to test some of the image processing
applications of the setup. Finally, we
constructed a hologram of a 3D object on film.
Fourier Optics and
Holograms
Allan
Jean-Pierre
Haldane
Auclair
260050293 260148852
Phys 359, March 13, 2006
Abstract
In this experiment, various properties of
light were confirmed, especially in relation to
Fourier Optics. First, we observed the fact that
plane waves of light passed through an aperture
will diffract to form the Fourier transform of the
aperture at an infinite distance away. We then
proved the bandwidth theorem using slits of
various widths as out aperture, and again
observed the frequency components present in
these slits. Next, we observed that we could
reverse the Fourier transformation to get back
the original image with a suitable setup of
lenses. Also, by blocking out part of the Fourier
transform of the image, and then reconstructing
an image from this modified Fourier transform,
we could selectively remove frequency
components from the image. We used this fact
1
1 – Introduction
T
2 2
T f ( x ) sin  x nx dx . (eq.2c)
T  2
The series can be simplified by the use of a
complex exponential to
This experiment explores the wave-like
properties of light and their implications.
Because of its wave-like properties, light
can be used to calculate the Fourier
transformation of functions and images. The
standard single-slit experiment is an example of
this, although it is not often emphasized,
especially to beginning physics students. It turns
out that the diffraction pattern obtained by
passing light though a single slit is actually the
Fourier transform of the slit, when viewed as a
step function.
The inverse transform can also be
calculated using light, allowing for image
processing to be done. The Fourier transform
decomposes a function into its frequency
components. The resulting beam of light can
then be masked, so that certain frequency
components are erased, and a modified image
can be reconstructed. Much work in computer
science has gone into writing fast, efficient
Fourier transform algorithms in order to do
image processing and other tasks. Yet here,
nature provides a way to calculate the Fourier
transform at the speed of light (but with
limitations, of course).
Finally, a hologram can be made by
taking advantage of these wave-like properties,
which was attempted here.

f ( x)   c n exp( in  x x) ,
where
an  ibn
.
(eq.6)
2
We can decompose non-periodic
functions in a similar way but we then have to
do it over a continuous spectrum of frequencies.
This is the idea behind Fourier transforms which
allows us to express the non-periodic functions
as
cn 

f ( x)   F ( x ) exp( i x x)d x ,

  a n sin  x nx 
where F(ω) is the Fourier transform given by

F ( x )   f ( x) exp( i x x)dx .
x 
2
an 
T
T
2
T

2

2
,
T
f ( x) cos x nx dx ,
(eq.8)

In summary, the Fourier transform of a function
gives a new function, which is like a plot of the
frequencies present in the original function.
Light can be used to calculate a Fourier
transform. The reason is that light is a wave, and
thus having a phase component, so that at a
distance x the phase of a beam of light may be
written as
Phase = exp( 2ix /  )
(eq.9)
for light of wavelength λ. If we integrate
spatially over the light from an aperture A(x,y),
we have
A' ( x' , y ' )   A( x, y ) exp( 2ix /  )dx
(eq.1)
(eq.10)
If we assume that r is much larger than the size
of A, then all the light from A will be
equidistant from any point at distance r, so that
A(x,y) is effectively constant (the light intensity
from all parts of A is constant). Then
n 1
where
(eq.7)

A function that is periodic with a period
T can be expanded using a Fourier series as

1
a 0   a n cos x nx 
2
n 1
(eq.5)

