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PHYS 408 HOMEWORK 8 SOLUTIONS

PHYS 408 HOMEWORK 8 SOLUTIONS - solution set 8 Contents 1 2...

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solution set 8 November 27, 2009 Contents 1 Inverse Fourier Transform ............................................ 1 2 optical switching ................................................. 2 3 phase contrast imaging .............................................. 3 3.1 bright±eld imaging .......................................... 3 3.2 phase contrast imaging, phase masking .............................. 3 3.3 matlab script ............................................. 4 3.4 plots .................................................. 5 List of Figures 1 p3.jpg ...................................................... 5 1 Inverse Fourier Transform From the previous homework, at the fourier plane the image is (without coordinate inversion at the input plane): g ( x, y )= ° ie 2 ikf λf ± F{ f ( x, y ) }| ° k =( k/f ) °x (1) (2) Now consider inversion of the input image coordinates. This will only a²ect the unevaluated fourier transform above, since only it contains the input image. This inversion amounts to f ( x, y ) f ( x, y ). The fourier transform above under this inversion is f ( x, y ) }| ° k =( k/f ) = °² dx ² dy f ( x, y ) e i ( k x x + k y y ) ± ° k =( k/f ) (3) = °² d ( x ) ² d ( y ) f ( x, y ) e i ( k x ( x )+ k y ( y )) ± ° k =( k/f ) (4) = °² dx ² dy f ( x, y ) e + i ( k x x + k y y ) ± ° k =( k/f ) (5) = F 1 { f ( x, y ) }| ° k =( k/f ) (6) (7) Thus, for inversion of the input image coordinates, the image at 2 f is the inverse fourier transform of the input image, i.e. g ( x, y ° ie 2 ikf λf ± F 1 { f ( x, y ) }| ° k =( k/f ) (8) 1
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2 optical switching The input function is f ( x, y )= ( x α ) δ ( y α ) forα =0 . 5cm (9) Propagate this through half of the 4 f system, to just before the mask. Here, the function will now take the form – according to a result from the previous homework – the fourier transform of f ( x, y ) mapped onto real space, a result further analyzed in problem 1 above. The result: p < ( x, y ° ie 2 ikf λf ± F{ f ( x, y ) }| ° k =( k/f ) °x (10) = ° ie 2 ikf λf ± ² Ae ik x α e ik y α ³ ° k =( k/f ) (11) = ° ie 2 ikf λf ± ´ Ae i ( kα/f )( x + y ) µ (12) The desired output function is g ( x, y ( x α ) δ ( y α ) (13) Propagate this “backwards” to just after the mask, which can be done by using the result from problem 1: p > ( x, y ° ie 2 ikf λf ± F 1 { f ( x, y ) }| ° k =( k/f ) (14) = ° ie 2 ikf λf ± ² Ae + ik x α e + ik y α ³ ° k =( k/f ) (15) = ° ie 2 ikf λf ± ´ Ae + i ( kα/f )( x + y ) µ (16)
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PHYS 408 HOMEWORK 8 SOLUTIONS - solution set 8 Contents 1 2...

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