Physics 3340 Spring Fourier Optics

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1 Physics 3340 Spring 011 Purpose Fourier Optics In this experiment we will show how the Fraunhofer diffraction pattern or spatial Fourier transform of an object can be observed within an optical system. We will construct a Fourier Optical Analyzer that can display both the real and Fourier image of any object. Introduction The optical system we will use is shown in Figure 4.1. In this experiment we will use commercial compound lenses instead of the simple lenses we have used so far. We will use a microscope objective to expand the laser beam and camera lenses when we need a large aperture converging optic. The beam is first focused at point P by the microscope objective. Microscope objectives are labeled by their magnification, m, which is related to their focal length, f, by f=(160 mm)/m. The second lens, L, is used to refocus the beam at the point P'. Since P' is conjugate to the initial focal point, P, P' is on a diffraction plane. A white screen placed at P' will show the diffraction pattern associated with the object slide, O. The object slide can be placed anywhere between L and P'. Microscope objective 40x Zeiss (+)10mm lens Object Diffraction plane Zeiss (+)10mm lens Diverging zoom lens (-)50 mm f.l. He-Ne laser f f P OA Laser L1 P L O D L3 L4 I I s3 s3 s s s4 s4 Figure 4.1. Converging beam Fourier optical analyzer. The diffraction pattern on the screen is given by the expression: I0 I u, v dxdyf x, yexp i ux vy (4.1) In this expression, I 0 is the intensity incident on the object, is the distance measured from the object to the diffraction plane, and u and v are x and y angular coordinates in the Fourier Optics 4.1 Spring 011

2 diffraction plane, measured from the point where the object intersects the optical axis. The final two lenses, L3 and L4, are used to create a magnified image, I, of the object. The combination of a converging lens, which creates an image I', and a diverging lens placed to the left of I', is called a zoom lens. This arrangement can provide a very large magnification. By alternately inserting and removing a screen at the diffraction plane, D, the direct and diffracted images can easily be compared. You can also block a portion of the diffraction pattern at D and see what the reconstructed image at I looks like with certain spatial Fourier components missing. This process is called spatial filtering. Fourier Optics 4. Spring 011

3 Introduction to Fourier Optics The Fourier Optics experiment that you are about to do builds upon knowledge that you might have gained about basic diffraction, perhaps via the Fraunhofer Diffraction Experiment. For completeness and to begin discussion of the Fourier Optics process, we repeat here the discussion of diffraction for the simpler Fraunhofer case: Fraunhofer diffraction discussion "Diffraction" refers to the spreading of waves and appearance of fringes that occur when a wave front is constricted by an aperture in a screen that is otherwise opaque. The light pattern changes as you move away from the aperture, being characterized by three regions. Plane waves Intensity profiles X Z OA Intensity Aperture Shadow Fresnel Diffraction Z ~ a / Fraunhofer Diffraction Figure 3.1 Diffraction of plane waves at an aperture. 1. In the shadow region, close to the aperture, the boundary of the transmitted light is sharp and resembles the aperture in shape.. As you move away into the Fresnel region, the beam width remains comparable to that in the aperture, but narrow fringes appear at the edges. 3. Far away, in the Fraunhofer region, the beam spreads to a width much greater than that of the aperture and is flanked by many weaker fringes. The Fraunhofer region is chosen for simple experiments because the broader fringes are easier to measure with an optical detector of finite aperture, and the calculations are more straightforward than in region. Fourier Optics 4.3 Spring 011

