Diffraction By An Array of Apertures

Diffraction By An Array of Apertures - Diffraction by an...

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Unformatted text preview: Diffraction by an Array of Apertures An infinite rectangular array of apertures is illuminated at normal incidence by a plane wave of freespace wavelength A. Each aperture has a transmittance of 100% and a diameter of d. The apertures are in a rectangular array with a center—to—center spacing of Am in the a: direction and a center—to—center spacing of Ag in the y direction as shown in the drawing. At all locations outside of the apertures, the transmittance is zero. J_ Av ...dT 1 To 000 Toooo ioooo oooo Calculate, showing all work, the radiant intensity of the far—field diffraction pattern of the above array of apertures. Express your answer as a function of Am, Ay, d, A, 0m, By, and I0 only where 01 is the far-field diffraction angle measured from the normal to the surface of the array in the a: direction, 6,, is the corresponding far—field diffraction angle measured in the y direction, and I0 is the far-field radiant intensity at 0x = 6y 2 0. Sim— plify your answer as much as possible and put your final answer in the space provided. I(03,0y) = Diffraction by an Array of Apertures The transmittance is 2 (2—D array of delta functions) * (circular aperture transmittance) Where “ * ” represents convolution. The 2—D array of delta functions itself can be repre- sented by a convolution. Thus The far—field diffraction pattern (in amplitude) is the Fourier transform of the transmit- tance. Thus H +00 +00 fl 2 6<m-mAx>* Z Max—mulfltcv/an m=—oo nz—oo F(jkx1jky) +00 +00 Puma) = [$2 6(w—mAa-IZ 6(y—nAy>]-fltc<r/a)1 +00 +00 F(jkx,jky) : 7-: 6(m—mAx)-}'Z 6(y—nAy)-}'[tc(r/a)] FUkmjky) = AL: 6(fx-—) 7A1; (my—Al) W (7rd 3m 0//\) IZ—OO ’LZ—OO “there A0 is a constant and since k2 + k2— — k2 and f2 + f2: f2 then sinzfl = 527120,; + sinzfiy. . . ” 1 1 +°° ' ‘ J1(7rd sinQ/A) FUkmjky) = AOEA—y.z 5(fx-‘—;) :2 5( (fa/"7) W . , 1 1 0° 32'on 2' +00 sinél i ’ J wdsinOA F(akz,gky) = AoK—K— Z 5( A WI). 2 5( A y‘K‘)'1—("—/_) m y (7rd sin H/A) The radiant intensity is Haney) oc |F(fl<:g,.-,jky)l2 and so +00 ’ +00 . . . . . 2 _ 3m 6w 2 3m 9y 2 J1(7rd sm B/A) New” ‘ “OLE M /\ Am) ,2 6( /\ Ag) (7rd sinB/A) 12—00 l=—(X) TWO - DIMENSIONAL FOURIER TRANSFORMS The two-dimensional Fourier transform is given by Fem): /: /: .flaykflfiflmmwdu The inverse two-dimensional Fourier transform is ”(mm /: 3/ pgmawwwwwn The Fraunhofer diffraction patterns (two-dimensional Fourier transform) due to three slits are shown in Fig. 1. Fig. 1 Narrow slits of various length (left) and their Fourier transforms (right). ...
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