1 00:00:00,000 --> 00:00:28,000 2 00:00:28,000 --> 00:00:32,207 Today we are going to talk about interference of 3 00:00:32,207 --> 00:00:36,056 electromagnetic radiation. And I will start, 4 00:00:36,056 --> 00:00:39,547 as a warm-up, with the famous historical 5 00:00:39,547 --> 00:00:43,843 experiment which was first done by young in 1801. 6 00:00:43,843 --> 00:00:48,767 By that time the issue of whether or not light was waves 7 00:00:48,767 --> 00:00:53,063 or whether it was particles was still unresolved. 8 00:00:53,063 --> 00:00:57,987 Newton always wanted light to be particles but the Dutch 9 00:00:57,987 --> 00:01:03,000 physicist Huygens wanted them to be waves. 10 00:01:03,000 --> 00:01:07,335 And the issue was unresolved. Let us agree that if light are 11 00:01:07,335 --> 00:01:11,892 particles and you have a screen that, say, has two openings and 12 00:01:11,892 --> 00:01:15,346 you throw particles through their like tomatoes, 13 00:01:15,346 --> 00:01:19,241 the tomatoes are particles, and you collect them here, 14 00:01:19,241 --> 00:01:23,797 then those tomatoes that don't get stuck on the screen but that 15 00:01:23,797 --> 00:01:28,280 make it would form a pile here of tomatoes and they would form 16 00:01:28,280 --> 00:01:32,784 a pile there of tomatoes. That is very typical for 17 00:01:32,784 --> 00:01:35,757 particles. But the situation is different 18 00:01:35,757 --> 00:01:40,441 when we deal with waves because the moment you have waves coming 19 00:01:40,441 --> 00:01:44,158 in, say, for instance, we have plan waves coming in 20 00:01:44,158 --> 00:01:48,023 like this, then the wave can go through both openings 21 00:01:48,023 --> 00:01:51,220 simultaneously. And that changes the picture 22 00:01:51,220 --> 00:01:53,599 quite dramatically. It was known, 23 00:01:53,599 --> 00:01:58,059 of course, already in the 17th century that if you have water 24 00:01:58,059 --> 00:02:02,000 waves going through a small opening -- 25 00:02:02,000 --> 00:02:05,111 Here are water waves moving in like so. 26 00:02:05,111 --> 00:02:09,286 And here is also water. That what you see coming out 27 00:02:09,286 --> 00:02:13,789 there are circular waves. The waves then look like this. 28 00:02:13,789 --> 00:02:17,883 And they are propagating out in a circular fashion. 29 00:02:17,883 --> 00:02:22,385 And, if the velocity of the wave here is the same as the 30 00:02:22,385 --> 00:02:27,298 velocity there then the wave length here would be the same as 31 00:02:27,298 --> 00:02:32,054 the wavelength there. But if the velocities are 32 00:02:32,054 --> 00:02:36,082 different, of course, you will see a difference in 33 00:02:36,082 --> 00:02:39,205 the wavelength. Huygens, my countryman, 34 00:02:39,205 --> 00:02:44,136 suggested in the 17th century the idea, which is now known as 35 00:02:44,136 --> 00:02:48,493 Huygens' Principle that we can think of this in a very 36 00:02:48,493 --> 00:02:51,287 different way. Huygens, by the way, 37 00:02:51,287 --> 00:02:55,972 is difficult to pronounce. Any one of you who knows how to 38 00:02:55,972 --> 00:03:00,000 say Huygens correctly is also Dutch. 39 00:03:00,000 --> 00:03:04,032 Because you miss the uh in your language and you miss the hua in 40 00:03:04,032 --> 00:03:06,783 your language. And the combination of uh and 41 00:03:06,783 --> 00:03:09,151 hua is a complete killer for you guys. 42 00:03:09,151 --> 00:03:12,927 You don't get extra 8.03 course credit if you can come to my 43 00:03:12,927 --> 00:03:16,256 office as say Huygens, but it would certainly put you 44 00:03:16,256 --> 00:03:19,199 in the Dutch category. In the 17th century then 45 00:03:19,199 --> 00:03:22,527 Huygens came with an idea, which was later amended by 46 00:03:22,527 --> 00:03:25,536 Fresnel in the 19th century and is now known as, 47 00:03:25,536 --> 00:03:29,376 I will pronounced it your way, the Huygens-Fresnel Principle, 48 00:03:29,376 --> 00:03:33,241 which works as follows. If we have a plane 49 00:03:33,241 --> 00:03:37,103 monochromatic wave and it is incident on a screen with an 50 00:03:37,103 --> 00:03:40,344 aperture, this would be an aperture, an opening. 51 00:03:40,344 --> 00:03:43,586 Here you would have two apertures, two openings. 52 00:03:43,586 --> 00:03:47,379 Then the Huygens-Fresnel Principle states the following. 53 00:03:47,379 --> 00:03:50,965 All points in the aperture plane may be thought of as 54 00:03:50,965 --> 00:03:53,862 secondary point sources of spherical waves. 55 00:03:53,862 --> 00:03:56,965 And the point sources replace the real source, 56 00:03:56,965 --> 00:04:01,752 which is floating the screen. And the screen itself is a 57 00:04:01,752 --> 00:04:05,056 perfect absorber of the radiation falling upon it. 58 00:04:05,056 --> 00:04:08,831 And so, I will repeat the Huygens-Fresnel Principle which 59 00:04:08,831 --> 00:04:11,797 I will need today. All points in the aperture 60 00:04:11,797 --> 00:04:15,303 plane may be thought of as secondary point sources of 61 00:04:15,303 --> 00:04:18,404 spherical waves. In the case that we had water, 62 00:04:18,404 --> 00:04:22,247 this is a two-dimensional surface, there would be circles. 63 00:04:22,247 --> 00:04:25,887 But, when you deal with light, you can think of them as 64 00:04:25,887 --> 00:04:30,000 three-dimensional. That means spherical waves. 