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4500HWF01

# 4500HWF01 - Fraunhofer Diffraction by...

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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 far-ﬁeld (as measured from the normal to the surface of the slit plate) and Io is the far-ﬁeld 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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