2 – General theory of the Fourier
analysis
f ( x) 
bn 
and
(eq.2a)
(eq.2b)
A'  A exp( 2ix /  )dx
2
(eq.11)
which is the same equation as the inverse
Fourier transform. Since A and x are in the
space domain, the function must be converting
to the frequency domain. Therefore, if r is large
enough, the pattern of light will be a frequency
decomposition of the original aperture, as is the
Fourier transform.
In this experiment, the image is the
original function f(x), and the Fourier transform
is the function F(νx). Instead of looking at the
transformed image at infinity, as required for
equation 11, we add a lens so r=infinity is
effectively at the lens’ focal point.
Slit #
EQ1574 #1
EQ1574 #2
EQ1574 #3
EQ1590
Description
Thinnest
Thin
Wide
Widest
Width (10-5 m)
15.8 ± 0.6
26.0 ± 0.9
55.4 ± 0.8
106.7 ± 0.7
Description
Coarse
Fine
Period (10-5 m)
116 ± 3
13.9 ± 0.8
Table 2: Slits
Grating #
EQ1579
EQ1581
Table 3: Gratings
Grid #
EQ1586
Period (10-5 m)
40 ± 0.5
3 – Apparatus
Table 4: Grid
3.1 – Equipment list
4 – Fraunhöfer diffraction as a way to
calculate the Fourier transform
The equipment used is shown in tables 1 to 4.
Lens #
EQ1567 and
EQ1568
EQ1566
Name
Lb
EQ2575
Lc
EQ0730
(1 and 2)
L1 and
L2
L0
4.1 – Study of the Fourier transform of
gratings (Activity 1)
Description
60x microscope
objective
10x microscope
objective
31.3 ± 0.5 cm focal
length lens
55.5 ± 0.5 cm focal
length lenses
To start our experiment, we use a
horizontal grating similar to the one shown in
figure 1 since it is the simplest periodic function
one can get.
Table 1: Lenses
The two Lb were used as beam
expanders in the holography part, whereas L0
was for the Fourier part of our experiments. The
other lenses were used only in the first part.
The focal length of Lc was found by
moving the lens to a certain distance so that the
beam coming out of the pinhole was parallel.
For L1 and L2, we measured the distances at
which the parallel beam coming from Lc was
focused in one point.
The widths and periods of the preceding
equipment were each calculated from 8 different
measurements with the travelling microscope
from which we calculated the mean and
standard deviation.
Otherwise we used different types of
mount and a laser with a wavelength of 632.8
nm.
Figure 1 - Horizontal grating
We can predict that the Fourier
transform in the horizontal direction will be a
series of equally spaced points, as the Fourier
transform of a periodic function shows the
frequencies of the function as points. On the
other hand, the image in the vertical direction is
a constant so the Fourier transform should yield
a delta function. Combining the two we can then
expect to have a horizontal line of equally
spaced dots. Of course the dots won’t be of
infinitesimal length since the edges of the lines
of our grating aren’t perfectly straight and
because we have a certain uncertainty for the
3
focal length of the lenses we are using. To get
this transform, we use the setup shown in figure
2. The first two lenses allow us to get a parallel
beam which we shine on the grating. We then
obtain the Fourier transform by placing another
lens after the object to simulate the infinite
distance where the diffraction pattern would
otherwise be observable.
cm
cm
Figure 3 – Fourier transform of the coarse grating
cm
Figure 2– Setup used for sections 4.1 and 4.2.
cm
We can measure the base frequency ν of
the slit using its period T since the two are
related by equation 2. We know that the
frequencies used to build the function
represented by our grid are integer multiples of
the base frequency. In the spatial frequency
domain, we can expect to observe regularly
spaced peaks in intensity for which we can
express the frequency as
x 
Figure 4 – Fourier transform of the fine grating
Grating used
Coarse
Fine
Calculated
distances
(mm)
0.303 ± 0.008
2.5 ± 0.1
Experimental
distances
(mm)
0.304 ± 0.004
2.53 ± 0.01
Table 5 – Distance between two dots of the Fourier
transform
2
d (eq.12)
f 
4.2 – Study of the Fourier transform of
different slits (Activity 2)
where f is the focal length f1 as shown in Figure
1, λ is the wavelength of the laser beam we are
using and d is the distance between two peaks.
Using equation 2 and 12, we can express the
distance between the peaks as
Using the setup shown in Figure 1 for
this part again, but now with a single slit as the
object, we can again compare the obtained
diffraction pattern to the expected Fourier
transform.
A slit can be expressed in percentage of
light transmitted as a function of space by
f 
(eq.13),
T
a function of the period T of the grating, giving
us a relation between the two variables we can
directly measure, d and T. We performed this
experiment with coarse and fine gratings, and
the transformed images on the screen are shown
in figures 3 and 4. Table 5 presents the results
we predicted using equation 13 compared to
those obtained from measuring the number of
pixels between dots in the transformed images.
d
f ( x)  0 for x  w/ 2
 1 for x  w/ 2
(eq.14)
for a slit of width w. This has a Fourier
transform given by
F ( x )  
w/ 2
w / 2
4
e i x x dx 
2 sin(  x w / 2)
x
(eq.15)
where νx is the spatial frequency. From this we
can see that the Fourier transform will be zero
whenever the argument of the sine function is an
integer multiple of π.
As shown in the lab manual, from
geometric considerations we can get
x 
2 sin 

