1 00:00:00,090 --> 00:00:02,430 The following content is provided under a Creative 2 00:00:02,430 --> 00:00:03,820 Commons license. 3 00:00:03,820 --> 00:00:06,030 Your support will help MIT OpenCourseWare 4 00:00:06,030 --> 00:00:10,120 continue to offer high-quality educational resources for free. 5 00:00:10,120 --> 00:00:12,660 To make a donation or to view additional materials 6 00:00:12,660 --> 00:00:16,620 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:16,620 --> 00:00:17,640 at ocw.mit.edu. 8 00:00:21,680 --> 00:00:24,770 GEORGE BARBASTATHIS: OK, good morning. 9 00:00:24,770 --> 00:00:28,250 So I'd like to pick up from where we left off last time. 10 00:00:28,250 --> 00:00:31,850 So we were discussing the various properties 11 00:00:31,850 --> 00:00:33,600 of the Fourier transform and what 12 00:00:33,600 --> 00:00:37,850 they mean in terms of the Fraunhofer diffraction 13 00:00:37,850 --> 00:00:39,032 patterns. 14 00:00:39,032 --> 00:00:41,240 So what we're doing last time when we ran out of time 15 00:00:41,240 --> 00:00:45,300 was we were proving the convolution theorem. 16 00:00:45,300 --> 00:00:47,090 So this is the proof here. 17 00:00:47,090 --> 00:00:49,610 And just to remind you very briefly, 18 00:00:49,610 --> 00:00:53,240 we start with an expression that looks like a convolution. 19 00:00:53,240 --> 00:00:55,610 We have an output function equals an input 20 00:00:55,610 --> 00:00:58,880 function times a kernel. 21 00:00:58,880 --> 00:01:04,010 Then what we did is we wrote each one of those, the input 22 00:01:04,010 --> 00:01:07,490 as well as the kernel, we wrote those as Fourier transforms. 23 00:01:07,490 --> 00:01:10,490 Here they are respectively. 24 00:01:10,490 --> 00:01:16,460 Then we rearranged the integrals and the order of integration. 25 00:01:16,460 --> 00:01:26,800 And we noticed that this expression here is also 26 00:01:26,800 --> 00:01:28,360 known as a delta function. 27 00:01:32,050 --> 00:01:35,590 The integral of an exponential, of a complex exponential 28 00:01:35,590 --> 00:01:38,395 from minus infinity to infinity is a delta function. 29 00:01:40,930 --> 00:01:43,295 Oh, too bright. 30 00:01:43,295 --> 00:01:44,920 OK, I think that's a little bit better. 31 00:01:50,950 --> 00:01:51,940 I can rewrite this. 32 00:01:51,940 --> 00:01:53,970 Let me do it one step at a time. 33 00:01:53,970 --> 00:01:57,850 So I can rewrite now this big integral. 34 00:02:02,560 --> 00:02:05,860 Unfortunately, the black marker has run out of steam, 35 00:02:05,860 --> 00:02:06,970 so I'll switch color. 36 00:02:32,420 --> 00:02:34,550 So since we have a double integral with respect 37 00:02:34,550 --> 00:02:39,590 to the two frequency variables and a delta function inside, 38 00:02:39,590 --> 00:02:43,280 the delta function will knock out one of the integrals. 39 00:02:43,280 --> 00:02:45,385 And we simply replace the-- 40 00:02:52,050 --> 00:02:54,470 well, let's pick-- we can pick whichever 41 00:02:54,470 --> 00:02:57,500 variable we want to integrate, so let's pick u1. 42 00:02:57,500 --> 00:03:05,730 So we'll simply get du1 times G sub in of u1 H of u1 one 43 00:03:05,730 --> 00:03:10,100 e to the i 2 pi u1 x prime. 44 00:03:10,100 --> 00:03:12,170 And now we realize and remember what this is. 45 00:03:12,170 --> 00:03:15,230 If you go back to the beginning of the derivation, 46 00:03:15,230 --> 00:03:18,060 this is the outcome of the convolution. 47 00:03:18,060 --> 00:03:20,120 So this is actually-- let me write it out again. 48 00:03:20,120 --> 00:03:24,290 This is G sub out of x prime. 49 00:03:27,950 --> 00:03:30,230 What we've got now is a Fourier transform. 50 00:03:30,230 --> 00:03:33,638 Because, you see, here is the Fourier transform kernel. 51 00:03:33,638 --> 00:03:35,180 Actually-- I'm sorry-- what we've got 52 00:03:35,180 --> 00:03:36,730 is a Fourier integral. 53 00:03:36,730 --> 00:03:40,400 We've got to write the outcome of the convolution 54 00:03:40,400 --> 00:03:42,170 as a Fourier integral, where this 55 00:03:42,170 --> 00:03:46,180 is the Fourier integral kernel or the inverse Fourier 56 00:03:46,180 --> 00:03:47,480 transform kernel. 57 00:03:47,480 --> 00:03:49,680 The two terms mean the same thing. 58 00:03:49,680 --> 00:03:52,580 And this is the actual inverse Fourier transform. 59 00:03:52,580 --> 00:03:56,090 So, therefore, this result is equivalent to G sub 60 00:03:56,090 --> 00:04:04,070 out of u equals G sub in of u times H of u. 61 00:04:04,070 --> 00:04:06,860 So this is, then, the proof of the convolution theorem, which 62 00:04:06,860 --> 00:04:13,700 says that if two functions are related 63 00:04:13,700 --> 00:04:16,970 as a convolution with a kernel, then 64 00:04:16,970 --> 00:04:20,390 the equivalent relationship in the Fourier domain 65 00:04:20,390 --> 00:04:23,570 is actually a product of the Fourier transform of the input 66 00:04:23,570 --> 00:04:26,990 function times the Fourier transform of the kernel. 67 00:04:26,990 --> 00:04:31,340 So this mathematical-- this analytical result is what 68 00:04:31,340 --> 00:04:34,440 you've seen in the simulations that I have on this screen 69 00:04:34,440 --> 00:04:37,820 here, where the two-- 70 00:04:37,820 --> 00:04:39,320 I will do this in one dimension. 71 00:04:39,320 --> 00:04:41,240 The simulations are obviously 2D, 72 00:04:41,240 --> 00:04:45,390 but it's a little bit easier to do everything in 1D here 73 00:04:45,390 --> 00:04:48,360 so we don't spend too much time writing. 74 00:04:48,360 --> 00:04:53,300 So the two functions that we have here are actually-- 75 00:04:53,300 --> 00:04:56,130 one of them is a sinusoid. 76 00:04:56,130 --> 00:04:58,025 So if I have a sinusoid-- 77 00:05:00,930 --> 00:05:04,980 I keep using the blue marker, [INAUDIBLE].. 78 00:05:04,980 --> 00:05:06,275 So the sinusoid is-- 79 00:05:10,980 --> 00:05:15,020 let me just write it with full contrast 80 00:05:15,020 --> 00:05:16,960 to save some writing here. 81 00:05:20,490 --> 00:05:22,470 This is a sinusoid. 82 00:05:22,470 --> 00:05:25,110 Then it's Fourier transform will actually-- 83 00:05:25,110 --> 00:05:27,480 as we have said several times, it 84 00:05:27,480 --> 00:05:29,430 will consist of three delta functions. 85 00:05:52,750 --> 00:05:56,130 Then we have another function which looks like a rectangle. 86 00:05:56,130 --> 00:05:58,720 That's on the right-hand side, top right. 87 00:05:58,720 --> 00:06:00,470 And we've already seen this one. 88 00:06:00,470 --> 00:06:03,880 So the rectangle is a rectangular function. 89 00:06:03,880 --> 00:06:10,750 If it has a width, let's say, a, then the Fourier transform 90 00:06:10,750 --> 00:06:11,865 will look like this. 91 00:06:20,992 --> 00:06:22,450 And we call this the sinc function. 92 00:06:27,860 --> 00:06:33,230 So the height of this function is a by virtue 93 00:06:33,230 --> 00:06:35,270 of the scaling theorem. 94 00:06:35,270 --> 00:06:41,690 And then the nulls, they go like 1/a, 2/a, and so on. 95 00:06:41,690 --> 00:06:43,970 And then, symmetrically, on the negative side, 96 00:06:43,970 --> 00:06:48,750 minus 1/a, minus 2/a, and so on and so forth. 97 00:06:48,750 --> 00:06:51,380 So these are two Fourier transform relationships. 98 00:06:51,380 --> 00:06:54,800 Now, what we see on the bottom side 99 00:06:54,800 --> 00:06:59,450 is basically a grating which is truncated. 100 00:06:59,450 --> 00:07:02,320 So I take my infinite grating. 101 00:07:09,040 --> 00:07:11,480 It continues on and on. 102 00:07:11,480 --> 00:07:12,590 And then I truncate it. 103 00:07:17,800 --> 00:07:19,260 That is, I multiply-- 104 00:07:32,924 --> 00:07:38,790 I multiply it by a rectangular function which sets the side-- 105 00:07:38,790 --> 00:07:43,510 which sets the aperture of the truncation. 106 00:07:43,510 --> 00:07:47,370 So the way I've written it, the width of this boxcar 107 00:07:47,370 --> 00:07:54,190 would be a, and the period of the grating itself 108 00:07:54,190 --> 00:07:55,890 would be uppercase lambda. 109 00:07:55,890 --> 00:07:57,640 So, of course, we can go ahead and compute 110 00:07:57,640 --> 00:08:00,317 this Fourier transform analytically, 111 00:08:00,317 --> 00:08:01,650 but that would be quite painful. 