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Unformatted text preview: Fraunhofer Diffraction by a Transmittance Function A spatial light modulator is adjusted such that the amplitude transmittance, f s , is as shown in the figure. At the center ( x = 0), the amplitude transmittance is 100% ( t = 1 . 00). The transmittance drops to 75% ( t = 0 . 75) at x = ± a/ 2. The transmittance drops again to 50% ( t = 0 . 50) at x = ± 3 a/ 2. The transmittance drops then to 25% ( t = 0 . 25) at x = ± 5 a/ 2. The transmittance drops to 0% ( t = 0 . 00) at x = ± 7 a/ 2. In addition, the transmittance is 25% ( t = 0 . 25) from x = ± 9 a/ 2 to ± 11 a/ 2. The spatial light modulator is illuminated in air at normal incidence by a plane wave of freespace wavelength λ . Derive, showing all work, the amplitude of the farfield (Fraunhofer) diffraction pattern resulting from the diffraction of the plane wave by this amplitude transmittance function, f s . Express your answer as A ( k x ) where A ( k x ) is a function of a and k x only , where k x is the xcomponent of the diffracted wavevector. Derive, showing all work,component of the diffracted wavevector....
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This note was uploaded on 04/29/2008 for the course ECE 4500 taught by Professor Gaylord during the Spring '08 term at Georgia Institute of Technology.
 Spring '08
 Gaylord

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