Difference between revisions of "Test in Plane for Theta at 20 degrees and Phi at 0"
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![X in detector plane.png](/./images/3/34/X_in_detector_plane.png)
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− | <center><math>\ | + | <center><math>\underline{\textbf{Navigation}}</math> |
[[In_the_Detector_Plane|<math>\vartriangleleft </math>]] | [[In_the_Detector_Plane|<math>\vartriangleleft </math>]] | ||
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− | + | =Test for <math>\theta=20</math> and <math>\phi=0</math>= | |
Substituting in the values found earlier for the case of <math>\theta=20^{\circ}</math> and <math>\phi=0</math> | Substituting in the values found earlier for the case of <math>\theta=20^{\circ}</math> and <math>\phi=0</math> | ||
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<center><math>x'=\frac{x_{lab}}{sin 65^{ \circ}}=.868=x'</math></center> | <center><math>x'=\frac{x_{lab}}{sin 65^{ \circ}}=.868=x'</math></center> | ||
+ | |||
+ | |||
+ | ---- | ||
+ | |||
+ | |||
+ | <center><math>\underline{\textbf{Navigation}}</math> | ||
+ | |||
+ | [[In_the_Detector_Plane|<math>\vartriangleleft </math>]] | ||
+ | [[VanWasshenova_Thesis#Determining_wire-theta_correspondence|<math>\triangle </math>]] | ||
+ | [[Test_in_Plane_for_Theta_at_20_degrees_and_Phi_at_1_degree|<math>\vartriangleright </math>]] | ||
+ | |||
+ | </center> |
Latest revision as of 20:28, 15 May 2018
Test for and
Substituting in the values found earlier for the case of
and
Since
The
distance from focal point 1 is:This is the radius from focal point 1, which is to be expected since the y component is equal to zero for
The focii are located at
This implies that with respect to the origin, x', we find
This is verified with CED
![X in detector plane.png](/./images/3/34/X_in_detector_plane.png)
Since the x' dimension is the hypotenuse in a right triangle of 65 degrees