65 00:04:30,000 --> 00:04:33,651 Now, the Huygens-Fresnel Principle is very powerful, 66 00:04:33,651 --> 00:04:37,732 though, there is a wide range of opinion to its scientific 67 00:04:37,732 --> 00:04:40,095 merit. And, in a very famous book, 68 00:04:40,095 --> 00:04:43,031 the principle of electrodynamics by Melvin 69 00:04:43,031 --> 00:04:47,326 Schwartz, I read the following. And I quote verbatim from his 70 00:04:47,326 --> 00:04:49,618 book. He says, Huygens' Principle 71 00:04:49,618 --> 00:04:54,057 tells us to consider each point on a wave front as a new source 72 00:04:54,057 --> 00:04:57,637 of radiation and add the radiation from all the new 73 00:04:57,637 --> 00:05:02,576 sources together. Physically, this makes no sense 74 00:05:02,576 --> 00:05:05,319 at all. Light does not emit light. 75 00:05:05,319 --> 00:05:08,312 Only accelerating charges emit light. 76 00:05:08,312 --> 00:05:12,717 Thus, we will begin by throwing out Huygens' Principle 77 00:05:12,717 --> 00:05:15,710 completely. Later we will see that it 78 00:05:15,710 --> 00:05:20,448 actually does give the right answer for the wrong reasons. 79 00:05:20,448 --> 00:05:25,435 I will proceed with my lecture today to get the right answer, 80 00:05:25,435 --> 00:05:28,095 but perhaps for the wrong reason. 81 00:05:28,095 --> 00:05:33,000 Suppose now we have an electromagnetic wave. 82 00:05:33,000 --> 00:05:38,200 I am particularly thinking of light. 83 00:05:38,200 --> 00:05:46,075 And I have here an opening and I have here an opening, 84 00:05:46,075 --> 00:05:54,693 and plane waves are coming in from the left moving with the 85 00:05:54,693 --> 00:06:01,231 speed of light. And let this opening be A and 86 00:06:01,231 --> 00:06:08,660 let this opening be B. And, here, the center of the 87 00:06:08,660 --> 00:06:13,563 two is O. This could be a circular 88 00:06:13,563 --> 00:06:18,318 opening. It could also be a slit. 89 00:06:18,318 --> 00:06:26,341 Most of the experiments I will do today are with slits, 90 00:06:26,341 --> 00:06:32,433 which are perpendicular to the blackboard. 91 00:06:32,433 --> 00:06:38,000 06:03 Then what you get is that the 92 00:06:38,000 --> 00:06:42,206 mountain of the E vector here will coincide with the valley of 93 00:06:42,206 --> 00:06:46,068 the E vector from the other one. And so you would get the 94 00:06:46,068 --> 00:06:49,517 situation that light plus light will give darkness. 95 00:06:49,517 --> 00:06:52,068 We call that destructive interference. 96 00:06:52,068 --> 00:06:56,344 The wave from here and the wave from there would be 180 degrees 97 00:06:56,344 --> 00:07:00,307 out of phase. And, if you take all the points 98 00:07:00,307 --> 00:07:03,508 for which the difference is one-half lambda, 99 00:07:03,508 --> 00:07:06,114 that is a high paraboloidal surface. 100 00:07:06,114 --> 00:07:09,017 It is like a bowl. It is not only in the 101 00:07:09,017 --> 00:07:12,813 blackboard, but it also comes out of the blackboard. 102 00:07:12,813 --> 00:07:16,387 It is like a bowl. Remember from your high school 103 00:07:16,387 --> 00:07:20,779 days that if the sum of the distance AP and BP is a constant 104 00:07:20,779 --> 00:07:24,724 then you get an ellipse, but if the difference is your 105 00:07:24,724 --> 00:07:29,969 constant you get a hyperbola. You will get a hyperboloidal 106 00:07:29,969 --> 00:07:34,189 surface and everywhere on that surface BP minus AP would then 107 00:07:34,189 --> 00:07:37,353 be one-half lambda. And then you would get the 108 00:07:37,353 --> 00:07:41,573 amazing consequence that light plus light will give darkness, 109 00:07:41,573 --> 00:07:44,949 destructive interference. Tomatoes don't do that. 110 00:07:44,949 --> 00:07:49,098 One tomato on top of another tomato does not give no tomato. 111 00:07:49,098 --> 00:07:52,123 That is distinctly different from particles. 112 00:07:52,123 --> 00:07:55,147 Now, of course, there are also hyperboloidal 113 00:07:55,147 --> 00:08:00,000 surfaces for which there is constructive interference. 114 00:08:00,000 --> 00:08:02,886 In other words, if I make this difference n 115 00:08:02,886 --> 00:08:05,636 times lambda, whereby n is either zero or 116 00:08:05,636 --> 00:08:09,348 plus or minus one or plus or minus two then I would get 117 00:08:09,348 --> 00:08:12,647 surfaces for which we get construct interference. 118 00:08:12,647 --> 00:08:16,496 Then the phase difference between the two waves is either 119 00:08:16,496 --> 00:08:19,246 zero or 2pi or 4pi. We have surfaces with 120 00:08:19,246 --> 00:08:23,026 constructive interference and then we have surfaces with 121 00:08:23,026 --> 00:08:26,807 destructive interference. And that is going to be at the 122 00:08:26,807 --> 00:08:31,000 heart of what I want to discuss with you now. 123 00:08:31,000 --> 00:08:37,239 I will start a new drawing. I think we do not need this 124 00:08:37,239 --> 00:08:41,398 anymore. And I will now have a screen 125 00:08:41,398 --> 00:08:46,597 with two narrow openings. They could be slits, 126 00:08:46,597 --> 00:08:52,143 as I said before. Here is one opening and here is 127 00:08:52,143 --> 00:08:57,689 the other opening. And let the separation between 128 00:08:57,689 --> 00:09:02,952 them be d. And the upper one we call 129 00:09:02,952 --> 00:09:08,740 source number one, and the other one source number 130 00:09:08,740 --> 00:09:12,874 two. Suppose now that we are looking 131 00:09:12,874 --> 00:09:17,952 at this at a distance that is very far away. 132 00:09:17,952 --> 00:09:23,622 Very, very far away. And where you are located in 133 00:09:23,622 --> 00:09:30,000 space seen from this point is an angle theta. 