1 00:00:02,550 --> 00:00:04,920 The following content is provided under a Creative 2 00:00:04,920 --> 00:00:06,310 Commons license. 3 00:00:06,310 --> 00:00:08,520 Your support will help MIT OpenCourseWare 4 00:00:08,520 --> 00:00:12,610 continue to offer high quality educational resources for free. 5 00:00:12,610 --> 00:00:15,150 To make a donation or to view additional materials 6 00:00:15,150 --> 00:00:19,110 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:19,110 --> 00:00:20,330 at ocw.mit.edu. 8 00:00:23,620 --> 00:00:27,580 YEN-JIE LEE: So I hope you can hear me. 9 00:00:27,580 --> 00:00:29,690 Welcome back, everybody. 10 00:00:29,690 --> 00:00:31,750 I hope you have the full energy back 11 00:00:31,750 --> 00:00:35,380 from Thanksgiving vacation. 12 00:00:35,380 --> 00:00:39,070 Everybody, welcome back. 13 00:00:39,070 --> 00:00:45,470 So before we start, we'll talk about where we are now. 14 00:00:45,470 --> 00:00:50,180 So this is actually the goal, which we said, for 8.03, 15 00:00:50,180 --> 00:00:51,580 at the beginning. 16 00:00:51,580 --> 00:00:54,520 So what we have been discussing is 17 00:00:54,520 --> 00:00:58,190 to learn about how to translate physical situations 18 00:00:58,190 --> 00:01:03,430 into mathematics, simple harmonic oscillator, coupled 19 00:01:03,430 --> 00:01:05,530 oscillators, et cetera. 20 00:01:05,530 --> 00:01:11,360 And we tried to put together infinite number of oscillators. 21 00:01:11,360 --> 00:01:17,380 And we found waves from this interesting exercise. 22 00:01:17,380 --> 00:01:21,280 And of course, we learned about Fourier decompositions of waves 23 00:01:21,280 --> 00:01:25,860 and also learned about how to put together physical systems. 24 00:01:25,860 --> 00:01:29,920 In order to do that, you need to define boundary conditions. 25 00:01:29,920 --> 00:01:32,350 And those conditions need to be satisfied, 26 00:01:32,350 --> 00:01:35,900 so that you can describe multiple physical systems 27 00:01:35,900 --> 00:01:37,550 all together. 28 00:01:37,550 --> 00:01:39,640 And the third part of the course, we 29 00:01:39,640 --> 00:01:44,320 have been focusing on many, many applications, 30 00:01:44,320 --> 00:01:46,840 for instance, for the phenomenon related 31 00:01:46,840 --> 00:01:50,320 to electromagnetic waves and also 32 00:01:50,320 --> 00:01:53,830 many practical applications in optics. 33 00:01:53,830 --> 00:01:59,020 And we are pretty close to the discussion between wave 34 00:01:59,020 --> 00:02:03,190 and vibrations and the future course, which 35 00:02:03,190 --> 00:02:06,900 is 8.04, the connections to quantum mechanics. 36 00:02:06,900 --> 00:02:10,520 And if we have time, we will talk about gravitational waves 37 00:02:10,520 --> 00:02:13,660 if we manage to do that. 38 00:02:13,660 --> 00:02:16,030 It depends on how fast we progress. 39 00:02:18,900 --> 00:02:21,380 So let's start the lecture today. 40 00:02:21,380 --> 00:02:25,250 So first, we will give you a short review. 41 00:02:25,250 --> 00:02:27,250 Before Thanksgiving, we were talking 42 00:02:27,250 --> 00:02:30,100 about polarizer filters. 43 00:02:30,100 --> 00:02:34,900 And we have been researching how to make a very good photo so 44 00:02:34,900 --> 00:02:37,070 that you can post it on Facebook, right? 45 00:02:37,070 --> 00:02:38,800 So that's essentially what we learned. 46 00:02:38,800 --> 00:02:44,230 So if you want to take a picture of the sky, which is deep blue, 47 00:02:44,230 --> 00:02:46,390 then you need to use a polarizer. 48 00:02:46,390 --> 00:02:47,620 And the reason? 49 00:02:47,620 --> 00:02:50,740 We also understand that it's because, if you 50 00:02:50,740 --> 00:02:56,380 look at the sky, which is actually, roughly a 45 to 90 51 00:02:56,380 --> 00:02:59,290 degree wave from the direction of the sun. 52 00:02:59,290 --> 00:03:03,870 Basically, what you get is that all those light fronts 53 00:03:03,870 --> 00:03:06,730 scattering, between the sunlight and the molecules 54 00:03:06,730 --> 00:03:12,170 or little dust in the sky, polarize the light. 55 00:03:12,170 --> 00:03:15,850 Therefore, you can actual filter them using the polarizer. 56 00:03:15,850 --> 00:03:18,670 Of course, you have to tune your polarizer carefully, 57 00:03:18,670 --> 00:03:21,140 so that you can actually minimize 58 00:03:21,140 --> 00:03:23,470 the light from the sky, so that you get 59 00:03:23,470 --> 00:03:26,710 a sharper image in your photo. 60 00:03:26,710 --> 00:03:31,630 And also, we discussed about, with a polarizer, polarization 61 00:03:31,630 --> 00:03:36,100 filter, we can actually filter out also the reflected light, 62 00:03:36,100 --> 00:03:42,100 for example, from the water or from the window of a car. 63 00:03:42,100 --> 00:03:46,090 And that is because something which is closely related 64 00:03:46,090 --> 00:03:51,250 to the boundary condition, which we learned about from Maxwell's 65 00:03:51,250 --> 00:03:52,960 equation in matter. 66 00:03:52,960 --> 00:03:54,950 And this is actually the four equations, 67 00:03:54,950 --> 00:03:56,670 which we discussed last time. 68 00:03:56,670 --> 00:04:01,370 And in that issue, we were using that to explain the incident 69 00:04:01,370 --> 00:04:07,840 light, from air to something which is denser, 70 00:04:07,840 --> 00:04:09,880 for example lighter gas. 71 00:04:09,880 --> 00:04:14,780 And then we found that, if we start with unpolarized light-- 72 00:04:14,780 --> 00:04:17,620 this incident wave unpolarized light-- 73 00:04:17,620 --> 00:04:20,589 what we found is that the transmitted wave, which 74 00:04:20,589 --> 00:04:25,070 is actually in the bottom of this diagram, 75 00:04:25,070 --> 00:04:30,250 is actually still pretty close to unpolarized 76 00:04:30,250 --> 00:04:34,900 light but slightly polarized because of the transmission 77 00:04:34,900 --> 00:04:36,880 and the boundary condition. 78 00:04:36,880 --> 00:04:40,900 And the reflected light, something very interesting 79 00:04:40,900 --> 00:04:43,060 happens. 80 00:04:43,060 --> 00:04:46,690 Only the component, which is actually 81 00:04:46,690 --> 00:04:51,430 polarizing the direction, such that the electric field is 82 00:04:51,430 --> 00:04:56,470 oscillating in a direction perpendicular 83 00:04:56,470 --> 00:05:01,210 to the surface of this light, will survive. 84 00:05:01,210 --> 00:05:07,010 And that actually gives you polarized light, 85 00:05:07,010 --> 00:05:09,260 which is actually reflected from the surface. 86 00:05:09,260 --> 00:05:14,220 And this interesting phenomenon reaches the maxima, 87 00:05:14,220 --> 00:05:17,860 where you get the fully polarized light, when 88 00:05:17,860 --> 00:05:23,020 you actually set the incident angle of the unpolarized light 89 00:05:23,020 --> 00:05:25,460 at so-called Brewster's angle. 90 00:05:25,460 --> 00:05:28,540 And this Brewster's angle is happening 91 00:05:28,540 --> 00:05:32,520 when the reflected light and this transmitted light 92 00:05:32,520 --> 00:05:37,030 direction actually are orthogonal to each other. 93 00:05:37,030 --> 00:05:39,940 And that actually gives you the maxima effect 94 00:05:39,940 --> 00:05:42,040 we are looking for. 95 00:05:42,040 --> 00:05:43,930 So that's actually what we have learned 96 00:05:43,930 --> 00:05:46,840 from electromagnetic wave in matter 97 00:05:46,840 --> 00:05:49,460 and, also, matching the boundary conditions 98 00:05:49,460 --> 00:05:57,670 between the electromagnetic waves inside the material 99 00:05:57,670 --> 00:06:01,300 and in the air, such that we actually learn about all 100 00:06:01,300 --> 00:06:03,250 those interesting phenomena. 101 00:06:03,250 --> 00:06:05,770 And basically, we have learned how 102 00:06:05,770 --> 00:06:08,830 to describe electromagnetic waves, how 103 00:06:08,830 --> 00:06:13,210 to add electromagnetic waves together, how they propagate 104 00:06:13,210 --> 00:06:16,120 from one position to the other position, 105 00:06:16,120 --> 00:06:19,090 and how the boundary condition works 106 00:06:19,090 --> 00:06:23,020 and your equation of motion, et cetera, and something 107 00:06:23,020 --> 00:06:26,230 related to dielectric material. 108 00:06:26,230 --> 00:06:28,990 And today what we're going to do is 109 00:06:28,990 --> 00:06:32,080 to put all the things we have learned together 110 00:06:32,080 --> 00:06:34,620 and see if we can actually explain 111 00:06:34,620 --> 00:06:38,090 a very interesting phenomenon. 112 00:06:38,090 --> 00:06:44,220 So before I start, I will show you 113 00:06:44,220 --> 00:06:48,580 a demonstration, which I'm not sure if I will be successful. 114 00:06:48,580 --> 00:06:50,250 It's very difficult, actually. 115 00:06:50,250 --> 00:06:52,890 And of course, during the break, you're 116 00:06:52,890 --> 00:06:55,970 welcome to come over and play with all those demos. 117 00:06:55,970 --> 00:06:59,710 And here, I have two sticks. 118 00:06:59,710 --> 00:07:04,690 And I am going to create a soap bubble from this soap 119 00:07:04,690 --> 00:07:05,670 and water. 120 00:07:05,670 --> 00:07:10,070 And let's see if I will make it or not. 121 00:07:10,070 --> 00:07:14,220 So basically, I put this into the soap water. 122 00:07:14,220 --> 00:07:19,450 And I will try to open it to see if I can create a bubble. 123 00:07:19,450 --> 00:07:20,470 Yeah. 124 00:07:20,470 --> 00:07:22,290 You can see. 125 00:07:22,290 --> 00:07:25,720 You can see that there's a colorful soap bubble created. 126 00:07:28,870 --> 00:07:30,770 You can see that it is not always easy. 127 00:07:35,260 --> 00:07:37,320 Oh, it's getting very messy now. 128 00:07:37,320 --> 00:07:40,110 I'm trying to destroy the classroom. 129 00:07:40,110 --> 00:07:44,880 But it's OK, because we are MIT. 130 00:07:44,880 --> 00:07:47,170 You can that it's really beautiful. 131 00:07:47,170 --> 00:07:49,385 It's colorful. 132 00:07:49,385 --> 00:07:53,630 It live for a while, then it breaks. 133 00:07:53,630 --> 00:07:57,060 And of course, during the break, you are welcome to do this. 134 00:07:57,060 --> 00:08:02,860 And it's actually non-trivial to create this size of bubble. 135 00:08:02,860 --> 00:08:04,380 So the success rate is like 50%. 136 00:08:07,080 --> 00:08:13,810 So as you can see from this demonstration, 137 00:08:13,810 --> 00:08:17,740 we see something really beautiful. 138 00:08:17,740 --> 00:08:19,735 This bubble is colorful. 139 00:08:23,420 --> 00:08:27,190 And I didn't actually shine this soap bubble 140 00:08:27,190 --> 00:08:31,570 by all kinds of different preset colors. 141 00:08:31,570 --> 00:08:34,659 So it appears automatically and just shiny 142 00:08:34,659 --> 00:08:36,740 with all kinds of different-- whatever, 143 00:08:36,740 --> 00:08:41,260 wavelengths I get from the lights that are in here. 144 00:08:41,260 --> 00:08:43,659 Pretty bright light, there, on my face. 145 00:08:43,659 --> 00:08:47,980 And you can see that it becomes colorful. 