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 size 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 size 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 above are together

All energy under this linse is good to run experiment for both conditions above

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