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MIT2_71S09_usol7 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

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�� �� �� MASSACHUSETTS INSTITUTE OF TECHNOLOGY 2.710 Optics Spring ’09 Solutions to Problem Set #7 Due Wednesday, Apr. 22, 2009 Problem 1: Zernicke phase mask For problem 1, general formulations for the 4– f system are presented here. As shown in Fig. A in problem 1, x , x , and x are the lateral coordinates at the input, Fourier, and output plane, respectively. The complex transparencies at the input and Fourier plane are denoted by t 1 ( x ) and t 2 ( x ), respectively. With on–axis plane illumination, we can formulate as follows: 1. field immediately after T1: t 1 ( x ) 2. field immediately before T2: F [ t 1 ( x )] x x λf 1 3. field immediately after T2: t 2 ( x ) F [ t 1 ( x )] x x λf 1 λf 1 4. field at the image plane: F t 2 ( x ) F [ t 1 ( x )] x x �� x x λf 2 f 2 x x x = F [ t 2 ( x )] F F [ t 1 ( x )] x = F [ t 2 ( x )] t 1 x , x �� x �� λf 2 λf 1 �� λf 2 f 1 x x λf 2 (1) where we use F [ F [ g ( x )]] = g ( x ). Note that the field at the image plane is a convo- lution of the scaled object field and the Fourier transform of the pupil function, where the FT of the pupil is the point spread function of the system. Next, it is important to model correctly the transparencies of the gratings. For T 1 , the phase delay caused by grooves is 2 λ π ( n 1) d 1 , where d 1 is the height of the groove (1 μ m), and the phase profile is shown in Fig. 1. Hence, the complex transparency of Figure 1: phase profile of the grating T 1 ( x ) 1
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( ) ) 1 T is written as 2 π x x 1 ( x ) t 1 ( x ) = e = exp i ( n 1) d 1 rect comb , (2) λ A A 1 2 where A 1 = 5 μ m and A 2 = 10 μ m. Hence, e (= 1) if | x | < A 1 / 2 , t 1 ( x ) = (3) 1 if A 1 / 2 < x < A 2 / 2 or A 2 / 2 < x < A 1 / 2 , for | x | < A 2 / 2. Using the Fourier series ( t 1 ( x ) is periodic) and A = A 2 = 2 A 1 , we find the Fourier series coefficients as A/ 2 1 c q = t 1 ( x ) e i 2 A A/ 2 A π qx d x A/ 4 A/ 4 A/ 2 1 A i 2 i 2 A 1 0 1 ( x A For q = 0, A A/ 2 A/ 4 A/ 4 For = 0, )d = 0 t q c = x . π π e π A i 2 qx d x (4) qx d x qx d x + = e e A/ 4 A/ 4 A/ 2 A A i 2 i 2 i 2 c q = + A i 2 π q i 2 π q i 2 π q A A A A/ 2 A/ 4 A/ 4 A π π π qx qx qx 1 e e e 1 i e π 2 iπq q e e i π 2 q e i π 2 q e iπq e i π q 2 = + A i 2 π q i 2 π q i 2 π q A A A π q e i π 2 iπq e iπq i q 2 e e q q = sinc ( q ) sinc = sinc = . i 2 π 2 i 2 πq 2 2 q Thus, c q = sinc 2 q + δ ( q ); all even orders disappear and only odd orders survive.
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