146 00:08:47,980 --> 00:08:50,780 And we are going to understand what is going on 147 00:08:50,780 --> 00:08:56,500 and where this color is coming from. 148 00:08:56,500 --> 00:09:00,280 And the good news is that, based on the knowledge we 149 00:09:00,280 --> 00:09:03,940 have learned, we are in a very good position 150 00:09:03,940 --> 00:09:06,770 to understand this phenomenon. 151 00:09:06,770 --> 00:09:11,870 So before we start to explain this phenomena, 152 00:09:11,870 --> 00:09:14,650 I would like to talk about a phenomenon, 153 00:09:14,650 --> 00:09:17,080 so-called interference. 154 00:09:17,080 --> 00:09:21,160 So suppose we have two electromagnetic waves. 155 00:09:21,160 --> 00:09:22,810 We can actually add them together 156 00:09:22,810 --> 00:09:25,160 because of superposition principle. 157 00:09:25,160 --> 00:09:27,160 So what we can do is that, suppose 158 00:09:27,160 --> 00:09:30,300 I have two electric fields. 159 00:09:30,300 --> 00:09:40,400 E1 is actually defined as A1 cosine omega t minus kz 160 00:09:40,400 --> 00:09:44,360 plus phi 1 in the x direction. 161 00:09:44,360 --> 00:09:47,340 So by now, you should know that this is actually 162 00:09:47,340 --> 00:09:52,710 the electric field propagating at angular frequency omega, 163 00:09:52,710 --> 00:10:00,580 with a wave number k, going toward positive z direction. 164 00:10:00,580 --> 00:10:04,180 And the electric field is perpendicular to the direction 165 00:10:04,180 --> 00:10:08,500 of propagation in the x direction. 166 00:10:08,500 --> 00:10:14,410 And also, this electric field have a phase of phi 1. 167 00:10:14,410 --> 00:10:18,310 So that's actually what we know already by now. 168 00:10:18,310 --> 00:10:20,710 What does this mean, this expression mean? 169 00:10:20,710 --> 00:10:25,540 And it's actually the harmonic oscillating electric field. 170 00:10:25,540 --> 00:10:28,960 And of course, since I am talking about interference, 171 00:10:28,960 --> 00:10:34,990 basically, I can add this, the first electric field 172 00:10:34,990 --> 00:10:37,660 and the second electric field together 173 00:10:37,660 --> 00:10:39,020 and see what will happen. 174 00:10:39,020 --> 00:10:42,300 So now, if you define the second electric field 175 00:10:42,300 --> 00:10:46,615 to be A2, which is the amplitude, cosine omega 176 00:10:46,615 --> 00:10:52,630 t minus kz, basically, they have the same wavelengths and also 177 00:10:52,630 --> 00:10:54,810 the angular frequency-- 178 00:10:54,810 --> 00:10:57,130 plus phi 2. 179 00:10:57,130 --> 00:10:59,710 But they have different phase. 180 00:10:59,710 --> 00:11:02,650 And of course, in this setup, I asked 181 00:11:02,650 --> 00:11:05,140 them to be pointing to the same direction, which 182 00:11:05,140 --> 00:11:07,240 is the x direction. 183 00:11:07,240 --> 00:11:08,920 So we were wondering, what is going 184 00:11:08,920 --> 00:11:15,510 to happen if I consider the superposition of these two 185 00:11:15,510 --> 00:11:18,600 electric fields. 186 00:11:18,600 --> 00:11:22,900 And the total electric field, which is called the E vector, 187 00:11:22,900 --> 00:11:26,110 is actually E1 plus E2. 188 00:11:29,380 --> 00:11:31,810 So before that, I would like to remind you 189 00:11:31,810 --> 00:11:37,970 about pointing vector and also the so-called intensity. 190 00:11:37,970 --> 00:11:40,510 So pointing factor is actually defined 191 00:11:40,510 --> 00:11:43,300 as this S vector, pointing vector, 192 00:11:43,300 --> 00:11:51,370 is equal to 1 over mu 0 E cross B. So this is actually 193 00:11:51,370 --> 00:11:56,290 the directional flux of energy per unit area. 194 00:11:56,290 --> 00:12:00,520 So that should be the pointing vector which 195 00:12:00,520 --> 00:12:03,720 we have been using for a while. 196 00:12:03,720 --> 00:12:09,630 And also, another reminder is that, given the electric field, 197 00:12:09,630 --> 00:12:13,780 which is actually a harmonic progressing wave, 198 00:12:13,780 --> 00:12:16,750 the corresponding B field would be 199 00:12:16,750 --> 00:12:23,800 equal to 1/v. v is actually the speed of the light 200 00:12:23,800 --> 00:12:26,800 in some specific material. 201 00:12:26,800 --> 00:12:30,180 And k hat is actually the direction 202 00:12:30,180 --> 00:12:35,940 of propagation across E. This will give you 203 00:12:35,940 --> 00:12:38,260 the magnitude and also the direction 204 00:12:38,260 --> 00:12:42,100 of the corresponding magnetic field 205 00:12:42,100 --> 00:12:46,720 for the electromagnetic waves. 206 00:12:46,720 --> 00:12:49,750 So now I'm interested in what would 207 00:12:49,750 --> 00:12:59,770 be the resulting intensity, I, if I try to superimpose this, 208 00:12:59,770 --> 00:13:05,140 to try to put together these two electric fields. 209 00:13:05,140 --> 00:13:08,200 As you can see, these two electric field 210 00:13:08,200 --> 00:13:09,370 have different phase. 211 00:13:09,370 --> 00:13:11,140 The first one has phase phi 1. 212 00:13:11,140 --> 00:13:14,270 The second one, has phase phi 2. 213 00:13:14,270 --> 00:13:18,920 So this means that they may reach maxima or minima 214 00:13:18,920 --> 00:13:23,000 at different position in space. 215 00:13:23,000 --> 00:13:25,970 In this case, it's in the z direction. 216 00:13:25,970 --> 00:13:33,710 And what I'm actually defining here is two plane waves. 217 00:13:33,710 --> 00:13:37,020 And so it really depends on what would 218 00:13:37,020 --> 00:13:41,230 be the relative phase for the first and second 219 00:13:41,230 --> 00:13:43,490 electromagnetic wave. 220 00:13:43,490 --> 00:13:48,920 In order to quantify how much they cancel each other 221 00:13:48,920 --> 00:13:53,000 or how much they enhance each other, what I'm going to do 222 00:13:53,000 --> 00:13:58,670 is to evaluate the intensity of the resulting 223 00:13:58,670 --> 00:14:00,470 electromagnetic waves. 224 00:14:00,470 --> 00:14:02,210 And what is actually intensity? 225 00:14:02,210 --> 00:14:05,310 Intensity is actually the amplitude 226 00:14:05,310 --> 00:14:07,490 of the pointing vector. 227 00:14:07,490 --> 00:14:14,170 So I write this as the length of the S vector, pointing vector. 228 00:14:14,170 --> 00:14:19,010 And I can, of course, calculate what will be the value of this 229 00:14:19,010 --> 00:14:21,440 or, say, the length of the pointing vector. 230 00:14:21,440 --> 00:14:24,530 This will be equal to 1 over mu 0, 231 00:14:24,530 --> 00:14:26,630 based on these equations here-- 232 00:14:26,630 --> 00:14:27,770 just a reminder. 233 00:14:27,770 --> 00:14:32,160 And I would like to know what would be the length of the E 234 00:14:32,160 --> 00:14:35,030 cross B field. 235 00:14:35,030 --> 00:14:38,600 That will give you the length of the pointing vector. 236 00:14:38,600 --> 00:14:47,900 And basically, what you're going to get is 1 over mu 0 times-- 237 00:14:47,900 --> 00:14:52,790 since the B field is actually highly related 238 00:14:52,790 --> 00:14:59,780 to the electric field, and it takes a hit of 1/v, right, 239 00:14:59,780 --> 00:15:01,670 in terms of the size of the amplitude. 240 00:15:01,670 --> 00:15:08,430 So basically, you're going to get 1/v, E squared. 241 00:15:12,280 --> 00:15:16,420 And this cross product actually is OK, 242 00:15:16,420 --> 00:15:21,550 because it become E squared, because B field and the E field 243 00:15:21,550 --> 00:15:24,930 are always orthogonal to each other. 244 00:15:24,930 --> 00:15:28,140 And I can now rewrite this v. v is actually 245 00:15:28,140 --> 00:15:32,830 the velocity of the speed of light in matter. 246 00:15:32,830 --> 00:15:35,120 So basically, what I can rewrite is 247 00:15:35,120 --> 00:15:42,790 that will become 1/v will become c/n, 248 00:15:42,790 --> 00:15:49,060 which is actually the refractive index of a specific material. 249 00:15:49,060 --> 00:15:52,560 And still I have E squared here. 250 00:15:52,560 --> 00:15:56,320 Finally, I can rewrite this formula, 251 00:15:56,320 --> 00:16:05,320 since c is equal to 1 over square root mu 0 epsilon 0. 252 00:16:05,320 --> 00:16:09,030 Therefore, I can rewrite this expression 253 00:16:09,030 --> 00:16:11,210 in terms of epsilon 0. 254 00:16:11,210 --> 00:16:16,190 And what I'm going to get is c times n time epsilon 0, 255 00:16:16,190 --> 00:16:19,120 There E squared. 256 00:16:19,120 --> 00:16:25,570 So basically, what I do is I multiply both numerator 257 00:16:25,570 --> 00:16:33,490 and the denominator by c, and also 258 00:16:33,490 --> 00:16:35,560 I actually use this expression. 259 00:16:35,560 --> 00:16:37,990 Then I can actually cancel the mu 0 260 00:16:37,990 --> 00:16:44,380 and then write everything in terms of c, n, and epsilon 0. 261 00:16:44,380 --> 00:16:47,750 So until here, there was basically no magic. 262 00:16:47,750 --> 00:16:52,290 Basically, it's just rewriting the length 263 00:16:52,290 --> 00:16:58,570 of the pointing vector in terms of n and also 264 00:16:58,570 --> 00:17:01,640 the electric field. 265 00:17:01,640 --> 00:17:06,829 So now, what I am going to do is to plug in this expression 266 00:17:06,829 --> 00:17:11,270 into that formula and see what we are going to get. 267 00:17:11,270 --> 00:17:16,163 So let me evaluate what would be the E squared, 268 00:17:16,163 --> 00:17:17,329 the length of the E squared. 269 00:17:17,329 --> 00:17:22,099 So basically, the definition of the E is shown here. 270 00:17:22,099 --> 00:17:25,760 It's the superposition of E1 and E2. 271 00:17:25,760 --> 00:17:28,910 Therefore, I can now quickly write down 272 00:17:28,910 --> 00:17:31,430 what would be E squared here. 273 00:17:35,450 --> 00:17:42,140 Basically, you are going to get A1 squared, cosine squared, 274 00:17:42,140 --> 00:17:47,660 omega t minus kz plus phi 1. 275 00:17:47,660 --> 00:17:53,300 Basically, that is coming from the E1 times E1. 276 00:17:53,300 --> 00:17:55,540 The second term, which I'm going to get, 277 00:17:55,540 --> 00:17:59,040 is actually E2 times E2. 278 00:17:59,040 --> 00:18:03,650 E2 times E3, you are going to get A2 squared, 279 00:18:03,650 --> 00:18:12,360 cosine squared, omega t minus kz plus phi 2. 280 00:18:12,360 --> 00:18:15,890 Basically based on this equation, and I square it, 281 00:18:15,890 --> 00:18:19,890 and then I get the second part here. 282 00:18:19,890 --> 00:18:22,340 And finally, what I am going to get 283 00:18:22,340 --> 00:18:25,360 is the third term, which is actually E1 times E2. 284 00:18:30,920 --> 00:18:36,995 Basically, you are going to get 2 times A1 A2 cosine omega 285 00:18:36,995 --> 00:18:47,201 t minus kz plus phi 1 cosine omega t minus kz 286 00:18:47,201 --> 00:18:53,150 plus phi 2, which is actually the cross term of this E 287 00:18:53,150 --> 00:18:55,420 vector squared. 