112 00:08:01,650 --> 00:08:05,420 We'd have to do a lot of changes of variable and so on. 113 00:08:05,420 --> 00:08:09,250 However, by virtue of the convolution theorem, 114 00:08:09,250 --> 00:08:11,620 because this is a product, this Fourier transform 115 00:08:11,620 --> 00:08:13,520 will be a convolution. 116 00:08:13,520 --> 00:08:15,700 So, basically, what we have to do in order-- 117 00:08:15,700 --> 00:08:23,480 when we multiply these two results in this domain, 118 00:08:23,480 --> 00:08:27,650 in this domain we have to convolve them. 119 00:08:27,650 --> 00:08:31,130 So what is now-- so the assumption here 120 00:08:31,130 --> 00:08:34,340 is that the size of the boxcar is actually bigger 121 00:08:34,340 --> 00:08:35,720 by the size of the period. 122 00:08:35,720 --> 00:08:37,250 It does not have to be much bigger, 123 00:08:37,250 --> 00:08:40,309 but it's nice if it is bigger for what I'm 124 00:08:40,309 --> 00:08:42,000 about to draw to be correct. 125 00:08:42,000 --> 00:08:46,220 So what we get if we can convolve this pattern 126 00:08:46,220 --> 00:08:50,210 with the three delta functions, each delta 127 00:08:50,210 --> 00:08:53,610 will produce a replica of this function centered 128 00:08:53,610 --> 00:08:55,610 at the location of the delta function. 129 00:08:55,610 --> 00:08:57,590 So, therefore, we'll get three replicas 130 00:08:57,590 --> 00:09:01,550 of this sinc pattern centered at the corresponding locations. 131 00:09:01,550 --> 00:09:03,650 So let me draw this a little bit carefully here. 132 00:09:10,050 --> 00:09:11,900 So I'll try to be as accurate as I can. 133 00:09:17,830 --> 00:09:18,630 OK. 134 00:09:18,630 --> 00:09:21,450 So here are the two-- 135 00:09:21,450 --> 00:09:22,960 well, there's three delta functions. 136 00:09:22,960 --> 00:09:24,150 One is at the origin. 137 00:09:24,150 --> 00:09:27,860 One is at minus 1 up on lambda, and the other 138 00:09:27,860 --> 00:09:30,000 is at plus 1 up on lambda. 139 00:09:30,000 --> 00:09:32,400 And then, on each one of those, I'm 140 00:09:32,400 --> 00:09:35,380 going to center one of the sinc functions. 141 00:09:35,380 --> 00:09:42,410 So here is one sinc function, and here 142 00:09:42,410 --> 00:09:50,160 is another sinc function, and another one. 143 00:09:55,500 --> 00:09:57,360 So I tried to do it carefully. 144 00:09:57,360 --> 00:09:58,385 Why is this taller? 145 00:09:58,385 --> 00:09:59,760 Because, if you recall, the delta 146 00:09:59,760 --> 00:10:02,130 function that was at the origin is actually 147 00:10:02,130 --> 00:10:05,590 twice the size of the other two delta functions. 148 00:10:05,590 --> 00:10:10,170 So the size of this one would be a/2 actually, 149 00:10:10,170 --> 00:10:11,610 because, if you recall, we're also 150 00:10:11,610 --> 00:10:15,810 picking up one factor of a from the sinc itself. 151 00:10:15,810 --> 00:10:20,490 Then the size of this would be a/4, and the size of this 152 00:10:20,490 --> 00:10:23,070 would also be a/4. 153 00:10:23,070 --> 00:10:26,580 When I say the size, I mean the height. 154 00:10:26,580 --> 00:10:28,350 And then where are the nulls located? 155 00:10:28,350 --> 00:10:29,830 Well, these nulls are simple. 156 00:10:29,830 --> 00:10:34,370 They're still at 1/a, minus 1/a. 157 00:10:34,370 --> 00:10:35,340 Where is this null? 158 00:10:35,340 --> 00:10:39,885 For example, this null would be at 1 over lambda plus 1/a. 159 00:10:39,885 --> 00:10:42,990 And then I have another null over here at 1 160 00:10:42,990 --> 00:10:46,900 over lambda minus 1/a, and so on and so forth. 161 00:10:46,900 --> 00:10:50,670 And then I have more nulls at 1 over lambda minus 2/a, 162 00:10:50,670 --> 00:10:53,850 minus 3/a, and so on and so forth. 163 00:10:53,850 --> 00:10:55,230 And now my assumption that I have 164 00:10:55,230 --> 00:10:57,710 several periods of the grating with the boxcar, 165 00:10:57,710 --> 00:11:00,150 you can see it is quite convenient because I can draw 166 00:11:00,150 --> 00:11:03,420 these sinc functions and I can kind of ignore what 167 00:11:03,420 --> 00:11:05,550 happens in between over here. 168 00:11:05,550 --> 00:11:07,260 As you can imagine, something complicated 169 00:11:07,260 --> 00:11:10,650 will happen over here because the sincs are actually 170 00:11:10,650 --> 00:11:14,190 adding coherently, so something funky will happen. 171 00:11:14,190 --> 00:11:16,170 But if they're pretty far apart, you 172 00:11:16,170 --> 00:11:18,240 can see that the envelope of the sinc function 173 00:11:18,240 --> 00:11:19,620 actually decays quite fast. 174 00:11:19,620 --> 00:11:22,640 So I can more or less ignore what is happening over here, 175 00:11:22,640 --> 00:11:25,460 and I can simply draw the sinc function. 176 00:11:25,460 --> 00:11:28,260 Of course, this is because I'm doing it by hand. 177 00:11:28,260 --> 00:11:31,920 If I use a computational tool, such as MATLAB or Mathematical, 178 00:11:31,920 --> 00:11:34,780 it will do it for me. 179 00:11:34,780 --> 00:11:36,730 So this is the one-dimensional calculation, 180 00:11:36,730 --> 00:11:38,660 that is easier to do by hand. 181 00:11:38,660 --> 00:11:41,420 As a bonus, I think you guys have already computed-- 182 00:11:41,420 --> 00:11:43,150 or, if you haven't already, you will 183 00:11:43,150 --> 00:11:46,300 compute some time between Tuesday midnight 184 00:11:46,300 --> 00:11:48,280 and Wednesday at 8:00 AM. 185 00:11:48,280 --> 00:11:53,380 Some time you will compute the same convolution 186 00:11:53,380 --> 00:11:55,640 for the two-dimensional case. 187 00:11:55,640 --> 00:11:57,350 And you can see the result here. 188 00:11:57,350 --> 00:11:59,330 So, in effect, this saves you from-- 189 00:12:02,290 --> 00:12:04,880 well, at least you know if your result is correct or wrong. 190 00:12:04,880 --> 00:12:07,090 If you derive something that looks like this, 191 00:12:07,090 --> 00:12:09,140 then you know you've done correctly. 192 00:12:09,140 --> 00:12:10,630 So you see what's happened. 193 00:12:10,630 --> 00:12:12,840 If you had the infinite grating, the infinite grating 194 00:12:12,840 --> 00:12:15,850 gives you three delta functions whose 195 00:12:15,850 --> 00:12:19,690 axis is kind of perpendicular to the fringes of the grating, 196 00:12:19,690 --> 00:12:22,270 and the one that the origin is stronger. 197 00:12:22,270 --> 00:12:23,950 Of course, the rectangular function 198 00:12:23,950 --> 00:12:25,768 produces a sinc pattern. 199 00:12:25,768 --> 00:12:27,310 And then when you convolve them, when 200 00:12:27,310 --> 00:12:35,600 you have a tilted grating now and multiply it by a rect, 201 00:12:35,600 --> 00:12:39,290 then you see that you get three sinc patterns oriented 202 00:12:39,290 --> 00:12:42,230 along the same axis, which is perpendicular to the fringes 203 00:12:42,230 --> 00:12:43,310 of the grating. 204 00:12:43,310 --> 00:12:45,290 But the sinc pattern itself is actually 205 00:12:45,290 --> 00:12:48,740 perpendicular to the rectangular aperture. 206 00:12:48,740 --> 00:12:51,110 So this is what the convolution theorem says. 207 00:12:51,110 --> 00:12:53,670 And, of course, you can turn it around. 208 00:12:53,670 --> 00:12:57,050 You can actually turn the grating around 209 00:12:57,050 --> 00:12:58,730 and rotate the aperture. 210 00:12:58,730 --> 00:13:02,990 In this case now, again, you will see the sincs oriented 211 00:13:02,990 --> 00:13:06,710 parallel to the fringes again, but now the rectangular pattern 212 00:13:06,710 --> 00:13:09,860 itself is rotated, and therefore the sinc patterns themselves 213 00:13:09,860 --> 00:13:12,730 are rotated. 214 00:13:12,730 --> 00:13:15,985 So the reason we use these properties is because-- 215 00:13:15,985 --> 00:13:17,360 again, as you can imagine, if you 216 00:13:17,360 --> 00:13:19,700 were to write this as an explicit integral, 217 00:13:19,700 --> 00:13:21,410 it can be done and you would-- of course 218 00:13:21,410 --> 00:13:23,300 you would still get the correct result, 219 00:13:23,300 --> 00:13:25,220 but it would be quite painful to compute. 220 00:13:30,447 --> 00:13:32,030 So now that we have all these results, 221 00:13:32,030 --> 00:13:33,790 we can actually apply them. 222 00:13:33,790 --> 00:13:35,670 So far, these were mathematically results. 