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Haverford College
Physics 211
Physical Optics: Interference and Diffraction, Abbe Theory of Imaging and Spatial
Filtering (updated 9/2008 SAK)
Introduction
The Physics 106 experiment on interference and diffraction demonstrated that geometrical ray optics is
not sufficient to describe the physical behavior of light waves. Similarly, the wave-like nature of light
must be considered in describing the details of image formation by optical instruments. In this lab, we
will see that a physical optics approach using solutions to Maxwell’s equations and Fourier transforms
provides a way to describe image formation that includes the wave behavior of light. By thinking in
terms of Fourier transforms, we will see how to remove noise from a laser beam, how to understand the
resolution limit of optical instruments and how to manipulate optical images. (Of course, even though
our discussion will focus here on visible light, our results will apply to a broader class of waves,
including sound waves, electrons, neutrons and invisible forms of electromagnetic radiation.)
First, however, let’s remember the lessons from the interference and diffraction lab. Consider a
diffraction grating consisting of equally spaced narrow absorbing and transmitting (black and white)
bands or slits. It is possible not only to determine mathematically the angles m of the diffracted orders
using:
sin m  m d
(1)
(with m as an integer), but one can also determine the relative intensities I() of the diffracted spots. If we
insert a lens behind the diffraction grating, the effect of the lens serves to refocus the interference pattern
so it reforms into an image of the diffraction grating. We will see shortly how this simple experiment can
be used to understand how imaging works from a wave perspective.
First, let’s recall how periodic waves are represented. Most of us are familiar with repetitions in time,
such as periodic oscillations associated with sound waves. Pure tones are made of sinusoidal sound
waves of a single frequency: I (t) = Aosin t. More complicated sound waveforms can be constructed
from the superposition of pure sine waves with different frequencies, amplitudes and phases.
The same concept can be applied to spatial frequencies describing the variation in space of a pattern. The
spatial frequency represents the rapidity with which a feature varies spatially across an object or image.
The idea is that you can have rapid variations (which represent high spatial frequencies) as well as slow
changes or fluctuations which occur on larger spatial scales (these represent low spatial frequencies).
Spatial frequencies, k, are usually measured in cycles/mm or mm-1. Just as we can define a frequency,
=2/T, for variations in time with period T, k=2/ for 1D spatial oscillations with wavelength Here
we’ve used a different symbol for the spatial wavelength because we’re reserving  for the wavelength of light.) A
sinusoidal variation in space would be represented as:
g ( x)  A sin kx 

A ikx ikx
e e
2i

(2)
All of this can be expressed mathematically in terms of Fourier series. Any periodic (repeating) function
can be expressed as a superposition of sine waves at the fundamental frequency (k for space, f for time)
and its higher harmonics at integer multiples of f or k (i.e. 2 k, 3 k, etc.) The amount that each frequency’s
contributes is different and can be calculated using integrals of the original function g(x). The
decomposition of the periodic function into its harmonic components is called Fourier analysis, and from
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this analysis, the amplitude of each harmonic contribution is determined as is its phase relative to the
fundamental (i.e., whether the harmonic is in
phase or 180 degrees out of phase with the
fundamental).
This procedure can be reversed. If a pattern at
the fundamental frequency is combined with
the appropriate number of higher harmonics, it
is possible to approximate any function with a
repetition frequency equal to that of the
fundamental. This is called the inverse Fourier
transform or Fourier synthesis.
Now let’s imagine how to represent the
function describing the diffraction gratings
used in the earlier experiments. If the grating
consisted of a sinusoidal variation only (see
Figure 1a), there would only be a zero
frequency component (for the constant offset)
and two first order peaks at frequencies ± f or k
(Why two peaks instead of one? Think of the
representation of sine in terms of complex exponentials.) Fig. 1(a) shows the function and its Fourier
coefficients.) If instead we had the actual pattern of alternating black and transparent bands, we would
have a function like that in Fig. 1(b). Then its
Fourier transform would have many
contributions from all values of odd integral f
or k. To completely synthesize a function such
as our idealized alternating black and
transparent grating, an infinite number of
harmonics would be needed. If only a finite
number of harmonics were used, the
synthesized function would resemble the
function, but it will have edges that are not as
sharp as the original (and it will have ringing
oscillations near the sharp edges). A simple
example is shown in Figure 2 in which only
the fundamental and two harmonics are used
to approximate the square wave function,
which is equivalent to our black and
transparent grating function.