288 00:18:55,420 --> 00:18:56,360 Any questions so far? 289 00:18:59,710 --> 00:19:02,830 I hope this is pretty straightforward to you. 290 00:19:02,830 --> 00:19:06,610 And of course, I can now rewrite this product 291 00:19:06,610 --> 00:19:09,740 with 2 cosine, cosine times cosine, right? 292 00:19:09,740 --> 00:19:11,920 Basically, I can rewrite this using 293 00:19:11,920 --> 00:19:14,960 the formula, which we have related 294 00:19:14,960 --> 00:19:16,730 to a cosine times a cosine. 295 00:19:16,730 --> 00:19:27,620 Basically, I can rewrite this as 1/2 cosine 2 omega t minus 2kz 296 00:19:27,620 --> 00:19:37,510 plus phi 1 plus phi 2 plus 1/2 cosine phi 1 minus phi 2. 297 00:19:37,510 --> 00:19:39,850 So basically, the first term is actually 298 00:19:39,850 --> 00:19:42,400 collecting the content of the two cosine 299 00:19:42,400 --> 00:19:44,560 and their length together. 300 00:19:44,560 --> 00:19:47,920 The second term is actually calculating 301 00:19:47,920 --> 00:19:52,240 the difference between the content of the cosine function. 302 00:19:52,240 --> 00:19:55,770 And what I'm going to get is actually phi 1 minus phi 2. 303 00:19:58,990 --> 00:20:03,910 Now, based on this definition, intensity, 304 00:20:03,910 --> 00:20:07,800 I is actually equal to the magnitude of that 305 00:20:07,800 --> 00:20:09,010 pointing vector. 306 00:20:09,010 --> 00:20:11,290 So remember, our goal is to evaluate 307 00:20:11,290 --> 00:20:15,860 what will be the resulting average intensity. 308 00:20:15,860 --> 00:20:20,740 So now I can calculate what would be the average intensity 309 00:20:20,740 --> 00:20:23,260 over one period. 310 00:20:23,260 --> 00:20:29,310 So this will be equal to 1/T, integration over 0 311 00:20:29,310 --> 00:20:38,830 to T, one period, and the instantaneous intensity, I, dt. 312 00:20:41,572 --> 00:20:44,170 And what I am going to get is that-- 313 00:20:44,170 --> 00:20:46,090 so we have three terms. 314 00:20:46,090 --> 00:20:49,480 The first term is here, which is actually A1 squared, 315 00:20:49,480 --> 00:20:51,200 cosine squared. 316 00:20:51,200 --> 00:20:54,400 It's actually related to omega t. 317 00:20:54,400 --> 00:20:56,530 And the second term is here. 318 00:20:56,530 --> 00:21:01,570 It's also proportional to cosine squared omega t. 319 00:21:01,570 --> 00:21:03,850 And finally, we have two terms here, 320 00:21:03,850 --> 00:21:09,780 which is actually proportional to cosine 2 omega t. 321 00:21:09,780 --> 00:21:16,210 And finally, the last term is actually independent of time. 322 00:21:16,210 --> 00:21:22,620 So what I'm going to do is to evaluate the individual terms. 323 00:21:22,620 --> 00:21:24,880 For the first term, basically, A1 324 00:21:24,880 --> 00:21:27,610 squared, cosine squared, omega t. 325 00:21:27,610 --> 00:21:32,290 So by now, it should be pretty straightforward for you 326 00:21:32,290 --> 00:21:36,560 if I integrate cosine squared over one period of time, 327 00:21:36,560 --> 00:21:40,210 basically, what you are going to get is 1/2. 328 00:21:40,210 --> 00:21:45,220 So is actually done several times in the p set. 329 00:21:45,220 --> 00:21:48,160 So basically, what you're going to get is-- 330 00:21:48,160 --> 00:21:51,550 I am going to collect all of those constants from here 331 00:21:51,550 --> 00:21:52,300 and copy here. 332 00:21:52,300 --> 00:21:57,500 So basically, you have c times n times epsilon 0, 333 00:21:57,500 --> 00:22:05,110 which is actually coming from the definition of intensity. 334 00:22:05,110 --> 00:22:10,630 And then for the first term, what I am going to get 335 00:22:10,630 --> 00:22:14,250 is A1 squared divided by 2. 336 00:22:14,250 --> 00:22:17,220 This 1/2 is actually just an integral 337 00:22:17,220 --> 00:22:19,630 related to cosine squared. 338 00:22:19,630 --> 00:22:23,960 Similarly, you are going to get the same result, a very similar 339 00:22:23,960 --> 00:22:26,350 result, for the second term. 340 00:22:26,350 --> 00:22:31,610 The second term is going to give you A2 squared divided by 2. 341 00:22:35,440 --> 00:22:38,850 Finally, you can have the third term. 342 00:22:38,850 --> 00:22:43,040 The third turn is going to give you what value? 343 00:22:43,040 --> 00:22:45,494 Can somebody help me? 344 00:22:45,494 --> 00:22:46,370 AUDIENCE: 0. 345 00:22:46,370 --> 00:22:47,500 YEN-JIE LEE: Yes, 0, right? 346 00:22:47,500 --> 00:22:50,450 Because this is actually cosine 2 omega t, right? 347 00:22:50,450 --> 00:22:52,900 So if you integrate over one period, 348 00:22:52,900 --> 00:22:54,540 you are going to get 0 plus 0. 349 00:22:54,540 --> 00:22:56,940 Each period will give you 0, right? 350 00:22:56,940 --> 00:22:59,410 So 0 plus 0 is 0, so therefore you get 0. 351 00:22:59,410 --> 00:23:01,030 Very good. 352 00:23:01,030 --> 00:23:07,042 How about the last term, anybody can help me? 353 00:23:07,042 --> 00:23:08,797 AUDIENCE: It should remain as it is. 354 00:23:08,797 --> 00:23:09,880 YEN-JIE LEE: That's right. 355 00:23:09,880 --> 00:23:11,680 Because it's a constant. 356 00:23:11,680 --> 00:23:16,040 So the average of a constant is a constant, which is actually 357 00:23:16,040 --> 00:23:27,080 giving you 1/2 times 2 times A1 times A2 cosine 358 00:23:27,080 --> 00:23:29,470 phi 1 minus phi 2. 359 00:23:29,470 --> 00:23:32,680 Of course, I can cancel this 1/2, which 360 00:23:32,680 --> 00:23:35,200 is actually coming from here, and the 2, 361 00:23:35,200 --> 00:23:36,742 which is coming from here. 362 00:23:36,742 --> 00:23:43,200 And basically, you are getting A1 A2 cosine phi 1 minus phi 2. 363 00:23:43,200 --> 00:23:47,250 And I need to close this bracket. 364 00:23:47,250 --> 00:23:49,790 Any questions so far? 365 00:23:49,790 --> 00:23:54,100 So what we have been doing is that I evaluated 366 00:23:54,100 --> 00:23:57,530 the total electric field. 367 00:23:57,530 --> 00:23:59,740 I basically calculated the superposition 368 00:23:59,740 --> 00:24:01,030 of the two fields. 369 00:24:01,030 --> 00:24:05,620 And then I am interested in what would be the average intensity 370 00:24:05,620 --> 00:24:07,630 coming from this field. 371 00:24:07,630 --> 00:24:11,000 And I write down E squared explicitly. 372 00:24:11,000 --> 00:24:15,490 There are four terms and only three of them actually survive. 373 00:24:15,490 --> 00:24:21,550 And basically, the expression I'm getting is like this. 374 00:24:21,550 --> 00:24:24,280 Basically, you have some constant multiplied 375 00:24:24,280 --> 00:24:27,790 by A1 squared over 2 plus A2 squared over 2 376 00:24:27,790 --> 00:24:32,570 plus A1 A2 cosine phi 1 minus phi 2. 377 00:24:32,570 --> 00:24:37,960 You can see that the intensity depends on phi 1 and phi 2, 378 00:24:37,960 --> 00:24:40,100 right? 379 00:24:40,100 --> 00:24:44,320 So this actually would change the resulting intensity. 380 00:24:44,320 --> 00:24:48,670 So in order to get some idea about what does that mean 381 00:24:48,670 --> 00:24:53,800 and also how does the average intensity change 382 00:24:53,800 --> 00:24:58,060 as a function of phi 1 minus phi2, what I'm going to do 383 00:24:58,060 --> 00:25:03,220 is to define phi 1 minus phi 2 to be delta, which 384 00:25:03,220 --> 00:25:06,100 I will call phase difference. 385 00:25:06,100 --> 00:25:11,890 Then I would like to plot the averaging intensity, I, 386 00:25:11,890 --> 00:25:17,280 as a function of delta and see what's going to happen. 387 00:25:17,280 --> 00:25:19,420 So this is actually the result. So if I 388 00:25:19,420 --> 00:25:22,840 have the x-axis to be delta, which 389 00:25:22,840 --> 00:25:29,780 is actually phi 1 minus phi 2, and the y-axis is intensity. 390 00:25:29,780 --> 00:25:33,190 Of course, I would like to take out the constant, which 391 00:25:33,190 --> 00:25:36,520 is c times n times epsilon 0. 392 00:25:36,520 --> 00:25:41,740 So I am plotting the y-axis' average intensity divided 393 00:25:41,740 --> 00:25:46,150 by c times n times epsilon 0. 394 00:25:46,150 --> 00:25:48,210 What I'm going to get is something 395 00:25:48,210 --> 00:25:52,690 which is actually oscillating up and down, like this. 396 00:26:00,070 --> 00:26:06,895 The maxima value happens when delta is equal to 0. 397 00:26:09,670 --> 00:26:13,840 When delta is equal to 0, what is going to happen? 398 00:26:13,840 --> 00:26:18,715 This means that cosine delta is equal to what? 399 00:26:18,715 --> 00:26:23,020 , Therefore what you are going to get is A1 squared over 2 400 00:26:23,020 --> 00:26:27,000 plus A2 squared over 2 plus A1 A2. 401 00:26:37,770 --> 00:26:41,620 This is actually when there delta is equal to 0. 402 00:26:41,620 --> 00:26:44,860 And the intensity, as you reach maxima, 403 00:26:44,860 --> 00:26:48,640 and the maxima values is actually 1/2 404 00:26:48,640 --> 00:26:55,870 A1 plus A2 squared, based on these calculations. 405 00:26:55,870 --> 00:26:58,860 On the other hand, you can expect 406 00:26:58,860 --> 00:27:02,790 that that intensity will reach a minima when 407 00:27:02,790 --> 00:27:06,220 delta is equal to which value? 408 00:27:06,220 --> 00:27:09,349 Anybody can help me. 409 00:27:09,349 --> 00:27:09,890 AUDIENCE: Pi. 410 00:27:09,890 --> 00:27:11,380 YEN-JIE LEE: Pi, yes. 411 00:27:11,380 --> 00:27:14,020 When delta is pi, what is going to happen? 412 00:27:14,020 --> 00:27:17,320 Cosine pi is minus 1. 413 00:27:17,320 --> 00:27:21,280 So therefore, what you are getting is 1/2 A1 414 00:27:21,280 --> 00:27:26,740 squared plus A2 squared minus 2 A1 A2. 415 00:27:26,740 --> 00:27:35,300 And that will give you 1/2 A1 minus A2 squared. 416 00:27:35,300 --> 00:27:40,960 You can see that, when the filter is equal to 0 417 00:27:40,960 --> 00:27:46,660 or when the filter is equal to 2 pi, for example, 418 00:27:46,660 --> 00:27:51,440 if you increase the phase difference large enough, 419 00:27:51,440 --> 00:27:56,890 or the filter is actually 4 pi, all of those number 420 00:27:56,890 --> 00:28:02,530 will keep you maxima constructive interference. 421 00:28:02,530 --> 00:28:03,490 So what does that mean? 422 00:28:03,490 --> 00:28:08,650 That means you are adding these two electric fields 423 00:28:08,650 --> 00:28:12,550 in the most efficient way. 424 00:28:12,550 --> 00:28:18,910 On the other hand, when the value of the filter 425 00:28:18,910 --> 00:28:26,800 is equal to pi or equal to 3 pi or equal to 5 pi, et cetera, 426 00:28:26,800 --> 00:28:31,390 the intensity, the average intensity reaches a minima. 427 00:28:31,390 --> 00:28:34,930 That means, instead of adding them, 428 00:28:34,930 --> 00:28:37,240 you are actually canceling them. 429 00:28:37,240 --> 00:28:40,330 You are canceling the electric field 430 00:28:40,330 --> 00:28:46,390 of the first and the second electromagnetic wave. 431 00:28:46,390 --> 00:28:51,730 And now I give you a maxima intensity, which is 1/2 A1 432 00:28:51,730 --> 00:28:53,860 minus A2 squared. 433 00:28:53,860 --> 00:28:57,610 Just a reminder, this A1 and A2 is actually 434 00:28:57,610 --> 00:29:02,410 the amplitude of the first and second electric field. 