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.