223 00:13:35,670 --> 00:13:38,700 I could pretend I'm teaching 18085, 224 00:13:38,700 --> 00:13:41,660 or whatever it is that you learn those things. 225 00:13:41,660 --> 00:13:45,420 But this is also-- of course, they actually become Fraunhofer 226 00:13:45,420 --> 00:13:49,200 diffraction patterns if you simply perform a scale-- 227 00:13:49,200 --> 00:13:51,420 a coordinate change. 228 00:13:51,420 --> 00:13:54,820 So the Fourier transforms, we compute them with respect 229 00:13:54,820 --> 00:13:59,730 to the frequency variables, u and v we call them. 230 00:13:59,730 --> 00:14:02,640 Goodman calls them F sub x, F sub y, 231 00:14:02,640 --> 00:14:05,010 but it is actually the same variable. 232 00:14:05,010 --> 00:14:08,520 Well, if you substitute a special variable 233 00:14:08,520 --> 00:14:11,093 and you multiply by the scaling factor lambda z, 234 00:14:11,093 --> 00:14:13,260 that we saw the derivation in the previous lecture-- 235 00:14:13,260 --> 00:14:14,550 I'm not going to do it again-- 236 00:14:14,550 --> 00:14:17,330 then you actually get the Fraunhofer diffraction pattern. 237 00:14:17,330 --> 00:14:19,200 So it's a simple scaling argument. 238 00:14:19,200 --> 00:14:21,630 And you can tell that-- 239 00:14:21,630 --> 00:14:23,540 well, at least it is plausibly correct. 240 00:14:23,540 --> 00:14:27,080 Here, the units are correct, because u and v 241 00:14:27,080 --> 00:14:28,500 are frequency variables. 242 00:14:28,500 --> 00:14:31,300 So, therefore, their units are inverse meters. 243 00:14:31,300 --> 00:14:33,260 And then multiply it by meter squared. 244 00:14:33,260 --> 00:14:35,490 The wavelength is meter, the distance is meter. 245 00:14:35,490 --> 00:14:37,380 So, therefore, I get meters. 246 00:14:37,380 --> 00:14:40,380 So the units are correct. 247 00:14:40,380 --> 00:14:42,640 And this is one example that we did. 248 00:14:42,640 --> 00:14:45,920 Then here's another one where I shrink 249 00:14:45,920 --> 00:14:47,240 the rectangular aperture. 250 00:14:47,240 --> 00:14:50,130 And, of course, the Fraunhofer diffraction pattern will grow. 251 00:14:50,130 --> 00:14:54,060 We call this a similarity or scaling theorem. 252 00:14:54,060 --> 00:14:57,090 Then I can, for example, use the shift theorem 253 00:14:57,090 --> 00:15:00,110 to calculate the Fraunhofer diffraction 254 00:15:00,110 --> 00:15:03,150 pattern from three rectangles. 255 00:15:03,150 --> 00:15:04,710 As we discussed the last time, this 256 00:15:04,710 --> 00:15:07,960 will give rise to a sinusoidal modulation in the Fraunhofer 257 00:15:07,960 --> 00:15:08,850 domain. 258 00:15:08,850 --> 00:15:11,460 And you can also do the convolution theorem 259 00:15:11,460 --> 00:15:13,362 in this case with a truncated aperture, 260 00:15:13,362 --> 00:15:14,320 and so on and so forth. 261 00:15:20,900 --> 00:15:26,360 So this basically concludes the discussion of the Fraunhofer 262 00:15:26,360 --> 00:15:29,100 diffraction pattern. 263 00:15:29,100 --> 00:15:34,640 But since we've been discussing Fourier transforms, 264 00:15:34,640 --> 00:15:37,880 there's another very basic property of Fourier 265 00:15:37,880 --> 00:15:41,600 transforms that I would like to introduce here, 266 00:15:41,600 --> 00:15:44,660 and then we will see it in full glory for the next two 267 00:15:44,660 --> 00:15:49,010 lectures, and that is spatial filtering. 268 00:15:49,010 --> 00:15:51,380 So spatial filtering is basically the following. 269 00:15:51,380 --> 00:15:54,070 It says if you go to this Fraunhofer 270 00:15:54,070 --> 00:15:57,440 domain, or, in general, in the transform domain-- 271 00:15:57,440 --> 00:15:59,120 which, we will see a little bit later, 272 00:15:59,120 --> 00:16:01,940 that we don't need to have to go very far actually. 273 00:16:01,940 --> 00:16:04,970 By using a lens, we can produce a Fraunhofer diffraction 274 00:16:04,970 --> 00:16:07,270 pattern at the back focal plane of the lens. 275 00:16:07,270 --> 00:16:08,870 That's very convenient. 276 00:16:08,870 --> 00:16:11,720 But if I go here and I do some modification, 277 00:16:11,720 --> 00:16:14,480 and then take another Fourier transform, then, of course, 278 00:16:14,480 --> 00:16:16,580 the signal I reconstruct is not identical 279 00:16:16,580 --> 00:16:18,260 to my original signal, but it will 280 00:16:18,260 --> 00:16:21,290 have changed because I've modified the frequency 281 00:16:21,290 --> 00:16:22,560 spectrum. 282 00:16:22,560 --> 00:16:24,960 So this is called spatial filtering. 283 00:16:24,960 --> 00:16:27,720 So here's an example that I have constructed. 284 00:16:27,720 --> 00:16:31,140 So in this case, I've contacted the signal 285 00:16:31,140 --> 00:16:34,690 in the space domain that looks like three sinusoids. 286 00:16:34,690 --> 00:16:38,150 Now, you cannot tell very clearly from this pattern that 287 00:16:38,150 --> 00:16:41,140 you have three sinusoids, but if you take the Fourier transform, 288 00:16:41,140 --> 00:16:44,090 then you see three spots here, three dots. 289 00:16:44,090 --> 00:16:45,740 Actually, you see six. 290 00:16:45,740 --> 00:16:48,410 But you recall that each sinusoid corresponds 291 00:16:48,410 --> 00:16:49,850 to two dots. 292 00:16:49,850 --> 00:16:52,130 So the conjugate dots here, this one and this one, 293 00:16:52,130 --> 00:16:53,330 they are one sinusoid. 294 00:16:53,330 --> 00:16:55,040 This and this one, another sinusoid. 295 00:16:55,040 --> 00:16:57,860 This and this one are yet another sinusoid. 296 00:16:57,860 --> 00:17:01,010 So spatial filtering, a very simple occurrence 297 00:17:01,010 --> 00:17:04,550 of spatial filtering is what happens if, for example, you 298 00:17:04,550 --> 00:17:09,369 go in with some black marker or some opaque screen 299 00:17:09,369 --> 00:17:12,859 in the case of optics and you remove one of these dots. 300 00:17:12,859 --> 00:17:15,770 If you do that in the transform domain, then you will see it. 301 00:17:15,770 --> 00:17:18,300 Now, watch as I transition the slide. 302 00:17:18,300 --> 00:17:21,139 You will see that the spatial pattern also changes. 303 00:17:24,079 --> 00:17:24,829 OK. 304 00:17:24,829 --> 00:17:27,920 So now it becomes kind of horizontal, 305 00:17:27,920 --> 00:17:30,950 and it is horizontal because the two dominant-- the two 306 00:17:30,950 --> 00:17:35,340 dominant sinusoids are actually along the horizontal axis. 307 00:17:35,340 --> 00:17:36,810 So, therefore, your grooves are-- 308 00:17:36,810 --> 00:17:37,310 I'm sorry. 309 00:17:37,310 --> 00:17:40,330 Your grooves are vertical because the two dots here 310 00:17:40,330 --> 00:17:42,980 are on the horizontal axis. 311 00:17:42,980 --> 00:17:44,840 But there's also a weaker sinusoid 312 00:17:44,840 --> 00:17:47,930 that gives rise to these weak diagonal 313 00:17:47,930 --> 00:17:49,730 fringes that you see over here. 314 00:17:49,730 --> 00:17:51,230 But, basically, you can see that one 315 00:17:51,230 --> 00:17:53,820 of the three spatial frequencies has vanished here. 316 00:17:53,820 --> 00:17:57,740 So this is the simplest case of spatial filtering. 317 00:17:57,740 --> 00:17:59,640 And, of course, you can generalize it. 318 00:17:59,640 --> 00:18:02,710 Here's again the Red Sox-- 319 00:18:02,710 --> 00:18:07,430 or I should say the GO SOX pattern on the Boston high rise 320 00:18:07,430 --> 00:18:09,110 that I showed last time. 321 00:18:09,110 --> 00:18:16,070 And this is the spatial frequency representation 322 00:18:16,070 --> 00:18:18,320 or the Fourier transform of this pattern. 323 00:18:18,320 --> 00:18:20,010 And then we can apply various filters. 324 00:18:20,010 --> 00:18:21,980 For example, if I go with a filter 325 00:18:21,980 --> 00:18:24,560 and I block all the high frequencies, 326 00:18:24,560 --> 00:18:27,260 then you can see that my pattern appears blurred. 327 00:18:27,260 --> 00:18:28,910 In fact, it is more than blurred. 328 00:18:28,910 --> 00:18:33,050 The windows have kind of disappeared of the building. 329 00:18:33,050 --> 00:18:35,730 And that is because the windows, if I go back, 330 00:18:35,730 --> 00:18:36,970 you will see that the-- 331 00:18:36,970 --> 00:18:37,730 hi, Colin. 332 00:18:37,730 --> 00:18:38,530 You're back. 333 00:18:38,530 --> 00:18:39,410 COLIN: Yes. 334 00:18:39,410 --> 00:18:40,160 GEORGE BARBASTATHIS: Welcome. 