435 00:29:02,410 --> 00:29:07,170 So what will happen if I set A1 equal to A2? 436 00:29:10,110 --> 00:29:14,130 If I set A1 equal to A2, that means the minima 437 00:29:14,130 --> 00:29:15,570 would be equal to what? 438 00:29:15,570 --> 00:29:16,350 AUDIENCE: 0. 439 00:29:16,350 --> 00:29:18,330 YEN-JIE LEE: 0, yeah, very good. 440 00:29:18,330 --> 00:29:20,956 How about the maxima? 441 00:29:20,956 --> 00:29:21,830 AUDIENCE: [INAUDIBLE] 442 00:29:21,830 --> 00:29:25,240 YEN-JIE LEE: It will be A1 plus A2, right? 443 00:29:25,240 --> 00:29:30,130 So basically, you are going to get four times larger value 444 00:29:30,130 --> 00:29:35,330 compared to the intensity before you add them together. 445 00:29:35,330 --> 00:29:41,030 So individual intensity is I. And after adding them together, 446 00:29:41,030 --> 00:29:43,120 with delta equal to 0, you are going 447 00:29:43,120 --> 00:29:46,660 to get four times larger intensity 448 00:29:46,660 --> 00:29:50,980 if the amplitude of the first and second electric field 449 00:29:50,980 --> 00:29:53,650 is the same. 450 00:29:53,650 --> 00:29:55,500 So very good. 451 00:29:55,500 --> 00:29:58,310 So that's actually the result of the calculation. 452 00:29:58,310 --> 00:30:03,370 And you can see that the amount of intensity we can get out 453 00:30:03,370 --> 00:30:07,770 of this highly depends on the filter, which is actually 454 00:30:07,770 --> 00:30:09,790 the phase difference between the first 455 00:30:09,790 --> 00:30:12,650 and the second electric field. 456 00:30:12,650 --> 00:30:17,290 Can we actually get some more feeling about this addition? 457 00:30:17,290 --> 00:30:20,980 So what I am going to do is to, again, 458 00:30:20,980 --> 00:30:27,460 write everything down in terms of imaginary number 459 00:30:27,460 --> 00:30:30,160 or, say, a complex number. 460 00:30:30,160 --> 00:30:38,220 So if I rewrite the electric field, E1, 461 00:30:38,220 --> 00:30:45,570 as a real part of A1 exponential i phi 1, 462 00:30:45,570 --> 00:30:51,400 exponential i omega t minus kz. 463 00:30:51,400 --> 00:30:54,340 In the x direction. 464 00:30:54,340 --> 00:30:58,240 And I can also rewrite the expression 465 00:30:58,240 --> 00:31:00,730 for the second electric field to be 466 00:31:00,730 --> 00:31:07,225 the real part of A2 exponential i phi 2, exponential omega 467 00:31:07,225 --> 00:31:13,180 t minus kz, again, in the x direction 468 00:31:13,180 --> 00:31:18,430 So if I add these two fields together, what I am doing 469 00:31:18,430 --> 00:31:24,520 is like in the complex plane I have an imaginary number 470 00:31:24,520 --> 00:31:27,310 contribution in the y direction. 471 00:31:27,310 --> 00:31:32,580 And the real part is actually in the x direction. 472 00:31:32,580 --> 00:31:36,890 Suppose omega t minus kz is 0. 473 00:31:36,890 --> 00:31:41,260 At some instant of time omega t minus kz is equal to 0. 474 00:31:41,260 --> 00:31:46,090 So what I'm doing is I have the first vector, which is actually 475 00:31:46,090 --> 00:31:52,810 presenting the contribution of the first electric field. 476 00:31:52,810 --> 00:31:58,460 And this electric field is going to be pointing to a direction 477 00:31:58,460 --> 00:32:05,200 phi 1 away from the real axis with amplitude equal to A1. 478 00:32:05,200 --> 00:32:08,860 This is actually what we learned from the first lecture. 479 00:32:08,860 --> 00:32:13,560 And if I add the second electric field, what I'm going to get 480 00:32:13,560 --> 00:32:21,820 is another vector, which is actually A2 in length. 481 00:32:21,820 --> 00:32:29,320 And the angle is phi 2, here. 482 00:32:29,320 --> 00:32:35,560 So the resulting amplitude is actually 483 00:32:35,560 --> 00:32:42,570 when I take the real part of the first and second expression, 484 00:32:42,570 --> 00:32:44,950 adding them together. 485 00:32:44,950 --> 00:32:51,220 Basically, I am taking a projection to the real axis. 486 00:32:51,220 --> 00:32:56,800 And that is actually the resulting amplitude 487 00:32:56,800 --> 00:33:02,050 of the electric field, which is actually the superposition 488 00:33:02,050 --> 00:33:04,330 of the first and second field. 489 00:33:04,330 --> 00:33:16,150 So you can see that, when phi 1 is 490 00:33:16,150 --> 00:33:21,710 equal to phi 2, what is going to happen is the following. 491 00:33:21,710 --> 00:33:24,640 So basically, what you're actually going to get 492 00:33:24,640 --> 00:33:29,260 is that you are increasing the length 493 00:33:29,260 --> 00:33:35,710 of the routing vector, which is the addition of the two 494 00:33:35,710 --> 00:33:37,580 vectors. 495 00:33:37,580 --> 00:33:42,310 You are actually getting a maxima out of this addition. 496 00:33:42,310 --> 00:33:45,790 Because phi 1 is equal to phi 2, therefore, these two vectors 497 00:33:45,790 --> 00:33:49,060 form a straight line, therefore, you 498 00:33:49,060 --> 00:33:53,950 can actually add and get maximum amount 499 00:33:53,950 --> 00:33:59,080 of the amplitude out of this. 500 00:33:59,080 --> 00:34:05,300 On the other hand, when phi 1 minus phi 2 501 00:34:05,300 --> 00:34:14,949 is pi, which I define as delta, when this happens, 502 00:34:14,949 --> 00:34:19,510 what we are doing is like addition of two vectors, 503 00:34:19,510 --> 00:34:23,409 in a complex plane, but they are pointing 504 00:34:23,409 --> 00:34:25,870 to the opposite direction. 505 00:34:25,870 --> 00:34:27,790 So the first one will be like this. 506 00:34:27,790 --> 00:34:31,630 And the second one will be looking like that. 507 00:34:31,630 --> 00:34:35,260 And they are actually trying to cancel each other. 508 00:34:35,260 --> 00:34:39,300 So that's actually how you can understand what is happening 509 00:34:39,300 --> 00:34:41,310 with different delta values. 510 00:34:41,310 --> 00:34:44,330 In this case, delta is equal to 0. 511 00:34:44,330 --> 00:34:46,960 The phase difference is equal to 0. 512 00:34:46,960 --> 00:34:51,159 And in the second case, phase difference is equal to pi. 513 00:34:51,159 --> 00:34:55,790 And then what happened in between is like this. 514 00:34:55,790 --> 00:34:58,730 You are adding them sort of together but not 515 00:34:58,730 --> 00:35:02,830 in the most efficient way or the most destructive way. 516 00:35:02,830 --> 00:35:05,290 And you are actually evaluating what 517 00:35:05,290 --> 00:35:08,830 will be the resulting amplitude by looking 518 00:35:08,830 --> 00:35:13,120 at the vector sum of the first and second field. 519 00:35:13,120 --> 00:35:16,180 So I hope that this will give you a some more intuition 520 00:35:16,180 --> 00:35:19,450 about what we have been doing. 521 00:35:19,450 --> 00:35:20,380 Any questions so far? 522 00:35:24,860 --> 00:35:29,150 So now, we are actually in a very good position 523 00:35:29,150 --> 00:35:32,960 once we understand this superposition 524 00:35:32,960 --> 00:35:36,410 of the two electric fields and the interference. 525 00:35:36,410 --> 00:35:40,160 Basically, the size of the resulting intensity 526 00:35:40,160 --> 00:35:43,070 will be highly dependent on the phase difference 527 00:35:43,070 --> 00:35:45,170 between the two fields. 528 00:35:45,170 --> 00:35:47,630 Then we are in a very good position 529 00:35:47,630 --> 00:35:53,270 to discuss the phenomenon which we just see in the demo. 530 00:35:53,270 --> 00:35:58,730 So before I actually perform the calculation 531 00:35:58,730 --> 00:36:00,590 and give you the explanation, I would like 532 00:36:00,590 --> 00:36:03,560 to take a vote, as of usual. 533 00:36:03,560 --> 00:36:08,390 So the question we are asking is, in addition 534 00:36:08,390 --> 00:36:11,630 to what we see in the demo-- 535 00:36:11,630 --> 00:36:16,310 we see a colorful bubble-- 536 00:36:16,310 --> 00:36:22,550 how thick is the soap film such that you can see color 537 00:36:22,550 --> 00:36:25,070 from the reflected light? 538 00:36:25,070 --> 00:36:28,850 The first option is maybe it's like 1 millimeter, which 539 00:36:28,850 --> 00:36:30,290 is possible. 540 00:36:30,290 --> 00:36:35,090 And that is about the size of the head of a pin. 541 00:36:35,090 --> 00:36:40,580 Or it can be 100 micron, so that's actually about the size, 542 00:36:40,580 --> 00:36:45,020 the thickness is about the size of the human hair. 543 00:36:45,020 --> 00:36:50,930 Or 100 nanometer, which is the size of the virus. 544 00:36:50,930 --> 00:36:54,410 How many of you think the thickness 545 00:36:54,410 --> 00:36:56,990 is roughly 1 millimeter? 546 00:36:56,990 --> 00:36:59,700 Raise your hand. 547 00:36:59,700 --> 00:37:01,447 Nobody thinks so. 548 00:37:01,447 --> 00:37:01,946 Really? 549 00:37:08,520 --> 00:37:11,460 Actually nobody think that's the case. 550 00:37:11,460 --> 00:37:14,090 How about 100 micron? 551 00:37:14,090 --> 00:37:15,680 How many of you think so? 552 00:37:28,680 --> 00:37:31,250 How about 100 nanometer? 553 00:37:31,250 --> 00:37:32,998 Me How many of you? 554 00:37:44,400 --> 00:37:47,140 So that is actually the vote. 555 00:37:47,140 --> 00:37:49,570 And we are going to know the result very soon. 556 00:37:49,570 --> 00:37:51,190 And how about the rest? 557 00:37:53,950 --> 00:37:54,770 Cool. 558 00:37:54,770 --> 00:37:58,900 So now we are going to solve the puzzle. 559 00:37:58,900 --> 00:38:03,670 So just a quick reminder about what 560 00:38:03,670 --> 00:38:06,620 we have learned from the last lecture. 561 00:38:06,620 --> 00:38:11,090 So there is a reason why we have the lecture first 562 00:38:11,090 --> 00:38:15,410 on the reflection of an electromagnetic wave before we 563 00:38:15,410 --> 00:38:19,560 discuss the color of the bubble. 564 00:38:19,560 --> 00:38:26,200 So from the last lecture, suppose 565 00:38:26,200 --> 00:38:33,080 I have two materials, which form an interface between material 566 00:38:33,080 --> 00:38:36,640 number 1, with refractive index n1, 567 00:38:36,640 --> 00:38:41,760 and the second material has a refractive index n2. 568 00:38:41,760 --> 00:38:45,020 If I have an incident wave, incident 569 00:38:45,020 --> 00:38:50,670 electromagnetic plane wave, and the incident angle 570 00:38:50,670 --> 00:38:57,960 is actually, in this case, 0, that means this incident plane 571 00:38:57,960 --> 00:39:02,680 wave is actually propagating in a direction which is actually 572 00:39:02,680 --> 00:39:05,800 hitting the surface directly. 573 00:39:05,800 --> 00:39:08,950 So if the initial amplitude is A, 574 00:39:08,950 --> 00:39:11,140 what we have learned from last time 575 00:39:11,140 --> 00:39:13,780 is that there will be a reflective wave, which 576 00:39:13,780 --> 00:39:17,620 is actually R times A. R is actually 577 00:39:17,620 --> 00:39:21,970 reflective coefficient. 578 00:39:21,970 --> 00:39:25,850 And finally, you have also the transmitted wave, 579 00:39:25,850 --> 00:39:34,510 which I call T times A, where is the transmission coefficient. 