134 00:09:30,000 --> 00:09:36,720 The wave from this source will reach you. 135 00:09:36,720 --> 00:09:46,802 And, since that point is very far away, these lines here are, 136 00:09:46,802 --> 00:09:56,211 of course, parallel to each other, very close to parallel 137 00:09:56,211 --> 00:10:05,788 because it is very far away. The wave that comes from this 138 00:10:05,788 --> 00:10:15,029 one has to travel over the larger distance than the wave 139 00:10:15,029 --> 00:10:24,271 that comes from this one. And that difference can easily 140 00:10:24,271 --> 00:10:32,000 be expressed algebraically. 10:02 141 00:10:32,000 --> 00:10:36,769 And that translates into a phase angle difference between 142 00:10:36,769 --> 00:10:38,473 the E vectors. Theta. 143 00:10:38,473 --> 00:10:40,687 First of all, d sine theta, 144 00:10:40,687 --> 00:10:45,542 which is the past difference, this is how many wavelengths 145 00:10:45,542 --> 00:10:49,886 you can fit on that past difference, so I divided by 146 00:10:49,886 --> 00:10:53,037 lambda. And for each wavelength that I 147 00:10:53,037 --> 00:10:56,955 can fit on there, the phase difference is going 148 00:10:56,955 --> 00:11:02,105 to be 2pi. And so, the phase angle, 149 00:11:02,105 --> 00:11:09,094 the difference between the wave from here and there is, 150 00:11:09,094 --> 00:11:16,470 therefore, 2pi times d divided by lambda times sine theta. 151 00:11:16,470 --> 00:11:23,070 And so, now I can set the condition for constructive 152 00:11:23,070 --> 00:11:28,376 interference. What I want is delta to be a 153 00:11:28,376 --> 00:11:35,439 multiple times 2pi. I already wrote that down 154 00:11:35,439 --> 00:11:39,980 earlier, but I will do that again. 155 00:11:39,980 --> 00:11:45,759 N can be zero, it can be plus or minus one, 156 00:11:45,759 --> 00:11:50,162 it can be plus or minus two, etc. 157 00:11:50,162 --> 00:11:55,803 If n is zero, by the way, you are right at 158 00:11:55,803 --> 00:12:01,995 this plane that comes perpendicular out of the 159 00:12:01,995 --> 00:12:07,498 blackboard. Clearly, when you are on this 160 00:12:07,498 --> 00:12:14,378 plane, when n is zero, the distance to any point on 161 00:12:14,378 --> 00:12:22,083 the plane form source number one and source number two is 162 00:12:22,083 --> 00:12:28,000 obviously the same. 12:00 163 00:12:28,000 --> 00:12:34,612 And now we can write down the condition for destructive 164 00:12:34,612 --> 00:12:39,877 interference. Now you want that delta is pi, 165 00:12:39,877 --> 00:12:44,408 3pi, 5pi. Now you have delta equals 2n 166 00:12:44,408 --> 00:12:50,163 minus one times pi. And, if you want to write it 167 00:12:50,163 --> 00:12:57,142 down in terms of the d sine theta then you get that d sine 168 00:12:57,142 --> 00:13:02,653 theta of n, which is the n that you see there, 169 00:13:02,653 --> 00:13:10,000 is now 2n minus one divided by two times lambda. 170 00:13:10,000 --> 00:13:14,477 And so, this is at the heart of the idea of the interference, 171 00:13:14,477 --> 00:13:17,238 in this case, of two of these sources, 172 00:13:17,238 --> 00:13:19,776 two slits or two circular openings. 173 00:13:19,776 --> 00:13:22,462 Let me check that we have that right. 174 00:13:22,462 --> 00:13:25,746 D sine theta, the n makes connection with the 175 00:13:25,746 --> 00:13:30,000 n, equals n lambda. Delta is a multiple of 2pi. 176 00:13:30,000 --> 00:13:34,641 Destructive interference, we get 2n minus one times pi. 177 00:13:34,641 --> 00:13:39,369 Yes, I can live with that. Now I want to be quantitative 178 00:13:39,369 --> 00:13:44,355 because I am going to do a demonstration very much with the 179 00:13:44,355 --> 00:13:49,169 numbers that you are going to see now on this blackboard. 180 00:13:49,169 --> 00:13:54,154 I will give this back to you, although you may not need it. 181 00:13:54,154 --> 00:14:00,000 We now have the two openings all the way on the left here. 182 00:14:00,000 --> 00:14:05,048 And they are so close together that you cannot even separate 183 00:14:05,048 --> 00:14:08,814 them anymore. This is the point where the two 184 00:14:08,814 --> 00:14:12,579 openings are. And we are going to look on the 185 00:14:12,579 --> 00:14:16,088 screen that is far away. Here is a screen, 186 00:14:16,088 --> 00:14:18,826 which is going to be that screen. 187 00:14:18,826 --> 00:14:23,276 And we want to see how the constructive interference, 188 00:14:23,276 --> 00:14:26,699 that means light, shows up on the screen, 189 00:14:26,699 --> 00:14:29,779 and how the destructive interference, 190 00:14:29,779 --> 00:14:35,000 that means darkness, shows up on that screen. 191 00:14:35,000 --> 00:14:38,601 And so here are these two openings. 192 00:14:38,601 --> 00:14:44,533 Light comes in from the left. And let this distance be l. 193 00:14:44,533 --> 00:14:50,360 And so, if I look under an angle theta, as seen from the 194 00:14:50,360 --> 00:14:55,974 slits, I want to know what kind of pattern I will see. 195 00:14:55,974 --> 00:15:02,068 Let's call this x equals zero. And we call this positive x. 196 00:15:02,068 --> 00:15:05,985 And that is negative x. You can also say that the sine 197 00:15:05,985 --> 00:15:10,123 of theta, if you want that, is approximately x divided by 198 00:15:10,123 --> 00:15:13,817 L, but that is only true for small angles of theta. 199 00:15:13,817 --> 00:15:16,403 And, with this kind of interference, 200 00:15:16,403 --> 00:15:20,467 we sometimes have angles of theta which are not small at 201 00:15:20,467 --> 00:15:23,054 all. I want to warn you that this is 202 00:15:23,054 --> 00:15:27,192 not something you can use universally, but today you will 203 00:15:27,192 --> 00:15:32,832 see that we can use it. I start now with red light. 