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2.710 Optics Problem Set 8 Solutions 1. (a) For a diffraction limited system the slopes of the OTF are constant. mu=25mm−1 = 68.75% = 0.6875 x i 1h Iin = 1 + cos 2π 2 Λ 1 1 1 1 1 ˆ + δ u+ ⇒ Iin = δ(u) + δ u − 2 4 Λ 4 Λ 1 a 1 1 a ˆ ˆ Iout = Iin · OTF = δ(u) + δ u − + δ u+ 2 4 Λ 4 Λ 0 1 x 1 + a cos 2π Iout (x0 ) = 2 Λ 1 a 1 a ( + ) − (2 − 2) =a mu= 1 = 21 a2 Λ ( 2 + 2 ) + ( 12 − a2 ) ∴ the contrastis the normalized value of the OTF at that frequency. Using similar triangles, if mu=25mm−1 = 0.6875 = (1 − 0.3125), then mu=50mm−1 = (1 − 0.6250) = 0.3750 = 37.5 % (b) The cut-off frequency for incoherent imaging is u0 = 80mm−1 . The cut-off frequency of the coherently illuminated system is 40mm−1 . Hence 50mm−1 frequencies do NOT go through if it is coherently illuminated. 2. 1 1 x 1 3x I(x) = 1 + cos 2π + cos 2π 2 2 40µm 2 40µm (a) The contrast m is given by: m= At the input, m = 1−0 1+0 = 1. 1 Imax − Imin Imax + Imin (b) The Fourier transform of I(x) is: 1 1 1 3 3 1 ˜ = I(u) δ u− +δ u+ +δ u− +δ u+ + δ(u) 8 40 40 40 40 2 The Fourier transform of the output intensity is: ˜ I˜0 (u) = (MTF) · I(u) 1 1 1 1 +δ u+ = δ(u) + (0.25) δ u − 2 8 40 40 1 1 1 1 1 1 = δ(u) + δ u− + δ u+ 2 16 2 40 2 40 0 1 x 1 cos 2π I0 (x0 ) = + 2 16 40 1 1 1 1 ( 2 + 16 ) − ( 2 − 16 ) 1 mout = 1 = = 0.125 1 1 1 8 ( 2 + 16 ) + ( 2 − 16 ) (c) The incoherent transfer function is an autocorrelation of the coherent transfer function. The coherent transfer function in this case is probably a triangle function with half the cut-off frequency. H(u) = F{h(x)} 2 3. h(x) = sinc2 x b (a) Incoherent iPSF 2 4 h̃(x) = |h(x)| = sinc x b (b) MTF = H̃(u) n o 2 x 2 x H̃(u) = F{h̃(x)} = F sinc · sinc n x o n b x o b ⊗ F sinc2 = F sinc2 b b = bΛ(bu) ⊗ bΛ(bu) bΛ(bu) MTF 3 2.710 Optics Solutions to Problem Set #8 Spring ’09 Due Wednesday, May 13, 2009 Problem 4: a) Consider the system shown in Figure 1. At the input we place a sinusoidal amplitude grating of perfect contrast, x i 1h 1 + cos 2 ; (1) gt (x) = 2 where = 0:5 m, f1 = f2 = f = 20cm, = 10 m, and the two apertures at the pupil plane have a diameter of 1cm and are separated by 1cm from the optical axis. We begin by computing the input intensity (assuming uniform intensity illumination), Iin = = = = x i2 1h 1 + cos 2 jgt (x)j = 4 h x x i 1 2 1 + cos 2 + 2 cos 2 4 1 x x 1 1 + 2 cos 2 1 + + cos 4 4 2 2 x 1 x 3 1 + cos 4 + cos 2 ; 8 8 2 2 (2) where we used the following trigonometric identity, cos2 ( ) = 1 1 + cos (2 ) : 2 2 Now we compute the frequency spectra of the input signal, Figure 1: Optical system for problem 4. 1 (3) Figure 2: Input signal spectrum. Figure 3: Graphical calculation of OTF. Gin = zfIin g 1 3 (u) + = 8 4 (4) (u 1 ) + (u + 1 ) + 1 16 2 (u ) + (u + 2 ) ; which can be represented graphically as shown in Figure 2. The ATF of the system is, H = rect x 00 00 xo x + xo + rect d d ; (5) where xo is the lateral shift (1cm) and d is the aperture diamter (also 1cm). We compute the OTF of the optical system graphically as shown in Figure 3. The OTF is the normalized cross-correlation of the ATF of the system. To compute the spectrum of the output signal we multiply the OTF by the input spectra, 2 ^ in Gout = HG 3 1 = (u) + 8 32 (6) 2 (u ) + (u + 2 ) 00 : u= x f The output intensity is, Iout = zfGout g 2 3 1 4e = + 8 16 i2 00 2x + ei2 2 1 2x 3 + cos 2 = 8 16 00 2x (7) 3 5 00 : (b) The constrast or visibility is given by, V = Imax Imin : Imax + Imin (8) For the output intensity of equation 7, the contrast is V = 0:167: (c) Finally we compare our results with the coherent case where the spectrum of the input …eld is, Gin = Ffgt (x)g = 1 1 (u) + 2 4 1 (u ) + (u + 1 ) : (9) The ATF of the system acts as a pass-band …lter as shown in Figure 4 and only allows the …rst di¤raction order to pass, Gout = 1 4 (u 1 ) + (u + 1 ) 00 : (10) u= x f The output …eld is, gout = zfGout g " x0 1 e i2 + ei2 = 2 2 = 1 x0 cos 2 2 3 ; x0 # (11) Figure 4: Band-pass …lter - coherent system. and the output intensity is, Iout = jgout j2 1 x0 2 = cos 2 4 2x0 1 1 1 + cos 2 = 4 2 2 1 1 2x0 = + cos 2 : 8 8 For the coherent case we note the the contrast is V = 1. 