# Holography 1 HOLOGRAPHY

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1 Holography 1 HOLOGRAPHY Introduction and Background The aesthetic appeal and commercial usefulness of holography are both related to the ability of a hologram to store a three-dimensional image. Unlike ordinary photographs, holograms record both phase and amplitude information. Because phase is a relative property, construction of a hologram requires a reference in addition to the light reflected from an object s surface. Additional requirements include a powerful, coherent, monochromatic light source (e.g. a laser, whose workings you should understand); a vibration-free flat surface; some common optical devices; and photographic plates, chemicals, and a darkroom. A hologram can be made by exposing a photographic plate to the interference pattern made by the reference and object s. The plate is then developed and dried. When illuminated by a reference similar to the original one, it recreates a three dimensional picture of the object. There is nothing mystical about this recreation, and you can understand how it works with the help of some Fourier transform mathematics. Mathematical Overview Consider the light waves reflected from an object (see Figure 1). The details of this object obviously depend upon the shape and nature of the particular object. We will take a general approach and say that the object is a superposition of plane waves, ψ 0 (r, t) = A 0 F 0 (k x, k y )e i(ωt kxx kyy kzz) dk x dk y, where (k 2 x + k 2 y + k 2 z) 1/2 = k = 2π/λ. Let s assume that we have permitted ψ 0 to interfere with a reference given by ψ 1 (r, t) = A 1 e i(ωt kz), and recorded the resulting intensity pattern on a photographic plate in the plane z = 0. recorded intensity will be I(x, y) = A 1 + A 0 F 0 (k x, k y )e i(kxx kyy) dk x dk y 2. When the right-hand side of this equation is expanded, it will have terms proportional to ψ 0 2, ψ 1 2, and ψ 0 ψ 1. We will assume that the reference is much more intense than the object, so that ψ 1 >> ψ 0. Then the term in ψ 0 2 can be ignored, and the intensity can be written I(x, y) a A 0 A 1 F 0 (k x, k y )e i(kxx+kyy) dk x dk y + A 1 A 0 F 0 (k x, k y )e +i(kxx+kyy) dk x dk y. The developed plate will have a transmission function T (x, y) = 1 γi(x, y), where γ is a function of the exposure and developing processes. The hologram will therefore have a transmission function T (x, y) = C 0 γa 0 A 1 F 0 (k x, k y )e i(kxx+kyy) dk x dk y γa 1 A 0 F 0 (k x, k y )e +i(kxx+kyy) dk x dk y, where C 0 is a constant. Put in words, the hologram contains scaled versions of the two-dimensional Fourier transform of the object, and the inverse two-dimensional Fourier transform of the The

2 Holography 2 Emulsion P2 Object Point Source Reference Point Source P1 Figure 1: A two dimensional projection of the interference patterns produced. A photographic plate at P1 would result in a transmission hologram, a plate at P2 in a reflection hologram. object, superimposed on each other. If you illuminate this hologram with a plane wave like the original reference, another Fourier transformation takes place, and what emerges is where, as is usual in Fourier optics, ψ(r, t) = F (k x, k y)e i(ωt k x x k y y k z z) dk xdk y F (k x, k y) = ( ) 1 2 2π T (x, y)e i(k xx+k yy) dxdy. The three terms constituting T (x, y) obligingly integrate as delta functions (the term we threw out above would not have been so obliging), and we find F (k x, k y) = γa 1A 0 F 0 (k x, k y) γa 1 A 0F 0 ( k x, k y) + C 0 δ(k x)δ(k y). Therefore the wave that emerges from the suitably illuminated hologram (see Figure 2) consists of three parts: (1) a term proportional to the original object, reconstructing the object in all details of amplitude and phase; (2) a so-called twin reconstruction that can be shown to be the original object reflected in the plane of the photographic plate and run backwards in time; and (3) an undeflected piece of the reference traveling along the z axis. Transmission and Reflection Holograms The first holograms were examples of transmission holography. Transmission holograms require a monochromatic reconstructive. In other words, the image is only visible when the hologram is illuminated by either a laser or other single frequency source. This limitation arises from the way a hologram stores information. The interference patterns produced from an object and reference, whose formalism has been derived above, are actually three dimensional hyperbolae of revolution[3]. The position of the photographic plate relative to the object determines how this interference pattern is recorded onto the photographic emulsion. Figure 1 shows two possible placements of the photographic emulsion relative to the object. Position one (P1) shows the development of a transmission hologram. Note how the interference pattern is recorded perpendicular to the plane of the plate. Consequently, any frequency of light

