1. INVERSE SQUARE LAW

2. The picture above demonstrates the typical x-ray tube used to produce a point source of x-rays. Then as radiation exits the tube it diverges to cover an increasingly larger area as the distance from the source increases. Notice that area "A" is smaller and the radiation is more concentrated than in an equal area "A1" which is some distance from "A." Each square A1 is the same size as "A" but only 1/4 the number of photons occupies it because of the divergence of the radiation with increasing distance.

5. In spite of the advances in radiation protection, such as collimators, cones, and positive beam limiting devices, distance is still the best tool for radiation protection and remains the most common method of protecting personnel, visitors, and adjacent patients from ionizing radiation use. But few persons in the health care environment understand why distance effectively protects them and therefore continuously question, “At what distance am I considered safe? The answer lies in understanding the relationship of ones distance from a source to exposure intensity. The type(s) of radiation one is exposed to as well as its energy content are also factors that affect personal dose. A safe distance can be accurately estimated from the vector of radiation exposure and its initial intensity using the inverse square law. The radiographer should note that this law applies only to a point source of radiation such as the primary beam. Additionally, the inverse square law applies only to electromagnetic radiation (x-rays and gamma rays), and does not apply to particulate ionizing radiation, or scatter radiation which is the major type of occupational radiation exposure personnel should encounter.

7. What is the inverse Square Law Formula? I2I2 I1I1 = (D 1 ) 2 (D 2 ) 2

8. Problem 1: For a given technique the x-ray intensity at 36 cm distance is 400 mR. What is the intensity if source of radiation tube was moved to 18 cm? old intensity I 1 In the preceeding problem identify: old distance D 1 new intensity I 2 new distance D 2 I2I2 I1I1 = (D 1 ) 2 (D 2 ) 2 36400 18 Nexthint ?

9. Animations http://highered.mcgraw- hill.com/sites/dl/free/007299181x/59233/6_2b.htm

10. Virtual Lab Link http://astro.unl.edu/classaction/animations/stellarprops/l ightdetector.html Demonstrates the inverse square law of light with a lightbulb and detector. The lightbulb's intensity and the detector's distance can be adjusted to see how they affect the reading. There are two bulbs and detectors to allow side-by-side comparisons