Inverse-Square Law
Lesson: Inverse-Square Law in Diagnostic Radiology
Introduction
The inverse-square law is a fundamental principle of radiation physics that describes how radiation intensity changes with distance from a source. When radiation spreads uniformly from a point source, its intensity decreases as the square of the distance increases.
This law is important in diagnostic radiology because it helps explain radiation exposure, the effect of distance on occupational exposure, and the adjustment of radiographic exposure factors when the source-to-image distance (SID) changes. It is an ideal relationship; actual measurements can also be affected by attenuation, scattering, and room geometry.
Definition
The inverse-square law states that the intensity of radiation at a point is inversely proportional to the square of the distance between that point and a point source, assuming constant source output and negligible attenuation.
Here, I represents radiation intensity and d represents the distance from the source. The symbol ∝ means “is proportional to.”
Physical Explanation
A point source emits radiation in different directions. As the radiation travels outward, it spreads over the surface of an expanding sphere. The surface area of a sphere is:
Because the surface area increases with the square of the radius, the same total radiation output is distributed over a larger area at greater distances. Therefore, the intensity received per unit area decreases according to the inverse square of the distance.
Diagram: Radiation Intensity and Distance
The diagram shows that when distance increases from r to 2r and 3r, the intensity falls to one-quarter and one-ninth, respectively.
Schematic illustration: the circles indicate increasing distances from the source. Radiation from an ideal point source spreads over spherical surfaces.
Mathematical Formula
Comparing Two Distances
When intensity at one distance is known, the intensity at another distance can be calculated using:
- I₁: Initial intensity.
- I₂: Intensity at the new distance.
- d₁: Initial distance from the source.
- d₂: New distance from the source.
Remember: In the intensity formula, the distance ratio is initial distance divided by new distance. Square the entire ratio.
Important Distance Rules
| Change in distance | New intensity | Intensity remaining |
|---|---|---|
| Distance stays the same | I | 100% |
| Distance doubles | I / 4 | 25% |
| Distance triples | I / 9 | 11.11% |
| Distance becomes four times greater | I / 16 | 6.25% |
| Distance is halved | 4I | 400% |
Applications in Diagnostic Radiology
1. Radiation Protection
Increasing distance from an X-ray source or a patient producing scatter radiation can reduce exposure. For an ideal point source, doubling the distance reduces intensity to one-quarter. Distance should be combined with reducing exposure time and using suitable shielding.
2. Source-to-Image Distance (SID)
When SID increases, the intensity reaching the image receptor decreases if the X-ray tube output remains unchanged. To maintain approximately the same receptor exposure, mAs can be adjusted using the distance compensation formula:
This calculation assumes other relevant exposure conditions remain unchanged. Follow local protocols and equipment guidance when selecting clinical exposure factors.
3. Occupational Exposure
Radiographers may reduce exposure by moving farther from the patient and X-ray tube during exposure when appropriate. Actual scatter radiation depends on patient size, beam direction, room layout, and shielding, so the inverse-square law is an approximation for real clinical environments.
Radiation intensity is 80 ฮผGy/h at a distance of 1 metre. Calculate the intensity at a distance of 2 metres.
- I₁ = 80 ฮผGy/h
- d₁ = 1 m
- d₂ = 2 m
Use the inverse-square law formula:
Final answer: 20 ฮผGy/h.
Explanation: The distance doubles, so intensity decreases to one-quarter of the original value.
Radiation intensity is 100 ฮผGy/h at 2 metres. At what distance will the intensity become 25 ฮผGy/h?
- I₁ = 100 ฮผGy/h
- I₂ = 25 ฮผGy/h
- d₁ = 2 m
Final answer: 4 metres.
Explanation: The intensity becomes one-quarter when the distance doubles from 2 metres to 4 metres.
An X-ray exposure uses 10 mAs at a SID of 100 cm. Calculate the mAs required at a SID of 150 cm to maintain approximately the same detector exposure.
- mAs₁ = 10
- d₁ = 100 cm
- d₂ = 150 cm
Final answer: Approximately 22.5 mAs.
Explanation: The distance ratio is 1.5, so the mAs is multiplied by 1.5², or 2.25.
Intensity is 120 ฮผGy/h at 3 metres. Calculate the intensity at 6 metres.
Solution:
Answer: 30 ฮผGy/h.
An intensity of 64 units is measured at 4 metres. Calculate the intensity at 2 metres.
Solution:
Answer: 256 units.
An X-ray exposure requires 8 mAs at 80 cm SID. Calculate the mAs required at 160 cm SID for approximately the same detector exposure.
Solution:
Answer: 32 mAs.
Conditions and Limitations
- Point-source approximation: The source should be small compared with the distance being considered.
- Constant source output: Source output must remain unchanged between measurements.
- Uniform spreading: Radiation should spread approximately uniformly in space.
- Attenuation: Absorption and scattering by air, tissue, shielding, or other materials can change measured intensity.
- Scatter radiation: Scattered radiation from patients and nearby surfaces can cause real measurements to differ from ideal predictions.
Conclusion
The inverse-square law explains the relationship between radiation intensity and distance. It is essential for understanding radiation protection and radiographic exposure calculations. Always remember that intensity varies inversely with the square of distance, and apply the formula carefully by squaring the distance ratio.
Educational note: This material is intended for learning radiation physics. Clinical exposure settings and radiation protection decisions should follow local protocols and equipment guidance.
2× farther → ¼ intensity · 3× farther → ⅑ intensity
Inverse-Square Law — 10 MCQs
Choose the best answer for each question. Answers and explanations are provided on the answer sheet below.