3 Holography 3 plate reference Object source plate reference Object source Hologram Primary image Hologram Conjugate Image reconstructing reconstructing Conjugate Image Transmission Reconstruction Primary image Reflection Reconstruction (a) (b) Figure 2: The recording and reconstruction setups for (a) transmission and (b) reflection holograms. The transmission figure was taken from [3]. can pass through the plate. If this hologram were viewed in white light, every frequency would create an image, the combined effect simply being a wash-out. This is why transmission holograms are not visible in white light. However if we place the photographic plate at P2, the interference pattern is recorded parallel to the plate and along the thickness of the emulsion. This thickness, irrelevant to transmission holograms is essential for reflection holograms. Reflection holograms get their name from the fact that instead of creating an image by transforming the reconstructive into a replica of the object as it passes through the plate, they reflect a reconstruction of the object. Figure 2 illustrates the difference between the two types of holograms. The unique characteristic of reflection holograms is that they can be viewed in white light. Therefore reflection holograms are commonly known as white light holograms. Exactly how does this happen? As we mentioned earlier, the thickness of the emulsion is crucial. In reflection holography, the interference pattern is recorded along the thickness of the emulsion, creating what are known as Bragg diffraction planes. These planes act similarly to half-silvered mirrors: some of the light used to illuminate the object reflects off each embedded interference fringe. As shown in figure 3, there is a path difference between light rays reflecting off different Bragg diffraction planes. These reflected rays then interfere with each other, and if they are not in phase, the interference will be destructive. Consequently, the only light rays reflected off a Bragg diffraction grating have path length differences equal to an integral number of wavelengths. The geometry shown in figure 3 allows us to derive a quantitative description of Bragg diffraction [5]. The passing through the top grating and reflecting off the lower layer travels a distance AB + BC longer than the reflected off the top layer. The angle of incidence θ is equal to the angle of reflection, so AB = BC = d sin θ. The path length difference must be equal to an integral multiple of wavelengths for constructive interference to occur. Therefore, the condition for

4 Holography 4 q q A q C d B Figure 3: Diagram representing Bragg diffraction. The path length difference is AB BC and the angle of reflection is θ. Figure taken from [5]. constructive interference is nλ = 2d sin θ, (1) which is known as Bragg s Law. Bragg s law allows only a particular wavelength to be reflected at a specific angle. Therefore, when viewed in white light, a reflection hologram will essentially select one particular frequency to reconstruct the image, preventing the wash-out effect seen in transmission holograms. Additional Details Following certain procedures will help you to ensure success in making your holograms. 1. Recall that in the derivation above we ignored the contribution of ψ 0 2 in comparison with ψ 1 2 ; that is, we assumed that the reference was more intense than the object. However, the derivation did not tell us how much more intense the reference should be. It has been found empirically that a reference/object intensity ratio between 3 and 10 yields good results. A photocell is provided to measure the intensities of the reference and object s. 2. Higher intensities (consistent with a suitable reference/object ratio), will need shorter exposure times, and, will, in general, produce better holograms. 3. A clean, uniform reference is needed for a high quality hologram. A pinhole apparatus can help you achieve this goal. Set it up carefully in order to minimize the diffraction and interference patterns from imperfect optical surfaces (mirrors, splitters, and lenses) in the path of the reference. In addition, you should make sure that the optical surfaces are clean and free of dust. 4. The path lengths of the reference and object s should differ by less that the coherence length of the laser. You should understand what this means about the purity of the laser light.

6 Holography 6 References [1] Gabor, D. 1972, Holography, , Science, 177, [2] Jeong, T. H. 1970, Holograms, in Physics Demonstration Experiments, ed. H. F. Meiners (New York: Ronald Press), vol. 2, chapter 37. [3] Beesley, M. J., Lasers and their Applications, (London: Taylor and Francis), chapter 12. [4] Lasers and Light, (San Francisco: W.H. Freeman and Company, 1969), especially Sections 6 and 7. [5] Schields, P. Bragg s Law and Diffraction. /ProjectJava/Bragg/home.html. 4/10/02. The paper by Gabor is the author s Nobel Prize address. Jeong gives a very helpful discussion of experimental approaches, and the last reference is a collection of articles from Scientific American. Revised January, 1990 J. H. Taylor Revised April, 2002 A. Erickcek, Z. Kermish

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