Minimum accelerator energy to run experiment

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general setup

fitting the collimator size into the hole through the concrete wall

I can express the distance [math]\displaystyle{ A_1D_1 }[/math] as function of collimator size [math]\displaystyle{ \Theta_C/m }[/math] and electron beam energy E:

[math]\displaystyle{ A_1D_1(E,\ \Theta_C/m) = \frac{469}{\sqrt{2}}\tan\left(\frac{0.511}{E}\right)
                              + \frac{469}{2}\tan\left(\frac{1}{m}\frac{0.511}{E}\right) }[/math]

To fit the collimator size into the hole through the concrete wall with radius R = 8.73 cm we need to solve equation:

[math]\displaystyle{ A_1D_1(E,\ \Theta_C/m) = 8.73\ cm }[/math]

1) some solutions of this equation for different collimator sizes m are:

[math]\displaystyle{ m = 1 \Rightarrow E_{min} = 33.1\ MeV   }[/math]

[math]\displaystyle{ m = 2 \Rightarrow E_{min} = 26.3\ MeV   }[/math]

[math]\displaystyle{ m = 4 \Rightarrow E_{min} = 22.8\ MeV   }[/math]

2) in general for arbitrary collimator size m the solutions are:

All energy under this line is good to run experiment for condition above

critical collimator line condition

Also I can express the distance [math]\displaystyle{ GH }[/math] as function of collimator size [math]\displaystyle{ \Theta_C/m }[/math] and electron beam energy E:

If I would like that nothing hitting the 4" box to go through the collimator I need to solve equation:

[math]\displaystyle{ GH(E,\ \Theta_C/m) = 5.08\ cm }[/math]

1) some solutions of this equation for different collimator sizes m are:

[math]\displaystyle{ m = 1 \Rightarrow E_{min} = 73.7\ MeV   }[/math]

[math]\displaystyle{ m = 2 \Rightarrow E_{min} = 36.9\ MeV   }[/math]

[math]\displaystyle{ m = 4 \Rightarrow E_{min} = 18.4\ MeV   }[/math]

2) in general for arbitrary collimator size m the solutions are:

All energy under this line is good to run experiment for condition above

both solutions together

All energies under this lines is good to run experiment for both conditions above

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