PHYS 408 HOMEWORK 8 SOLUTIONS

# 5 cm 9 propagate this through half of the 4f system

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Unformatted text preview: (x, y )}|￿ ⊥ =(k/f )￿ ⊥ k x 1 (8) 2 optical switching The input function is f (x, y ) = Aδ (x − α)δ (y − α)f orα = 0.5 cm (9) Propagate this through half of the 4f 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 ) = = = ￿ ￿ ￿ −ie2ikf λf −ie2ikf λf −ie2ikf λf The desired output function is ￿ ￿ F{f (x, y )}|￿ ⊥ =(k/f )￿ ⊥ k x ￿ Ae−ikx α e−iky α ￿￿ ￿ ￿ ⊥ =(k/f )￿ ⊥ k x Ae−i(kα/f )(x+y) ￿ g (x, y ) = Aδ (x − α)δ (y − α) (10) (11) (12) (13) Propagate this “backwards” to just after the mask, which can be done by using the result from problem 1: p> (x, y ) = ￿ ￿ −ie2ikf λf ￿ F −1 {f (x, y )}|￿ ⊥ =(k/f )￿ ⊥ k x ￿ −ie2ikf ￿ +ikx α +iky α ￿ = Ae e ￿ ⊥ =(k/f )￿ ⊥ k x λf ￿ ￿￿ ￿ 2ikf −ie = Ae+i(kα/f )(x+y) λf (14) (15) (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(x, y ) = p> (x, y ) e+i(kα/f )(x+y) = −i(kα/f )(x+y) = e+2i(kα/f )(x+y) p< (x, y ) e (17) Recalling that the transmission of a thin ﬁlm of is just t(x, y ) ≈ eikd0 eik(n−1)d(x,y) , we can – within an overall phase which can be adjusted by adjusting the maximum thickness d0 – achieve a mask with the derived transmission above with a thin ﬁlm of thickness according to d(x, y ) = α (x + y ) 2(n − 1)f (18) This is just a prism with increasing thickness in both x and y , which can be fabricated directly (or be composed of a succession of a prism in x and a prism in y ). 2 3 phase contrast imaging 3.1 brightﬁeld imaging 3.2 phase contrast imaging, phase masking √ See section 1.8 of the lab manual, especially page 13. Plane wave I0 eikz illumination, p...
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