Transcript Slide 1

Autocollimator Calibration
Calibration : interpretation of output from Autocollimator Controller
What we have:
Output from controller i.e. numbers that
quantify the lateral displacment (Δx, Δ y)
caused by tilts in arbitrary unit
What we search:
A coefficient that transform the lateral
displacment arbitrary unit in µm
Autocollimator Calibration
The idea
• Generate a tilt of a known
angle…
• …measure it…
• …calibrate the output!
Autocollimator Calibration
Generate a tilt of a known angle
Response spring
=
Micrometric
translation stage
Micrometric
translation
+
Rail arm
autocollimator
Known angle
Autocollimator Calibration
Procedure
Step of T.S.
20 μm
Min angle
Δθ = 0,0571 mrad
y  f  tan(2 )
i.e. Δy=100 a.u. read  Δy = 56.5 μm
Lateral displacement
Δy = 32 µm
Average of conv coeff: 0,565082 μ/a.u.
Error:
ACCURACY [%]
0,068163 μ/a.u.
12,06248
Why this error?
Average Δy read
74 measure of 20 μm step
Δy = 56,783 a.u.
Autocollimator Calibration
Focal length
F  280 mm
Rail length
R  350 mm
 R  1mm
Positioning (per step)
P  20m
 P  2m
Data Acquisition (per step)
X
X
Y
Average on100 meas
St. dev. on 100 meas
Y
Difference between the two Y values
Shift Angle
L  F  tan(2   )
2
 

2 F  2
2 F  1
  2

2
 cos (2  ( 2  1 ))   cos (2  ( 2  1 )) 

 L  
 Y
P
  arctan 
R
Linear Shift


P
  
 P 2
 1  2
  R
2






P  R
 

2




 P 
  R 
 1  2   R 2 
  R 

 
Y
  Y1   Y 2
2
Average of  Y
Conversion Coeff
Y  56.78
C  L / Y
 
 L *  Y 
 C   L   
2

yY 

 Y


Difference between the two next angles
   2  1
2
 0.48
2
Autocollimator Calibration
Next Step
Reduce Min. angle Δθ
+
Improve Accurancy
•Using Exapod
•Using Theodolite+ large mirror