4 y x P(x,y) R r x y Z axis Incident waves Aperture da Figure 3.. Geometry for the Kirchoff-Fresnel diffraction integral. z Observation plane The Huygens-Fresnel principle governs diffraction phenomena: "Every unobstructed element of a wavefront acts as a source of spherical waves with the same frequency as the primary wave. The amplitude of the optical field beyond is a superposition of all these wavelets taking account of their amplitudes and phases." The Kirchoff-Fresnel diffraction integral gives quantitative expression to these ideas. Consider plane waves incident on an aperture from the left, as shown in Figure 3.. The incident field is described via: E z, t E e INC 0 i kz t The field in the aperture i.e., where z=0, is then INC 0 the wave front of area, da', and at position, r x, y,0 i t E 0, t E e. A typical element of, then acts as a source of Huygens wavelets. Our light detector sits at a point P in the observation screen, at a vector distance, R, from the origin of the aperture. Note that the distance of the detector, P, from the element da', is given by: r R r The field at P due to the element da' is then equal to: it ikr Ee 0 e de P da r Source strength Huygens spherical wave The field at P due to the entire aperture is then a superposition of the wavelets from all the elemental areas: 0 E P Aperture Area it ikr Ee e da r The detector measures the light intensity at P, rather than the electric field strength. Fourier Optics 4.4 Spring 011

5 Intensity is given by the magnitude of the time averaged Poynting vector, ˆ 0 0 S E B E Z z. Therefore, the detector measures: 0 I P E P Z Where Z is the characteristic impedance (the ratio of the electric field magnitude to the magnetic field magnitude try it out, go ahead, take the ratio of an electric field to a magnetic field and see how the units work out. It s Ohms.) of free space. In the Fraunhofer experiment, we study the diffraction through a single-slit aperture. The aperture is a slit of width, a, while the detector is a photodiode at position, P. Evaluation of the Kirchoff-Fresnel integral for the slit gives the following prediction for the diffracted intensity: xa sin a z I P x, z I INC z xa z This prediction is subject to the condition that the observation point is far enough away so that z a. From a practical perspective, z 10a is sufficiently far away for the theory to be quite accurate. Notice that the formula applies to a situation in which plane waves of uniform intensity are incident normal to a long narrow slit of uniform width. Further, the rays from different parts of the slit to a given observing point are effectively parallel (the Fraunhofer condition). To observe Fraunhofer diffraction, the design of the experiment must mimic these conditions as closely as is possible. The condition on parallelism of rays is adequately satisfied at a distance of z 10a or more, since the error depends on the square of this quantity. The case of Fourier Optics. Now that you ve been reminded of how Fraunhofer diffraction works, let s see how Fourier Optics works. The Fraunhofer case assumes that observations are made far enough away so that the light rays that cause the diffraction pattern are well treated as being effectively parallel. In Fourier optics, we use a converging lens to cause these parallel rays that leave a diffracting object to produce the diffraction pattern at a more convenient location (the focal plane of the lens), rather than requiring that we put detectors very far away from the object, where the parallel rays would eventually interfere to produce the Fraunhofer pattern. To understand how the lens functions to create the diffraction pattern, just think about what a converging lens does to parallel light rays. In the simplest case of a lens illuminated by Fourier Optics 4.5 Spring 011

6 rays that travel parallel to the optical axis, the lens focuses the rays to a point on the optical axis. This position is the so-called focal point of the lens. In this case, the lens takes a set of parallel light rays traveling at a particular angle to the optical axis, and converges them to a point. What does the lens do to a beam of light composed of parallel rays, but where the rays travel at some angle to the optical axis? For that case, consider the diagram in Figure 4.. Focal point, f Figure 4.. Parallel rays incident on the lens from the left are focused to a point, but not a point on the optical axis. The position of the point for the case of thin lenses is one focal length past the lens, and f tan above the optical axis. As the figure shows, for two of the parallel rays, the rules of geometrical optics tell us what to do: For the ray passing through the focal point, we know that it leaves the lens parallel to the optical axis. For the ray passing through the center of the lens, in the thin lens limit, there is no deflection of the ray. These two rays cross at a common point. In fact, ALL the rays converge at this point. Therefore, a lens will take a parallel beam at some incident angle to the optical axis, and converge it to a point, but the point is located off the axis. Now you are in a position to see how Fourier Optics works: A lens maps the angle incident parallel beams onto distinct spatial points. For every plane wave incident at some angle, the lens creates a delta-function intensity at some position on the focal plane. The statement, The Fourier Transform of a plane wave is a delta-function, should be triggered at this point. If you have several different incident plane waves, you will have several different bright spots at the focal plane. You are building the Fourier Transform as in Equation 4.1. If you diffract a plane wave through an object, like you did for the single slit in the Fraunhofer experiment, you will get a set of diffracted plane waves travelling in different directions. You can then either let Fourier Optics 4.6 Spring 011

