuable for surveying sources and planning safe working distances under the ALARA principle (As Low As Reasonably Achievable). The calculation is fast, requires only two values (I₁ and d₁), and gives an order-of-magnitude result useful for initial planning.
However, for real shielding calculations, this tool is not sufficient. Shielding design requires the full attenuation equation: I₂ = I₁ × e^(−μx) / (d₂/d₁)², where μ is the linear attenuation coefficient of the shielding material. The half-value layer (HVL) concept lets you determine the thickness of a specific material (lead, concrete, water) needed to reduce intensity by half. Multiple HVLs give multiplicative reduction. Neither of these is handled by this calculator.
References: ICRP Publication 103 (2007); NCRP — National Council on Radiation Protection and Measurements; Cember H, Johnson TE — Introduction to Health Physics, 4th ed.; ACGIH TLV/BEI Booklet — Physical Agents — Ionizing Radiation. For regulatory compliance, always verify applicable dose limits with the relevant authority (Health Canada, NRC, IRSN, etc.).
Frequently Asked Questions
Is this the same as the noise distance attenuation tool?
Does this work for radiation, light, and sound?
What is a point source?
Why doesn't this apply at very short distances?
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