580 00:39:34,510 --> 00:39:38,380 From the exercise, which we actually already 581 00:39:38,380 --> 00:39:47,270 done last time, R is equal to n1 minus n2 divided by n1 plus n2. 582 00:39:47,270 --> 00:39:51,010 And the transmission coefficient is actually 583 00:39:51,010 --> 00:39:58,630 T equal to 2n1 divided by n1 plus n2. 584 00:39:58,630 --> 00:40:01,870 So basically, what I am actually talking about here 585 00:40:01,870 --> 00:40:04,150 is a conclusion from the exercise 586 00:40:04,150 --> 00:40:10,960 we have done in the last lecture, just a quick reminder. 587 00:40:10,960 --> 00:40:17,500 So I would like to discuss with you various situation related 588 00:40:17,500 --> 00:40:19,970 to R value. 589 00:40:19,970 --> 00:40:24,730 So the n1 and the n2 are related to the property 590 00:40:24,730 --> 00:40:27,790 of the first medium and the second medium. 591 00:40:27,790 --> 00:40:36,700 So it could be that n1 is actually greater than n2. 592 00:40:36,700 --> 00:40:42,760 So if n1 is greater than n2 in the experimental setup, 593 00:40:42,760 --> 00:40:48,370 that means that R will be greater than 0. 594 00:40:51,050 --> 00:40:52,940 Because R is actually n1 minus n2 595 00:40:52,940 --> 00:40:57,090 divided by the sum of n1 and n2. 596 00:40:57,090 --> 00:41:01,700 Therefore, what I'm going to get is something like this. 597 00:41:01,700 --> 00:41:04,750 So basically, I'm going to have an incident 598 00:41:04,750 --> 00:41:09,580 wave like this, where, say, I use the notation pointing 599 00:41:09,580 --> 00:41:12,250 upwards. 600 00:41:12,250 --> 00:41:19,690 Once they got reflected, it is actually still like this, 601 00:41:19,690 --> 00:41:25,070 pointing upward, because the R is actually greater than 0. 602 00:41:25,070 --> 00:41:29,830 There's no changing sign in the amplitude. 603 00:41:29,830 --> 00:41:41,582 Therefore, there's an no flip in amplitude 604 00:41:41,582 --> 00:41:46,960 if n1 is actually greater than n2. 605 00:41:46,960 --> 00:41:50,080 On the other hand, the transmitted wave, 606 00:41:50,080 --> 00:41:54,530 if you look at the functional form of the transmitted wave, 607 00:41:54,530 --> 00:41:57,400 and transmission coefficient, T is actually 608 00:41:57,400 --> 00:42:02,050 equal to 2n1 divided by n1 plus n2. 609 00:42:02,050 --> 00:42:05,900 It's always positive. 610 00:42:05,900 --> 00:42:11,370 Therefore, will there be any possibility to flip the sign? 611 00:42:11,370 --> 00:42:12,400 No. 612 00:42:12,400 --> 00:42:14,110 You are absolutely right. 613 00:42:14,110 --> 00:42:16,750 So therefore, what is going to happen 614 00:42:16,750 --> 00:42:20,020 is that I will use this little arrow to keep 615 00:42:20,020 --> 00:42:23,260 track of the sign change. 616 00:42:23,260 --> 00:42:27,340 Basically, you'll see that after it pass through the boundary, 617 00:42:27,340 --> 00:42:32,290 there will be no change in sign in amplitude no matter 618 00:42:32,290 --> 00:42:33,910 what happens. 619 00:42:33,910 --> 00:42:41,230 On the other hand, if I have the situation n1 smaller than n2, 620 00:42:41,230 --> 00:42:43,140 what is going to happen? 621 00:42:43,140 --> 00:42:51,520 If you calculate the R value, it will be negative, right? 622 00:42:51,520 --> 00:42:54,890 In this case, R will be smaller than 0. 623 00:42:54,890 --> 00:43:03,380 So what is going to happen is that, initially, the incident 624 00:43:03,380 --> 00:43:06,130 wave has positive amplitude. 625 00:43:06,130 --> 00:43:08,710 And I keep track of the sign of this amplitude 626 00:43:08,710 --> 00:43:13,120 by this arrow pointing up. 627 00:43:13,120 --> 00:43:17,080 Because the R is actually smaller than 0, 628 00:43:17,080 --> 00:43:23,080 therefore, there is flip in sign in the amplitude. 629 00:43:23,080 --> 00:43:24,770 So what is going to happen? 630 00:43:24,770 --> 00:43:26,620 So the reflective wave will look like this. 631 00:43:26,620 --> 00:43:32,470 And I use this arrow to keep track of the flip in amplitude. 632 00:43:32,470 --> 00:43:35,540 And finally, as I mentioned before, 633 00:43:35,540 --> 00:43:39,890 the transmitted wave, the T, is always positive. 634 00:43:39,890 --> 00:43:46,780 Therefore, there will be no change in sign in amplitude. 635 00:43:46,780 --> 00:43:49,390 Finally, the third example is, if I 636 00:43:49,390 --> 00:43:51,690 have somehow two different materials, 637 00:43:51,690 --> 00:43:55,220 but they have the same refractive index, what 638 00:43:55,220 --> 00:43:58,540 is going to happen is that there will be no reflection, 639 00:43:58,540 --> 00:44:01,240 and everything goes through. 640 00:44:01,240 --> 00:44:05,560 Even if you have two different kinds of material, but if they 641 00:44:05,560 --> 00:44:09,430 have the same refractive index, then what is going to happen 642 00:44:09,430 --> 00:44:11,240 is that everything will pass through. 643 00:44:11,240 --> 00:44:13,450 And what you are going to get is that you 644 00:44:13,450 --> 00:44:16,230 will have no reflected light. 645 00:44:16,230 --> 00:44:20,610 Meaning R is actually equal to 0. 646 00:44:20,610 --> 00:44:21,850 Any questions so far? 647 00:44:26,300 --> 00:44:29,840 I would like to make sure that everybody 648 00:44:29,840 --> 00:44:34,440 understands the consequence of this calculation. 649 00:44:34,440 --> 00:44:39,590 So if I introduce no flip in amplitude, 650 00:44:39,590 --> 00:44:42,350 this means that this contribution will 651 00:44:42,350 --> 00:44:46,310 introduce a filter equal to 0. 652 00:44:46,310 --> 00:44:51,140 So basically, there will be no change in the phase, 653 00:44:51,140 --> 00:44:54,830 because there's no flip in amplitude. 654 00:44:54,830 --> 00:45:02,120 On the other hand, if there's a changing sign in amplitude, 655 00:45:02,120 --> 00:45:05,050 what would be the resulting filter value? 656 00:45:05,050 --> 00:45:06,870 Can somebody actually tell me? 657 00:45:06,870 --> 00:45:07,810 AUDIENCE: Pi. 658 00:45:07,810 --> 00:45:09,240 YEN-JIE LEE: It would be pi. 659 00:45:09,240 --> 00:45:10,030 Very good. 660 00:45:10,030 --> 00:45:13,600 So that means you are getting hit 661 00:45:13,600 --> 00:45:16,660 by a phase difference of pi. 662 00:45:16,660 --> 00:45:20,650 Therefore, the amplitude changes by a factor 663 00:45:20,650 --> 00:45:25,300 of cosine pi, which is minus 1. 664 00:45:25,300 --> 00:45:28,450 So that is actually something pretty important 665 00:45:28,450 --> 00:45:38,990 when we have the discussion of the soap bubble reflection. 666 00:45:38,990 --> 00:45:43,450 So let me give you a quick example 667 00:45:43,450 --> 00:45:48,160 about why is actually the amount of the reflected light and also 668 00:45:48,160 --> 00:45:51,670 what is the amount of transmitted light. 669 00:45:51,670 --> 00:45:53,980 Let me give you a concrete example. 670 00:45:53,980 --> 00:46:01,660 For example, if I have n1 equal to 1, which is actually 671 00:46:01,660 --> 00:46:12,070 the refractive index of the air, and n2 equal to 1.5 672 00:46:12,070 --> 00:46:19,210 If that happens what is the resulting intensity? 673 00:46:19,210 --> 00:46:25,360 Just a quick reminder, average I, the average intensity 674 00:46:25,360 --> 00:46:33,680 will be equal to c times n times epsilon 0 A 675 00:46:33,680 --> 00:46:38,470 squared divided by 2, where A is the amplitude 676 00:46:38,470 --> 00:46:42,190 of the electric field. 677 00:46:42,190 --> 00:46:43,420 Just a quick reminder. 678 00:46:43,420 --> 00:46:48,370 And this 1/2 is coming from the time average, just a reminder. 679 00:46:48,370 --> 00:46:53,470 So now I can go ahead and use these two formula, R and T, 680 00:46:53,470 --> 00:46:57,790 to calculate the reflection coefficient 681 00:46:57,790 --> 00:46:59,650 and this transmission coefficient. 682 00:46:59,650 --> 00:47:07,510 So R will equal to 1 minus 1.5 divided by 1 plus 1.5. 683 00:47:07,510 --> 00:47:15,100 So basically, you get minus 0.5 divided by 2.5. 684 00:47:15,100 --> 00:47:23,560 And that is actually going to give you minus 0.2. 685 00:47:23,560 --> 00:47:25,410 Of course, I can also calculate what 686 00:47:25,410 --> 00:47:33,220 will be the T, which would be a 2 divided 2.5. 687 00:47:33,220 --> 00:47:39,970 So basically what you are getting is 0.8. 688 00:47:39,970 --> 00:47:47,020 So I can now calculate what will be the resulting intensity 689 00:47:47,020 --> 00:47:49,190 of the reflected light. 690 00:47:49,190 --> 00:47:52,150 Everybody's following? 691 00:47:52,150 --> 00:47:58,340 So what would be the intensity of the reflected light? 692 00:47:58,340 --> 00:48:06,090 This will be equal to minus 0.2 squared, right? 693 00:48:06,090 --> 00:48:12,330 Because the average intensity is proportional to A squared. 694 00:48:12,330 --> 00:48:16,410 A is actually the amplitude of the electric field. 695 00:48:16,410 --> 00:48:19,020 R should tell you what is actually 696 00:48:19,020 --> 00:48:22,390 the relative amplitude between the reflected light 697 00:48:22,390 --> 00:48:25,660 and the incident light. 698 00:48:25,660 --> 00:48:31,480 Therefore, you are getting hit by 0.2 squared multiplied 699 00:48:31,480 --> 00:48:37,130 by the initial intensity. 700 00:48:37,130 --> 00:48:39,190 Basically what you are going to get 701 00:48:39,190 --> 00:48:47,210 is 0.04 I, initial intensity. 702 00:48:47,210 --> 00:48:55,040 So basically 4% of the light is reflected. 703 00:48:55,040 --> 00:48:57,260 That may surprise you a bit, right? 704 00:48:57,260 --> 00:49:01,610 Because when you see, for example, the soap bubble, 705 00:49:01,610 --> 00:49:04,040 you see that it is still pretty bright, right? 706 00:49:04,040 --> 00:49:08,615 But in reality, only 4% of the light or 4% of the intensity 707 00:49:08,615 --> 00:49:10,610 got reflected. 708 00:49:10,610 --> 00:49:15,090 That is because your eyes is actually having nonlinear. 709 00:49:15,090 --> 00:49:17,315 Your eye responds to the-- 710 00:49:17,315 --> 00:49:22,130 or, say, receiving or interpreting the intensity 711 00:49:22,130 --> 00:49:24,380 is really highly nonlinear. 712 00:49:24,380 --> 00:49:28,210 So basically, you get 4% reflected. 713 00:49:28,210 --> 00:49:31,500 And the rest actually goes through. 714 00:49:31,500 --> 00:49:37,450 And just to convince you that the total intensity is 100%, 715 00:49:37,450 --> 00:49:42,320 we can calculate what would be the intensity 716 00:49:42,320 --> 00:49:43,890 of the transmitted light. 717 00:49:43,890 --> 00:49:47,050 This will be equal to 1.5-- 718 00:49:47,050 --> 00:49:54,470 this is actually related to n2, because the intensity 719 00:49:54,470 --> 00:50:00,470 is proportional to c n epsilon 0 A squared over 2-- 720 00:50:00,470 --> 00:50:03,110 times T squared. 721 00:50:03,110 --> 00:50:09,530 So basically you have a 0.8 squared 722 00:50:09,530 --> 00:50:13,680 and the I initial intensity. 