Spatio-Spectral Imaging with the RSI
Abstract
We use a rotational shearing interferometer to determine the location and spectra of a
sparce array of point sources in only two CCD “snapshots” without dithering the path
length, but by physically rotating the interferometer itself a small known amount in
between snapshots.
Introduction
The ultimate goal of imaging is to capture a full five dimensional record of a scene –
three spatial dimensions, one spectral dimension, and one time dimension.
Itoh, Ihoue, and Ichioka constructed a rotational shear interferometer for the purpose of
taking a 64x64 image with 64 channels of spectral information. This was done by
dithering one arm of the interferometer and capturing 64 two dimensional 64x64 images
and performing a 3D Fourier Transform on the whole thing.
With a Rotatioal Shear Interferometer, both the spectral and spatial information in the
optical field can be measured, but by capturing a single interference pattern, the two
kinds of information are coupled. This must be so, since three dimensions of spatial
information plus one of spectral information must be projected on to two dimensions of
CCD measurement.
One way of dealing with this is to use a narrow spectral filter and only attempt to find the
angular direction of sources. This eliminates the spectral dimension and reduces the three
spatial dimensions of x, y, and z to x/z and y/z where the z dimension is along the optical
axis. This measurement can be made by taking the Fourier transform of a single frame as
discussed below.
To add the third spatial dimension, measurements could be made of an object at several
angles around a circle with the object at its center and cone beam tomography used to
reconstruct the three dimensional object.
To add spectral information, one use the Wiener-Khintchine Theorem and the idea of
Fourier Transform Spectroscopy where one dithers the path length of one arm of the
interferometer and the variation of measured intensity values is the Fourier Transform of
the power spectral density of the object.
Dithering the mirror, however, requires expensive sub-wavelength translation. In one
sense, moving the object to a different spatial location has the same effect on the
interference pattern as shifting its spectrum by a certain amount. Given this, it is possible
to determine the power spectral density by translating or rotating the interferometer a
known amount and taking a second image.
Equations for RSI:
Here, we consider a simple point source far away from the interferometer emitting at
wavelength  . The equation for the complex analytic field emitted by this point source
where r is the distance to the point source is given by
r
A i 2
E e 
r
The field at the CCD is the sum of the fields reflected off the two arms. We assume the
amplitudes of the two fields are approximately the same at the CCD and from here
forward, we will neglect this overall amplitude. With r1 and r2 as the distances from a
pixel of the CCD to the point source through the first and second arms of the
interferometer, the field is given by:
i 2
r1
i 2
r2
Ee
e 
The intensity measured is then given by the magnitude squared of this field – a constant
plus a co-sinusoidal modulation term given by
Ie
i 2
r1

e
i 2
r2 2

r
i 2 2
 i 2 r1

 e
e 


r
i 2 2
 i 2 r1
 e
e 


2
2
i  r1  r2 
i  r1  r2 

 2
  2e 
r1  r2 
e 
 2  2 cos

 


The geometry of the RSI determines r1 and r2 . Instead of an RSI, to begin with, consider
a Michalson interferometer with flat mirrors instead of the RSI’s right angle mirrors.
[Picture of 3D michaelson with (xs, ys) going to (xc, yc) with a bar]
[Top view of michaelson with z and delta-z labeled]
Specifically, we want to calculate the intensity measured on a CCD pixel at xc , y c  for a
point source at x s , y s  a distance z away where we’ll let z include the distance through
the fixed arm of the interferometer. The difference in path length between the fixed arm
(#1) and the moveable arm (#2) we’ll call z .
By “unfolding” the interferometer as shown in Fig ##, one can see that this distance is the
given by:
r1 
xc  xs 2   yc  y s 2  z 2
and r2 
xc  xs 2   yc  y s 2  z  z 2
Now we’ll consider the effect of replacing the flat mirrors with right angle prisms. First
we consider having no sheer – the right angle runs along the y-axis as shown. This is an
RSI with sheer angle   0 . If you look through either arm of the interferometer, it has
the effect of making the source at x s , y s  appear at the same place as a point at
 xs , y s  would be through the flat mirror interferometer. We define x1 , y1  in the
following way: When we look at a point source at x s , y s  through arm #1 of the RSI, it
looks like it would be at x1 , y1  if we were looking through a flat mirror interferometer.
Similarly, if we were looking through arm #2, the point source looks like it’s at x2 , y2  .
For no sheer, x1 , y1   x2 , y 2    x s , y s  .
Now we rotate right angle prism #1 by angle  and right angle prism #2 by angle   .
When we look through arm #1, the source looks like it’s been flipped across the line
that’s the right angle vertex. If you looked at yourself through one of these right angle
prism with the right angle vertical (   0  ), your image will the flipped right-to-left, but
you will still look right-side-up: x1 , y1    x s , y s . If you looked at yourself through
right angle prisms with the right angle horizontal (   90  ), you would look upside-down
(rotated by 180  ): x1 , y1   x s , y s  . For an arbitrary angle:
 x1    cos2  sin 2   x s 
 x    cos2   sin 2  x s 
   