4 (12) 2.710 Optics Solutions to Problem Set #8 Spring ’09 Due Wednesday, May 13, 2009 Problem 5: a) Consider the system shown in Figure 1. The input transparency is a binary amplitude grating with the following parameters: m = 1; duty cycle = = 1=3; = 10 m. The operating wavelength is = 0:5 m, and the focal lengths are f1 = f2 = f = 20cm. At the Fourier plane, the pupil mask has two apertures with a diameter of 1cm shifted 2cm from the optical axis. We begin by expressing the binary amplitude grating in a Fourier Series, 1 X gt (x) = sinc ( q) ei2 qx (1) : q= 1 Since equation 1 has a binary amplitude dependence that goes from 0 to 1, the intensity is also binary, Iin = jg(t)j2 . The spectrum of the input signal is given by, Gin (u) = " 1 X sinc ( q) u q q= 1 00 ! Gin (x ) = 1 X sinc ( q) # (2) 00 u= x f x fq 00 ; q= 1 and is shown in Figure 2. ^ of the optical system is computed in the same way as in problem 4 and The OTF, H, it is shown in Figure 3. The output …eld spectrum is given by, ^ in Gout = HG 1 00 (x ) = 3 p 3 16 (3) x 00 The output intensity is, 1 f4 + 00 x + f4 : Figure 1: Optical system for problem 5. Figure 2: Input signal spectrum. Figure 3: System’s OTF. 2 Iout = zfGout g 2 0 p x 4 1 3 4 e i2 + ei2 = 3 8 2 p 0 3 x4 1 cos 2 : = 3 8 0 x 4 3 (4) 5 (b) The resulting contrast is, V = Imax Imin = 0:2067: Imax + Imin (5) (c) Comparying with the results from Lecture 19, p. 24 , for the coherent case, the output intensity is, Iout = 3 8 2 + 0 3 8 2 and the contrast V = 1. 3 cos 2 x4 ; 2.710 Optics Solutions to Problem Set #8 Spring ’09 Due Wednesday, May 13, 2009 Problem 6: Consider the optical system shown in Figure 1. The input transparency is a binary amplitude grating with the same parameters as in problem 5. The operating wavelength is = 0:5 m, and the focal lengths are f1 = f2 = f = 20cm. The pupil mask is given by, 00 00 00 gP M (x ) = x rect a + (i x 1) rect b ! H(u) = gP M (u) u = rect + (i 1) rect u (1) x00 =u f ; where = a= f , = b= f , a = 3cm and b = 1cm. The magnitude and phase of the pupil mask are shown in Figure 2. We now compute the coherent PSF, h(x) = F 1 fH(u)g = sinc ( x) + (i 1) sinc ( x) = sinc ( x) sinc ( x) + i sinc ( x) : (2) The incoherent PSF is given by, ^ h(x) = jh(x)j2 = h h = [ sinc ( x) sinc ( x)]2 + 2 sinc2 ( x) + 2 2 sinc2 ( x) (3) 2 2 sinc ( x) 2 sinc ( x) sinc ( x) : The OTF of the system is found by computing the Fourier transform of equation 3, u ^ H(u) = triag + 2 triag u and is shown in Figure 3. 1 2 rect u rect u ; (4) Figure 1: Optical system for problem 6. Figure 2: Magnitud and phase of the pupil mask. 2 Figure 3: OTF and spectrum of the input signal. From problem 5, the spectrum of the input signal is, Gin (u) = 1 X sinc ( q) u q : (5) q= 1 To compute the spectrum of the output signal we multiply equations 5 and 6, p 1 3 2 2 Gout (u) = (u) + u + u+ : 3 12 (6) The output intensity is given by, Iout = FfGout g 2 0 p x 2 1 3 4 e i2 + ei2 = + 3 6 2 p 0 1 3 x2 = + cos 2 : 3 6 0 x 2 3 (7) 5 (b) The resulting contrast is, V = Imax Imin = 0:2757: Imax + Imin 3 (8) (c) Finally, we compare our results to those from the coherent case as discussed in Lecture 19, p. 27, where the output intensity is, 0 Iout (x ) = 1 3 2 + 3 2 2 and the contrast is V = 0:2548: 4 + 0 3 2 2 cos 2 x2 ; (9) 2.710 Optics Solutions to Problem Set #8 Spring ’09 Due Wednesday, May 13, 2009 Problem 7: In this problem we are required to show that the intensity at the Fourier plane of the system shown in Figure 1 is proportional to the autocorrelation of the input …eld, gin = gillum (x) gt (x). From the supplement to lecture 18 , we can see that the optical …eld at the Fourier plane for the special case of z = f is, 2f ei2 G gF (x ; y ) = i f 00 00 00 00 x y ; f f : (1) The intensity at that plane is, IF = gF gF _ G G ; (2) which is proportional to the autocorrelation of the input …eld as they are related by a Fourier transform, =fgin gin g = G G : Figure 1: Fourier transforming lens. 1 (3) 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.