4500PF13 - Fraunhofer Diffraction by a Three-Slit Uneven...

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Unformatted text preview: Fraunhofer Diffraction by a Three-Slit Uneven Transmittance A spatial light modulator is adjusted such that the amplitude transmittance, t, is a three-slit uneven transmittance function, f3“, as shown in the figure. Atthe center (:1: = 0), the slit amplitude transmittance is 100% (t : 1). The amplitude transmittance drops to zero (t = 0) at a: 2 21:1. The left and right slits are a in width. The amplitude transmittance of the left and right slits is 100% (t = 1). The spatial light modulator is " illuminated at normal incidence by a plane wave of freespace wavelength A. f3u |‘-2°—>H°-l* Calculate, showing all work, the radiant intensity of the far—field (Fraunhofer) diffraction pattern resulting‘ifrom the diffraction of the plane wave by this three—slit uneven amplitude transmittance function, f3“. Express your answer as I (k1) where I (k1) is a function of I0, 0., and k,c only where kglc is the x-component of the diffracted wavevector and I0 is the far-field radiant intensity at kg,- = 0. Also express your answer as I (0) where 1(0) is a function of I0, 0., A, and 0 only where 0 is the angle of propa- gation in the far-field (as measured from the normal to the surface of the spatial light modulator) and I0 is the far-field radiant intensity at (9 = O. Simplify your answers as ‘ much as possible and put your final answers in the spaces provided. Make an accurate (computer-based not hand-drawn) plot of your result I (km) for —41r/a < k:ac < +47r/a. Attach your plot to your solution. I (k3) = [(9) = Fraunhofer Diffraction by a Three-Slit Uneven Transmittance The transmittance is we) = mac) + was) Where t1(:r)=:1 —a<:1:<a 5a 3a 3a 5a t2(:r) — 1 ——2—<m<—? and +§<m<+3 The Fourier transforms are obtained using t —> ac, w ——> km, and T —> a and where km = 2—)? sin0. F1 (3' km a) = 2a EETEIE—f—ZSQ center slit F2 (j km a) = 2a cos(2 kgE a) left and right slits 2 F(jkx a) 2 F1 + F2 F(jkx a) = 2a[ W + 003(2 lsac (1)] F2(0) = ID = (4a)2 = 16a2 The radiant intensity is thus 10%) 2 [—0 <3in(kma) 4 W+7€~ZL°TCOSWW> The radiant intensity is also 19 = — () 4 (2303mm + $sm6) co“ ,\ 2 I0 (smegma-n9) sin("T“sin6) 47ra8m0)> Alternatively, the transmittance is t($) = t1w(x) — t1m(-’E) + Mm) Where 5a 5a t1w($):1 3a 3a t1(a:)=1 —-a<m<a F1“, (j kac a) = 5 a Wide slit 5kza 2 3kg“). 2 F1m(jkx a) = 3 a medium slit sin( Is:z a) F1(jk$ a) = 2a (km a) center slit : Flw _‘ Flm + F1} . sin( km a) ngw) = 5a——2—— — 3a——2—— + 2a——— ( (—lecz a) (—Lskz a) (kxa) 172(0) 2 I0 = (4a)2 2 16a2 The radiant intensity is thus 2 I0 sin( kz a)) “k” : KG (3%) ‘3 (3%“) H Um) 8.7.55.4.32 0000000 £225 362:5 1 0.9 0.1 ...
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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 Tech.

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4500PF13 - Fraunhofer Diffraction by a Three-Slit Uneven...

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