723 00:50:13,680 --> 00:50:15,860 And if I calculate this value, basically you 724 00:50:15,860 --> 00:50:21,450 are going to get 96% of the initial intensity. 725 00:50:21,450 --> 00:50:25,490 So 96% of the initial intensity actually passes 726 00:50:25,490 --> 00:50:30,470 through the boundary and continues and propagates 727 00:50:30,470 --> 00:50:34,150 in the second medium, which, actually, in this case, 728 00:50:34,150 --> 00:50:35,214 is the soap. 729 00:50:42,770 --> 00:50:45,530 So the picture is the following. 730 00:50:45,530 --> 00:50:53,970 When 100% of light intensity going 731 00:50:53,970 --> 00:50:58,280 towards the boundary, what is going to happen 732 00:50:58,280 --> 00:51:06,220 is that 4% of the light got reflected. 733 00:51:06,220 --> 00:51:09,030 4% of intensity got reflected. 734 00:51:09,030 --> 00:51:14,140 And also, because n1 is smaller than n2, 735 00:51:14,140 --> 00:51:21,750 therefore, there is a flip in sign in the amplitude. 736 00:51:21,750 --> 00:51:27,680 And the rest continues, 96% of them. 737 00:51:27,680 --> 00:51:34,400 And there is no flipping sign in the amplitude. 738 00:51:34,400 --> 00:51:36,180 Any questions so far? 739 00:51:36,180 --> 00:51:37,515 We're really pretty close. 740 00:51:41,040 --> 00:51:45,480 So now we are in a position to discuss what is actually really 741 00:51:45,480 --> 00:51:50,290 happening to this soap bubble. 742 00:51:50,290 --> 00:51:55,070 So I'm going to keep this result here. 743 00:51:58,740 --> 00:52:04,330 And I will now discuss a situation in which 744 00:52:04,330 --> 00:52:09,002 you have two interfaces. 745 00:52:12,860 --> 00:52:19,150 So suppose I zoom in, zoom, and zoom in this soap bubble 746 00:52:19,150 --> 00:52:21,100 and put it on the board. 747 00:52:21,100 --> 00:52:29,990 So this is actually the soap film. 748 00:52:29,990 --> 00:52:35,090 And I have now an incident wave, which is actually 749 00:52:35,090 --> 00:52:40,310 going into this bubble. 750 00:52:40,310 --> 00:52:43,700 So now I have 100%, which is actually 751 00:52:43,700 --> 00:52:48,110 going toward this film. 752 00:52:48,110 --> 00:52:54,170 So after this light, this plane wave 753 00:52:54,170 --> 00:52:58,370 hits the film, what is going to happen? 754 00:52:58,370 --> 00:53:01,750 The first thing which happens is that there will be 755 00:53:01,750 --> 00:53:06,306 4% of the light got reflected. 756 00:53:09,420 --> 00:53:11,050 n2 is equal to 1.5. 757 00:53:11,050 --> 00:53:12,704 It's the same setup, just a reminder, 758 00:53:12,704 --> 00:53:17,060 just to make sure everybody is on the same page. 759 00:53:17,060 --> 00:53:20,350 So 4% of the light got reflected. 760 00:53:20,350 --> 00:53:23,225 Of course, the sign changed. 761 00:53:27,880 --> 00:53:33,880 96% of the intensity actually continue. 762 00:53:33,880 --> 00:53:37,570 And what is actually happening is 763 00:53:37,570 --> 00:53:45,010 that there will be no change in amplitude in sign. 764 00:53:45,010 --> 00:53:47,830 And this is actually not the end of the story, right? 765 00:53:47,830 --> 00:53:51,850 Because the light will continue and continue to propagate. 766 00:53:51,850 --> 00:53:53,680 What is going to happen is that it 767 00:53:53,680 --> 00:54:02,120 will reach another boundary, where the incident light is 768 00:54:02,120 --> 00:54:06,630 you're traveling, from n2 refractive index material, 769 00:54:06,630 --> 00:54:09,520 to n1, which is actually the air. 770 00:54:09,520 --> 00:54:13,600 Now I have a situation where the light is actually 771 00:54:13,600 --> 00:54:18,080 going through the boundary and going out of the air. 772 00:54:18,080 --> 00:54:22,690 That means the light is actually going into the bubble. 773 00:54:22,690 --> 00:54:25,280 So this is actually inside the bubble. 774 00:54:31,130 --> 00:54:33,910 So what is going to happen is the following. 775 00:54:33,910 --> 00:54:40,130 Basically, the calculation is the same, except that now 776 00:54:40,130 --> 00:54:46,340 the R is actually 0.2 instead of minus 0.2, right? 777 00:54:46,340 --> 00:54:51,110 Because now n2 minus n2 is actually 0.5. 778 00:54:51,110 --> 00:54:55,280 0.5 divided by 2.5 is positive 0.2. 779 00:54:55,280 --> 00:54:57,200 So basically, what you are going to get 780 00:54:57,200 --> 00:55:03,486 is a reflected light, which actually doesn't change. 781 00:55:03,486 --> 00:55:08,240 It doesn't change the sign of the amplitude. 782 00:55:08,240 --> 00:55:11,270 And what is actually the intensity? 783 00:55:11,270 --> 00:55:18,280 The intensity will be 96% times 4%, because only 4% 784 00:55:18,280 --> 00:55:20,650 of the light got reflected. 785 00:55:20,650 --> 00:55:23,680 And of course, a large fraction of the light 786 00:55:23,680 --> 00:55:28,580 actually pass through the bubble, 96% times 96%. 787 00:55:28,580 --> 00:55:32,350 And this would be, again, pointing upward, 788 00:55:32,350 --> 00:55:36,460 because T is always positive. 789 00:55:36,460 --> 00:55:39,437 Any questions so far. 790 00:55:39,437 --> 00:55:41,520 You can see that this actually really interesting, 791 00:55:41,520 --> 00:55:43,670 because most of the light actually 792 00:55:43,670 --> 00:55:47,990 pass through the bubble. 793 00:55:47,990 --> 00:55:49,490 So that's actually already one thing 794 00:55:49,490 --> 00:55:52,010 we've learned from this exercise. 795 00:55:52,010 --> 00:55:55,860 Now, what is going to happen to this light if I continue 796 00:55:55,860 --> 00:55:58,220 and increase the time? 797 00:55:58,220 --> 00:56:00,640 What is going to happen is that this reflected 798 00:56:00,640 --> 00:56:06,650 light, from the second surface or second boundary, 799 00:56:06,650 --> 00:56:10,580 will go backward and pass through the first boundary 800 00:56:10,580 --> 00:56:13,070 again. 801 00:56:13,070 --> 00:56:14,930 What is going to happen is the following. 802 00:56:14,930 --> 00:56:20,150 So basically, we're going to get, again, transmitted light 803 00:56:20,150 --> 00:56:21,393 and the reflected light. 804 00:56:26,090 --> 00:56:29,360 What will be the sign of the reflected light? 805 00:56:29,360 --> 00:56:31,769 Will the arrow be pointing up or down? 806 00:56:31,769 --> 00:56:32,310 AUDIENCE: Up. 807 00:56:32,310 --> 00:56:34,055 YEN-JIE LEE: Up, yeah, very good. 808 00:56:34,055 --> 00:56:36,470 Right now, if you are bored, then that 809 00:56:36,470 --> 00:56:39,560 means I am very successful. 810 00:56:39,560 --> 00:56:44,840 So that means I'm getting 4% times 96% times 4%. 811 00:56:47,870 --> 00:56:51,080 What would be the sign for the transmitted light? 812 00:56:51,080 --> 00:56:52,019 Pointing up or down? 813 00:56:52,019 --> 00:56:52,560 AUDIENCE: Up. 814 00:56:52,560 --> 00:56:53,226 YEN-JIE LEE: Up. 815 00:56:53,226 --> 00:56:56,060 Very good, so everybody gets it. 816 00:56:56,060 --> 00:57:06,151 And 96% pass through, 96% times 96% times 4% will pass. 817 00:57:09,330 --> 00:57:11,820 And of course, I can now continue and continue. 818 00:57:11,820 --> 00:57:15,580 What is going to happen is that now you have learned 8.03. 819 00:57:15,580 --> 00:57:20,350 You will see that this is a crazy phenomenon. 820 00:57:20,350 --> 00:57:22,050 What is going to happen is that there 821 00:57:22,050 --> 00:57:28,710 will be a tiny fraction of the light which is trapped forever 822 00:57:28,710 --> 00:57:30,620 between the two surfaces. 823 00:57:30,620 --> 00:57:33,630 They are going to be bouncing back and forth, 824 00:57:33,630 --> 00:57:36,870 boo, boo, boo, boo, boo, boo, boo, boo, forever. 825 00:57:36,870 --> 00:57:40,290 Of course, the fraction of the intensity 826 00:57:40,290 --> 00:57:44,010 is really, really small. 827 00:57:44,010 --> 00:57:47,130 Because every time you've got the reflection actually 828 00:57:47,130 --> 00:57:50,660 happening, you take a hit of 4%. 829 00:57:50,660 --> 00:57:55,800 But since we are talking about theoretical physics, 830 00:57:55,800 --> 00:57:58,650 so, theoretically, that would continue forever. 831 00:57:58,650 --> 00:58:01,900 That's actually pretty interesting. 832 00:58:01,900 --> 00:58:06,030 And going back to practical situation, 833 00:58:06,030 --> 00:58:11,490 basically, I can safely ignore any further reflection, 834 00:58:11,490 --> 00:58:13,410 because they are hitting so hard, 835 00:58:13,410 --> 00:58:16,830 because every time I get a 4% hit, right? 836 00:58:16,830 --> 00:58:20,200 Therefore, I can ignore all the other contribution. 837 00:58:20,200 --> 00:58:25,710 And what we are actually seeing is what? 838 00:58:25,710 --> 00:58:29,200 Our eye is here. 839 00:58:29,200 --> 00:58:34,050 We see the contribution of the first pass, 840 00:58:34,050 --> 00:58:38,430 which is actually reflected from the first surface. 841 00:58:38,430 --> 00:58:41,560 The second pass, OK, it pass through the first surface, 842 00:58:41,560 --> 00:58:46,200 got reflected from the second boundary, 843 00:58:46,200 --> 00:58:54,090 and pass through the first boundary, in the second round, 844 00:58:54,090 --> 00:58:58,500 and then, also, reaching your eye. 845 00:58:58,500 --> 00:59:00,960 So what are we looking at ? 846 00:59:00,960 --> 00:59:04,590 We are looking at the superposition 847 00:59:04,590 --> 00:59:10,776 of two electromagnetic waves coming from one, 848 00:59:10,776 --> 00:59:15,935 which is like this, and two, which is actually like this. 849 00:59:23,310 --> 00:59:31,200 The question now is what is the thickness of the film? 850 00:59:31,200 --> 00:59:34,580 Now I can define the thickness or, say, 851 00:59:34,580 --> 00:59:39,060 the width of this film to be d. 852 00:59:39,060 --> 00:59:42,720 Now the question we are actually asking 853 00:59:42,720 --> 00:59:46,970 is, what would be the thickness, d, 854 00:59:46,970 --> 00:59:53,820 which is needed such that I can have constructive interference? 855 00:59:53,820 --> 00:59:58,010 Now the question becomes really clear. 856 00:59:58,010 --> 01:00:01,770 And we can actually calculate that 857 01:00:01,770 --> 01:00:06,840 by evaluating the phase difference between the path 858 01:00:06,840 --> 01:00:09,780 number one and the path number two. 859 01:00:09,780 --> 01:00:15,030 So now, in order to have constructive interference, 860 01:00:15,030 --> 01:00:18,420 I need a specific phase difference. 861 01:00:18,420 --> 01:00:22,310 But before that, I need to calculate the phase difference 862 01:00:22,310 --> 01:00:32,100 first between path number one and path number two. 863 01:00:32,100 --> 01:00:35,430 What will be the phase difference? 864 01:00:35,430 --> 01:00:42,190 The phase difference delta will be equal to, of course, pi. 865 01:00:42,190 --> 01:00:49,560 This pi contribution is coming from the flip of the amplitude. 866 01:00:49,560 --> 01:00:53,810 That will actually give you an pi phase difference. 867 01:00:53,810 --> 01:00:58,140 The second phase difference is coming from the difference 868 01:00:58,140 --> 01:01:01,900 in the optical path length. 