Calculating Intensity with the Inverse Square Law I_{1}/ I_{2} = D_{2}^{2}/ D_{1}^{2}
Where: 
I_{1} 
= 
Intensity 1 at D_{1} 
I_{2} 
= 
Intensity 2 at D_{2} 
D_{1} 
= 
Distance 1 from source 
D_{2} 
= 
Distance 2 from source 
Example Calculation 1 The intensity of radiation is 530 R/h at 5 feet away from a source. What is the intensity of the radiation at 10 feet? Rework the equation to solve for the intensity at distance 2 I_{2} = I_{1} x D_{1}^{2} / D_{2}^{2} Plug in the known values I_{2} = 530R/h x (5ft)^{2 }/ (10ft)^{2} Solve for I _{2} I_{2} = 132.5 R/h In this instance the distance has been doubled and the intensity at that point has decreased by a factor of four. Example Calculation 2 A source is producing an intensity of 456 R/h at one foot from the source. What would be the distance in feet to the 100, 5, and 2 mR/h boundaries. Convert R/hour to mR/hour 456R/h x 1000 = 456,000 mR/h Rework the equation to solve for D_{2} Plug in the known values and solve D_{2}= 67.5 feet Using this equation the 100mR/h boundary would be at 68 feet, the 5mR/h boundary would be at 301.99 feet, and the 2mR/h boundary would be at 477.5 feet. Sources are seldom operated for an entire hour, and collimators are often used which reduce these distances considerably.
