Dan
Review of the optics elements: Pinhole (“GAP”), FZP, Coded Aperture
Extracting information from the GAP what is the GAP width? introduction to diffraction calculations
Extracting information from the FZP how is the in-focus image separated from the out-of-focus?
is there an intrinsic image width?
Extracting information from the Coded Aperture how can this image be fitted?
Comparison of the “5μm” and “10μm” coded apertures
Results from a representative set of runs

Pin Hole (GAP)
The height is about 50 μm and is optimized for the minimum size image.
Fresnel Zone Plate (FZP)
Ours has 239 rings,
1200 μm diameter, diam. of center opaque: 77.8 μm, width of last ring: 1.26 μm.
Focus if by diffraction; paths through successive rings differ by one wavelength.
Oddly, it is optimized for 2.5 keV.
Coded Aperture (CA)
The height is about 310 μm.
(Yellow is transmitting.)
Transmitting slits range in size from
40 μm (begin diffraction) to
10 μm (very diffracting).
The image from a single 10 μm slit is much wider than that of the 40 μm

The pinhole is a simple camera
– similar to what I used to look at a solar eclipse in
1959*.
The image size (
σ i
) is a convolution of components due to the source (or object) size ( σ
S
) and a due to the pinhole (or GAP) size ( σ
G
).
There are variables relevant to the discussion:
σ i
σ
S is the image size as observed on the detector, is the beam size at the source (our product),
G is the full width of the pinhole,
σ
G
σ iG is the
σ of the pinhole, =G/sqrt(12) , is the pinhole image size as observed on the detector and
σ
SG is the pinhole image size “referred to the source”
σ
SG
=
σ iG
/M, where
M is the magnification, i/o
We have been using the formula:
σ
S
= [(
σ i
/M) 2 -
σ
SG
2 ] 1/2 (which is equivalent to [
σ i
2 -
σ iG
2 ] 1/2 /M )
* It was July 20, 1963 when a total eclipse was visible from Nova Scotia.
iG
SG
SG

For large pinhole size, the image size is a projection:
σ iG
~ G.
For small pinhole size, the image is a diffraction pattern,
σ iG
~ 1/G.
There is a pinhole size, G, which provides the minimum image size σ iG
.
We measure the image width while varying the pinhole size,
G. Above are the measurements for Cline from 20100803 and 20101223 where the minimum was found at
TASGAP=4.55 and 4.57.
JimA has done calculations using the x-ray spectrum, in
December 2009, where he finds
σ
SG between 14.1 and 17.0
μm; we are using 16.

I wrote a program (an Excel spreadsheet) to perform the numerical integration:
Σ exp( i 2 π P(x) /λ T (x) ) where x is a position along the optics element,
P(x) is the path length from the source, to the position on the optic, to the detector pixel,
T (x) is the transparency of the optics element at that position (1 or 0).
Pulse height is calculated for 64 pixels of size 25
μm.
Limitations: the paths are separated by 0.5 μm at the optic.
It does not integrate over x-ray energy; I use one energy at a time.
G= 10
μm 20 30 80 100 240
The minimum image is found with G =45 μm.
However, extracting
σ iG from this distribution is not simple.
The diffraction bumps will be interpreted as background when fitting the data and must be treated as background when determining the width of the central peak.

A cut-off distance must be chosen for the image.
At any candidate cut-off, the next 7 bins are used to define the background.
The cut-off is chosen where background subtracted distribution crosses zero.
The RMS calculated within the cut-off.
Note that the RMS of the full distribution is much larger.
The minimum RMS image size, for x-ray energy about 2keV, is
σ iG
= 49 μm.
σ
SG
=
σ iG
/ 2.34 = 21
μm

The Fresnel Zone Plate produces a true focus.
The image, for σ
S
=0, is so small that it can be lost with the 25 μm bins.
I model a FZP with 7 transmitting rings.
This is the image at the optimization x-ray energy, 2.5keV, with 2 μm bins .
Distributions below show that the chromatic aberration is severe, with a half width of only about 0.1keV.
E x-ray
= 2.5
E x-ray
= 2.1 2.3 2.4 2.6 2.7 2.9
The distributions also shows that the “wide part” is due to the off-energy x-rays.
The width of the central peak, in the range 2.4 to 2.6keV, is only 5 μm, so the width “referred to the source” is negligible.
The only width contribution that must be corrected is due to the pixel size: 50 μm / 12 1/2 / 2.34 = 6.2
μm.

The coded aperture is an array of slits ranging is width from 10 μm to 40μm.
Shown is the image for σ
S
=0. (blue points)
With a goal of developing a parameter-based, rather than template-based fitting procedure, the red line is the sum of 12 gaussian distributions:
F =
Σ
J=1,12
A
J exp ( -0.5 ( x-x
J0
) 2 /(
σ
J
2 ) )
Return to this after discussing some points about the Coded Aperture.

The peaks of the Coded Aperture are not direct images of individual slits. This is very much a diffractive device.
The first 10
μm slit , alone, produces a broad diffraction peak.
The first two 10
μm slits produce a diffraction pattern with 4 easily visible peaks.
The widest slit, at 40 μm, produces a peak similar to the optimized GAP image, but does not directly correspond to the large feature in the image.

A parameter-based fitting function with
12 gaussian distributions:
The
σ
S
=0 distribution is “fit” to the sum:
F =
Σ
J=1,12
A
J exp ( -0.5 ( x-x
J0
) 2 /( σ
J
2 ) )
The 36 parameters: A
J
, x
J0
,
σ
J
, are fixed.
For non-zero source size , contributions from the source add non-coherently.
There are 3 floating variables:
C, an overall normalization,
X
0
, an overall position offset, and
σ
I
, the image width due to the beam size.
Then,
F = C
Σ
J=1,12
A
J
( σ
J
) /( σ
J
2 exp ( -0.5 ( x-x
+ σ
I
2 ) 1/2
J0
- X
0
) 2 /( σ
J
2 + σ
I
2 ) )
Shown at right is the calculation and function for
σ
S
= 15
μm.

The original reason for creating the numerical integration was to look at the image KEK Coded Aperture.
The KEK Coded Aperture has all features reduced to half size. The image for 2keV,
σ
S
= 15
μm, does not have wellresolved peaks. This is why we replaced it with the
Canadian Coded Aperture in the C-line.
At 4keV, the half size Coded Aperture produces resolved peaks.

So, now there is a function for fitting each of the images.
For the PinHole, use 1 gaussian + background.
Extract the beam size:
σ
S
= [(
σ i
/M) 2 – 21μm 2 ] 1/2
For the FZP, use 2 gaussians + background.
The wide component has fixed width.
The narrow component is the measurement.
Extract the beam size:
σ
S
= [(
σ i
/M) 2 – 6μm 2 ] 1/2
For the Coded Aperture, use
F = C
Σ
J=1,12
A
J
(
σ
J
) /(
σ
J
2 +
σ
I
2 ) 1/2 exp ( -0.5 ( x-x
J0
- X
0
) 2 /(
σ
J
2 +
σ
I
2 ) )
Extract the beam size: σ
S
= σ
I
/M ; there are no other corrections.

The beam position is measured for individual turns.
The peak is located. Then the average is measured within a window about the peak.
(The window is different for each optics element.)
The individual turn is shifted to the average beam position and accumulated in an average distribution. Only the average distribution is fitted.

And this is a comparison of a set of runs, using the 3 optics elements, taken during beam dynamics studies, 2010-12-20