204 00:15:32,832 --> 00:15:38,183 Lambda 600 nanometers. That is the light I will use. 205 00:15:38,183 --> 00:15:44,059 I use monochromatic light. And we will have a distance to 206 00:15:44,059 --> 00:15:48,045 the screen, which is about five meters. 207 00:15:48,045 --> 00:15:52,662 The setup is here. This is about five meters. 208 00:15:52,662 --> 00:15:58,747 And we have light from a red laser which is a little larger 209 00:15:58,747 --> 00:16:04,600 than 600 nanometers. But I just want some wrong 210 00:16:04,600 --> 00:16:07,799 numbers. And the separation here, 211 00:16:07,799 --> 00:16:13,299 d, between the two slits is one-quarter of a millimeter. 212 00:16:13,299 --> 00:16:17,500 Now, you may think that it is rather small. 213 00:16:17,500 --> 00:16:22,000 Well, you will see that that is what you need. 214 00:16:22,000 --> 00:16:27,899 You need very small values of d to even see this phenomenon. 215 00:16:27,899 --> 00:16:33,700 Now, let's evaluate what n is, let's evaluate what delta is 216 00:16:33,700 --> 00:16:40,000 and let's evaluate what the sine is of theta of n. 217 00:16:40,000 --> 00:16:42,989 And then we will evaluate what x of n is. 218 00:16:42,989 --> 00:16:46,651 And we will do that for constructive interference. 219 00:16:46,651 --> 00:16:51,284 The numbers that you are going to see here is for constructive. 220 00:16:51,284 --> 00:16:53,526 And, of course, if you want to, 221 00:16:53,526 --> 00:16:57,786 you can do the same exercise for destructive interference. 222 00:16:57,786 --> 00:17:01,000 I need constructive interference. 223 00:17:01,000 --> 00:17:05,121 So I have to turn to delta. And I make them n times 2pi. 224 00:17:05,121 --> 00:17:08,569 I start with n equals zero. Well, that is easy. 225 00:17:08,569 --> 00:17:12,690 N is zero, delta is zero, the sine theta n is zero and x 226 00:17:12,690 --> 00:17:15,613 of n is zero. That is where you will see 227 00:17:15,613 --> 00:17:18,611 maximum light, constructive interference. 228 00:17:18,611 --> 00:17:22,733 Obviously, that is the plane that goes right through the 229 00:17:22,733 --> 00:17:27,229 middle of those two openings. And the distance to each one of 230 00:17:27,229 --> 00:17:31,126 those sources everywhere on that plane is, of course, 231 00:17:31,126 --> 00:17:36,258 the same. And so, you expect here to see 232 00:17:36,258 --> 00:17:41,037 light maximum. Now I go to plus or minus one. 233 00:17:41,037 --> 00:17:44,621 And I calculate now what delta is. 234 00:17:44,621 --> 00:17:50,595 Then I get plus or minus 2pi. And so the sine is now 2.4 235 00:17:50,595 --> 00:17:55,265 times 10 to the minus 3. You can check that. 236 00:17:55,265 --> 00:18:02,000 Because you know that d sine theta is n times lambda. 237 00:18:02,000 --> 00:18:06,441 And I told you what lambda is. D sine theta is now n times 238 00:18:06,441 --> 00:18:08,935 lambda. And we take n equals one, 239 00:18:08,935 --> 00:18:12,597 and you know what lambda is, you know what d is, 240 00:18:12,597 --> 00:18:15,792 so you can calculate the angle sine theta. 241 00:18:15,792 --> 00:18:20,155 And that translates then into an x value of about plus or 242 00:18:20,155 --> 00:18:22,805 minus plus n minus 1.2 centimeters. 243 00:18:22,805 --> 00:18:27,168 Now I know what the sine of theta is, and I use the small 244 00:18:27,168 --> 00:18:32,000 angle approximation which is more than adequate. 245 00:18:32,000 --> 00:18:36,142 And I know now where on the this screen I get again maxima. 246 00:18:36,142 --> 00:18:39,714 And that is then here, 1.2 centimeters on this side 247 00:18:39,714 --> 00:18:43,500 and 1.2 centimeters on the other side because you get, 248 00:18:43,500 --> 00:18:45,785 of course, plus as well as minus. 249 00:18:45,785 --> 00:18:48,714 And then I can continue. For n equals one, 250 00:18:48,714 --> 00:18:51,642 n equals two, I will just do plus or minus 251 00:18:51,642 --> 00:18:54,142 ten. Then you get here plus or minus 252 00:18:54,142 --> 00:18:56,571 20pi. Then you get here 24 times 10 253 00:18:56,571 --> 00:18:59,857 to the minus 3. And so you get plus or minus 12 254 00:18:59,857 --> 00:19:05,269 centimeters. It is almost linear because the 255 00:19:05,269 --> 00:19:09,170 sine of theta, for these small angles, 256 00:19:09,170 --> 00:19:12,967 is still the same as theta in radius. 257 00:19:12,967 --> 00:19:16,763 Now, before I show the demonstration, 258 00:19:16,763 --> 00:19:22,563 I want a little bit more information on the shape of the 259 00:19:22,563 --> 00:19:27,098 dark and light areas that I am going to see. 260 00:19:27,098 --> 00:19:33,942 What I plot here now is -- I will lower this and I will 261 00:19:33,942 --> 00:19:40,456 raise it later because I want to work above my head so that you 262 00:19:40,456 --> 00:19:46,130 can see what I am doing. I am plotting here the sine of 263 00:19:46,130 --> 00:19:50,018 theta. And keep in mind it is linearly 264 00:19:50,018 --> 00:19:55,376 proportional with x. But I always prefer to plot the 265 00:19:55,376 --> 00:20:00,000 sine of theta. And so here is zero. 266 00:20:00,000 --> 00:20:02,687 Let here be lambda divided by d. 267 00:20:02,687 --> 00:20:08,063 That is where my first maximum will come, and you can see that. 268 00:20:08,063 --> 00:20:11,098 And then here, at the same distance, 269 00:20:11,098 --> 00:20:15,780 I have two lambda divided by d. And, on the other side, 270 00:20:15,780 --> 00:20:18,554 I have minus lambda divided by d. 271 00:20:18,554 --> 00:20:23,150 And all of these are maxima constructive interference. 