 k sin  ,
Slit width
Calculated
Measured
w
distances
distances
(mm)
(mm)
(mm)
Thinnest
2.22 ± 0.09
2.43 ± 0.02
Thin
1.35 ± 0.05
1.41 ± 0.02
Wide
0.63 ± 0.01
0.64 ± 0.01
Widest
0.329 ± 0.006
0.324± 0.04
Table 6 – Calculated and experimental values of the
distance between the zeros of the Fourier transform,
found by measuring pixel distances images.
(eq.16)
Slit width
Calculated
Fitted
w
distances
distances
(mm)
(mm)
(mm)
Thinnest
2.22 ± 0.09
2.42 ± 0.01
Thin
1.35 ± 0.05
1.45 ± 0.06
Wide
0.63 ± 0.01
0.634 ± 0.007
Widest
0.329 ± 0.006
N/A
Table 7 – Calculated and experimental values of the
distance between the zeros of the Fourier transform,
found by fitting the intensity in matlab.
where we have defined a new constant
k  2  , and θ is the angle of incidence, λ is
the wavelength of the light. It is then convenient
to define a variable α for the argument of the
sine function as

1
kwsin    x w / 2 .
2
(eq.17)
We used two different systems of
measurement to get the experimental values for
d shown in tables 6 and 7. The first was to
simply count pixels in the scanned images, and
convert these counts to distances using a
conversion factor. The other was to fit equation
8 to the data. The transformed images are shown
in figure 5 to 8.
The intensity of the diffraction pattern is then
 sin  2 
 .
I  I 0 
2
  
(eq.18)
y
,
f
(eq.19)
cm
d
f 
,
w
cm
Figure 6 – Fourier transform of the thin slit
cm
where y is the distance from the center of the
image, and f is the focal length. We define d to
be the distance from the center of the image to
the first node in intensity (at α = π), which is
also the distance between two successive nodes.
Solving equations 17 and 19 (where y=d) for d,
we get
Figure 5 – Fourier transform of the thinnest slit
cm
sin   tan  
cm
This will equal 0 (a node) anywhere α = nπ.
Also from geometric consideration, at the focal
plane, for small θ,
(eq.20)
cm
Figure 7 – Fourier transform of the wide slit
cm
a relation between the two measurable values d
and w similar to the one found in equation 13.
From equation 20 we calculated the expected
values for d shown in tables 6 and 7.
Figure 8 – Fourier transform of the widest slit
cm
5
As can be seen in figure 9, the fits to the
data are not great. To do the fit, we excluded the
central peak, because it was much too small
compared to the other peaks. The intensity was
smaller than expected both in the central peak
and in the outlying peaks. This has mostly to do
with the properties of the film being used. There
is a maximum exposure possible on film, since
there is a limited amount of pigment per unit
area. This is what causes the central peak to be
much lower in intensity than expected, since the
intensity is actually very high there. The film
also has a lower intensity threshold, so that light
will not cause a reaction in the pigment unless it
is above some minimum intensity. This causes
the outlying peaks, which are at low light
intensity, to be lower than expected.
or in an easier way to see for table 8,
Intensity (0 to 255)
 xw
 1.
2
(eq.22)
cm
cm
Intensity (0 to 255)
Figure 10 - Failed fit to the Fourier transform of the
widest slit.
Slit width (w)
(mm)
Thinnest
Thin
Wide
Widest
 xw
2
1.09 ± 0.04
1.07 ± 0.06
1.00 ± 0.02
0.98 ± 0.01
Table 8– Experimental values for the bandwidth
theorem. Note that they were calculated with the fitted
values when possible.
cm
cm
We can see in table 8 that the values are
close to the one we expected but that it does fall
out of the error range for most. This is probably
caused by some systematic errors in the slit
width since the exact position of the slit’s edges
is a bit vague under the traveling microscope.
Figure 9 – Fit to the intensity of the Fourier transform
for the thin slit.
Due to overexposure and to the small
distance between two successive dots, our
picture of the Fourier transform of the widest
slit couldn’t be fit correctly, as the fitting
produced a single large peak instead of many
small peaks as shown in figure 10.
These results also allow us to prove the
bandwidth theorem. As t  1 holds in
time/frequency domain, an equivalent must exist
in the space/space frequency domain, which is
that x w  ct . From equations 17 and 19 (with y
= d) we get
 x  k sin  
2 d 2