To see what all this has to do with optics and
image formation, now let’s apply this
reasoning to what you observed in the earlier
interference and diffraction experiments. First
it’s important to realize that the effect of diffraction is to Fourier transform the object diffracting the light.
(The derivation of this result for the general case is not that hard, but it’s beyond the scope of our course.
ikx
The e factors in the Fourier transform result from keeping track of the varying pathlengths of each
wave and its resulting phase difference; the sums or integrals from adding all the lightwaves together as
they superimpose to form the interference pattern.) To be perfectly clear, the Fourier transform of a
diffraction grating’s spatial pattern is the electric field at each point in its interference pattern. What’s
being transformed mathematically? The function, g(x), that describes the spatial variation of the
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diffraction pattern. In the Fourier transform, what plays the role of the spatial frequencies? The values of
 at which diffraction peaks occur for a given value of wavelength of light. What is the amplitude of the
Fourier transform? The electric field amplitude, which is the square root of the light intensity at each
point in the diffraction pattern. Similarly, the phase of the electric field at each point in the diffraction
corresponds to the phase of that Fourier component.
This simple idea that diffraction provides a probe of the Fourier transform of an object is a powerful one
that underlies many crucially important experimental techniques, including x-ray and neutron
crystallography, electron microscopy and diffraction, static and dynamic light scattering and many forms
of spectroscopy.
If we were to diffract light from a sinusoidal diffraction grating like that in Fig. 1(a), we would observe a
diffraction pattern with only a zeroth and the first orders m=±1. As the repetitive patterns in the grating
sharpened up and departed from sinusoidal, approaching the situation in Fig. 1(b), additional diffraction
orders would appear and, in the idealized case of the black and transparent grating (with alternating
perfectly transmitting and absorbing slits), the infinite series of diffraction orders would be present (Fig.
3(a)). Thus, the theoretical interference patterns you learned in Physics 106 lecture and lab merely
corresponded to special cases of a Fourier transform! You could compute the interference pattern of any
sort of diffraction grating using this method. For example, if instead of a diffraction grating, a circular
aperture is used, then the resulting interference pattern looks like that in Fig. 3(b).
Figure 3
Furthermore, the action of a lens acts to perform a Fourier transform as well, allowing you to recover the
original spatial pattern of the diffracting object: in other words, to form its image. To start, consider Fig.
4 , which shows how a lens takes all the rays of light with a particular angle (in other words, all those
corresponding to a particular spatial frequency) and focuses them down to a point at the back focal plane
one focal length f from the lens. In this geometry, the lens maps angle (the rays with the same helpfully
shown as having a single color in the figure) onto x at the focal plane, a nice experimental trick that you might
wish to file away for future reference! Next, we see from Fig. 3 that the action of the lens is to take all the
light diffracted through all angles from each point on the diffracting object and refocus it down to a single
point at the image plane. Summing all the light at various angles indeed corresponds to summing all
contributions of a Fourier series over all k, which is the inverse Fourier transform. Thus, the action of a
lens is in effect to undo the Fourier transform effected by the diffraction of light.
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Fig. 4) http://www.doitpoms.ac.uk/tlplib/DD1-6/image.php
One can see that the larger the lens used, the more of the widely diffracted (i.e., high spatial frequency)
components will be retained and focused to make the image. Although very little light is collected at the
edge of the lens, it is this light that contributes to the fine detail in high-resolution images. This same idea
can be used to explain why large diameter mirrors are used to provide fine resolution in modern
telescopes.
Since the light in the Fourier transform plane (Figure 3) is arranged according to increasing spatial
frequency with radius, blocking light at that location with a mask which selectively obstructs and
transmits the light will change the distribution of spatial frequencies present. This will in turn change the
content of the image in a predictable way: if higher frequencies are removed, the image will be blurred,
for example. This method of modifying an image by changing the spatial frequencies that are present is
called spatial filtering. It is used widely in electron microscopy and other imaging methods.
