From Ohm

Attenuation of the radiation intensity $I$ of monochromatic (mono-energetic) rays while travelling through a homogeneous material with thickness d and attenuation coefficient $\mu$:

$ I(d) = I_0 \cdot e^{−\mu \cdot d} $

What is the thickness of a material that absorbs exactly half of the rays intensity?

$ I/I_0 = e^{−\mu \cdot d} = 0.5 $

$ D_\frac12 = \frac{ln 2}{\mu} $

$ ln 2 \approx 0.7 $

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Page last modified on October 13, 2013, at 03:13 PM