This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Fraunhofer Diffraction by a ThreeSlit Triangular Transmittance A spatial light modulator is adjusted such that the amplitude transmittance, t, is
a threeslit triangular transmittance function, f at, as shown in the ﬁgure. At the center
(a: = 0), the slit amplitude transmittance is 100% (t = 1). The amplitude transmittance
drops to zero (t = 0) at :1: = ia/ 2. The left and right slits are also a in width. The
amplitude transmittance of the left and right slits is 50% (t = 0.5). The spatial light modulator is illuminated at normal incidence by a plane wave of freespace wavelength A. Calculate, showing all work, the radiant intensity of the far—ﬁeld (Fraunhofer) diffraction pattern resulting from the diffraction of the plane wave by this threeslit tra
iangular amplitude transmittance function, f3t. Express your answer as I (km) Where I (km) is a function of Io, a, and kI only where km mcomponent of the diffracted wavevec
tor and Io is the far—ﬁeld radiant intensity at k1; = 0. Also express your answer as 1(0)
where 1(6) is a function of IO, a, A, and 0 only where 0 is the angle of propagation in the far—ﬁeld (as measured from the normal to the surface of the spatial light modula
tor) and Io is the farﬁeld radiant intensity at 9 = 0. Simplify your answers as much as possible and put your ﬁnal answers in the space provided. Plot your result I (kx) for —47r/a, < krt < +47r/a. Attach your plot to your solution.
[(1%) = 1(9) = Fraunhofer Diffraction by a ThreeSlit Triangular 'D‘ansmittance The transmittance is t0”) = t1(9'5) + M33) where
t1(a:)=1 —%<x<g 1 5a 3a 3a 5a
t2($)=§ ——2—<:I:<——2— and +—2—<m<+? The Fourier transforms are obtained using t ——> m, w ——> kz, and 7' —> a and where k; = 277'sin0. F1 (j kl a) = a center slit [clEa. 2  M
F2(jk$ a) = a 8m( 2 ) 003(2 [cat a) left and right slits 16:0, 2 2 F1 + F2
 1222 ' m F(jk$a) = (1% +aiz—T—LiT2)—cos(2k$a)
‘2L ‘32"
~ 1212 F(jkz a) = aw [l + cos(2kx a)] 2 The radiant intensity is thus I(kx) : [1 + 003(2kz a)]>2 2 where ;Q_2
4—61. The radiant intensity is also 1(6) 2 E <————sm(ﬂxqsm6) [1 + cos(2wasin0)])2 4 ﬁ—asinﬁ) A
where
If 2 (12.
Alternatively,
1533,5019ac a) = g center slit, 0.50 transmittance
2
F3,0_5(jl<:31c a) : % swig2) [1 + 2003(2 km (1)] three slits, 0.50 transmittance
2 16.1.9. M F019,, a) : garb—ISL: ) + Egan—IE1; )[1 + 2cos(2kza)]
2 2
m F(jkm a) = a mix; )[1 + cos(2kxa)] as before. 7. 6. 5. A 3.
0 O 0 0 0 E295 869:5 ...
View
Full
Document
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

Click to edit the document details