Circular apertures and the Rayleigh resolution limit
The interference pattern for a circular aperture, shown in Fig. 3(b), is a particularly interesting case
because many results from this situation also apply to imaging with either a spherical lens or a mirror
with the same diameter. In particular, the interference pattern for this case can be used to derive the
spatial resolution limit for both the mirror and lens cases. The Fourier transform of a filled circular region
is called the Airy function, and consists of a series of circularly-symmetrical intensity maxima and nodes,
and an amplitude envelope that decrease with increasing radius.
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Figure 6. The Rayleigh criterion. The plot of intensity along a line
between the centers of two diffraction patterns from close-spaced
circular apertures is shown below a photo of two sources barely
resolved as specified by the Rayleigh criterion.
The region within the first dark ring is called the Airy disk, and
the angular separation  between the central maximum of the Airy
disk and the first dark ring is given by:
 sin   1.22 D
(3)
where D is the aperture diameter and the first relationship is the
small angle approximation (since in practice D>>) The
measurement of the diameter D of different size pinholes using
the observed Airy disk pattern is part of this lab.
One good approximation of a point source is a bright star. A pair
of stars close to one another can give a measure of the diffraction
limits of a system which is called the Rayleigh resolution. The
resolution of a system can be determined by the smallest angular
separation between such sources that still allows them to be distinguished. A limit of the resolution that
has been used in many instances is that two point sources are considered just resolvable if the maximum
of the diffraction pattern of one point source falls on the first dark ring of the pattern of the second point
source (Fig. 6), giving the same result as equation (3).
Spatial Filtering
Another important application of this idea is spatial filtering, the manipulation of an interference pattern
to remove selected frequency components. One application of spatial filtering is to remove high
frequency noise from an image in electron diffraction or other imaging techniques. Another is to “clean
up” a laser beam by removing the effects of scattering (see Figure 6). The intensity distribution of the
beam in many lasers is a Gaussian function of radius. However, dust and small imperfections in the air,
lenses, windows, and surfaces it encounters can produce irregularities in the intensity pattern that show
up as mostly high frequency contributions to the Fourier transform. By contrast, the Fourier transform of
a Gaussian beam is another Gaussian, centered at zero frequency and with a width determined by the
reciprocal of the original beam’s diameter. When the laser beam is focused with a lens, the Fourier
transform intensity distribution has a distribution in angle arranged according to spatial frequency, with
the Gaussian components peaked near zero angle and the high frequency noise at high angles (Fig. 6) If
a small pinhole is placed at the lens’s focal point, then it passes only the low frequency components. If
the pinhole’s diameter is chosen to pass only the low-frequency Gaussian beam contributions, then the
irregularities will be removed from the emerging beam and a “clean” beam will result.
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Figure 6: Spatial filtering of a laser beam
Prelab Problem: Design a “beam expander”: an optical device to take a collimated (parallel)
laser beam with one diameter and output a collimated beam with a larger diameter. Using
lenses of your choice and using a ray optics diagram, show how to construct this device and
compute its magnification (the factor by which the diameters have been expanded.) There are
several designs possible!
Experiment 1: Abbe imaging & spatial filtering
In this lab the frequency distribution of an object illuminated with a laser beam will be examined with a
single lens placed after a picture slide. The light distribution formed at the focal plane tells us the
frequency content of the image, and by manipulating the light in that plane we can control the quality
and content of the image to be displayed. The optical setup is shown in Fig. 7.
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The laser beam has a smooth profile (i.e., it does not have appreciable high-frequency components), and
when it is focused it produces a well defined Gaussian spot. If we pass this beam through a grating or
screen, then in the focal plane of the lens we will see higher angle spots indicating that additional spatial
frequency components have been added. We will build and experimental set up which will allow us to
examine this feature.
1.
Mount a laser assembly (item A) on the far side of the optical table, and adjust the position such
that the laser beam is parallel to the edge. Tape an index card with a small (2 mm) hole in it to
the front of the laser so that the laser beam can pass through it. This card will be used to monitor
the reflections from the optical components for alignment purposes as they are inserted into the
beam.