,
 f
w
4.3 – Computing inverse Fourier transform
(Activity 3)
By adding one more lens at the end of
the setup used in activities 1 and 2, we can
calculate the Fourier transform of the “object”
with regard to this new lens, which is actually
the Fourier transform of the actual object. The
setup shown in figure 10 thus allows us to
retrieve the actual object from its Fourier
transform.
(eq.21)
6
cm
Figure 11 - Setup used for sections 4.3, 4.4 and 4.5.
Using the medium grid, as the object, we
expected something similar to figure 12, the
Fourier transform of a grid computed using
matlab. We got the image shown in figure 13
and the inverse of the Fourier transform as show
in figure 14. The periods of the original and the
reconstructed images are shown in table 9. All
periods were measured by counting pixels in the
scanned images.
cm
Figure 14 – Inverse Fourier transform of the Fourier
transform of the grid
4.4 – Spatial filtering as a way to alter images
(Activity 5)
Using the setup from activity 3, we can
also filter some part of the Fourier transform
before the reconstruction of the image so that
only some features of the image are
reconstructed. Still using the medium grid, we
filtered out all but the central horizontal line of
the Fourier transform. By doing so, we got rid
of all the vertical frequencies which are the
constituents of the horizontal lines of the grid
and got the grating shown in figure 15 with a
period shown in table 10 again measured by
counting pixels and converting to distance
afterward.
cm
cm
Figure 12 – Expected Fourier transform of a grid
calculated in matlab.
cm
cm
Figure 15 – Reconstructed image
of the grid show in
Figure 13 with all but the central horizontal line of
frequency filtered out.
cm
Figure 13 – Fourier transform of the grid.
Original period (mm)
0.400 ± 0.005
Original period (mm)
0.400 ± 0.005
Image period (mm)
0.406 ± 0.003
Image period (mm)
0.400 ± 0.005
Table 10 – Periods of the original grid and
reconstructed grating.
Table 9 – Periods of the original and reconstructed
grids.
7
We then filtered out all but the diagonal
starting from the lower right corner to the upper
left one, which changed the reconstructed image
to a diagonal grating shown in figure 16 with a
period that should be smaller by a factor of
square root of two. This is due to the fact that
the diagonal distance between two points in the
Fourier transform plane is bigger by a factor of
root two then from each other then between two
points in a horizontal or vertical line. The
frequencies used to reconstruct the image being
bigger by a factor of root two, we can see from
equation 2a that the period of the reconstructed
image must be smaller by a factor of root two,
as shown in table 11.
reconstruct an image in which only the nonhorizontal paths appear. The downside of this
treatment is that we then filter out the horizontal
parts of the interaction paths as well. Starting
from the picture shown in figure 17a and using a
very thin vertical filter to get rid of the
frequencies in the middle vertical line, which
are the most important ones in creating
horizontal lines, we could the reconstruct the
interaction paths free from the incoming ones as
shown in figure 17b.
(a)
(b)
cm
Figure 17 – Cloud chamber picture before and after
filtering.
Although we did not record images, we also
observed that by filtering out the parts of the
Fourier transform near the center, the low
frequency components of the image were lost.
Masking the central dot in the transform of the
grid created another grid with half the original
spacing. The high frequency components were
lost when the outlying parts of the Fourier
transform were masked, although this was hard
to see, since it just caused the ‘on/off’ wave of
the grid to become smoother.
cm
Figure 16 – Diagonal grating obtained from filtering
out all but a diagonal line in the frequency space when
using the grid as an image.
Predicted period
p
(mm)
2
0.282 ± 0.004
Measured period
(mm)
5 – Using Fourier optics to create a
Hologram
0.288 ± 0.004
Table 11 – Predicted and experimental periods for the
image with only the diagonal frequency components
retained.
Starting from the principles of Fourier
optics, one can understand the basic of the
creation of a hologram. The light coming from
an object, after going through a lens, computes
its Fourier transform, which we can then
reconstruct by calculating the inverse Fourier