869 01:01:01,900 --> 01:01:07,140 You can see that the first path, it doesn't go into the film. 870 01:01:07,140 --> 01:01:10,270 It got reflected, directly. 871 01:01:10,270 --> 01:01:15,300 And the second path, which is path number two, 872 01:01:15,300 --> 01:01:20,120 it takes more effort or more time 873 01:01:20,120 --> 01:01:23,740 for the light to go back and reach your eye. 874 01:01:23,740 --> 01:01:27,320 How big is the path length difference? 875 01:01:27,320 --> 01:01:33,030 The size of the path length difference is 2 times d, right? 876 01:01:36,540 --> 01:01:40,500 Of course, I need to actually translate that back 877 01:01:40,500 --> 01:01:45,330 to the phase. 878 01:01:45,330 --> 01:01:49,020 So first, I need to actually calculate how many period. 879 01:01:49,020 --> 01:01:54,080 So the length divided by lambda will be the period. 880 01:01:54,080 --> 01:02:00,510 So lambda is actually the wavelength 881 01:02:00,510 --> 01:02:03,780 of the incident light. 882 01:02:03,780 --> 01:02:06,100 But I am missing a factor here. 883 01:02:06,100 --> 01:02:07,410 And can somebody help me? 884 01:02:07,410 --> 01:02:10,110 Because this lambda is actually inside the material, right? 885 01:02:10,110 --> 01:02:11,948 So which factor, I'm missing? 886 01:02:11,948 --> 01:02:13,232 AUDIENCE: n2. 887 01:02:13,232 --> 01:02:14,230 YEN-JIE LEE: n2, right? 888 01:02:14,230 --> 01:02:15,580 Yeah, thank you very much. 889 01:02:15,580 --> 01:02:19,060 So basically, inside the material, 890 01:02:19,060 --> 01:02:23,440 since the speed of light is 1.5 times smaller 891 01:02:23,440 --> 01:02:25,870 than the speed of light in vacuum, 892 01:02:25,870 --> 01:02:34,840 therefore, the wavelength is actually lambda divided by n2. 893 01:02:34,840 --> 01:02:37,030 And this is actually the number of period. 894 01:02:37,030 --> 01:02:40,780 And now, I need to translate that to phase difference. 895 01:02:40,780 --> 01:02:44,620 Therefore, I multiply this by 2pi. 896 01:02:44,620 --> 01:02:49,080 So you can see that now I have successfully evaluated 897 01:02:49,080 --> 01:02:52,510 or quantified the phase difference between path number 898 01:02:52,510 --> 01:02:54,370 one and two. 899 01:02:54,370 --> 01:02:56,830 That is there are two contributions. 900 01:02:56,830 --> 01:02:59,170 The first one is pi. 901 01:02:59,170 --> 01:03:04,750 It's related to the flip in amplitude. 902 01:03:04,750 --> 01:03:08,280 The second contribution, the blue one, 903 01:03:08,280 --> 01:03:13,720 is actually coming from the optical path length difference. 904 01:03:13,720 --> 01:03:19,600 And of course, we can evaluate that really precisely. 905 01:03:19,600 --> 01:03:22,240 Therefore, we can now quickly conclude 906 01:03:22,240 --> 01:03:36,170 that, in order to have constructive interference, 907 01:03:36,170 --> 01:03:43,190 I need to have filter equal to 2N pi, where N is an integer. 908 01:03:48,080 --> 01:03:51,410 And in order to have destructive interference, 909 01:03:51,410 --> 01:04:02,870 I need to have filter equal to 2N 910 01:04:02,870 --> 01:04:07,420 plus 1, pi, which is actually the result 911 01:04:07,420 --> 01:04:12,070 of the calculation which we have done, I think, before. 912 01:04:12,070 --> 01:04:12,880 Yeah, there. 913 01:04:12,880 --> 01:04:15,250 So this is actually based on the calculation 914 01:04:15,250 --> 01:04:18,730 we have done in the beginning. 915 01:04:18,730 --> 01:04:22,120 So we are really close. 916 01:04:22,120 --> 01:04:28,660 So now we have this result, delta is equal to pi plus 2d, 917 01:04:28,660 --> 01:04:32,670 times 2 pi divided by lambda divided by n2, right, 918 01:04:32,670 --> 01:04:35,680 so this complicated formula? 919 01:04:35,680 --> 01:04:39,010 Now we are in the position to evaluate what 920 01:04:39,010 --> 01:04:43,030 would be the phase difference. 921 01:04:43,030 --> 01:04:45,310 So the first thing which I would like to discuss 922 01:04:45,310 --> 01:04:51,140 is that, when d goes to 0, what does is actually the limit? 923 01:04:51,140 --> 01:04:57,780 The limit is when the width of the film 924 01:04:57,780 --> 01:05:01,030 is really, really small, it goes to 0. 925 01:05:01,030 --> 01:05:02,770 What is going to happen? 926 01:05:02,770 --> 01:05:10,621 You are going to have destructive interference. 927 01:05:14,310 --> 01:05:15,530 Why is that? 928 01:05:15,530 --> 01:05:19,450 That is because, even when you have d equal to 0, 929 01:05:19,450 --> 01:05:27,120 the filter is pi because of the flip in sign in path number 930 01:05:27,120 --> 01:05:29,640 one. 931 01:05:29,640 --> 01:05:31,490 The second thing is that now I can 932 01:05:31,490 --> 01:05:35,316 calculate what would be the constructive interference 933 01:05:35,316 --> 01:05:35,816 width. 934 01:05:44,650 --> 01:05:50,805 So this will happen when d is equal to 2N minus 1 lambda, 935 01:05:50,805 --> 01:05:53,620 divided by 4 n2. 936 01:05:53,620 --> 01:06:01,430 So basically, you can use that formula there and solve d. 937 01:06:01,430 --> 01:06:04,390 Then basically that's the formula we are going to get. 938 01:06:04,390 --> 01:06:07,750 And I will not go into detail with this. 939 01:06:10,420 --> 01:06:11,310 Any questions so far? 940 01:06:15,600 --> 01:06:25,750 So now, the third conclusion is that, if I 941 01:06:25,750 --> 01:06:35,890 fix d and the change in lambda, that is actually 942 01:06:35,890 --> 01:06:39,100 the more practical situation. 943 01:06:39,100 --> 01:06:40,900 Because I have the soap bubble. 944 01:06:40,900 --> 01:06:44,120 And it have a well-defined width, which is d. 945 01:06:44,120 --> 01:06:46,180 And what is happening is that I am trying 946 01:06:46,180 --> 01:06:50,285 to shine this soap bubble with light 947 01:06:50,285 --> 01:06:53,140 with different wavelengths, right? 948 01:06:53,140 --> 01:06:55,900 So that is actually the third situation. 949 01:06:55,900 --> 01:07:02,050 If I fix the width of the film, and the change the wavelength, 950 01:07:02,050 --> 01:07:04,640 lambda, what I am going to get is 951 01:07:04,640 --> 01:07:11,170 that the lambda max, which is the wavelength needed 952 01:07:11,170 --> 01:07:29,720 to have constructive interference, 953 01:07:29,720 --> 01:07:37,710 will be equal to 4d n2 divided by 2N minus 1. 954 01:07:37,710 --> 01:07:42,170 So basically, I can solve the lambda if I am given a d value. 955 01:07:45,180 --> 01:07:47,550 So actually, we already get the answer 956 01:07:47,550 --> 01:07:50,520 we are asking in the beginning. 957 01:07:50,520 --> 01:07:54,990 The first question is, why do we see color? 958 01:07:54,990 --> 01:07:57,650 The second question is, when I see color, 959 01:07:57,650 --> 01:08:01,080 what is actually the width of the soap film? 960 01:08:01,080 --> 01:08:03,945 We are going to know the result in a moment. 961 01:08:07,860 --> 01:08:11,170 So now, I have this formula in hand. 962 01:08:11,170 --> 01:08:24,979 If I have d roughly equal to 100 nanometer, which 963 01:08:24,979 --> 01:08:37,790 is the third option we were discussing, 964 01:08:37,790 --> 01:08:45,892 that is going to give you lambda maxima equal to 4 times 100 965 01:08:45,892 --> 01:08:49,910 nanometer times 1.5-- 966 01:08:49,910 --> 01:08:56,180 n2 is 1.5-- divided by 2N minus 1. 967 01:08:56,180 --> 01:09:00,450 So that is actually 600 nanometer 968 01:09:00,450 --> 01:09:03,000 divided by 2N minus 1. 969 01:09:06,050 --> 01:09:10,489 Suppose I have N equal to 1, basically I 970 01:09:10,489 --> 01:09:15,319 am getting 600 nanometer. 971 01:09:15,319 --> 01:09:22,220 Suppose I have N equal to 2, 2N minus 1 is actually 4 minus 1 972 01:09:22,220 --> 01:09:23,029 is 3. 973 01:09:23,029 --> 01:09:29,330 Therefore, you get 200 nanometer and 120 nanometer, 974 01:09:29,330 --> 01:09:32,050 et cetera, et cetera, which are the required 975 01:09:32,050 --> 01:09:38,870 wavelengths in order to have constructive interference 976 01:09:38,870 --> 01:09:42,930 between path number one and path number two. 977 01:09:42,930 --> 01:09:46,010 Everybody is following? 978 01:09:46,010 --> 01:09:50,450 If I plot the spectra of this lambda max, 979 01:09:50,450 --> 01:09:56,720 assuming d is 100 nanometer, what I am getting is like this. 980 01:10:00,740 --> 01:10:06,380 So this is a situation of very thin film. 981 01:10:06,380 --> 01:10:09,320 So this is the lambda. 982 01:10:09,320 --> 01:10:13,865 What I am getting is that there will be a maxima here, which 983 01:10:13,865 --> 01:10:17,700 is actually 600 nanometer. 984 01:10:20,580 --> 01:10:27,350 Red color is actually roughly 650 nanometer. 985 01:10:27,350 --> 01:10:31,130 This is red light. 986 01:10:31,130 --> 01:10:35,680 And this is actually roughly the range 987 01:10:35,680 --> 01:10:39,780 of the visible light, which is actually between lambda 988 01:10:39,780 --> 01:10:44,290 equal to lambda violet-- 989 01:10:44,290 --> 01:10:48,270 violet is equal to 400 nanometer. 990 01:10:48,270 --> 01:10:52,640 So you can see that the first maxima, lambda maxima, 991 01:10:52,640 --> 01:10:55,580 where you have constructive interference 992 01:10:55,580 --> 01:10:58,640 is at 600 nanometer. 993 01:10:58,640 --> 01:11:02,400 So that means you are going to see what kind of color 994 01:11:02,400 --> 01:11:03,623 in your soap bubble? 995 01:11:03,623 --> 01:11:04,490 AUDIENCE: Red. 996 01:11:04,490 --> 01:11:07,400 YEN-JIE LEE: You are going to see red, right? 997 01:11:07,400 --> 01:11:10,760 And then the next wavelength which 998 01:11:10,760 --> 01:11:15,190 you can have constructive interference is 200 nanometer. 999 01:11:15,190 --> 01:11:22,130 That is actually shorter than the wavelength of the violet 1000 01:11:22,130 --> 01:11:23,390 light. 1001 01:11:23,390 --> 01:11:26,690 It's out of the range of the visible light. 1002 01:11:26,690 --> 01:11:27,820 What is going to happen? 1003 01:11:27,820 --> 01:11:30,050 Your eye will not see it. 1004 01:11:32,900 --> 01:11:36,680 So the next one would be here, whatever, blah, blah, blah, 1005 01:11:36,680 --> 01:11:38,980 blah, which I don't care, because they 1006 01:11:38,980 --> 01:11:41,070 are so short in wavelength. 1007 01:11:41,070 --> 01:11:43,280 And you cannot see them. 1008 01:11:43,280 --> 01:11:47,770 So you can see that, if I have a width which 1009 01:11:47,770 --> 01:11:54,560 is roughly 100 nanometer, very same situation, what 1010 01:11:54,560 --> 01:11:55,580 is going to happen? 1011 01:11:55,580 --> 01:11:58,430 What is going to happen is that you 1012 01:11:58,430 --> 01:12:04,540 are going to get only one maxima in the visible light range. 1013 01:12:04,540 --> 01:12:06,205 And therefore, you can see color. 