7 them travel to infinity, where they interfere to produce the Fraunhofer pattern, or you can use a lens to produce the same pattern of bright spots at the focal plane of the lens. The Converging Beam Fourier System The discussion above emphasizes the way a lens maps the plane waves leaving a diffracting object onto bright spots at the focal plane of the lens. However, in this experiment, you are constructing a converging beam system. How does it work? One way to understand it is to think of the Zeiss camera lens as effectively acting like two lenses. One of the messages of geometrical optics is that any pair of lenses can be replaced by a single lens with a focal length whose inverse is the sum of the inverse focal lengths of the partner lenses (or if you like, the diopters add!). Another message is that the order of optical components does not matter if they are closely spaced. OK, so split the Zeiss into two lenses and swap the second lens and the diffracting object, so the object is between the two lenses. Set the focal length of the first so that the point source, P (see Figure 4.1), is located at the focal point of that lens. The point source then results in plane wave parallel rays exiting that first lens and illuminating the object of interest. The object then diffracts beams in various directions. The second lens then converts the diffracted beams back into bright points at its focal plane. If you work through the geometrical optics, the diffracted points will be found to be at point, P, as in Figure 4.1. That s it! Fourier Optics 4.7 Spring 011

8 Outline of the Experiment 1. Using your results from the prelab as a guide set up the Fourier Optical Analyzer. Make whatever adjustments or changes that are needed to get good direct and Fourier images.. Examine the formation of diffraction patterns. Try to understand in as much detail as you can the relationship between the direct and the Fourier image. Can you verify the idea that parallel rays from different directions lead to spots at different spatial positions off the optical axis? Can you understand diffraction from simple shapes like round and square holes of different sizes? If so, then look at regular arrays of the same objects. Your goal is to connect the spatial shape to the Fourier Transform, and then observe whether the intensity at the point P bears any resemblance to that transform. 3. Ronchi grating. The diffraction pattern of the Ronchi grating is closely related to the Fourier transform of a square wave. Which Fourier components are visible? Recall (or derive) that a square wave has only the odd harmonic components. How is this visible in the diffraction image? 4. Spatial filtering. First look at the Ronchi. How does the reconstructed image look if you a) remove the central (zeroth order) diffraction spot, b) let only the third order spots go through, c) let only the zeroth order spot pass? Next, use crossed Ronchi Gratings to create a checkerboard object. Can you find a way to filter the light at the diffraction plane so that all of the horizontal lines are removed from the reconstructed image? 5. Explore Look at some of the other objects available in the lab. What features can be removed or emphasized by spatial filtering? Fourier Optics 4.8 Spring 011

9 Problems These problems will give you a starting design for the Fourier Optical Analyzer. Do not be afraid to make changes once you get to the lab. 1. Refer to Figure 4.1. We will carry out our design using thin lens formulas. A 40x microscope objective has a 4 mm focal length. So the point P is 4 mm to the right of L1. Suppose we want the distance from P to P' to be 10 cm. A) Where should we put the 10 mm focal length camera lens L? You will find that this problem has two solutions. Choose the one with L closest to L1. b) Assuming that the laser beam begins with a 1 mm diameter, what will the beam diameter be at L?. Suppose now that the object, O, is placed 5 cm to the right of L, and L3 is placed 8 cm to the right of P'. a) Where is the image point I'? b) Now suppose that we want the final image, I, to be 100 cm to the right of L3. Where should we put the diverging lens, L4? 3. Make a sketch of the system showing the positions of all lenses and images. Remember: Converging f>0 Diverging f<0 Object to left s>0 Object to right s<0 Fourier Optics 4.9 Spring 011

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