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Unformatted text preview: Fraunhofer Diffraction by a ThreeSlit Uneven Transmittance A spatial light modulator is adjusted such that the amplitude transmittance, t,
is a threeslit 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 xcomponent of the diffracted
wavevector and I0 is the farﬁeld 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 farﬁeld (as measured from the normal to the surface of the spatial light
modulator) and I0 is the farﬁeld radiant intensity at (9 = O. Simplify your answers as ‘
much as possible and put your ﬁnal answers in the spaces provided. Make an accurate
(computerbased not handdrawn) 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 ThreeSlit 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 + $sm6) co“ ,\ 2
I0 (smegman9) sin("T“sin6) 47ra8m0)> Alternatively, the transmittance is t($) = 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.
 Spring '08
 Gaylord

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