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 brightfield 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 λ f F{ f ( x, y ) }| k =( k/f ) x (1) (2) Now consider inversion of the input image coordinates. This will only a ff ect 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{ f ( x, y ) }| k =( k/f ) x = dx dy f ( x, y ) e i ( k x x + k y y ) k =( k/f ) x (3) = d ( x ) d ( y ) f ( x, y ) e i ( k x ( x )+ k y ( y )) k =( k/f ) x (4) = dx dy f ( x, y ) e + i ( k x x + k y y ) k =( k/f ) x (5) = F 1 { f ( x, y ) }| k =( k/f ) x (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 λ f F 1 { f ( x, y ) }| k =( k/f ) x (8) 1
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2 optical switching The input function is f ( x, y ) = A δ ( x α ) δ ( y α ) for α = 0 . 5 cm (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 λ f F{ f ( x, y ) }| k =( k/f ) x (10) = ie 2 ikf λ f Ae ik x α e ik y α k =( k/f ) x (11) = ie 2 ikf λ f Ae i ( k α /f )( x + y ) (12) The desired output function is g ( x, y ) = A δ ( 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 λ f F 1 { f ( x, y ) }| k =( k/f ) x (14) = ie 2 ikf λ f Ae + ik x α e + ik y α k =( k/f ) x (15) = ie 2 ikf λ f Ae + i ( k α /f )( x + y ) (16) Now that we have explicit forms for the function just before and just after the mask, we can write down an explicit form for the mask itself: m (
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