335 00:18:40,160 --> 00:18:41,077 COLIN: Sorry I'm late. 336 00:18:41,077 --> 00:18:42,932 GEORGE BARBASTATHIS: Oh, no problem. 337 00:18:42,932 --> 00:18:44,390 I thought you were still in Poland. 338 00:18:44,390 --> 00:18:45,480 COLIN: No, I got back. 339 00:18:45,480 --> 00:18:46,480 GEORGE BARBASTATHIS: Oh. 340 00:18:46,480 --> 00:18:48,920 Oh, OK, welcome back. 341 00:18:48,920 --> 00:18:51,530 And so the windows, if you recall, 342 00:18:51,530 --> 00:18:55,740 the windows are kind of periodic in this high rise here. 343 00:18:55,740 --> 00:18:59,210 So they correspond to these dots in the frequency domain, 344 00:18:59,210 --> 00:19:01,160 kind of like delta functions. 345 00:19:01,160 --> 00:19:04,670 And because, in this case, I have blocked the dots, 346 00:19:04,670 --> 00:19:08,898 you see that the windows disappear from the high rise. 347 00:19:08,898 --> 00:19:11,190 And, of course, you can do other funky kind of filters. 348 00:19:11,190 --> 00:19:13,290 This is called a band pass filter. 349 00:19:13,290 --> 00:19:15,200 And, in this case, the windows reappear 350 00:19:15,200 --> 00:19:19,280 because now I center this doughnut, this annulus. 351 00:19:19,280 --> 00:19:22,940 I centered it so that some of these dots in the frequency 352 00:19:22,940 --> 00:19:24,210 domain, they survive. 353 00:19:24,210 --> 00:19:26,480 And you can see that, of course, the-- 354 00:19:26,480 --> 00:19:28,715 well, it is not fully reconstructed, 355 00:19:28,715 --> 00:19:30,090 the original building, of course, 356 00:19:30,090 --> 00:19:32,480 because there's still spatial frequencies missing. 357 00:19:32,480 --> 00:19:35,810 But you can see that the pattern of windows of the high rise 358 00:19:35,810 --> 00:19:37,610 has kind of reappeared. 359 00:19:37,610 --> 00:19:40,340 Now, what happened to the sign GO SOX? 360 00:19:40,340 --> 00:19:43,100 It vanished, and it vanished because, in this case, 361 00:19:43,100 --> 00:19:45,440 I have blocked the lower frequencies. 362 00:19:45,440 --> 00:19:50,970 The GO SOX sign, it has survived the low pass filter 363 00:19:50,970 --> 00:19:55,440 because this is a relatively slow-varying signal, right? 364 00:19:55,440 --> 00:20:00,930 So its frequency content, you expect it to be centered-- 365 00:20:00,930 --> 00:20:03,960 I'm sorry-- to be concentrated near the center of the Fourier 366 00:20:03,960 --> 00:20:05,580 domain where the frequencies are low. 367 00:20:05,580 --> 00:20:08,550 When we do the band pass filter, the GO SOX vanishes. 368 00:20:11,480 --> 00:20:14,030 And that is because I blocked the low frequencies where 369 00:20:14,030 --> 00:20:16,610 this signal was represented. 370 00:20:16,610 --> 00:20:20,750 So you can see that you can do quite interesting manipulations 371 00:20:20,750 --> 00:20:24,620 on images using this concept of spatial frequency. 372 00:20:24,620 --> 00:20:29,480 And, actually, the GO SOX signal has not quite vanished. 373 00:20:29,480 --> 00:20:33,470 If you look carefully, there is a little bit of evidence of it 374 00:20:33,470 --> 00:20:35,070 here, but it's quite hard to see. 375 00:20:35,070 --> 00:20:37,070 And, of course, there's a little bit of evidence 376 00:20:37,070 --> 00:20:41,063 because there is a little bit of the frequency content leaking 377 00:20:41,063 --> 00:20:42,480 into the intermediate frequencies. 378 00:20:42,480 --> 00:20:44,330 So, therefore, some of it has survived, 379 00:20:44,330 --> 00:20:46,280 but mostly it is gone. 380 00:20:46,280 --> 00:20:48,080 So that is the-- 381 00:20:48,080 --> 00:20:52,130 so it is not a perfect filter, but it works quite well. 382 00:20:52,130 --> 00:20:54,650 Of course, the other thing that vanished is the average. 383 00:20:54,650 --> 00:20:56,640 You can see that the sky, that used 384 00:20:56,640 --> 00:20:59,750 to be kind of an average gray, it is gone also. 385 00:20:59,750 --> 00:21:02,120 Because the average-- of course, the average 386 00:21:02,120 --> 00:21:05,610 is presented at the zero spatial frequency, and I blocked it. 387 00:21:05,610 --> 00:21:08,180 So, therefore, the average is gone. 388 00:21:08,180 --> 00:21:13,080 So this is called a spatial filter. 389 00:21:15,860 --> 00:21:17,240 OK. 390 00:21:17,240 --> 00:21:20,630 So we're still one lecture behind. 391 00:21:20,630 --> 00:21:25,040 So this is what I was supposed to have done last Wednesday. 392 00:21:25,040 --> 00:21:29,800 And if you look ahead, there is some discussion of the transfer 393 00:21:29,800 --> 00:21:31,750 function of a Fresnel propagation, 394 00:21:31,750 --> 00:21:34,570 and then something called the Talbot effect. 395 00:21:34,570 --> 00:21:36,370 So I will not do this right now. 396 00:21:36,370 --> 00:21:39,115 I will postpone it, if I may, for next week. 397 00:21:39,115 --> 00:21:40,660 What I would like to do is I would 398 00:21:40,660 --> 00:21:45,740 like to switch to the lecture that I posted today online, 399 00:21:45,740 --> 00:21:48,550 and that is Lecture 10A. 400 00:22:00,995 --> 00:22:02,120 The reason I'm doing that-- 401 00:22:02,120 --> 00:22:04,280 I will go back and talk about the Talbot effect. 402 00:22:04,280 --> 00:22:05,150 Don't worry. 403 00:22:05,150 --> 00:22:06,980 But the reason I'm doing that now 404 00:22:06,980 --> 00:22:09,260 is because I would like to press on 405 00:22:09,260 --> 00:22:11,390 with the concept of spatial frequencies 406 00:22:11,390 --> 00:22:13,100 and spatial filtering. 407 00:22:13,100 --> 00:22:15,770 Because it is quite an important one, and I think the sooner 408 00:22:15,770 --> 00:22:17,340 you learn it, the better. 409 00:22:17,340 --> 00:22:20,870 Talbot effect, well you can learn later. 410 00:22:20,870 --> 00:22:24,140 But this business of spatial filtering, in my experience, 411 00:22:24,140 --> 00:22:26,150 it takes quite a bit of time to digest, 412 00:22:26,150 --> 00:22:29,202 so I would like to do it sooner rather than later. 413 00:22:29,202 --> 00:22:30,410 So I already alluded to that. 414 00:22:30,410 --> 00:22:34,090 I said that this Fraunhofer diffraction pattern 415 00:22:34,090 --> 00:22:36,320 is a Fourier transform, but we don't 416 00:22:36,320 --> 00:22:39,320 have to go to infinity to watch the Fraunhofer diffraction 417 00:22:39,320 --> 00:22:42,740 pattern if we want to generate a Fourier transform optically. 418 00:22:42,740 --> 00:22:44,887 We can also do it by using a lens. 419 00:22:44,887 --> 00:22:46,220 So this is what I will do today. 420 00:22:46,220 --> 00:22:48,410 For the rest of the lecture, I will show you 421 00:22:48,410 --> 00:22:53,330 how a lens can produce a spatial Fourier transform, 422 00:22:53,330 --> 00:22:56,792 and what can we do with it. 423 00:22:56,792 --> 00:22:59,900 So, very briefly, to remind you, from geometrical optics, 424 00:22:59,900 --> 00:23:00,770 this is-- 425 00:23:00,770 --> 00:23:04,450 we did this some time ago, maybe about a month ago. 426 00:23:04,450 --> 00:23:07,040 So, to remind you, a lens is a device 427 00:23:07,040 --> 00:23:13,820 that looks like a, well, at least one curved glass surface, 428 00:23:13,820 --> 00:23:17,150 typically more than one. 429 00:23:17,150 --> 00:23:19,880 And it is a device that we can use 430 00:23:19,880 --> 00:23:22,430 to focus or collimate light. 431 00:23:22,430 --> 00:23:25,100 So, for example, if you illuminate a lens with a plane 432 00:23:25,100 --> 00:23:28,400 wave, then the lens will focus that plane wave 433 00:23:28,400 --> 00:23:31,240 at one focal distance to the right. 434 00:23:31,240 --> 00:23:34,340 On the other hand, if you place a point source 435 00:23:34,340 --> 00:23:36,620 at one focal distance to the left, 436 00:23:36,620 --> 00:23:38,810 the lens will collimate it, will produce 437 00:23:38,810 --> 00:23:41,180 a plane wave, which we also refer 438 00:23:41,180 --> 00:23:42,742 to as an image at infinity. 439 00:23:42,742 --> 00:23:44,450 And, of course, what I am discussing here 440 00:23:44,450 --> 00:23:48,305 is for the case of a positive spherical lens. 441 00:23:48,305 --> 00:23:49,680 There's other lenses that we saw, 442 00:23:49,680 --> 00:23:53,270 negative lenses that would do something slightly different. 