Warning: This experiment will be performed in a darkened room. Extreme care should be
taken when using the laser. Your pupils will be expanded and will let in 60 times more light
than a lighted room. Do not look at direct specular reflections or the direct laser beam. Please
wear the appropriate eye protection when the laser is in use.
2.
Mount a steering mirror assembly (item B) at the far corner of the table. Adjust the mirror so that
the beam intersects the center of the mirror and reflects the beam at a right angle.
3.
Place a second mirror steering assembly (item C) in line with the laser beam at the lower left
corner of the table. Again adjust the mirror so that the reflected beam is at a right angle.
4.
Set up a beam expander (item D) between the two mirror assemblies.
5.
Mount the transform lens (focal length 150 mm) in a lens assembly (item E) at a distance 12
inches (305 mm) from the second steering mirror (item C). Center the lens in the laser beam path.
6.
Set up a slide holder and place a focus index card (item F) at one focal length (150 mm) behind
the transform lens (item E).
7.
Mount another slide holder (item G) at a distance 225 mm in front of the transform lens (item E).
This slide holder will be used to hold the picture slides in the path of the laser beam.
8.
The set up is now ready for the examination of slides. The image will be observed on an
observation index card mounted about 450 mm behind the lens (300 mm behind the focus card)
and will give an image about twice the size of the object slide image when the focus card is
removed.
9.
With no slide present, the laser beam produces a focused Gaussian spot. The center of this spot
corresponds to the zero of spatial frequency. The Gaussian laser beam’s Fourier transform has
only low frequency components very close in angle to this zero spatial frequency. Any other
spots appearing on the card indicate that other spatial frequencies are present. Remove the focus
card and you will see uniform (low spatial frequency) illumination at the observation card.
10. Place the square mesh in the slide holder and replace the index card to the focal plane of the lens.
On the focus card, you should see a square grid pattern of dots representing the frequency
content of the mesh in both the horizontal and vertical directions. Mark on the card with a pencil
the location of these axes. The dots on the x and y axes represent the frequencies present in the
horizontal and vertical directions respectively in the slide.
11. Remove the focus card and adjust the observation index card distance to achieve the sharpest
image. This image can be manipulated by blocking certain frequencies with the focus card. To
illustrate this, cut out a vertical slit in the focus card which will pass through only the vertical
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frequency components (only the center horizontal component, the DC, gets through). Examine
the image on the index card at the observation plane and record what you see.
12. Rotate the focus card so that the dots on the x-axis (horizontal direction) are passed. Record what
you see.
13. Using another focus card, cut out a hole so that only the central focus spot (DC) is transmitted.
You should see only uniform (flat) illumination on the observation card. This is the principle of
the spatial filter and the “cleaning” of optical beams described above. The high spatial
frequencies are removed by obstructing the unwanted harmonics, resulting in a low-pass filtered
signal.
14. Make another filter in the form of a larger circular hole that passes only the central spot and the
two adjacent spots. Record what you see on the observation card. Note that this removes sharp
edges and produces a soft image (this is another low-pass filter at work). Alternately, make a
special filter that will obstruct only the central spot by putting a drop of ink on a microscope slide
and locating it at the center spot on the focal plane. Record your observations. This will remove
the uniform illumination (DC) component and will leave the edges enhanced. This is a high-pass
filter.
15. Replace the square mesh slide with a picture slide. Notice that this slide has lots of horizontal
lines superimposed on it. This will be the original image which will be filtered. On the focus
card the light distribution consists of an irregular distribution of light representing the many
spatial frequencies present in the picture. Superimposed on this distribution is a set of faint spots
aligned along the vertical axis and passing through the central spot. This line of spots represents
the frequency content of the horizontal lines on the slide.
16. Attach with tape to the slide two pins (or toothpicks or thin objects) so that when it is placed back
in the focal plane, the objects will obstruct the vertical line of spots. The central spot should not
be obscured. Record your observations.
17. Other cutouts may be used at the focal plane. For example, cut a hole in the index card so it
obstructs the outside spots. These spots contain the high frequency information. Record your
observations. Obstructing them results in a fuzzier picture at the observation plane.
