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 using:
 m of the diffracted orders 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) = A o sin  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:
( )

A sin kx

A
2 i
 e ikx  e
 ikx

(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

2 Physical Optics 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.
The e ikx 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

3 Physical Optics 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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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.

5 Physical Optics
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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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

8 Physical Optics 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!

9 Physical Optics
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.

10 Physical Optics
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 4 th 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.