272 00:20:23,150 --> 00:20:27,572 And the destructive interference clearly falls smack 273 00:20:27,572 --> 00:20:32,245 in the middle. I did not calculate them. 274 00:20:32,245 --> 00:20:36,378 And so, the destructive interferences are here. 275 00:20:36,378 --> 00:20:41,768 And so, you get light curves, light intensity just like this. 276 00:20:41,768 --> 00:20:45,271 Light intensity, remember, is always the 277 00:20:45,271 --> 00:20:48,865 pointing vector. This is light intensity. 278 00:20:48,865 --> 00:20:52,189 And that is in watts per square meter. 279 00:20:52,189 --> 00:20:56,321 Now, the first thing that I want you to notice, 280 00:20:56,321 --> 00:21:01,082 and you will see that today in various demonstrations, 281 00:21:01,082 --> 00:21:05,484 is that the location of the maximum depends on the 282 00:21:05,484 --> 00:21:10,552 wavelength. So that means if red light, 283 00:21:10,552 --> 00:21:14,712 650 nanometers, fall here as a maximum and it 284 00:21:14,712 --> 00:21:20,290 falls here as a maximum and it has here a maximum and it has 285 00:21:20,290 --> 00:21:26,152 there a maximum that blue light, which would have a wavelength, 286 00:21:26,152 --> 00:21:31,447 say, I pick 400 nanometers, would not have its maximum at 287 00:21:31,447 --> 00:21:36,091 the same location. Therefore, not the minimum 288 00:21:36,091 --> 00:21:38,262 either. Because, look at the 289 00:21:38,262 --> 00:21:41,801 relationship, d sine theta is n times lambda. 290 00:21:41,801 --> 00:21:45,179 And so, the angles of theta are difference. 291 00:21:45,179 --> 00:21:49,844 And so, you would get here, roughly, it is about two-thirds 292 00:21:49,844 --> 00:21:52,579 down the way, the maximum for blue. 293 00:21:52,579 --> 00:21:55,715 You would get here the maximum for blue. 294 00:21:55,715 --> 00:22:00,461 Notice that for n equals zero, the red and the blue have the 295 00:22:00,461 --> 00:22:06,207 maximum at the same location. But you would get the first 296 00:22:06,207 --> 00:22:10,050 maximum, the blue here, you would get the second 297 00:22:10,050 --> 00:22:15,119 maximum here and you would get the third maximum almost exactly 298 00:22:15,119 --> 00:22:19,698 at the location where you get the second maximum for red. 299 00:22:19,698 --> 00:22:24,849 The reason being that two times 650 is very close to three times 300 00:22:24,849 --> 00:22:26,729 400. That is the reason. 301 00:22:26,729 --> 00:22:32,153 Each lives a life of its own. The light intensity that you 302 00:22:32,153 --> 00:22:36,384 see is a cosine square function of delta divided by two. 303 00:22:36,384 --> 00:22:41,153 And this is so fundamental that first I thought I would ask you 304 00:22:41,153 --> 00:22:44,538 to derive it, but then I decided to spend one 305 00:22:44,538 --> 00:22:47,153 minute on it and divide it for you. 306 00:22:47,153 --> 00:22:51,230 Because it is extremely important as we use that later 307 00:22:51,230 --> 00:22:54,615 in our predictions. Let us assume that source 308 00:22:54,615 --> 00:23:00,000 number one, and the other one I call source number two -- 309 00:23:00,000 --> 00:23:04,444 At source number one, E one has an amplitude E zero 310 00:23:04,444 --> 00:23:08,977 one when it reaches the screen. And then, of course, 311 00:23:08,977 --> 00:23:13,599 you have cosine omega t. And this is the frequency of 312 00:23:13,599 --> 00:23:16,888 the light. That is the electromagnetic 313 00:23:16,888 --> 00:23:19,911 radiation. E two, that is the other 314 00:23:19,911 --> 00:23:23,733 source, has E zero two times cosine omega t. 315 00:23:23,733 --> 00:23:29,244 And here is that crucial delta, which is the difference between 316 00:23:29,244 --> 00:23:33,968 the two in phase. That is why you get maxima and 317 00:23:33,968 --> 00:23:37,642 that is why you get minima. That is the whole idea of the 318 00:23:37,642 --> 00:23:40,988 interference pattern. Now, if you were very far away 319 00:23:40,988 --> 00:23:44,005 from the two openings then these E vectors are, 320 00:23:44,005 --> 00:23:46,105 of course, very closely the same. 321 00:23:46,105 --> 00:23:50,041 Remember E is proportional to one over r, but if the distance 322 00:23:50,041 --> 00:23:53,453 r is very closely the same then we can assume to good 323 00:23:53,453 --> 00:23:57,389 approximation that the E vector, the amplitudes are the same. 324 00:23:57,389 --> 00:24:00,145 That means the total E vector that you get, 325 00:24:00,145 --> 00:24:05,000 the vectorial sum of the two is then going to be 2E zero. 326 00:24:05,000 --> 00:24:09,909 I simply call it E zero now. And then I get the cosine of 327 00:24:09,909 --> 00:24:13,415 half the sum. I get the cosine of omega t 328 00:24:13,415 --> 00:24:18,149 minus delta divided by two times the cosine of half the 329 00:24:18,149 --> 00:24:23,584 difference, so that becomes the cosine of delta divided by two. 330 00:24:23,584 --> 00:24:28,581 And now you see why the light intensity is proportional to 331 00:24:28,581 --> 00:24:34,333 cosine square delta over two. Because, when you calculate the 332 00:24:34,333 --> 00:24:38,698 mean value of the pointing vector, you have to take E cross 333 00:24:38,698 --> 00:24:40,956 b. And so, b also has this term 334 00:24:40,956 --> 00:24:44,870 and also has this term. And so, you get the square of 335 00:24:44,870 --> 00:24:47,655 this term and the square of this term. 336 00:24:47,655 --> 00:24:51,720 But the average of the square of this term is one-half. 337 00:24:51,720 --> 00:24:55,408 Forget that for now. And so you see it is going to 338 00:24:55,408 --> 00:25:00,000 be proportional to the square of delta over two. 339 00:25:00,000 --> 00:25:03,923 And so this function that you see, delta is linearly 340 00:25:03,923 --> 00:25:08,153 proportional to sine theta. This curve that you see is a 341 00:25:08,153 --> 00:25:10,000 cosine square curve. 