MIT2_71S09_usol7

MIT2_71S09_usol7 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

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Unformatted text preview: MASSACHUSETTS INSTITUTE OF TECHNOLOGY 2.710 Optics Solutions to Problem Set #7 Spring '09 Due Wednesday, Apr. 22, 2009 Problem 1: Zernicke phase mask For problem 1, general formulations for the 4f 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 t1 (x) and t2 (x ), respectively. With onaxis plane illumination, we can formulate as follows: 1. field immediately after T1: t1 (x) 2. field immediately before T2: F [t1 (x)]x x f1 3. field immediately after T2: t2 (x )F [t1 (x)]x x f1 4. field at the image plane: F t2 (x )F [t1 (x)]x x f1 x x f 2 = F [t2 (x )]x x f2 F F [t1 (x)]x x f1 x x f 2 = F [t2 (x )]x x f2 t1 - f2 x f1 , (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 T1 , the phase delay caused by grooves is 2 (n - 1)d1 , where d1 is the height of the groove (1 m), and the phase profile is shown in Fig. 1. Hence, the complex transparency of Figure 1: phase profile of the grating T1 (x) 1 T1 is written as t1 (x) = ei1 (x) = exp i 2 (n - 1)d1 rect x A1 comb x A2 , (2) where A1 = 5 m and A2 = 10 m. Hence, t1 (x) = ei (= -1) if |x| < A1 /2, 1 if A1 /2 < x < A2 /2 or - A2 /2 < x < -A1 /2, (3) for |x| < A2 /2. Using the Fourier series ( t1 (x) is periodic) and A = A2 = 2A1 , we find the Fourier series coefficients as cq = 1 A A/2 -A/2 t1 (x)e-i A qx dx e -i 2 qx A 2 1 = A For q = 0, c0 = For q = 0, cq = 1 A -A/4 -A/2 dx - A/4 -A/4 e -i 2 qx A A/2 dx + A/4 e -i 2 qx A dx (4) t1 (x)dx = 0. -i 2 qx A 1 e A -i 2 q A iq 2 -A/4 - -A/2 e -i 2 q A -i 2 qx A A/4 + -A/4 iq 2 e -i 2 q A -i 2 qx A A/2 A/4 1 e - eiq e -e = - 2 A -i 2 q -i A q A -i q 2 i iq e - e-iq q q e 2 q - e-i 2 q = - = sinc (q) - sinc = -sinc . i2q 2 2 i2 2 q -iq - e-i 2 q e + -i 2 q A q Thus, cq = -sinc 2 + (q); all even orders disappear and only odd orders survive. For the grating T2 , the phase profile is shown in Fig. 2. (a) phase profile (b) real part (c) real part Figure 2: complex transparency of the grating T2 (x ) 2 Figure 3: the field immediately before T2 . The complex transparency can be written as t2 (x ) = rect x b - rect x a + irect x a . (5) 1.a) the intensity immediately after T1 is 1 because |t1 (x)|2 = 1. Since T1 is a pure phase object and there is no intensity variation. 1.b) the field immediately before T2 can be computed from the Fourier series co efficients of t1 (x). Since the period of T1 is A, the diffraction angle of the order q is q = q A , and the diffraction order q is focused at f1 q on the Fourier plane. Hence, the field immediately before T2 is f1 ((q) - sinc (q/2)) x - q = ((q) - sinc (q/2)) (x - q cm) . (6) A q=- q=- 1.c) Since b (the width of the grating T2 ) is 7 cm, the diffraction orders passing through the grating T2 are q = -3, -1, +1, +3, where -1 and +1 orders get phase delay of /2. The field immediately after the grating is 1 3 -sinc [(x - 3) + (x + 3)] - ei 2 sinc [(x - 1) + (x + 1)] . (7) 2 2 The field at the image plane is the Fourier transform of the field immediately after the grating T2 , which is computed as 3 1 F -sinc [(x - 3) + (x + 3)] - isinc [(x - 1) + (x + 1)] = 2 2 x u f 3 (x - q) + (x + q) 1 (x - 1) + (x + 1) - 2sinc F - 2isinc F = 2 2 2 2 -2 3x 2 x 3 1 -2sinc cos (23u)-2isinc cos (2u) = -2 cos 2 -2i cos 2 2 2 3 f2 f2 4 4 2 2 cos (0.3)x - i cos (0.1)x . (8) 3 3 = The intensity at the