  and  2   
 
 y1   sin 2  cos2  y s 
 y 2    sin 2  cos2   y s 
The fact that this looks similar to a rotation of 2 is somewhat intuitive since when we
turn the prism from   0  to   90  , we have to go from right-side-up to up-side-down
in some continuous way.
The quantities we are interested in are r1 and r2 . Given in equation ## but with x s , y s 
replaced by x1 , y1  due to the effect of the sheering right angle mirrors:
r1 
xc  x1 2   yc  y1 2  z 2
and r2 
xc  x2 2   yc  y 2 2  z  z 2
To find the intensity of light on the CCD, we substitute these values for r1 and r2 into
equation ##:
 2
r1  r2   2  2 cos 2 
I  2  2 cos



  
x c  x1 2   y c  y1 2  z 2

x c  x 2 2   y c  y 2 2  z  z 2  
At this point, we make the approximation that the distance to the source z is much
farther than the distance the object is off axis and the distance we dither the mirror z :
xc  xs 2   yc  ys 2  z
and z  z
By Taylor expanding the square roots in equation ## for this large z :
 2
I  2  2 cos
 


x  x1 2   y c  y1 2
x  x 2 2   y c  y 2 2
z  c
z c

2z
2z





Using equation ## to substitute for x1 , y1  and x2 , y2  and doing nightmare trig
simplification, we arrive at the RSI point source equation for the intensity measured on a
CCD pixel at xc , y c  for a point source at x s , y s  a distance z away radiating with
wavelength  and for a known shear angle of  :
 4 sin 2 
x s y c  y s x c 
I  2  2 cos
z


Since we are only interested in the direction to the source and not it’s absolute position,
we consider  x and  y (where tan  x   x s / z ), the angle from the z-axis of the RSI out
to the point source in the x-direction. Similarly  y as the angle to the point source in the
y-direction.

The expression for intensity becomes
 4 sin 2 
tan  x yc  tan  y xc 
I  2  cos



For example, for a point source located 10m away at (20mm, 30mm), Mirror Shear Angle
= 10deg, Wavelength = 500nm, what you’d measure on a 1cm x 1cm CCD and the
positive part of its 2D Fourier Transform would look like:
200
100
80
150
60
100
40
50
20
0
0
0
50
100
150
200
0
20
40
60
80
100
[How do we know if the source is top-right or bottom left? There isn’t any complex
phase data to do this, but after the rotation, it’s obvious which way the point sources
move.]
As discussed elsewhere, this RSI pattern is a measurement of the mutual coherence. If
the goal was to reconstruct an image via the van Cittert-Zernikie Theorem, we would take
the inverse Fourier transform of this measurement to recover projections through the
image. This is why the Fourier transform pictured is a single point (and a DC spot from
the 1+)
Effects of changing angular position and wavelength are shown. It’s apparent from the
intensity equation that angular positions are indistinguishable from wavelength  in a
single measurement. Moving the object twice as far away (thus reducing tan  x  and
tan  y  by a factor of two) has the same effect on the fringe pattern as doubling the
wavelength. To break this degeneracy and find the location and wavelength a
monochromatic point source, first take a measurement with the RSI at some unknown,
but desired angle  x with respect to the source. Then rotate the RSI a known angle  x
in the “x-direction” and take another measurement where the RSI is at angle  x   x .
[2D top view with all the angles labeled would be better.]
The two dimensional fourier transform of the CCD measurements will be a single peak
centered around one point in fourier space measured to be say u1 ,v1  for the unrotated
measurement and u2 ,v2  for the rotated one. From RSI point source equation, these
“spatial frequencies” correspond to measurements of the angle and wavelength:
v1 
4 sin   tan  

and v 2 
4 sin   tan    

This systems of two equations and two unknowns ( tan   and  ) can be solved by
expanding tan     and substitution:
v 2  v1 

tan   
v1  v2 2  4 v1 tan 2  
v2
and
v tan  
2 1
v2 4 sin  
v 2  v1 
v1  v2 2  4v1v2 tan 2  
2v 2 tan  
Ugly, but this is a closed form for both the wavelength and the direction of the point
source using no approximations other than the initial Fresnel approximation in deriving
the RSI point source equation.
Continuing the numerical example above, treating it as the first measurement, if we rotate
the RSI by  x  0.2 , we get the following simulated CCD Fourier Transform:
100
80
60
40
20
0
0
20
40
60
80
100
Using our laboratory RSI setup and the same point source as above, we got the following
experimental results…
If you had a relatively sparse array of point sources at well-defined frequencies, by
rotating the RSI in steps and tracking the positions of the point sources, an accurate
measurement could be made of the location and wavelength of all of the sources.