342 00:25:10,000 --> 00:25:15,000 343 00:25:15,000 --> 00:25:20,446 The experiment that I am going to show you has a separation of 344 00:25:20,446 --> 00:25:24,196 the slits, which is one-quarter millimeter. 345 00:25:24,196 --> 00:25:28,660 My wavelength is 633 nanometers, which is very much 346 00:25:28,660 --> 00:25:32,571 the same. And the distance to the screen 347 00:25:32,571 --> 00:25:36,685 is about five meters. And so the way that we have 348 00:25:36,685 --> 00:25:41,314 this arranged is that the two slits, they are vertical, 349 00:25:41,314 --> 00:25:45,428 they are like this. And so this separation is 1.4 350 00:25:45,428 --> 00:25:48,857 millimeters. And then we put over there a 351 00:25:48,857 --> 00:25:52,457 laser beam. And then the result is what you 352 00:25:52,457 --> 00:25:56,657 are going to see there. And so I now have to think 353 00:25:56,657 --> 00:26:01,836 about this is where it is. You are going to see it there. 354 00:26:01,836 --> 00:26:05,510 And, in order for you to see it, I have to make it completely 355 00:26:05,510 --> 00:26:06,000 dark. 356 00:26:06,000 --> 00:26:16,000 357 00:26:16,000 --> 00:26:20,200 When you look here on the screen, you will see areas which 358 00:26:20,200 --> 00:26:23,884 are distinctly dark, and you will see areas whereby 359 00:26:23,884 --> 00:26:26,315 you see the light. In other words, 360 00:26:26,315 --> 00:26:30,000 you see the pattern that we just calculated. 361 00:26:30,000 --> 00:26:33,718 You see constructive interference and you see 362 00:26:33,718 --> 00:26:38,366 destructive interference. And you are looking here at an 363 00:26:38,366 --> 00:26:42,676 experiment that is historically of great importance. 364 00:26:42,676 --> 00:26:46,985 It is one of the most mysterious and puzzling issues 365 00:26:46,985 --> 00:26:50,704 in all of physics. This interference pattern, 366 00:26:50,704 --> 00:26:55,521 it was first shown by Young in 1801, is clearly convincing 367 00:26:55,521 --> 00:27:00,000 evidence that light is a wave phenomenon. 368 00:27:00,000 --> 00:27:04,943 However, 20th century physics has shown that light comes in 369 00:27:04,943 --> 00:27:09,801 the form of individual photons with well-defined energies. 370 00:27:09,801 --> 00:27:12,784 And photons can behave like bullets. 371 00:27:12,784 --> 00:27:17,045 They have momentum, the produce radiation pressure, 372 00:27:17,045 --> 00:27:19,943 and they can be localized in space. 373 00:27:19,943 --> 00:27:24,630 We can detect individual photons in a way that we detect 374 00:27:24,630 --> 00:27:30,000 individual tomatoes. Photons behave like particles. 375 00:27:30,000 --> 00:27:35,565 However, the interference pattern that you were staring at 376 00:27:35,565 --> 00:27:41,228 on that screen can only be explained if we assume that each 377 00:27:41,228 --> 00:27:46,599 photon went through both slits. But how on earth can one 378 00:27:46,599 --> 00:27:52,555 particle go through both slits? It will have to be one slit or 379 00:27:52,555 --> 00:27:55,777 the other. And here lies the great 380 00:27:55,777 --> 00:27:59,000 mystery. Light is both. 381 00:27:59,000 --> 00:28:01,837 It is a wave, if you want it to be a wave, 382 00:28:01,837 --> 00:28:05,435 and it is a particle if you want it to be a particle. 383 00:28:05,435 --> 00:28:09,517 If you manage to determine for every photon through which of 384 00:28:09,517 --> 00:28:13,116 the two slits it went, and that is possible if you do 385 00:28:13,116 --> 00:28:16,921 it at very low light intensity, you can really determine 386 00:28:16,921 --> 00:28:20,174 through which of the two slits each photon went. 387 00:28:20,174 --> 00:28:23,357 Then you will not see any interference anymore. 388 00:28:23,357 --> 00:28:27,301 The pattern on the screen will be like that of the tomato. 389 00:28:27,301 --> 00:28:32,191 You will get two piles. The moment that you establish 390 00:28:32,191 --> 00:28:36,498 the particle character of the light by determining through 391 00:28:36,498 --> 00:28:39,974 which slit it went, you have destroyed its wave 392 00:28:39,974 --> 00:28:42,921 character. And it will not behave like a 393 00:28:42,921 --> 00:28:45,642 wave. It will behave like a particle. 394 00:28:45,642 --> 00:28:50,327 However, as long as you do not determine through which slit the 395 00:28:50,327 --> 00:28:54,710 photon went, as long as it remains your guess through which 396 00:28:54,710 --> 00:29:00,000 one it went, light will reveal to you its wave character. 397 00:29:00,000 --> 00:29:04,145 And you will see interference. The choice is yours, 398 00:29:04,145 --> 00:29:06,797 but you cannot have it both ways. 399 00:29:06,797 --> 00:29:11,854 And so you are looking now at something that had a huge impact 400 00:29:11,854 --> 00:29:15,253 on physics. Clearly, at the time that this 401 00:29:15,253 --> 00:29:19,398 experiment was done, it was accepted that light are 402 00:29:19,398 --> 00:29:22,466 waves. But we now look at it in a very 403 00:29:22,466 --> 00:29:26,031 different way. I want to show you a slide of 404 00:29:26,031 --> 00:29:32,000 this what we call double-slit interference in white light. 405 00:29:32,000 --> 00:29:34,467 We did it in red light. That is the easiest. 406 00:29:34,467 --> 00:29:37,337 And you get very sharply defined maxima and minima. 407 00:29:37,337 --> 00:29:40,781 And the reason why you get them so sharply defined is because 408 00:29:40,781 --> 00:29:43,708 you do not have the interference of the other color. 