Calculating Intensity with the Inverse Square Law

I1/ I2 = D22/ D1

2

Where:

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 = I1 x D1

2 / 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.

-Radiographic Inspection - Formula Based on 

Newton's Inverse Square LawIn radiographic inspection, the radiation spreads out as it travels away from the gamma or X-ray source.  Therefore, the intensity of the radiation follows Newton's Inverse Square Law.  As shown in the image to the right, this law accounts for the fact that the intensity of radiation becomes weaker as it spreads out from the source since the same about of radiation becomes spread over a larger area.  The intensity is inversely proportional to the distance from the source. 

In industrial radiography, the intensity at one distance is typically known and it is necessary to calculate the intensity at a second distance.  Therefore, the equation takes on the form of:

Where:

I1 = Intensity 1 at D1

I2 = Intensity 2 at D2

D1 = Distance 1 from source

D2 = Distance 2 from source

Note: This is the commonly found form of the equation.  However, for some it is easier to remember that the intensity time the distance squared at one location is equal to the intensity time the distance squared at another location.  The equation in this form is:

I1 x d12 = I2 x d2

2

Example 1)  Use Newton's Inverse Square Law to calculate the intensity of a radioactive source at a different distance than the distance it was originally measured.  If the intensity of a Iridium 192 source was found to be 62 milliroentgen/hour 100 feet, what is the exposure at a distance of 1 foot.

   

Where:I1 = Intensity at D1

I2 = Intensity at D2

D1 = Distance 1

D2= Distance 2

Reworking the equation to solve for I2

Substitute in the known values and solve for I2

 

Example 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 Rem per hour to mRem per 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 68 feet, the 5mR/h boundary would be 301.99 feet, and the 2mR/h boundary would be 477.5 feet.