4500HWF04 - Fraunhofer Diffraction by a Three-Slit...

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Unformatted text preview: Fraunhofer Diffraction by a Three-Slit Triangular Transmittance A spatial light modulator is adjusted such that the amplitude transmittance, t, is a three-slit 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 three-slit 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 m-component 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 far-field 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 Three-Slit Triangular 'D‘ansmittance The transmittance is t0”) = t1(9'5) + M33) where t1(a:)=1 —%<x<g 1 5a 3a 3a 5a t2($)=§ ——2—<:I:<——-2— and +—2—<m<+? The Fourier transforms are obtained using t ——> m, w ——> kz, and 7' —> a and where k; = 277'sin0. F1 (j kl a) = a center slit [cl-Ea. 2 - M F2(jk$ 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) : % swig-2) [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 ...
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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.

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4500HWF04 - Fraunhofer Diffraction by a Three-Slit...

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