A parallel bundle of monoenergetic photons is used for imaging in the configuration below. The bundle passes through a layer of water with a thickness of 1 cm, then through a zone where a cube with a thickness of 1 mm is placed, and then through another layer of water with a thickness of 1 cm. Upon exiting the second water screen, the bundle reaches a panel composed of individual detectors, identical and cubic with a side length of 1 mm. Each detector has an efficiency ε, which may depend on the energy of the incident photons.
If the intensity of the photon beam at the entrance to the detector is $I$ (photons/s), then the measured value by the detector is $S=\varepsilon I \pm \sqrt{\varepsilon I}$ (pulses/s). The effect of scattering on screens and objects on the detector signal is neglected. Uncertainties regarding the constructive thicknesses of the objects used are neglected.
- Assuming that the detectors' efficiency does not depend on energy ($\varepsilon$ = 1.0), what is the intensity (measured in photons/(cm² s)) of a photon beam with an energy of 50 keV needed to determine (based on the detector response) that the object has been placed between the two screens?
My method is wrong because it doesn't add up at the end. Given that the intensity of the photon beam at the material's exit depends on the relationship
$I=I_0e^{-d\cdot \mu}(J/(cm^2 s))$
The scattering effect is neglected. Knowing that the luminous radiant flux is proportional to energy per unit area per unit time, and since the beam's area remains constant, and the illumination time is the same for the entire system, they can be simplified in the following relationships.
Where $I_0$ is the initial intensity, $I$ is the intensity at the material's exit, $d$ is the distance traveled by the beam through the material, and $\mu$ is the attenuation coefficient.
Knowing that luminous intensity is proportional to energy per unit area and unit time ($I=\frac{E}{A\cdot t}$). Using the relationship $E_1=E_0e^{-d\cdot \mu_{H_2O_{\{1\}}}}(keV)$ to determine the energy $E_1$ exiting the screen.
$E_1=50e^{-0.227}=39.8\,(keV)$
I will approximate the input energy into the iron block with $E_1\approx40\,keV$ and from the table \cite{NISTfier}, extract the value $\mu_{Fe}/\rho_{Fe}=3.629(cm^2/g)$, resulting in $\mu_{Fe}=0.461(cm^{-1})$. I determine that the energy $E_2$ exiting the iron block will be $E_2=E_1e^{-d\cdot \mu_{Fe}}(keV)$.
$E_2=40e^{-10\cdot0.461}=38.19\,(keV)$
I will approximate the input energy into the iron block with $E_2\approx40\,keV$ and from the table nist.gov for $H_2O$, extract the value $\mu_{H_2O_{\{2\}}}/\rho_{H_2O}=0.268(cm^2/g)$, resulting in $\mu_{H_2O}=0.268(cm^{-1})$. Since the energy exiting the iron block is similar to the input energy, I will use the same attenuation coefficient for the second water screen for both cases, where the beam passes through the iron block and where it does not.
The measured values by the detectors will be:
$S_{1,3,4}=\epsilon I_0e^{-\mu_{H_2O_{\{1\}}}}e^{-\mu_{H_2O_{\{2\}}}}\pm\sqrt{\epsilon I_0e^{-\mu_{H_2O_{\{1\}}}}e^{-\mu_{H_2O_{\{2\}}}}}(pulses/s)$
For detector 2, the light beam passes through the iron block and both water screens. The luminous intensity $I_4$ will be:
$I_4=I_0e^{-\mu_{H_2O_{\{1\}}}}e^{-10\mu_{Fe}}e^{-\mu_{H_2O_{\{2\}}}}(J/(cm^2 s))$
And the measured value by the detector will be:
$S_2=\epsilon I_0e^{-\mu_{H_2O_{\{1\}}}}e^{-10\mu_{Fe}}e^{-\mu_{H_2O_{\{2\}}}}\pm\sqrt{\epsilon I_0e^{-\mu_{H_2O_{\{1\}}}}e^{-10\mu_{Fe}}e^{-\mu_{H_2O_{\{2\}}}}}(pulses/s)$
We will obtain the following solutions:
$S_{1,3,4}=I_0e^{-0.227}e^{-0.268}\pm\sqrt{I_0e^{-0.227}e^{-0.268}}(pulses/s)$
$S_2=I_0e^{-0.227}e^{-0.046}e^{-0.268}\pm\sqrt{I_0e^{-0.227}e^{-0.046}e^{-0.268}}(pulses/s)$
Create a system of equations using the determined relationships above:
$\begin{cases} S_{1,3,4}=0.609\cdot I_0\pm\sqrt{0.609\ cdot I_0} \\ S_2=0.582\cdot I_0\pm\sqrt{0.582\cdot I_0} \end{cases}$
We know that on detectors 1, 3, and 4, more pulses will be measured, i.e., $S_{1,3,4}>S_2$. We will solve the equation for extreme cases of uncertainties:
$\begin{cases} 0.609\cdot I_0-\sqrt{0.609\cdot I_0}>0.582\cdot I_0+\sqrt{0.582\cdot I_0} \\ 0.609\cdot I_0+\sqrt{0.609\cdot I_0}>0.582\cdot I_0-\sqrt{0.582\cdot I_0}\\ 0.609\cdot I_0+\sqrt{0.609\cdot I_0}>0.582\cdot I_0+\sqrt{0.582\cdot I_0} \\ 0.609\cdot I_0-\sqrt{0.609\cdot I_0}>0.582\cdot I_0-\sqrt{0.582\cdot I_0} \end{cases}\Rightarrow\begin{cases} I_0>57.158\sqrt{I_0} \\ I_0>-57.158\sqrt{I_0}\\ I_0>-0.648\sqrt{I_0} \\ I_0>0.419 \end{cases}$
As a result, the light intensity needed to determine that the object was placed between the two screens should be $I_0<3267.07\,(J/(cm^2 s))$.
I don't understand if I interpreted the problem wrong, messed up some units of measure, or used incorrect formulas.
