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i've got a bit of a problem with the inverse square law (I1/I2=D2 squared/D1 squared)(Where I=intensity and D=distance)

I need to change a distance from 1000mm to 400mm (I'm a Radiographer). Most of our current techniques are set at 1000mm, but one of our clients has asked if we can reduce the distance we use down to 400mm. I'm having a bit of trouble converting our old techniques from 1000mm to 400mm, if I reduced the distance from 1000mm to 500mm i'd simply 1/4 the time per exposure to get similar results.

I was just wondering if there is any way of changing the equation so that it deals with 1/5 reductions in distance as opposed to 1/2, it would make my job a tiny bit easier. Thanks in advance.

RandomRadiographer.
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1 Answers1

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$$ \frac{I_1}{I_2} = \frac{D_2^2}{D_1^2} $$

so if we have $D_0$ and $I_0$ as our default values then we want to reduce the distance by $1/5$ for example thus $$ D_{\text{new}} = \frac{1}{5}D_0 = D_2 $$

we then have

$$ \frac{I_0}{I_{\text{new}}} = \frac{D_{\text{new}}^2}{D_0^2} = \frac{\left(\frac{1}{5}D_0\right)^2}{D_0^2} = \frac{\left(\frac{1}{5}\right)^2D_0^2}{D_0^2} = \frac{1}{25} $$ thus

$$ 25I_0 = I_{\text{new}} $$

similarly if we only halved the distance then it would be

$$ I_{\text{half distance}} = 4I_0 $$ so you would decrease the time of exposure by a factor of 4 here, since it is 4 times the intensity. Also, for my previous example, you will get the same exposure for $1/25$th of the time.

Chinny84
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