Optics, Problem Set Solutions 7

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 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 t1 (x) and t2 (x ), respectively. With on–axis plane illumination, we can formulate as follows: 1. field immediately after T1: t1 (x) 2. field immediately before T2: F [t1 (x)]x→ x�� λf1 3. field immediately after T2: t2 (x )F [t1 (x)]x→ x�� λf1   4. field at the image plane: F t2 (x )F [t1 (x)]x→ x�� λf1  = F [t2 (x )]x�� → x� λf2  ⊗F F [t1 (x)]x→  x� λf1 � x x�� → λf 2   f2  = F [t2 (x )]x�� → x� ⊗t1 − x , λf2 f1  � 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 T1 , the phase delay caused by grooves is 2λπ (n − 1)d1 , where d1 is the height of the groove (1 μm), 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      2π x x ⊗ comb , = exp i (n − 1)d1 rect λ A1 A2  t1 (x) = e iφ1 (x) where A1 = 5 μm and A2 = 10 μm. Hence,  iπ e (= −1) if |x| < A1 /2, t1 (x) = 1 if A1 /2 < x < A2 /2 or − A2 /2 < x < −A1 /2, (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  2π 1 A/2 t1 (x)e−i A qx dx cq = A −A/2  −A/4  A/4  A/2 2π 1 −i 2Aπ qx −i 2Aπ qx dx − e dx + e−i A qx dx (4) = e A −A/2 −A/4 A/4 For q = 0, c0 = For q =  0, 1 A ⎡ t1 (x)dx = 0. A/2 ⎤ 1 e e e    ⎦ − + cq = ⎣ 2π   2π 2π    A −i A q −i A q −i A q −A/2 −A/4 A/4  iπq π π π  1 e 2 − eiπq e−i 2 q − ei 2 q e−iπq − e−i 2 q = − + A −i 2Aπ q −i 2Aπ q −i 2Aπ q   π π q  q  eiπq − e−iπq ei 2 q − e−i 2 q = − = sinc (q) − sinc = −sinc . i2πq i2 π2 q 2 2   Thus, cq = −sinc 2q + δ(q); all even orders disappear and only odd orders survive. For the grating T2 , the phase profile is shown in Fig. 2. −A/4 −i 2Aπ qx  (a) phase profile A/4 −i 2Aπ qx  (b) real part −i 2Aπ qx  (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           x x x  t2 (x ) = rect − rect + irect . b a 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 θq = q Aλ , and the diffraction order q is focused at f1 θq on the Fourier plane. Hence, the field immediately before T2 is   ∞ ∞   f1 λ  = (δ(q) − sinc (q/2)) δ (x − q cm) . (6) (δ(q) − sinc (q/2)) δ x − q 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 π/2. The field immediately after the grating is     1 3   i π2 [δ(x − 3) + δ(x + 3)] − e sinc [δ(x − 1) + δ(x + 1)] . (7) −sinc 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     [δ(x − 3) + δ(x + 3)] − isinc [δ(x − 1) + δ(x + 1)] F −sinc = 2 2 x� u�� → λf         3 δ(x − q) + δ(x + q) δ(x − 1) + δ(x + 1) 1 − 2sinc F − 2isinc F = 2 2 2 2         −2 3x 2 x 3 1 −2i cos 2π cos (2πu) = −2 cos 2π = −2sinc cos (2π3u)−2isinc 2 2 3π λf2 π λf2     4 4 2π 2π cos (0.3)x − i cos (0.1)x . (8) 3π λ π λ 3 The intensity at the image plane is     2 1  2π 2π  (0.3)x − i cos (0.1)x  = I(x) ∼  cos 3 λ λ     1 2π 2π 2 2 cos (0.3)x + cos (0.1)x (9) 9 λ λ 0.5 0.4 0.45 0.35 0.4 0.3 intensity [a.u.] intensity [a.u.] 0.35 0.3 0.25 0.2 0.25 0.2 0.15 0.15 0.1 0.1 0.05 0.05 0