18. Do not take apart your optical setup—we will use most of it in the next experiment.
Experiment 2: Diffraction by a circular aperture and the Rayleigh resolution limit (OPTIONAL)
In this experiment you will measure the diffraction effects of circular apertures and will quantitatively
determine the aperture sizes using the diffraction pattern. The diffraction associated with the size of the
aperture determines the resolving power of all optical instruments, from the electron microscope to giant
radio telescope dishes used in astronomy. We will see this effect at work in the Optical Instruments lab.
In addition, you will discover that a solid object not only casts a shadow, but it is possible for a bright
spot to appear in the center of the shadow!
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1.
Using your earlier optical setup, make the following changes. See Fig. 8. Remove the beam
expander, first slide holder and transform lens, but retain all of the other components in their
previous configurations.
2.
Mount a lens chuck assembly (item E), which will serve as a the circular aperture holder, in the
path of the beam five inches after the second mirror assembly (item C) (i.e., the beam will hit item
E after item C.
3.
Place the first pinhole target P1 into the lens chuck assembly (item E). Adjust it so that the beam
strikes the target at the center. The pinhole target will reflect a large percentage of the beam.
4.
Adjust the second mirror steering assembly (item C) so that the laser beam fills the pinhole (item
E). Look for a bright red glow from the back side of the pinhole target.
5.
Carefully adjust the second mirror steering assembly (item C) to produce the brightest image on
the target screen (item D). You should see a bright central circle surrounded by dark and light
circular bands. This is the Airy disk pattern.
6.
Measure the distance d from the pinhole target (item E) to the target screen (item D). Mark and
measure the diameter of the first dark circular band around the bright central circle in the
diffraction image. This is a measure of the amount of diffraction caused by the pinhole.
7.
Since the angular diffraction is small, the sine and tangent of the diffraction angle are about
equal:   tan   sin   1.22 D . The tangent of  can be found by dividing the radius (be
careful: not the diameter!) of the first dark band by the distance d from the pinhole to the target
screen. Calculate the diameter D of the pinhole using the wavelength of the He-NE laser, = 633
nm. Be sure to include the uncertainties in each of your measurements and derive an uncertainty
in your final derived value for the pinhole diameter.
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8.
Replace the pinhole target P1 with the next pinhole target P2. Repeat the necessary
measurements and calculations to determine the pinhole diameter (again with uncertainty) for
P2.
9.
Assemble a 6:1 beam expander between the first and second mirror steering assemblies (between
items B and C). Now replace the pinhole target P2 with the Fresnel target. (This device has a
solid spot surrounded by a transparent region, rather than a circular aperture.) Look at the
diffraction pattern on the target screen. Note that the center of the image has several bright and
dark rings. This is Fresnel diffraction. Depending on the distance between the Fresnel target and
the target screen, the center of the pattern may be light or dark. Although the Fresnel target has a
central obscuring circle, note that there is still light at the center of the pattern. The bright spot at
the center is sometimes called the Poisson Spot or the Spot of Arago. (In an amusing story retold
in Hecht’s Optics text [pg. 494 in the 2002 4th edition], we learn that Fresnel (then a 30-year-old
engineering) entered a paper on the wave theory of diffraction into a competition sponsored by
the French Academy of sciences. The judges included Poisson, a famous scientist strongly
opposed to the wave theory of light. Poisson thought he dealt a fatal blow to Fresnel’s entry
when he showed that Fresnel’s work implied that there should be a bright spot at the center of
the diffraction pattern of a solid target—a result the scientist Arago immediately found
experimentally! Section 10.3 of this text also derives diffraction by circular apertures and shows
why the bright spot appears in detail.)
10. Now examine the shadows of other objects put in the expanded laser beam. Pencil points, wires,
and small beads on a string are all good objects that give interesting Fresnel patterns. Note how
the patterns change as you move the objects along the beam direction. Sketch some of the
patterns.
Turn In
Please turn in explanations of your setup, observations, measurements, and uncertainties. Include your
derived pinhole diameters and a comparison of the real pinhole diameter values with their measured
values. Make statements about their consistency. If the values were not consistent, what went wrong?
What could account for the difference? What were the largest sources of uncertainty? Etc. Please also
include some sketches of your Fresnel patterns from other kinds of objects.