443 00:23:53,270 --> 00:23:55,460 But I don't want to do a full review of lenses here, 444 00:23:55,460 --> 00:23:58,280 just to remind you what is relevant to our discussion 445 00:23:58,280 --> 00:23:58,943 here. 446 00:23:58,943 --> 00:24:00,860 And, of course, the other thing that lenses do 447 00:24:00,860 --> 00:24:04,010 is they can produce images as finite conjugates. 448 00:24:04,010 --> 00:24:08,750 If you place an object at some distance s sub o, 449 00:24:08,750 --> 00:24:11,120 then the lens will form an image at a distance 450 00:24:11,120 --> 00:24:16,340 s sub i, which is related to s sub o by the lens law. 451 00:24:16,340 --> 00:24:19,170 So we did the stuff to that when we did geometrical optics, 452 00:24:19,170 --> 00:24:26,070 so I don't want to produce recurrent nightmares to you 453 00:24:26,070 --> 00:24:27,800 by repeating it here. 454 00:24:27,800 --> 00:24:30,080 So what I will do now is I will describe the lens 455 00:24:30,080 --> 00:24:33,030 in the context of wave optics. 456 00:24:33,030 --> 00:24:34,960 So, of course, in the context of wave optics, 457 00:24:34,960 --> 00:24:39,060 we have to describe the lens as some kind of a transparency, 458 00:24:39,060 --> 00:24:41,380 as some kind of a phase function that is 459 00:24:41,380 --> 00:24:43,970 applied to the optical field. 460 00:24:43,970 --> 00:24:46,818 So I don't want to go into the details of this one. 461 00:24:46,818 --> 00:24:47,860 Is there a question, or-- 462 00:24:51,720 --> 00:24:53,430 someone has a microphone on, so I can 463 00:24:53,430 --> 00:24:54,867 hear you shifting on your seat. 464 00:24:54,867 --> 00:24:56,284 Anyway, it doesn't matter, though. 465 00:24:59,297 --> 00:25:01,130 If you have a question, please interrupt me. 466 00:25:01,130 --> 00:25:02,570 Of course, this applies always. 467 00:25:08,977 --> 00:25:09,810 So what is this now? 468 00:25:09,810 --> 00:25:15,310 So we will do a very crude approximation here. 469 00:25:15,310 --> 00:25:18,310 We will actually neglect the thickness of the lens. 470 00:25:18,310 --> 00:25:20,760 We did this again when we did geometrical optics. 471 00:25:20,760 --> 00:25:23,730 And that is, of course, because it is not, strictly speaking, 472 00:25:23,730 --> 00:25:27,420 correct, but the results that we get are pretty good, 473 00:25:27,420 --> 00:25:29,800 and it makes our mathematics pretty simple. 474 00:25:29,800 --> 00:25:34,800 So the combination of the two is a good justification 475 00:25:34,800 --> 00:25:35,850 to make an approximation. 476 00:25:35,850 --> 00:25:39,330 If you have one of the two reasons, 477 00:25:39,330 --> 00:25:41,160 it is not good to make an approximation. 478 00:25:41,160 --> 00:25:43,810 For example, if it makes your math simple 479 00:25:43,810 --> 00:25:47,820 but the answer is wrong, then you don't do the approximation. 480 00:25:47,820 --> 00:25:52,477 If you get the correct answer but the math is not simplified, 481 00:25:52,477 --> 00:25:54,060 again we don't make the approximation. 482 00:25:54,060 --> 00:25:56,940 You might as well go with the accurate calculation. 483 00:25:56,940 --> 00:25:58,590 But, in this case, we get two bonuses. 484 00:25:58,590 --> 00:26:01,950 And if we have both bonuses, then we do the approximation. 485 00:26:04,670 --> 00:26:10,310 So what is happening here, if you take a field-- 486 00:26:10,310 --> 00:26:13,730 imagine like Huygens wavelets impinging 487 00:26:13,730 --> 00:26:15,650 on the lens from the left. 488 00:26:15,650 --> 00:26:19,720 The wavelet that impinges in the center 489 00:26:19,720 --> 00:26:23,300 will actually see the thickest part of the lens, 490 00:26:23,300 --> 00:26:27,450 so it will sustain the longer phase delay because it 491 00:26:27,450 --> 00:26:29,040 propagates through glass. 492 00:26:29,040 --> 00:26:32,670 If you take a Huygens wavelet that actually impinges away 493 00:26:32,670 --> 00:26:36,670 from the axis, it will see a thinner portion of the lens. 494 00:26:36,670 --> 00:26:38,640 Therefore, it will have less phase delay 495 00:26:38,640 --> 00:26:42,360 because it propagates a shorter distance in glass. 496 00:26:42,360 --> 00:26:45,390 It will still propagate some distance in air 497 00:26:45,390 --> 00:26:48,610 on the left and the right of the lens. 498 00:26:48,610 --> 00:26:53,750 So if you compute that now, the difference is, 499 00:26:53,750 --> 00:26:58,940 of course, it is given by the spherical calculation. 500 00:26:58,940 --> 00:27:00,540 I don't want to go through this. 501 00:27:00,540 --> 00:27:02,590 You can go back and do it yourselves. 502 00:27:02,590 --> 00:27:04,970 It's a very simple geometrical calculation, 503 00:27:04,970 --> 00:27:08,880 with the addition of the paraxial approximation. 504 00:27:08,880 --> 00:27:11,420 So even if you glance at this here, you see that-- 505 00:27:11,420 --> 00:27:13,790 actually, this I copied from Goodman, 506 00:27:13,790 --> 00:27:16,533 so the equations are verbatim from Goodman's book. 507 00:27:16,533 --> 00:27:17,450 It's a scan, actually. 508 00:27:19,800 --> 00:27:20,300 I'm sorry. 509 00:27:20,300 --> 00:27:21,140 It is not a scan. 510 00:27:21,140 --> 00:27:23,060 It is a scan from my own notes from last year. 511 00:27:23,060 --> 00:27:25,960 But, anyway, it is verbatim copied from Goodman. 512 00:27:25,960 --> 00:27:28,730 And you can see that I replaced the square root with a Taylor 513 00:27:28,730 --> 00:27:29,570 expansion. 514 00:27:29,570 --> 00:27:31,573 So it is a paraxial approximation. 515 00:27:31,573 --> 00:27:32,990 And the result that you get, which 516 00:27:32,990 --> 00:27:35,630 is what I really wanted to do, is 517 00:27:35,630 --> 00:27:37,250 something that looks like this. 518 00:27:58,510 --> 00:28:00,210 OK. 519 00:28:00,210 --> 00:28:01,130 This is what we get. 520 00:28:01,130 --> 00:28:05,910 So we'll get the complex amplitude transmittance 521 00:28:05,910 --> 00:28:09,360 of the lens if you express it in wave optics. 522 00:28:09,360 --> 00:28:11,350 It is actually a quadratic phase delay. 523 00:28:11,350 --> 00:28:11,850 That's if. 524 00:28:15,466 --> 00:28:18,600 And in this quadratic phase delay, 525 00:28:18,600 --> 00:28:20,610 a magical distance happens. 526 00:28:20,610 --> 00:28:24,770 A magical distance appears, which we recognize 527 00:28:24,770 --> 00:28:26,080 to be the focal length. 528 00:28:26,080 --> 00:28:27,930 This, if we recall from geometrical optics, 529 00:28:27,930 --> 00:28:32,840 we used to call this the lensmaker's formula. 530 00:28:36,640 --> 00:28:38,540 So, basically, we recover the expression 531 00:28:38,540 --> 00:28:43,805 for the focal length of the lens, but now we have a wave-- 532 00:28:46,723 --> 00:28:48,140 I don't want to say wave function. 533 00:28:48,140 --> 00:28:49,920 That sounds like quantum mechanics. 534 00:28:49,920 --> 00:28:54,500 But now we have actually-- we have a complex amplitude 535 00:28:54,500 --> 00:28:55,610 transmittance. 536 00:28:55,610 --> 00:28:59,130 That's what we have associated with that one. 537 00:28:59,130 --> 00:29:04,280 So now let's see why this is the same result as we had before. 538 00:29:07,500 --> 00:29:10,410 So the trick here is that we replace the lens-- 539 00:29:10,410 --> 00:29:17,710 when we have a situation like this one, 540 00:29:17,710 --> 00:29:20,640 we replace the lens with its amplitude-- 541 00:29:20,640 --> 00:29:22,662 complex amplitude transmittance. 542 00:29:34,150 --> 00:29:35,438 So forget about the curvature. 543 00:29:35,438 --> 00:29:37,230 Forget about the glass and everything else. 544 00:29:37,230 --> 00:29:40,480 We just replace it with a thin transparency. 545 00:29:40,480 --> 00:29:42,520 And then we illuminate it with something. 546 00:29:45,350 --> 00:29:48,440 Let's start by choosing this something to be a plane wave. 547 00:29:52,660 --> 00:29:55,500 So here's the wave vector of this plane wave. 548 00:29:55,500 --> 00:30:00,640 And because it is propagating at an angle, we write it as g sub 549 00:30:00,640 --> 00:30:01,140 in. 550 00:30:04,382 --> 00:30:09,470 Actually, we write it g sub t sub minus, 551 00:30:09,470 --> 00:30:11,630 because it is the field immediately 552 00:30:11,630 --> 00:30:21,760 to the left of the transparency, of x comma y. 553 00:30:21,760 --> 00:30:22,820 It is a plane wave. 554 00:30:22,820 --> 00:30:25,130 Let's call this angle theta-- 555 00:30:25,130 --> 00:30:26,120 what did they call it? 556 00:30:26,120 --> 00:30:26,620 Theta 0. 