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Unformatted text preview: Fraunhofer Diffraction by Triangular 'I‘ransmittance Slit A photographic plate in air contains a triangular amplitude transmittance slit of
width a. The slit plate is illuminated at normal incidence by a plane wave of freespace
wavelength x\. The amplitude transmittance of the slit plate is linear with position as shown in the diagram below. There is 100% transmittance at the center of the slit and 0% transmittance at the edges. L—«—»l Calculate, by taking the Fourier transform and showing all steps, the radiant in
tensity of the far—ﬁeld (Fraunhofer) diffraction pattern resulting from the diffraction of
the plane wave by this slit. Express your answer as a function of ID, a, A, and 0 only
where 0 is the angle of propagation in the farﬁeld (as measured from the normal to the
surface of the slit plate) and Io is the farﬁeld radiant intensity at 0 = 0. Simplify your answer as much as possible and put your ﬁnal answer in the space provided. Fraunhofer Diffraction by Triangular Transmittance Slit For a triangular amplitude transmittance f (m) is %a:+1 —%<x<0
f0”): —%m+1 0<m<+%
0 otherwise Fourier Transform Method The Fourier transform of an even function is given by +00
F(jkm) = / f($) cos kmx dx, $=—OO and therefore 0 2 W2 2 = (—1: + 1) cos kzx d1: + (— —:l: + 1) cos kxa: dz:
z=~a/2 0’ 23:0 0’ =
2 1 +a./2 +a/2
=__2[_ / (kxx)cos(kzw)d(km:r) — / (kxm)cos(kxx)d(kwx)]
akin 32:0 w=0
2 +a/2
+—/ cos(kzm) d(kz:r)
k$ :1:=0
_4 +kxa/2 2 +k1a/2 2 k2 / (kmx)cos(kxm)d(kmx)+—k—/ cos(kmx) d(kzx)
:ca’ kmcczO 1E kzz=0
Using
/§cos§d§ E c035 + £3in§ foosédé _=_ sing then —4 km kac
F (1W) ‘ m[ws(‘2—a)+(i
Using
1 =  2 €
— cosé ._ 23m then
, — km , kxa
F(szw) = k :2[(Ta)3m(7
(s)
s'n2 m
FUkmx) 2% Z 13:42)
(42)
and so
2  4 m
onm) = “; ———“"k(a44)
(2) Since kw = 2T"sin0 and I0 = (12/4 sintt wasinO)
_ _____L — Io (wasin9)4
2A
Convolution Method
Deﬁne
Ct ( x ) { 1 —% < ac < +5:
re ————— =
(1/2 0 otherwise
and
g3: + 1 —% < a: < 0
t ' m — 2 a
“(m)— —;x+1 0<x<+§ 0 otherwise Then tri(ﬁ§) can be repesented by the convolution tri(&%) = ﬁreciai/z) * ﬁrect<aiﬂ)
Then the Fourier transform is f[tri($)] = §f[rect(%)]  f[rect(a%)] Since
sin 1332
f[rect(a7§)] = E (15:24))
Then
2 sin2 1322
F(jkza:) = .7:[tri(a—/2)] = (Leggy) Fem) = (— ...
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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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