409 00:29:43,708 --> 00:29:46,233 But, of course, if you do it with white light 410 00:29:46,233 --> 00:29:49,620 then you will the blue light has its own spacing and the red 411 00:29:49,620 --> 00:29:52,203 light has its own spacing. And the net result, 412 00:29:52,203 --> 00:29:55,359 of course, is that you see a somewhat different pattern. 413 00:29:55,359 --> 00:30:00,142 And that I will show you here. I will make it a little darker 414 00:30:00,142 --> 00:30:02,714 for that. The upper panel is then the 415 00:30:02,714 --> 00:30:05,428 double-slit interference in blue light. 416 00:30:05,428 --> 00:30:09,714 And you see the clearly dark lines, which are the destructive 417 00:30:09,714 --> 00:30:12,000 inference locations. And you see, 418 00:30:12,000 --> 00:30:16,285 as I already indicated on the blackboard, that the separation 419 00:30:16,285 --> 00:30:20,214 of the red is larger because it is wavelength dependent. 420 00:30:20,214 --> 00:30:23,928 And, if you then do it with white light, then you see 421 00:30:23,928 --> 00:30:27,000 something like this. It is more difficult to 422 00:30:27,000 --> 00:30:31,000 actually see the maxima and the minima. 423 00:30:31,000 --> 00:30:36,374 Now, what you can do with light, if you can turn light 424 00:30:36,374 --> 00:30:42,153 plus light into darkness then obviously you should also be 425 00:30:42,153 --> 00:30:46,412 able to turn sound plus sound into silence. 426 00:30:46,412 --> 00:30:51,888 It is just a matter of scaling up the whole experiment. 427 00:30:51,888 --> 00:30:56,349 And that is what we have set up for you here. 428 00:30:56,349 --> 00:31:02,757 We have two loud speakers. And so all the equations that 429 00:31:02,757 --> 00:31:07,893 you have here apply verbatim, except that we scale it a 430 00:31:07,893 --> 00:31:11,887 little bit. So our d is no longer a quarter 431 00:31:11,887 --> 00:31:15,596 millimeter, but our d is now 1.5 meters. 432 00:31:15,596 --> 00:31:20,827 Here is a source of sound and here is a source of sound, 433 00:31:20,827 --> 00:31:24,631 and they are fed by the same electronics. 434 00:31:24,631 --> 00:31:28,720 And so, the d, the separation is 1.5 meters. 435 00:31:28,720 --> 00:31:33,000 And the frequency is 3000 hertz. 436 00:31:33,000 --> 00:31:38,000 437 00:31:38,000 --> 00:31:40,864 We do this experiment now for sound. 438 00:31:40,864 --> 00:31:45,366 And so, lambda is about 11.3 centimeters if you take 340 439 00:31:45,366 --> 00:31:48,640 meters per second for the speed of sound. 440 00:31:48,640 --> 00:31:51,832 We have, in a way, monochromatic lights, 441 00:31:51,832 --> 00:31:55,843 monochromatic sounds, more or less one wavelength. 442 00:31:55,843 --> 00:32:01,000 And so you can calculate what lambda divided by d is. 443 00:32:01,000 --> 00:32:04,444 It is about 0.075. And so you can calculate where 444 00:32:04,444 --> 00:32:08,177 now those surfaces are. They are really hyperboloidal 445 00:32:08,177 --> 00:32:12,339 surfaces, they are coming out from the center here and they 446 00:32:12,339 --> 00:32:16,071 bend in your direction. Where all those hyperboloidal 447 00:32:16,071 --> 00:32:18,727 surfaces are, where there is no sound. 448 00:32:18,727 --> 00:32:22,028 And then you can calculate where all those high 449 00:32:22,028 --> 00:32:25,401 hyperboloidal surfaces are where there is sound. 450 00:32:25,401 --> 00:32:29,564 Let's first calculate how many hyperboloidal surfaces there 451 00:32:29,564 --> 00:32:32,803 are. In order to do that, 452 00:32:32,803 --> 00:32:38,214 we simply have to stick in our equation d sine theta equals n 453 00:32:38,214 --> 00:32:41,281 lambda. We just put in theta is the 454 00:32:41,281 --> 00:32:45,429 maximum value possible, and that is 90 degrees. 455 00:32:45,429 --> 00:32:50,029 You put in theta max, and that is 90 degrees pi over 456 00:32:50,029 --> 00:32:53,456 two radiance. And so, when you do that, 457 00:32:53,456 --> 00:32:58,506 you will find that n max, that is the highest number of n 458 00:32:58,506 --> 00:33:01,302 in that series 0, 1, 2, 3, 4, 5, 459 00:33:01,302 --> 00:33:06,494 is then divided by lambda. Which, in our case, 460 00:33:06,494 --> 00:33:09,561 is about 13. It is one over this number. 461 00:33:09,561 --> 00:33:13,808 That means there are 13 of those surfaces going in this 462 00:33:13,808 --> 00:33:18,213 direction, but there are also 13 going in this direction. 463 00:33:18,213 --> 00:33:20,730 Remember, we have plus and minus? 464 00:33:20,730 --> 00:33:23,719 We really have to multiply this by two. 465 00:33:23,719 --> 00:33:28,359 And so, there are 26 surfaces, no more, and probably no less 466 00:33:28,359 --> 00:33:33,000 in this audience of constructive interference. 467 00:33:33,000 --> 00:33:39,948 And then there are probably also around 26 destructive 468 00:33:39,948 --> 00:33:45,848 interferences. Now I want to be as qualitative 469 00:33:45,848 --> 00:33:51,092 as I can be. And I would like to know the 470 00:33:51,092 --> 00:33:58,827 students who are sitting about five meters away from me what 471 00:33:58,827 --> 00:34:06,431 the distance is between one maximum, the zero order maximum 472 00:34:06,431 --> 00:34:11,676 right at the center and the next maximum. 473 00:34:11,676 --> 00:34:19,542 We get an idea where to search for those maxima and where for 474 00:34:19,542 --> 00:34:24,000 those minima. 34:00 475 00:34:24,000 --> 00:35:04,555 34:29 That separation x1, 476 00:35:04,555 --> 00:35:08,076 I call it x1, is roughly L times that 0.075. 477 00:35:08,076 --> 00:35:10,695 And that is about 38 centimeters. 