557 00:30:30,700 --> 00:30:34,160 So since it is a plane wave, the proper expression for it 558 00:30:34,160 --> 00:30:43,165 is e to i 2 pi sine theta 0 up on lambda x plus cosine theta 559 00:30:43,165 --> 00:30:45,430 0 up on lambda z. 560 00:30:48,790 --> 00:30:50,360 And I'm going to do two things here. 561 00:30:50,360 --> 00:30:54,540 First of all, I'm going to place the transparency at z equal 0. 562 00:30:54,540 --> 00:30:58,550 So that knocks out this factor, because equal 0. 563 00:30:58,550 --> 00:31:00,900 And then I'm going to make the paraxial approximation, 564 00:31:00,900 --> 00:31:02,940 so that knocks out the sine. 565 00:31:02,940 --> 00:31:05,760 So, basically, the field incident 566 00:31:05,760 --> 00:31:09,090 upon the transparency, upon the lens, that this, 567 00:31:09,090 --> 00:31:16,370 is simply e to the i 2 pi theta 0 x up on lambda. 568 00:31:19,720 --> 00:31:22,850 So what I would like to do now is 569 00:31:22,850 --> 00:31:29,870 to compute the field after the transparency, g sub t plus. 570 00:31:29,870 --> 00:31:39,040 So the rule that we described when we did thin transparencies 571 00:31:39,040 --> 00:31:40,580 is that we multiply. 572 00:31:40,580 --> 00:31:41,080 I'm sorry. 573 00:31:41,080 --> 00:31:44,360 I'm using a slightly different notation on the whiteboard 574 00:31:44,360 --> 00:31:45,410 than in the notes. 575 00:31:45,410 --> 00:31:47,505 You don't have a minus and a plus, 576 00:31:47,505 --> 00:31:51,330 but it's basically the same thing. 577 00:31:51,330 --> 00:32:06,170 So g sub t plus multiplied by the transparency itself. 578 00:32:11,770 --> 00:32:16,240 So the plus stands for after, the minus stands for before, 579 00:32:16,240 --> 00:32:19,055 and the nothing stands for the transparency itself. 580 00:32:19,055 --> 00:32:23,755 Oh, and another thing that I did in the notes is I defined-- 581 00:32:30,900 --> 00:32:37,220 I defined this quantity u0 equals theta 0 up on lambda. 582 00:32:37,220 --> 00:32:40,340 And this now we recognize as a spatial frequency 583 00:32:40,340 --> 00:32:43,010 because it has units of inverse meters. 584 00:32:46,530 --> 00:32:48,000 So what do we get now? 585 00:32:48,000 --> 00:32:50,970 What do we do is a little bit of algebra, 586 00:32:50,970 --> 00:32:55,470 but it actually results in an physically 587 00:32:55,470 --> 00:32:58,320 intuitive, physically meaningful result. 588 00:32:58,320 --> 00:33:01,080 So that justifies the algebra, I suppose. 589 00:33:01,080 --> 00:33:02,110 So let me write it out. 590 00:33:02,110 --> 00:33:05,370 Let me write out this product over here. 591 00:33:20,370 --> 00:33:22,800 OK, that's it. 592 00:33:22,800 --> 00:33:25,400 So now I have to do something that-- 593 00:33:25,400 --> 00:33:27,155 let me leave it here so you can see. 594 00:33:27,155 --> 00:33:29,660 We have to do something that you may remember from horror, 595 00:33:29,660 --> 00:33:32,300 from your high school or elementary school. 596 00:33:32,300 --> 00:33:34,202 I don't know where you learned these things. 597 00:33:34,202 --> 00:33:35,660 It's called to complete the square. 598 00:33:38,778 --> 00:33:40,570 What is the square that I want to complete? 599 00:33:40,570 --> 00:33:42,640 If you look at the exponents, you 600 00:33:42,640 --> 00:33:44,910 have an expression that looks like this. 601 00:33:44,910 --> 00:33:49,400 I'll knock out the minus sign. 602 00:33:49,400 --> 00:33:58,022 x square over lambda f minus 2 u0 x. 603 00:33:58,022 --> 00:33:58,730 Can you see that? 604 00:33:58,730 --> 00:34:00,880 I've neglected the y's and everything else, 605 00:34:00,880 --> 00:34:02,800 and the pi's, and so on. 606 00:34:02,800 --> 00:34:06,530 If I can do that, then I can take care of the rest. 607 00:34:06,530 --> 00:34:09,020 So how can I complete the sign? 608 00:34:09,020 --> 00:34:11,449 Well, I tend to get confused with this, 609 00:34:11,449 --> 00:34:13,690 so let me knock out the 1 over lambda f also. 610 00:34:23,889 --> 00:34:24,800 Now it looks better. 611 00:34:24,800 --> 00:34:25,467 Now I can do it. 612 00:34:35,510 --> 00:34:37,000 Basically, to complete the sign, I 613 00:34:37,000 --> 00:34:39,889 have to add and subtract the square of this business here. 614 00:34:52,260 --> 00:34:54,340 And now I can write it as-- 615 00:35:08,810 --> 00:35:10,790 OK, now I can go back and substitute 616 00:35:10,790 --> 00:35:12,140 into my original expression. 617 00:35:12,140 --> 00:35:15,860 I'm done with manipulating the exponent. 618 00:35:15,860 --> 00:35:18,980 And my original expression was this one. 619 00:35:18,980 --> 00:35:25,235 So I can rewrite out now, g sub t plus of x comma y equals-- 620 00:35:28,466 --> 00:35:32,400 first of all, I have this constant term. 621 00:35:32,400 --> 00:35:33,250 That is constant. 622 00:35:33,250 --> 00:35:37,520 It means it does not depend on x, the spatial variable. 623 00:35:37,520 --> 00:35:38,750 So I'll just take it out. 624 00:35:43,110 --> 00:35:44,570 I should not forget my pi's. 625 00:35:44,570 --> 00:35:45,950 So there's a pi over here. 626 00:35:56,310 --> 00:36:01,810 So-- no, I don't like the way this came out. 627 00:36:01,810 --> 00:36:04,378 I was expecting to divide, but I had multiplied. 628 00:36:09,856 --> 00:36:12,100 OK, that looks better. 629 00:36:12,100 --> 00:36:15,280 So I'm doing OK here, because what are the units? 630 00:36:15,280 --> 00:36:15,860 No units. 631 00:36:15,860 --> 00:36:18,130 I have a spatial frequency squared times 632 00:36:18,130 --> 00:36:21,520 distance squared, so no units. 633 00:36:21,520 --> 00:36:23,358 And what I have left is something 634 00:36:23,358 --> 00:36:24,400 that looks like this now. 635 00:36:39,804 --> 00:36:40,304 OK. 636 00:36:42,820 --> 00:36:45,240 So the first part, I don't have to worry too much about. 637 00:36:45,240 --> 00:36:47,750 This is just a constant factor, as I said. 638 00:36:47,750 --> 00:36:52,460 But this one, now I recognize that's a spherical wave. 639 00:36:52,460 --> 00:36:57,050 It is a spherical wave because it contains quadratic phases 640 00:36:57,050 --> 00:36:58,700 in the exponent. 641 00:36:58,700 --> 00:37:01,690 It is a converging spherical wave because of the minus sign 642 00:37:01,690 --> 00:37:02,190 here. 643 00:37:07,790 --> 00:37:12,410 And it's not quite its origin but its sink. 644 00:37:12,410 --> 00:37:15,880 The location where the spherical wave becomes a point 645 00:37:15,880 --> 00:37:21,420 is actually shifted by this factor over here. 646 00:37:24,890 --> 00:37:27,330 This is displacement. 647 00:37:27,330 --> 00:37:27,830 OK. 648 00:37:31,198 --> 00:37:32,740 So, basically, this is what I've got. 649 00:37:32,740 --> 00:37:35,550 I've got a spherical wave which converges. 650 00:37:35,550 --> 00:37:37,610 Oh, and where does it converge? 651 00:37:37,610 --> 00:37:40,900 Well, the distance that a spherical wave converges 652 00:37:40,900 --> 00:37:44,630 is what multiplies the wavelength in the denominator. 653 00:37:44,630 --> 00:37:48,652 So this is where it converges. 654 00:37:51,960 --> 00:37:54,050 So this is basically what you see here. 655 00:37:54,050 --> 00:37:56,270 The spherical wave after the lens 656 00:37:56,270 --> 00:37:59,800 converges to a distance u0 lambda f. 657 00:37:59,800 --> 00:38:04,890 If you substitute the definition for u0, it is theta 0 times 658 00:38:04,890 --> 00:38:07,250 f away from the axis. 659 00:38:07,250 --> 00:38:10,250 And the distance between the lens and the focus 660 00:38:10,250 --> 00:38:11,730 is one focal distance. 661 00:38:11,730 --> 00:38:12,860 So this is not news. 662 00:38:12,860 --> 00:38:14,590 We knew this from geometrical optics. 663 00:38:14,590 --> 00:38:19,238 We just rederived it using the thin transparency. 664 00:38:19,238 --> 00:38:20,780 So this, I guess, gives us conviction 665 00:38:20,780 --> 00:38:24,410 that our approach is correct, because we rederived something 666 00:38:24,410 --> 00:38:28,550 from geometrical optics. 667 00:38:28,550 --> 00:38:30,800 I will not do the next one. 668 00:38:30,800 --> 00:38:33,020 The next one, I'll let you do by yourselves. 