image plane is I(x) 1 cos 3 2 (0.3)x - i cos 2 (0.1)x = 1 2 cos2 (0.3)x + cos2 9 2 2 (0.1)x (9) 0.5 0.45 0.4 0.4 0.35 0.3 0.35 intensity [a.u.] intensity [a.u.] 15 10 5 0.3 0.25 0.2 0.15 0.25 0.2 0.15 0.1 0.1 0.05 0 20 0 5 10 15 20 0.05 0 20 15 10 5 0 5 10 15 20 x [m] x [m] (a) with the phase mask (b) with the phase mask Figure 4: intensity pattern at the image plane Figure 4 shows the intensity pattern at the image plane with and without the phase mask. 1.d) In Fig. 4, the phase mask introduces more dramatic intensity contrast, whose frequency is proportional to the twice of the spatial frequency of the object grating. In Fig. 4(b), there is a intensity variation but the contrast is smaller. This phase mask is particularly useful for imaging phase object because phase variation is converted into intensity variation. 1.e) In Fig. 4(a), although all the orders are recovered, the field signal is not identical as the input field (the field immediately after T1). Hence, we may still able to observe some intensity variation although the contrast could be very limited (but still better than the case without the phase mask). 1.f) If a = 0.5 cm, then the first order does not get the phase delay, and all the orders are imaged at the image plane. The output field is identical to the input field (the field immediately after T1); no intensity variation is produced. Intuitively, in Fig. 4(b), as all the order contribute, the valleys of the intensity pattern is filled and eventually uniform intensity pattern is produced. 4 (A Another solution with an alternative definition of the grating T1 if*II I of - ltw to - PrrLupra , I f") 1!- ie "-if,oi I t/ s ctnr q^L "^^sb;+A, -' / vt-wlet 1l4L a"t rMs., Tt do+t f**e-"ff*'^o*'t ty ,l \uru'gA^tro-kui; "AA"rU&-1t^e- I{il l t (u*'{-) - i+ - fX.4( 2{ltn1r(fpor = ?-'S tlr !4 pwioeieJfu -*fe'"fr 0t J,^sIL,/* rt, fu Ie+utoJ= l,z yxl +u!- R*aq. t,o"d"; - "+tl-i*fkdft r t f . I ula J Nnn cI, tl;try +1^* ^1 t,Zt I p t?-6T-' J L h-ad'rrusn' , g*r< =4 rr / Lr , ?zrnz Z-fF-e-'v_=n(z) '7lrr , Wc obtqin .t siJ$l- L - i?r^* c A Vl'r I pherc lr roii Problem Set 3 - page 20 Practice ?1to lto fn e" T - * a]aa*4n^dtrtg I f *nr*1s&* -dLaappgJw-3) -y-- ry&^v f c-"?e,np O* r"laralfi dtrw . -.l_.. X Practice Problem Set 3 - page 21 4tt|],a'i lsU'u o t 0^ I I I ,T{vr/\ T ,fr ) n,'s f,l-____ ry 7L nlt^t t-tfu - *+{t+-+}, z S,o=1f5 + =-ur# --5r -Wqttat- yftTsrirp :lr-Tryilw Practice Problem Set 3 - page 22 #l -- -'- ytt ry i tt I = -t'#F rLl -a-^,,r6J 7nqffi-"HW VM$f / wq ?ry\ -Zf%r-{-ZIffi r lti-ii 6ry "'Td w " I , I- 7n4ol .1r --t \ | Al..W-ut, 1 :I __2 J4 l^nrrff"ry] Practice Problem Set 3 - page 23 rrtv l vt ^ - a t11, .} - ---i rl.,_, i ,t i \rLt Problem 2: Signum phase mask 2.a) The phase shift induced by the mask is 2 2 s(n - 1) = (1 m)(0.5) = . 1 m Hence, the pupil function can be written as x - a/4 x + a/4 i + rect e . P (x ) = rect a/2 a/2 ATF is a scaled pupil function, which is found to be f u + a/4 i f u - a/4 + rect e . H(u) = P (f u) = rect a/2 a/2 (10) (11) (12) (a) magnitude (b) phase Figure 5: ATF of the system 1 Note that a/(2f ) = 20 m-1 . 2.b) The magnitude and phase of the transparency are shown as below. (a) magnitude (b) phase Figure 6: Transparency 2.c) The optical field at the image plane can be obtained from two ways: 1) direct forward computation or 2) frequency analysis. 