Calculating Intensity with the Inverse Square Law
I1/ I2 = D22/ D12
||Intensity 1 at D1
||Intensity 2 at D2
||Distance 1 from source
||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
I2 = I1 x D12 / D22
Plug in the known values
I2 = 530R/h x (5ft)2 / (10ft)2
Solve for I 2
I2 = 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 D2
Plug in the known values and solve
D2= 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.