669 00:38:33,020 --> 00:38:37,340 You can repeat a similar procedure of completing squares 670 00:38:37,340 --> 00:38:39,850 in order to see what happens to a diverging 671 00:38:39,850 --> 00:38:42,860 spherical wave placed at one focal distance 672 00:38:42,860 --> 00:38:44,240 to the left of the lens. 673 00:38:44,240 --> 00:38:46,220 And you can convince yourselves easily 674 00:38:46,220 --> 00:38:49,040 that this becomes a plane wave propagating 675 00:38:49,040 --> 00:38:53,270 at an angle equal to the ratio of the displacement 676 00:38:53,270 --> 00:38:54,628 over the focal length. 677 00:38:54,628 --> 00:38:57,170 So this is again something that we saw in geometrical optics. 678 00:38:57,170 --> 00:38:59,880 It is not new. 679 00:38:59,880 --> 00:39:00,380 OK. 680 00:39:06,440 --> 00:39:08,600 The real result that I want to derive here-- 681 00:39:08,600 --> 00:39:11,300 and I will try to do it carefully in the time that we 682 00:39:11,300 --> 00:39:12,800 have left-- 683 00:39:12,800 --> 00:39:16,230 is the Fourier transform property which 684 00:39:16,230 --> 00:39:17,483 I will do for a special case. 685 00:39:17,483 --> 00:39:19,400 Actually, I will not do it for a special case. 686 00:39:19,400 --> 00:39:20,190 I'm going to-- 687 00:39:20,190 --> 00:39:21,340 I take it back. 688 00:39:21,340 --> 00:39:25,480 I will do it for the general case of a lens-- 689 00:39:25,480 --> 00:39:28,010 I'm sorry-- of a thin transparency 690 00:39:28,010 --> 00:39:32,830 placed at a distance z to the left of the lens. 691 00:39:32,830 --> 00:39:34,230 Goodman does three cases. 692 00:39:34,230 --> 00:39:37,090 First, he does the case z equals 0. 693 00:39:37,090 --> 00:39:40,860 Then he does the case z equals f, and then another case. 694 00:39:40,860 --> 00:39:41,710 It doesn't matter. 695 00:39:41,710 --> 00:39:44,085 We'll just do it for the general case, and we're covered. 696 00:39:49,180 --> 00:39:52,000 So what-- first of all, let me do 697 00:39:52,000 --> 00:39:54,790 the derivation in one variable. 698 00:39:54,790 --> 00:39:56,110 So don't write too much. 699 00:39:56,110 --> 00:39:58,310 So we'll basically skip y. 700 00:39:58,310 --> 00:40:03,570 All of the derivations will be just with x. 701 00:40:06,260 --> 00:40:08,240 So I have a thin transparency g of x. 702 00:40:13,600 --> 00:40:15,550 And then what I will do is I will propagate 703 00:40:15,550 --> 00:40:24,910 it distance z to the lens. 704 00:40:31,390 --> 00:40:37,420 Now, on the lens, my coordinate is x prime. 705 00:40:37,420 --> 00:40:39,925 And since I'm propagating a field-- 706 00:40:42,850 --> 00:40:45,070 also, I forgot to say-- that's quite 707 00:40:45,070 --> 00:40:47,170 important-- the implicit assumption here 708 00:40:47,170 --> 00:40:51,340 is that the illumination is an on-axis plane wave coming 709 00:40:51,340 --> 00:40:52,540 from the left. 710 00:40:52,540 --> 00:40:54,730 So that, if you recall, we said a couple 711 00:40:54,730 --> 00:40:56,500 of times, that is simply-- 712 00:40:56,500 --> 00:41:01,480 its complex amplitude is 1, because I can choose that-- 713 00:41:01,480 --> 00:41:02,680 a constant phase. 714 00:41:02,680 --> 00:41:04,690 And there's no x variations with this one. 715 00:41:08,640 --> 00:41:12,209 So the Fresnel propagation kernel, if I go from z-- 716 00:41:16,041 --> 00:41:21,640 So g is-- let me maintain my notation consistent here. 717 00:41:21,640 --> 00:41:31,220 So to the left of the lens, L minus of x prime-- 718 00:41:31,220 --> 00:41:34,340 so that is the field to the left of the lens-- 719 00:41:34,340 --> 00:41:38,730 is going to be given by a Fresnel diffraction integral. 720 00:41:38,730 --> 00:41:45,195 And, actually, in my derivation, I skipped the constant. 721 00:41:59,940 --> 00:42:02,390 And what is the constant that I skipped? 722 00:42:02,390 --> 00:42:03,300 It is this one. 723 00:42:11,240 --> 00:42:13,160 And this constant should be there, 724 00:42:13,160 --> 00:42:15,382 but it is not doing anything significant for us 725 00:42:15,382 --> 00:42:17,090 in this case, so that's why I skipped it. 726 00:42:17,090 --> 00:42:19,160 To save writing, basically. 727 00:42:19,160 --> 00:42:22,490 So from now on, we will basically neglect this. 728 00:42:22,490 --> 00:42:25,410 Even though it is there, we will simply neglect. 729 00:42:28,230 --> 00:42:36,470 Now, the field after the lens equals the field 730 00:42:36,470 --> 00:42:40,570 before the lens times the lens itself. 731 00:42:40,570 --> 00:42:42,480 And the lens itself is something of the form 732 00:42:42,480 --> 00:42:47,200 e to the minus i pi x squared up on lambda f. 733 00:42:50,350 --> 00:43:01,410 And, finally, I have this field, and I have to propagate it. 734 00:43:07,280 --> 00:43:08,100 How long? 735 00:43:08,100 --> 00:43:10,160 Now I have to do this part, which 736 00:43:10,160 --> 00:43:13,730 means I have to propagate by distance f until I 737 00:43:13,730 --> 00:43:15,250 reach the back focal plane. 738 00:43:22,170 --> 00:43:27,010 And that is another Fresnel integral. 739 00:43:27,010 --> 00:43:30,930 I will call it g sub f, I guess. 740 00:43:36,250 --> 00:43:39,120 Again, there is a factor here which I will neglect. 741 00:43:56,530 --> 00:43:59,010 OK. 742 00:43:59,010 --> 00:44:00,645 So now let's put everything together. 743 00:44:06,150 --> 00:44:12,590 I have two Fresnel convolution integrals-- 744 00:44:12,590 --> 00:44:15,140 one with respect to the input coordinates, 745 00:44:15,140 --> 00:44:20,310 one with respect to the lens coordinates. 746 00:44:20,310 --> 00:44:24,300 And what is left inside, I will simply substitute all the rest. 747 00:44:24,300 --> 00:44:25,425 I have the input itself. 748 00:44:28,260 --> 00:44:30,990 Then I have the propagation kernel 749 00:44:30,990 --> 00:44:33,120 from the input to the lens. 750 00:44:38,410 --> 00:44:39,420 Then I have the lens. 751 00:44:44,100 --> 00:44:46,140 And then I have the propagation kernel 752 00:44:46,140 --> 00:44:48,100 from the lens to the back focal plane. 753 00:44:56,884 --> 00:44:59,330 OK. 754 00:44:59,330 --> 00:45:00,990 That's what it is. 755 00:45:00,990 --> 00:45:02,660 This looks a little scary, but part 756 00:45:02,660 --> 00:45:05,420 of the purpose of this class is to teach you 757 00:45:05,420 --> 00:45:09,100 how to not be scared by this kind of integrals. 758 00:45:09,100 --> 00:45:10,940 So the way you know this, you don't 759 00:45:10,940 --> 00:45:12,910 get scared by this sort of integral 760 00:45:12,910 --> 00:45:14,810 is you manipulate the exponents here. 761 00:45:14,810 --> 00:45:16,340 And you try to-- the first thing you 762 00:45:16,340 --> 00:45:19,130 do when you reach an integral of this kind 763 00:45:19,130 --> 00:45:22,220 is you try to collect terms. 764 00:45:22,220 --> 00:45:23,660 So I'll write the exponents here. 765 00:45:23,660 --> 00:45:29,520 I will expand the exponents and collect terms then. 766 00:45:29,520 --> 00:45:35,140 So if I expand the exponents, I will get x prime squared 767 00:45:35,140 --> 00:45:41,650 plus x squared minus 2x x prime over lambda z. 768 00:45:41,650 --> 00:45:44,320 This came from this term over here. 769 00:45:44,320 --> 00:45:51,300 Then I have minus x square over lambda f. 770 00:45:51,300 --> 00:45:52,335 And then I have plus-- 771 00:46:01,030 --> 00:46:03,650 and I think I missed a prime here. 772 00:46:03,650 --> 00:46:04,900 That should have been x prime. 773 00:46:07,810 --> 00:46:09,070 Yes, correct. 774 00:46:09,070 --> 00:46:11,180 The lens should also be with an x prime. 775 00:46:11,180 --> 00:46:11,680 Thank you. 776 00:46:24,938 --> 00:46:26,730 It is very fortunate that you corrected me, 777 00:46:26,730 --> 00:46:29,630 because if you hadn't I would be kind of stuck here. 778 00:46:29,630 --> 00:46:30,130 OK. 779 00:46:32,902 --> 00:46:34,750 So the thing that you notice first 780 00:46:34,750 --> 00:46:39,090 is that some of these exponents get knocked out. 781 00:46:42,180 --> 00:46:46,100 This one kills this one. 782 00:46:46,100 --> 00:46:47,560 That's very pleasant. 783 00:46:53,140 --> 00:46:54,550 Now what do I have to do? 784 00:47:07,175 --> 00:47:09,750 I still need to make an integration with respect 785 00:47:09,750 --> 00:47:10,490 to x prime. 