1014 01:12:09,550 --> 01:12:10,580 Any questions so far? 1015 01:12:14,120 --> 01:12:20,090 Now, what I'm going to do is take the same formula here, 1016 01:12:20,090 --> 01:12:28,580 but now I would like to change this d. 1017 01:12:28,580 --> 01:12:30,530 So now I would like to change the d 1018 01:12:30,530 --> 01:12:36,930 to consider a situation where you have a very thick layer. 1019 01:12:36,930 --> 01:12:40,280 So now I would like to change the situation to a very thick 1020 01:12:40,280 --> 01:12:46,165 layer, so maybe I need to erase this part of the board 1021 01:12:46,165 --> 01:12:49,320 to make some space. 1022 01:12:49,320 --> 01:12:58,130 So now if I have d equal to 100 micron, what 1023 01:12:58,130 --> 01:12:59,930 is going to happen? 1024 01:12:59,930 --> 01:13:03,590 So I can now still use this formula 1025 01:13:03,590 --> 01:13:07,320 to calculate what would be the lambda maxima. 1026 01:13:07,320 --> 01:13:15,020 So lambda maxima will be equal to 600 micron, which 1027 01:13:15,020 --> 01:13:18,190 is when you have N equal to 1. 1028 01:13:18,190 --> 01:13:22,400 But this wavelength is way, way larger, much, much larger 1029 01:13:22,400 --> 01:13:26,840 than the wavelength of the visible light. 1030 01:13:26,840 --> 01:13:29,510 So it's not going to work. 1031 01:13:29,510 --> 01:13:32,940 Therefore, you have to be patient. 1032 01:13:32,940 --> 01:13:41,780 You have to increase the N value until N is equal to 500. 1033 01:13:41,780 --> 01:13:46,520 So I am calculation 1, 2, 3, 4, 5, 6, until 500. 1034 01:13:46,520 --> 01:13:51,360 Ahh, we are in the visible light range, right? 1035 01:13:51,360 --> 01:13:57,147 Now, 5000 will give you 600.6 nanometer. 1036 01:13:57,147 --> 01:13:58,430 Phew. 1037 01:13:58,430 --> 01:14:02,210 Suddenly, your eye can see it. 1038 01:14:02,210 --> 01:14:03,070 Very good. 1039 01:14:03,070 --> 01:14:04,520 That's very nice, right? 1040 01:14:04,520 --> 01:14:07,610 So I can now put it in my diagram. 1041 01:14:07,610 --> 01:14:10,130 This is actually the wavelength, again. 1042 01:14:10,130 --> 01:14:14,480 And, ah, I get one line here. 1043 01:14:14,480 --> 01:14:18,980 How about the next one, N equal to 501? 1044 01:14:18,980 --> 01:14:23,770 I'm going to get 599.4 nanometer. 1045 01:14:23,770 --> 01:14:26,210 It's pretty close to this one. 1046 01:14:26,210 --> 01:14:37,010 And the next one would be 598.2 nanometer if N is equal to 502. 1047 01:14:37,010 --> 01:14:38,540 And what you are getting is that you 1048 01:14:38,540 --> 01:14:42,380 can see that, no, things are not going very well. 1049 01:14:42,380 --> 01:14:49,962 They are full spectra, all very, very narrow. 1050 01:14:49,962 --> 01:14:53,600 Very, very large number of wavelengths 1051 01:14:53,600 --> 01:14:58,150 can give you constructive interference. 1052 01:14:58,150 --> 01:15:00,600 So what is going to happen? 1053 01:15:00,600 --> 01:15:06,540 What is going to happen is that your eye will see reflection 1054 01:15:06,540 --> 01:15:10,210 with all kinds of different wavelengths. 1055 01:15:10,210 --> 01:15:11,884 And what color is that? 1056 01:15:11,884 --> 01:15:12,800 AUDIENCE: White light. 1057 01:15:12,800 --> 01:15:13,430 YEN-JIE LEE: White! 1058 01:15:13,430 --> 01:15:15,346 You are going to see something which is white. 1059 01:15:18,590 --> 01:15:22,620 So that is actually the answer to our question. 1060 01:15:22,620 --> 01:15:26,690 So what would be the required thickness? 1061 01:15:26,690 --> 01:15:32,410 The required thickness is something like 100 nanometer. 1062 01:15:32,410 --> 01:15:37,190 So we can see how thin is the water bubble. 1063 01:15:37,190 --> 01:15:39,080 That may surprise some of you, right? 1064 01:15:39,080 --> 01:15:41,230 Most of you actually didn't think 1065 01:15:41,230 --> 01:15:44,120 that's actually that thin. 1066 01:15:44,120 --> 01:15:47,510 Secondly, if d equal to 0, you are 1067 01:15:47,510 --> 01:15:49,710 going to have destructive interference. 1068 01:15:49,710 --> 01:15:52,150 That means there will be no reflected light. 1069 01:15:52,150 --> 01:15:54,400 Everything is going to pass through. 1070 01:15:54,400 --> 01:15:58,610 And the bubble is like transparent. 1071 01:15:58,610 --> 01:16:03,710 Finally, when the bubble is really thick, 1072 01:16:03,710 --> 01:16:06,920 you are going to see white. 1073 01:16:06,920 --> 01:16:13,670 So let me finish this lecture with a demonstration here. 1074 01:16:13,670 --> 01:16:15,462 First, before I turn off the light, 1075 01:16:15,462 --> 01:16:16,670 I would like to turn this on. 1076 01:16:20,330 --> 01:16:28,340 So what I have here is a very complicated machine. 1077 01:16:28,340 --> 01:16:31,220 It's not that complicated, actually. 1078 01:16:31,220 --> 01:16:33,920 So basically, I have a light source here, 1079 01:16:33,920 --> 01:16:41,540 which emits light with all kinds of different wavelengths. 1080 01:16:41,540 --> 01:16:43,810 And I have this little device here. 1081 01:16:43,810 --> 01:16:46,340 There's soap solution inside. 1082 01:16:46,340 --> 01:16:48,830 And I can actually rotate from outside. 1083 01:16:48,830 --> 01:16:52,790 You see how sweet is this setup. 1084 01:16:52,790 --> 01:16:57,560 And I can actually create a soap film out of this. 1085 01:16:57,560 --> 01:17:02,670 You can see that I am rotating and trying to actually project 1086 01:17:02,670 --> 01:17:06,190 the result on the wall. 1087 01:17:06,190 --> 01:17:14,480 You can see that, initially, there's nothing really 1088 01:17:14,480 --> 01:17:17,160 striking in the beginning. 1089 01:17:17,160 --> 01:17:21,584 You can see that the light is which color? 1090 01:17:21,584 --> 01:17:22,390 AUDIENCE: White. 1091 01:17:22,390 --> 01:17:24,750 YEN-JIE LEE: It's white, right? 1092 01:17:24,750 --> 01:17:29,610 Remember, because of the optical setup, 1093 01:17:29,610 --> 01:17:33,440 this image is actually upside down. 1094 01:17:33,440 --> 01:17:39,190 So the upper edge of that image is actually 1095 01:17:39,190 --> 01:17:43,990 the lower edge of my setup, which is 1096 01:17:43,990 --> 01:17:46,295 the lower edge of my soap film. 1097 01:17:51,830 --> 01:17:55,390 We have gravity, right, so that I can walk around. 1098 01:17:55,390 --> 01:18:02,514 Due to gravity, you can see that it 1099 01:18:02,514 --> 01:18:06,160 will form a thicker and thicker layer 1100 01:18:06,160 --> 01:18:09,700 in the bottom of my experimental setup 1101 01:18:09,700 --> 01:18:13,720 or in the upper edge of the image. 1102 01:18:13,720 --> 01:18:16,490 On the other hand, due to gravity, 1103 01:18:16,490 --> 01:18:21,700 the upper edge or the lower edge of the image 1104 01:18:21,700 --> 01:18:26,060 will become thinner and thinner as a function of time. 1105 01:18:26,060 --> 01:18:31,840 At some point, the color will start to show up. 1106 01:18:31,840 --> 01:18:39,270 As you can see now, since we wait long enough, 1107 01:18:39,270 --> 01:18:41,950 the soap film becomes thinner and thinner. 1108 01:18:41,950 --> 01:18:46,011 And you can see that there are colors popping up. 1109 01:18:46,011 --> 01:18:47,910 It's like a rainbow. 1110 01:18:47,910 --> 01:18:48,610 Why is that? 1111 01:18:48,610 --> 01:18:54,020 Because I am varying the thickness of the film 1112 01:18:54,020 --> 01:18:58,180 as a function of the vertical distance. 1113 01:18:58,180 --> 01:19:01,160 So therefore, you can see that this is actually showing you 1114 01:19:01,160 --> 01:19:07,760 that different d value will give you very different colors. 1115 01:19:07,760 --> 01:19:10,900 And if we wait long enough, basically what 1116 01:19:10,900 --> 01:19:15,730 we are going to get is that the whole film will 1117 01:19:15,730 --> 01:19:18,850 become more and more colorful. 1118 01:19:18,850 --> 01:19:21,520 And I am sure that after this class, 1119 01:19:21,520 --> 01:19:23,140 you can walk out of the classroom 1120 01:19:23,140 --> 01:19:29,320 and explain to your friend why the soap bubble is colorful. 1121 01:19:29,320 --> 01:19:30,530 Thank you very much. 1122 01:19:30,530 --> 01:19:34,480 And if you have any questions, I will be around. 1123 01:19:34,480 --> 01:19:37,750 And of course, if you want to make your own soap bubble, 1124 01:19:37,750 --> 01:19:40,610 you can actually go ahead and play the demo here. 1125 01:19:48,310 --> 01:19:49,870 So hello, everybody. 1126 01:19:49,870 --> 01:19:54,250 So we are going to show you a demonstration, which 1127 01:19:54,250 --> 01:20:00,060 we can see colorful interference pattern from a soap film. 1128 01:20:00,060 --> 01:20:04,270 So basically, the experimental setup up is like this. 1129 01:20:04,270 --> 01:20:06,040 Basically, we have light, which is 1130 01:20:06,040 --> 01:20:11,710 trying to shine this thin layer of soap film. 1131 01:20:11,710 --> 01:20:17,030 And the result is actually projected on the screen. 1132 01:20:17,030 --> 01:20:20,170 And you can see, at first, you don't really 1133 01:20:20,170 --> 01:20:26,640 see a lot of colorful pattern, because the thickness 1134 01:20:26,640 --> 01:20:33,430 of the film is still rather large, rather thick. 1135 01:20:33,430 --> 01:20:36,060 Therefore, you don't really see a lot of pattern. 1136 01:20:36,060 --> 01:20:38,560 But as a function of time, you can 1137 01:20:38,560 --> 01:20:40,960 see that this pattern is actually changing. 1138 01:20:40,960 --> 01:20:46,220 Because of gravitational force, you 1139 01:20:46,220 --> 01:20:51,940 will be able to see that the lower part of the film 1140 01:20:51,940 --> 01:20:53,740 becomes thicker and thicker. 1141 01:20:53,740 --> 01:20:58,540 And the upper part of the film, which you see that upside down 1142 01:20:58,540 --> 01:21:03,490 on the screen, is actually becoming thinner and thinner. 1143 01:21:03,490 --> 01:21:08,530 As we actually discussed during the class, when soap film is 1144 01:21:08,530 --> 01:21:16,375 thin enough, there will be only one or only a few maximas 1145 01:21:16,375 --> 01:21:23,500 in the interference pattern as a function of wavelengths, 1146 01:21:23,500 --> 01:21:28,600 which happened to be inside the visible light range. 1147 01:21:28,600 --> 01:21:32,910 And you can see it now, already, this colorful pattern 1148 01:21:32,910 --> 01:21:33,820 really develops. 1149 01:21:33,820 --> 01:21:35,470 It's really beautiful. 1150 01:21:35,470 --> 01:21:42,620 And you can see that, in the lower part of the experiment, 1151 01:21:42,620 --> 01:21:45,280 really have authentic color, because there 1152 01:21:45,280 --> 01:21:51,260 are multiple maximas in the visible light range. 1153 01:21:51,260 --> 01:21:55,400 On the other hand, in the upper part of the film, 1154 01:21:55,400 --> 01:21:59,800 you typically have very little number of maximas 1155 01:21:59,800 --> 01:22:04,430 or only have one maxima, in the visible light range, 1156 01:22:04,430 --> 01:22:06,110 as a function of wavelength. 1157 01:22:06,110 --> 01:22:08,570 Therefore, you see really, really dramatic 1158 01:22:08,570 --> 01:22:14,310 and very, very colorful pattern develop from this experiment.