PHYS 408 HOMEWORK 9 SOLUTIONS

# PHYS 408 HOMEWORK 9 SOLUTIONS

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Unformatted text preview: left structure add large transparent square x = -D/2 - L/2 to y = -L/2 to +L/2 -D/2 + L/2 subtract small opaque square x = -D/2 - l/2 to -D/2 + l/2 y = -l/2 to +l/2 right structure add large transparent square x = +D/2 - L/2 to y = -L/2 to +L/2 +D/2 + L/2 subtract small opaque square x = +D/2 - l/2 to +D/2 + l/2 y = -l/2 to +l/2 5 The fourier transform is of the input image is: F {f (x, y )} = = ￿ ￿ dx ￿ dy f (x, y )e−i(kx x+ky y) −D/2+L/2 dx −D/2−L/2 + ￿ ￿ +L/2 −L/2 +D/2+L/2 dx +D/2−L/2 − ￿ = E0 dy E0 e−i(kx x+ky y) (12) ￿ (13) −D/2+￿/2 +L/2 −L/2 dx −D/2−￿/2 − ￿ ￿ (11) ￿ +D/2+￿/2 dy E0 e−i(kx x+ky y) +￿/2 −￿/2 dx +D/2−￿/2 ￿ dy E0 e−i(kx x+ky y) +￿/2 dy E0 e−i(kx x+ky y) −￿/2 ￿−D/2+L/2 ￿ ￿+L/2 1 −ikx x 1 −iky y e e −ikx −iky −D/2−L/2 −L/2 ￿ ￿+D/2+L/2 ￿ ￿+L/2 1 −ikx x 1 −iky y +E0 e e −ikx −iky +D/2−L/2 −L/2 ￿ ￿−D/2+￿/2 ￿ ￿+￿/2 1 −ikx x 1 −iky y −E0 e e −ikx −iky −D/2−￿/2 −￿/2 ￿ ￿+D/2+￿/2 ￿ ￿+￿/2 1 −ikx x 1 −iky y −E0 e e −ikx −iky +D/2−￿/2 −￿/2 = E0 L e 2 +ikx D/2 sinc(kx L/2)sinc(ky L/2) +E0 L e 2 −ikx D/2 −E0 ￿ e sinc(kx ￿/2)sinc(ky ￿/2) 2 −ikx D/2 −E0 ￿ e = 2E0 cos(kx D/2) ￿ (15) (16) (17) (18) (19) (20) (21) (22) sinc(kx L/2)sinc(ky L/2) 2 +ikx D/2 (14) sinc(kx ￿/2)sinc(ky ￿/2) ￿ 2 2 L sinc(kx L/2)sinc(ky L/2) − ￿ sinc(kx ￿/2)sinc(ky ￿/2) (23) (24) (25) (26) (27) Evaluate at ￿ ⊥ = (k/d)￿ ⊥ = (2π /λd)￿ ⊥ to calculate the quantity the Fraunhofer far-ﬁeld is proportional to: k x x g (x, y ) ∝ 2E0 cos(π xD/λ) ￿ ￿ L sinc(π xL/λ)sinc(π yL/λ) − ￿ sinc(π x￿/λ)sinc(π y ￿/λ) 2 2 (28) The far-ﬁeld intensity is proportional to |g (x, y )|2 , which is plotted below. As for validity of the Fraunhofer approximation, the input image is conﬁned to a circle of radius squared ((L + D)2 + (L)2 )/4, thus the Fresnel number for the input image is (L + D)2 + (L)2 /λd, for an output image a distance d away from the input image. Thus we need d￿ (L + D)2 + (L)2 = 417 m 4λ (29) And – although not asked in the problem – we would need to make sure the Fresnel number for the output pattern of interest satisﬁes the Fraunhofer approximation for a given input-output distance. As can be seen in the plot of the image, the “interesting stuﬀ ” can be seen within a radius of approximately 3 tenths of a mm from the center of the pattern and, at a distance of 104 meters, the corresponding Fresnel number 3 · 10−4 /(6 · 10−7 · 104 ) = 5 · 10−2 satisﬁes the Fraunhofer approximation. 6 3.1 ﬁeld, interpreted contour Figure 1: p3a.jpg 7 3.2 intensity, interpreted contour Figure 2: p3b.jpg 8 3.3 intensity, 3D Figure 3: p3c.jpg 9 3.4 scripts 3.4.1 p3.m smax = 3e-4; N = 1e3; s = linspace(-smax,smax,N); scale = 1e-3; % mm pcolor(s/scale,s/scale,g(s,s)); shading interp xlabel(’x, mm’) ylabel(’y, mm’) title(’g(x,y)’) saveas(gcf,’p3a.jpg’); pcolor(s/scale,s/scale,abs(g(s,s)).^2); shading interp xlabel(’x, mm’) ylabel(’y, mm’) title(’|g(x,y)|^{2}’) saveas(gcf,’p3b.jpg’); meshz(s/scale,s/scale,abs(g(s,s)).^2); shading interp xlabel(’x, mm’) ylabel(’y, mm’) title(’|g(x,y)|^{2}’) saveas(gcf,’p3c.jpg’); 3.4.2 g.m function g = g(x,y) D = 2e-2; L = 1e-2; ell = 2e-3; lambda = 600e-9; g = (L^2)*( cos(pi*x’*D/(2*lambda)).*sinc(x’*L/(2*lambda)) )*sinc(y*L/(2*lambda)) ... - (ell^2)*( cos(pi*x’*D/(2*lambda)).*sinc(x’*ell/(2*lambda)) )*sinc(y*ell/(2*lambda)); g = g’; 10...
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## This document was uploaded on 03/11/2014 for the course PHYS 408 at University of British Columbia.

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