786 00:47:15,250 --> 00:47:19,050 So x prime appears here and here. 787 00:47:19,050 --> 00:47:21,810 And I have to make an integration with respect to x. 788 00:47:24,740 --> 00:47:27,570 Well, here is x. 789 00:47:27,570 --> 00:47:28,880 Here is x. 790 00:47:32,060 --> 00:47:32,560 OK. 791 00:47:35,270 --> 00:47:37,945 So what-- any ideas? 792 00:47:37,945 --> 00:47:40,070 Anybody want to speculate on what I should do here? 793 00:47:48,180 --> 00:47:50,157 Let me write the integral. 794 00:47:50,157 --> 00:47:51,990 That's a little bit confusing the way it is. 795 00:47:51,990 --> 00:47:55,260 Let me rewrite it so you can see what the integral looks like. 796 00:49:05,834 --> 00:49:07,050 Let me do it carefully. 797 00:49:07,050 --> 00:49:12,282 So I have e to the minus i 2 pi x prime up on lambda. 798 00:49:12,282 --> 00:49:13,740 That's common in the two exponents. 799 00:49:13,740 --> 00:49:20,400 Inside, I have x up on z plus x double prime up on f. 800 00:49:31,228 --> 00:49:32,270 So what should I do next? 801 00:50:03,220 --> 00:50:08,500 Is there any glaring sort of integral that popped up here? 802 00:50:39,850 --> 00:50:42,190 What is-- here, we have an integral 803 00:50:42,190 --> 00:50:45,790 of another exponential, right? 804 00:50:45,790 --> 00:50:58,490 And they-- let me rewrite it like this. 805 00:51:40,960 --> 00:51:43,460 OK. 806 00:51:43,460 --> 00:51:45,470 So the glaring integral that I was referring 807 00:51:45,470 --> 00:51:48,930 to before is this one. 808 00:51:48,930 --> 00:51:52,440 That's a Fourier transform. 809 00:51:52,440 --> 00:51:54,316 Whose Fourier transform? 810 00:51:54,316 --> 00:51:56,580 The Fourier transform of whomever 811 00:51:56,580 --> 00:51:58,290 appears in this location over here. 812 00:52:07,690 --> 00:52:09,940 And where is the Fourier transform computed? 813 00:52:09,940 --> 00:52:15,580 Well, it is computed in this spatial frequency, right? 814 00:52:15,580 --> 00:52:19,630 It is computed in whatever spatial frequency multiplies 815 00:52:19,630 --> 00:52:23,880 the dummy variable in the exponent. 816 00:52:23,880 --> 00:52:25,380 Now, what is this Fourier transform? 817 00:52:25,380 --> 00:52:28,170 We don't know, but we have our notes, or we have Goodman, 818 00:52:28,170 --> 00:52:30,270 or we have the tables of formulas. 819 00:52:30,270 --> 00:52:35,910 So switching to Lecture 9B. 820 00:52:52,960 --> 00:52:56,690 This is our Fourier transform pairs. 821 00:52:56,690 --> 00:53:01,130 I recognize this integral, recognize 822 00:53:01,130 --> 00:53:02,620 this Fourier transform. 823 00:53:02,620 --> 00:53:06,380 It is the second row from the bottom. 824 00:53:06,380 --> 00:53:13,910 If you look at this expression and the expression over here, 825 00:53:13,910 --> 00:53:16,270 it is actually the same Fourier transform. 826 00:53:16,270 --> 00:53:18,370 It is the Fourier trans-- what you 827 00:53:18,370 --> 00:53:20,380 see here is the Fourier transform 828 00:53:20,380 --> 00:53:24,220 of the quadratic phase in the exponential. 829 00:53:24,220 --> 00:53:26,770 So we have the answer. 830 00:53:26,770 --> 00:53:28,740 The answer is right here. 831 00:53:28,740 --> 00:53:29,980 Again, I will neglect-- 832 00:53:29,980 --> 00:53:31,810 actually, this constant is quite important, 833 00:53:31,810 --> 00:53:33,800 but I will be neglect it nevertheless. 834 00:53:36,580 --> 00:53:42,850 So, basically, the way to get a one-to-one correspondence is 835 00:53:42,850 --> 00:53:45,990 to simply substitute what would-- 836 00:53:45,990 --> 00:53:49,840 what in the table is denoted as a square is actually 837 00:53:49,840 --> 00:53:55,630 identical to 1 over lambda z in our case. 838 00:53:55,630 --> 00:54:00,050 So I can write out now the answer. 839 00:54:00,050 --> 00:54:10,850 This thing equals-- first of all, before I do any further, 840 00:54:10,850 --> 00:54:13,930 we recognize that this does not play in the integration. 841 00:54:13,930 --> 00:54:15,710 This has the output plane coordinate, 842 00:54:15,710 --> 00:54:17,120 so I will simply take it outside. 843 00:54:35,310 --> 00:54:36,620 And then I will write out the-- 844 00:54:36,620 --> 00:54:39,260 in one shot, I will write out the outcome 845 00:54:39,260 --> 00:54:42,390 of this Fourier transform. 846 00:54:42,390 --> 00:54:50,060 So that a square that I have in the original function, 847 00:54:50,060 --> 00:54:52,840 it will go inverse in the other one. 848 00:54:52,840 --> 00:54:56,560 So we'll get, then, e to the what? 849 00:54:56,560 --> 00:54:58,210 I will get an extra minus sign. 850 00:54:58,210 --> 00:55:03,140 If I have plus j here, I have minus j here. 851 00:55:03,140 --> 00:55:05,946 So this will then become e to the minus-- 852 00:55:05,946 --> 00:55:11,470 we'll have all the pi's and so on-- minus i pi lambda z. 853 00:55:11,470 --> 00:55:15,520 And then I will get the square of the spatial frequency. 854 00:55:15,520 --> 00:55:21,400 So it will be 1 over lambda square x up on z plus x double 855 00:55:21,400 --> 00:55:25,370 prime up on f squared. 856 00:55:25,370 --> 00:55:27,496 OK, that's it. 857 00:55:27,496 --> 00:55:30,770 So now I can manipulate it a little bit further. 858 00:55:46,150 --> 00:55:49,240 And now let me write out all these exponents 859 00:55:49,240 --> 00:55:51,910 that come out of this square. 860 00:55:51,910 --> 00:55:53,590 So I will get-- 861 00:55:53,590 --> 00:55:57,640 the first one will be x square up on z square. 862 00:55:57,640 --> 00:55:59,320 So one z will be killed. 863 00:55:59,320 --> 00:56:01,540 One lambda has already been killed. 864 00:56:01,540 --> 00:56:05,530 So I will get e to the minus i pi x 865 00:56:05,530 --> 00:56:08,610 square over lambda z, right? 866 00:56:11,130 --> 00:56:12,390 Then I will get this term. 867 00:56:12,390 --> 00:56:16,960 That will be e to the minus i pi-- 868 00:56:16,960 --> 00:56:23,100 this is tricky-- z x double prime square over lambda 869 00:56:23,100 --> 00:56:25,110 f square. 870 00:56:25,110 --> 00:56:27,130 This came out of the square of this one. 871 00:56:27,130 --> 00:56:29,040 And I will also get the cross term. 872 00:56:29,040 --> 00:56:33,562 So that will be e to the minus i. 873 00:56:33,562 --> 00:56:35,170 And then we'll do it carefully. 874 00:56:35,170 --> 00:56:37,790 I will get 2 pi. 875 00:56:37,790 --> 00:56:40,070 And what is left here-- 876 00:56:40,070 --> 00:56:41,670 one z will cancel. 877 00:56:41,670 --> 00:56:49,130 I will get x x double prime up on lambda f. 878 00:56:49,130 --> 00:56:52,560 And now, happily, we see that this additional quadratic that 879 00:56:52,560 --> 00:56:56,070 was very annoying over here, this one, it got 880 00:56:56,070 --> 00:56:58,950 killed by this one. 881 00:56:58,950 --> 00:57:02,460 This one is not playing in the integration either, 882 00:57:02,460 --> 00:57:04,490 so I can actually take it out of here. 883 00:57:38,260 --> 00:57:40,770 And this is the result that I was after. 884 00:57:40,770 --> 00:57:43,620 You see that I actually got another Fourier transform. 885 00:57:43,620 --> 00:57:46,590 This is well recognizable as a Fourier transform kernel. 886 00:57:50,590 --> 00:57:54,090 So what I have in this part over here, 887 00:57:54,090 --> 00:57:56,250 it is actually the Fourier transform 888 00:57:56,250 --> 00:58:02,615 of the transparency calculated at these coordinates, that is 889 00:58:02,615 --> 00:58:05,250 the argument of the integral. 890 00:58:09,160 --> 00:58:10,930 Now, something funny happened here, 891 00:58:10,930 --> 00:58:14,820 and this doesn't look quite right to me. 892 00:58:14,820 --> 00:58:16,570 That should be f, right? 893 00:58:16,570 --> 00:58:19,110 OK. 894 00:58:19,110 --> 00:58:20,420 I don't know how this became z. 895 00:58:24,300 --> 00:58:25,050 Oh, yes, yes, yes. 896 00:58:25,050 --> 00:58:25,883 OK, I know now, yes. 897 00:58:25,883 --> 00:58:29,250 That should be f, yes. 898 00:58:29,250 --> 00:58:31,200 Somewhere in my notes I converted 899 00:58:31,200 --> 00:58:34,560 this to f, but, thankfully, not the physics. 900 00:58:34,560 --> 00:58:35,802 So this should not be z. 901 00:58:35,802 --> 00:58:36,510 This should be f. 902 00:58:43,340 --> 00:58:46,510 OK, so now it looks right. 903 00:58:46,510 --> 00:58:48,060 OK.