# Distance-Intensity Calculation

**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}

$D_{2}=\sqrt{\frac{I_{1}D_{1}^{2}}{I_{2}}}$

Plug in the known values and solve

$D_{2}=\sqrt{\frac{456,000mR/h \times (1ft)^{2}}{100mR/h}}$

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.