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Calculating Intensity with the Inverse Square Law
 I1/ I2 = D22/ D12
| Where: |
| I1 |
= |
Intensity 1 at D1 |
| I2 |
= |
Intensity 2 at D2 |
| D1 |
= |
Distance 1 from source |
| D2 |
= |
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.
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