Calculating Intensity with the Inverse Square LawI1/ I2= D22/ D12Where:

I1=Intensity 1 at D1

I2=Intensity 2 at D2

D1=Distance 1 from source

D2=Distance 2 from source

Example Calculation 1The 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 I2= I1x D12/ D22Plug in the known values I2= 530R/h x (5ft)2/ (10ft)2Solve for I2 I2= 132.5 R/hIn this instance the distance has been doubled and the intensity at that point has decreased by a factor of four.Example Calculation 2A 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/hRework the equation to solve for D2Plug in the known values and solveD2= 67.5 feetUsing 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.