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5 0 5 10 15 20 x [μm] (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 1. Another solution with an alternative definition of the grating T1 (a) T1 is a phase mask; therefore, under the scalar and paraxial approximations, T1 does not modify the incident intensity. � I(x)� just = |eiφ(x) |2 = 1 (uniform) after T1 where φ(x) = ���� one period � 0 2π (n λ − 1)t = 2π (1.5 1µm − 1) × 1µm = π if |x| < 2.5µm if 2.5µm < |x| < 5µm (periodically repeating with period = 10µm) (b) Just before T2, the optical field is the Fourier transform of T1, scaled by λf1 . Using the formula [Goodman p.126, Figure 5.21] ∞ � sin( πn ) i2πnx/x 2 = e ≡ a(x) πn n=−∞ We obtain e iφ(x) = 2[a(2) − 1 ] 2 ∞ � sin( πn ) i2πnx/x 2 = e , where x = 10µm. πn n=−∞ n=0 � Note sin( nπ ) 2 nπ values: n sin(nπ/2) nπ . . . − 5 −4 −3 −2 −1 0 1 2 3 . . . 1 1 1 + 5π 0 − 3π 0 + π1 1 + π1 0 + 3π These are the amplitudes of the diffracted orders. 5 Lens L1 focuses each diffracted order to a focal point just before T2: The spacing between diffracted orders is: 1µm × 10cm λf1 = = 1cm grating period 10µm The phase shift applied by the elevated portion (center) of the T2 (pupil plane) mask is: 2π 2π π (n − 1)tT 2 = (1.5 − 1) × 0.5µm = λ 1µm 2 So, before mask T2 the field is: � � 1 1 1 1 1 1 1 �� �� �� �� �� �� · · · δ(x + 5) − δ(x + 3) + δ(x + 1) + δ(x − 1) − δ(x − 3) + δ(x − 5) · · · 2 5π 3π π π 3π 5π (c) The mask T2 cuts off all orders beyond the ±4th , and phase shifts the ±1th order with respect to the ±3rd orders. So just after mask T2 the field is: � � � � π π 1 1 e−i 2 e−i 2 1 �� �� �� �� − δ(x + 3) + δ(x + 1) + δ(x − 1) + − δ(x + 3) 2 3π π π 3π At the output plane, the field is the Fourier transform scaled by λf12 of the field π just after T2, i.e. [recall e−i 2 = −i] � � 3x� x� x� 3x� 1 1 −i2π 10µm i −i2π 10µm i +i2π 10µm 1 +i2π 10µm − e − e − e − e 2 3π π π 3π � � � � 1 3x� i x� =− cos 2π − cos 2π 3π 10µm π 10µm The intensity is: � � � � � � 1 3x� 1 x� � � �2 2 2 � Iout (x ) = fieldout (x ) = 2 cos 2π + 2 cos 2π 3π 10µm π 10µm (d) We recognize this as a Zernike phase mask. The original phase object T1 would have been invisible without the mask T2. With T2, we can observe intensity variation in the output plane. 6 Problem 2: Signum phase mask 2.a) The phase shift induced by the mask is 2π 2π s(n − 1) = (1 μm)(0.5) = π. λ 1 μm Hence, the pupil function can be written as �  � �  � x + a/4 iπ x − a/4  P (x ) = rect + rect e . a/2 a/2 ATF is a scaled pupil function, which is found to be � � � � λf u + a/4 iπ λf u − a/4 + rect e . H(u) = P (λf u) = rect a/2 a/2 (a) magnitude (10) (11) (12) (b) phase Figure 5: ATF of the system 1 μm−1 . Note that a/(2λf ) = 20 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 2π − + δ + = δ(x −5 mm)+ δ(x +5 mm). = δ �� x Λ λf 2 λf Λ 2 λf Λ 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 δ(x − 