transform. Then by simulating the Fourier
transform of an object, we can reconstruct it and
get an image of the object. If we wanted to
remake a sine function which is a truly periodic
function, we would then need to simulate the
Fourier transform made of two dots, which,
when the light they emit interfere, allows us to
get the sine function. This is shown in figure 18,
4.5 – Image processing (Activity 6)
The goal in this activity is to filter out
certain parts of an image to have a clearer view
of the interesting parts. In this case, we used a
picture from a cloud chamber showing multiple
paths of incoming particles, which are
horizontal lines in this case, as well as some
paths from interacting particles which are more
chaotic. By filtering out the components making
the horizontal lines in the image, we can then
8
where the dashed lines represent the transition
between finite and infinite distance from the
source, the thick line is the screen and the other
lines represent the wave, changing from
spherical waves at finite distance to plane waves
at infinite distance.
the light coming towards it was brighter then the
light of the reference beam, creating an
imbalance that prevented the possibility of
having a usable interference pattern.
6 - Conclusion
In this experiment, we applied Fourier
analysis to different setups used to compute
Fourier transforms and their inverses. From the
values obtained in the first activity, we can see
that this treatment is legitimate and that lenses
can be a powerful tool to make such
computations as long as one is interested only in
specific parts of the Fourier transform.
The second activity shows us the
limitation of such a treatment as the film used,
and probably the scanning process as well, have
distorted the intensity graphs away from what
they would theoretically have been supposed to
be. Even if the intensity of the Fourier transform
was altered during the process, we can see from
the result of the three widest slits that we can
still identify the spacing between the
frequencies, thus identify which ones are used in
the Fourier series corresponding to our slit. The
result from the thinnest one is quite off,
probably because of a systematic error in the
measurement of the width of the slit of the order
of 10-5 meters, which is more important
compared to the actual size of the smallest slit
then to the others, causing the bigger deviation
of the experimental value from our expected
value.
Optically computing the inverse Fourier
transform of an optically computed Fourier
transform also proved to be successful since our
values for the period of the grating used and its
reconstructed image agree well. Preventing
some part of the Fourier transform to participate
in the reconstruction of the image also worked
well since our period for the diagonal grating
agrees with what was expected. This can then be
applied without risk of distortions to real life
problems such as filtering out the paths of
incoming particles from a cloud chamber
photograph. It worked well beside the fact that it
also filtered out the horizontal parts of the
interaction paths, which was expected as this
Figure 18 – Setup to get a picture of a sine function.
When the lines representing the crests of
the plane waves intersect, we get a maximum as
well as when two minima intersect, with zeros
in between since what we get on the
photographic paper is the intensity of the light,
which is equal to the square of the wave
function.
To make our hologram, we used
a somewhat similar process but replaced one of
the light source by the object of which we
wanted the hologram, in this case the statue of a
swan, which is the setup shown in figure 19.
Everything has to be fixed in order to prevent
vibrations and that lenses needed to be as clean
as we could get them in order for it to work.
Figure 19 - Setup used to make a hologram.
The hologram itself is a success in only
some sections of the film, half of it not working
probably because it was closer than the other
half to the object and farther from the mirror
reflecting the reference beam, which means that
9
treatment took away all horizontal lines in the
photograph.
The hologram part was the most
demanding part of the experiment in term of the
precision needed for the setup and the
sensitivity of the film. It was nevertheless
successful and reproducible since we have two
holograms made in the same conditions that
show similar behaviour.
10