5 [1] direct forward computation: The incident field to the Fourier plane is x 1 1 1 1 1 1 x x F cos 2 = - + + = (x -5 mm)+ (x +5 mm). x f 2 f 2 f 2 2 (13) Since the width of the pupil is 10 mm, both delta functions pass through the pupil with a phase delay. The field immediately after the phase mask is 1 1 1 1 (14) (x - 5) + ei (x + 5). = (x - 5) - (x + 5). 2 2 2 2 The field at the image plane is 1 1 1 1 F (x - 5) - (x + 5) = iF (x - 5) - (x + 5) 2i 2i 2 2 x x f f 1x x = i sin 2 = i sin 2 . (15) 5 f 20 m [2] frequency analysis: The Fourier transform of the output field is a multiplication of the Fourier transformof the input fieldand the ATF. Since the Fourier transform of the input signal is 1 u - 20 1m + 1 u + 20 1m , the FT of the output field is 2 2 1 1 1 1 u- - u+ , (16) 2 20 m 2 20 m and the output field is i sin 2 20xm . 2.d) Similarly, still there are two ways to analyze. [1] If the incident wave is tilted, then one of the two delta functions at the Fourier plane is blocked by the pupil. The other delta function still gets phase delay and propagates to the image plane. The field immediately after the grating is x 2 . (17) exp i x cos 2 20 m Then the field incident to the Fourier plane is 2 x 1 1 1 x 1 x x F exp i x cos 2 = - - + + 20 m x f 2 f 2 f f f 1 f 1 x - f - + x - f + 2 2 1 1 = (x - 10 mm - 5 mm) + (x - 10 mm + 5 mm) , (18) 2 2 where the second delta function does not get phase delay now. The field at the output plane is 1 x 1 2 1 = exp -i2 (5 mm) = exp i (-0.05)x . (19) F (x - 5 mm) 2 2 f 2 x f 6 thus, the output field is a tilted plane wave with an angle of -0.05 rad. The factor of 1 indicates that the amplitude of the output field is half of the amplitude of the input 2 field. [2] The spatial frequency of the tilted plane wave is u = = 0.1 m-1 . Then, the field immediately after the grating is 1 1 1 1 1 1 u- - + u- + . (20) 2 10 m 20 m 2 10 m 20 m Then, the FT of the output field is 1 1 u- , 2 20 m and the output field is 1 2 1 1 exp -i2 x = exp i (-0.05)x . 2 20 2 (22) (21) 2.e) At the Fourier plane, a signum function with phase shift is multiplied. It is equivalent to the Hilbert transform. ( http://en.wikipedia.org/wiki/Hilbert transform) Note that the Hilbert transform converts cos to sin and visa versa. 7 Problem 3 Plot of the Original Image and its Fourier Transform High-pass Filter: Edges get enhanced Low-pass Filter: Soften the Edges (only the general shape of the mountain can be detected) Problem 4: Fourier transform 4.a) If the Fourier transform is linear, then it should satisfy two conditions 1. F[f (x) + g(x)] = F[f (x)] + F[g(x)], where f (x) and g(x) are input functions to the Fourier transform 2. F[af (x)] = aF[f (x)], where a is a constant. For condition 1, F[f (x) + g(x)] = {f (x) + g(x)} e-i2xu dx = -i2xu dx + g(x)e-i2xu dx = F[f (x)] + F[g(x)]. (23) f (x)e For condition 2, F[af (x)] = {af (x)} e -i2xu dx = a -i2xu f (x)e dx = aF[f (x)]. (24) Hence, the Fourier transform is a linear transform. 4.b) The transfer function is not defined in this case because the Fourier integral is not a form of convolution. In other words, the relation is not shift invariant. 8 MIT OpenCourseWare http://ocw.mit.edu 2.71 / 2.710 Optics Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. ...
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This note was uploaded on 02/24/2012 for the course MECHANICAL 2.710 taught by Professor Sebaekoh during the Spring '09 term at MIT.

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