5) + eiπ δ(x + 5). = δ(x − 5) − δ(x + 5). (14) 2 2 2 2 The field at the image plane is � � � � 1 1 1 1     δ(x − 5) − δ(x + 5) F δ(x − 5) − δ(x + 5) = iF 2 2 2i 2i x� x� λf λf � � � � x 1 x . (15) = i sin 2π = i sin 2π 5 λf 20 μm [2] frequency analysis: The Fourier transform of the output field is a multiplication of the Fourier transform�of the input Since the Fourier transform � field�and the ATF. � 1 1 1 1 of the input signal is 2 δ u − 20 μm + 2 δ u + 20 μm , the FT of the output field is � � � � 1 1 1 1 δ u− − δ u+ , (16) 2 20 μm 2 20 μm � � � and the output field is i sin 2π 20xμm . 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 2π . (17) exp i θx cos 2π λ 20 μm Then the field incident to the Fourier plane is �� � � � � �� �  � � �  � �  2π x θ 1 1 1 x x 1 x F exp i θx cos 2π =δ − ⊗ δ − + δ + λ 20 μm x�� λf λ 2 λf Λ 2 λf Λ λf � � � � λf 1 λf 1 δ x − θf − + δ x − θf + 2 Λ 2 Λ 1 1 = δ (x − 10 mm − 5 mm) + δ (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 � � � � � � x 1 2π 1 1   = exp −i2π (5 mm) = exp i (−0.05)x . (19) F δ (x − 5 mm) 2 2 λf 2 λ x� λf 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 = λθ = 0.1 μm−1 . Then, the field immediately after the grating is � � � � 1 1 1 1 1 1 δ u− − + δ u− + . (20) 2 10 μm 20 μm 2 10 μm 20 μm Then, the FT of the output field is � � 1 1 δ u− , 2 20 μm (21) �� � � � � 1 2π 1 1   exp −i2π x = exp i (−0.05)x . 2 20 2 λ (22) and the output field is 2.e) At the Fourier plane, a signum function with π 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 33 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−i2πxu dx = � � −i2πxu dx + g(x)e−i2πxu dx = F[f (x)] + F[g(x)]. (23) f (x)e For condition 2, � F[af (x)] = −i2πxu {af (x)} e � dx = a f (x)e−i2πxu 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 5. g(x� , y � ) = f (x, y) ⊗ h(x, y) g(x� ) = F −1 {F{g}} G(u) = F{g} = F{f } · F{h} � � � � �� 1 1 1 1 1 = δ(u) + δ u − + δ u+ · x0 rect(x0 , u) 2 4 Λ 4 Λ � 1 δ(u) 2 1 δ(u) 2 � 1 iφ e 2 � 1 1+ 2 G(u) = � g(x ) = if 1 1 1 1 + 4 δ(u − Λ ) + 4 δ(u + Λ ) if if � x �� cos 2π Λ if 1 2x0 1 2x0 < > 1 Λ 1 Λ 1 2x0 1 2x0 < > 1 Λ 1 Λ (plane wave) (all the input gets imaged at the output) 6. First create a labeled sketch of the problem. (a) � x �� 1� g1 (x) = t1 (x) = 1 + cos 2π Λ � � 2 � � � � 1 1 λf 1 λf � � � � = δ(x ) + δ x − g2 (x ) = G1 (u)�� + δ x + 2 4 4 Λ Λ x� u= λf � � � � �� � �� 1 1 λf 1 λf x � � � � � � g3 (x ) = g2 (x ) · t2 (x ) = δ(x ) + δ x − + δ x + · rect 2 4 Λ 4 Λ a � λf 1 δ(x� ) if a2 < Λ � � � � = 21 δ(x� ) + 14 δ x� − λf + 14 δ x� + λf if a2 > λf 2 Λ Λ Λ 9 (b) In this problem, we take the Fourier transform of t1 (x) and multiply it with a rect() function. This is equivalent to taking the inverse Fourier of the rect() function, which is a sinc() function, convolving it with t1 (x) and taking the Fourier of the convolution, which is what we did in problem # 5. 10