478 00:35:10,695 --> 00:35:14,134 Right there those guys I am looking at now, 479 00:35:14,134 --> 00:35:16,918 you will have a maximum sound here. 480 00:35:16,918 --> 00:35:20,766 And, when you move 38 centimeters, you will have 481 00:35:20,766 --> 00:35:24,122 another maximum. And there is a minimum in 482 00:35:24,122 --> 00:35:27,315 between. The minima and the maxima right 483 00:35:27,315 --> 00:35:32,980 there is about 19 centimeters. Now, for you there in the back 484 00:35:32,980 --> 00:35:35,885 it is larger. And for you here in front it is 485 00:35:35,885 --> 00:35:38,063 smaller. Of course, remember those 486 00:35:38,063 --> 00:35:41,430 surfaces go like this. And so here the separation is 487 00:35:41,430 --> 00:35:43,674 smaller. I just give you this as an 488 00:35:43,674 --> 00:35:47,767 example that you know roughly an order of magnitude of what you 489 00:35:47,767 --> 00:35:50,540 are looking for. I am going to turn on this 490 00:35:50,540 --> 00:35:54,171 sound now, and then I want you to wiggle back and forth, 491 00:35:54,171 --> 00:35:58,000 as you did when you were in nursery school. 492 00:35:58,000 --> 00:36:01,321 I want you to find the locations of maxima and minima. 493 00:36:01,321 --> 00:36:05,206 And I can assure that the ones with minima, the ones with sound 494 00:36:05,206 --> 00:36:07,838 silence are extremely well-defined because, 495 00:36:07,838 --> 00:36:11,096 of course, we have done this several times ourselves. 496 00:36:11,096 --> 00:36:13,415 Don't move too fast. Move very slowly. 497 00:36:13,415 --> 00:36:17,300 And there will be points where you hear no sound and there will 498 00:36:17,300 --> 00:36:19,744 be points where you hear a lot of sound. 499 00:36:19,744 --> 00:36:22,000 Are we ready for this? 500 00:36:22,000 --> 00:36:30,000 501 00:36:30,000 --> 00:36:34,000 Keep moving. Most of you move too fast. 502 00:36:34,000 --> 00:36:40,000 503 00:36:40,000 --> 00:36:44,678 Who was able to clearly distinguish the maxima from the 504 00:36:44,678 --> 00:36:46,411 minima? Not too many. 505 00:36:46,411 --> 00:36:50,743 The reason is that all of you are lousy scientists. 506 00:36:50,743 --> 00:36:56,115 [LAUGHTER] If the difference in space between maxima and minima 507 00:36:56,115 --> 00:36:59,234 is 19 centimeters, that is this much, 508 00:36:59,234 --> 00:37:04,000 what is the separation between your two ears? 509 00:37:04,000 --> 00:37:07,864 It is about 20 centimeters. It is excellent that I chose 510 00:37:07,864 --> 00:37:10,113 that. If you have a maximum here, 511 00:37:10,113 --> 00:37:13,978 you would have a minimum there. What would I have to do? 512 00:37:13,978 --> 00:37:17,562 You have to close one ear, and then you will be real 513 00:37:17,562 --> 00:37:20,372 scientists. I am going to turn it back on 514 00:37:20,372 --> 00:37:22,762 again. I want you to close one ear. 515 00:37:22,762 --> 00:37:26,627 And then you will be able to find those minima and those 516 00:37:26,627 --> 00:37:30,000 maxima. Move again back and forth. 517 00:37:30,000 --> 00:37:33,533 Now, there may be some reflections from the wall which 518 00:37:33,533 --> 00:37:37,000 may interfere with his experiment, no pun implied. 519 00:37:37,000 --> 00:37:43,000 520 00:37:43,000 --> 00:37:46,000 Go slowly. Go very slowly. 521 00:37:46,000 --> 00:37:51,000 522 00:37:51,000 --> 00:37:54,253 It is remarkable. Unbelievable. 523 00:37:54,253 --> 00:38:00,000 I hear nothing here and here I hear a lot of noise. 524 00:38:00,000 --> 00:38:03,344 So who heard it now? There you go. 525 00:38:03,344 --> 00:38:07,195 All right. This is the great moment for 526 00:38:07,195 --> 00:38:11,047 the mini-break. There is the mini-quiz. 527 00:38:11,047 --> 00:38:15,000 If someone is going to help me -- 528 00:38:15,000 --> 00:38:20,000 529 00:38:20,000 --> 00:38:23,382 I owe you something regarding Exam 2. 530 00:38:23,382 --> 00:38:26,671 Here you see a histogram off Exam 2. 531 00:38:26,671 --> 00:38:31,463 And, what is interesting and in a way helpful to me, 532 00:38:31,463 --> 00:38:36,531 it is almost bimodal. Now, we have 47% to go in the 533 00:38:36,531 --> 00:38:40,198 course of which 40% is for the final alone. 534 00:38:40,198 --> 00:38:43,603 And so, clearly, I cannot say much about 535 00:38:43,603 --> 00:38:47,007 dividing lines between a, b, c, d and f. 536 00:38:47,007 --> 00:38:50,325 I want to make only a global statement. 537 00:38:50,325 --> 00:38:55,039 And, I only warn those of you I put in the danger zone. 538 00:38:55,039 --> 00:38:59,579 The danger zone for me are students whose cumulative, 539 00:38:59,579 --> 00:39:03,333 assuming that they have taken all the exams, 540 00:39:03,333 --> 00:39:08,802 is less than 30 or near 30. And those are really the 541 00:39:08,802 --> 00:39:13,233 students, almost exclusively, who have less than 45 for Exam 542 00:39:13,233 --> 00:39:17,064 2, or approximately 45. And this is the danger zone. 543 00:39:17,064 --> 00:39:21,045 Now, that does not mean that you will fail the course. 544 00:39:21,045 --> 00:39:24,725 It is just the matter of probability, I would say, 545 00:39:24,725 --> 00:39:27,504 is high. And then those of you who are 546 00:39:27,504 --> 00:39:31,109 dying to know whether they are going to get an A, 547 00:39:31,109 --> 00:39:35,240 all I can say is if your cumulative now is more than 45, 548 00:39:35,240 --> 00:39:39,446 assuming you have taken all exams, then you will probably 549 00:39:39,446 --> 00:39:44,614 end up with an A. Will you end up with an A? 550 00:39:44,614 --> 00:39:48,134 I do not know. It is all a matter of 551 00:39:48,134 --> 00:39:52,156 probability, but the chances are you may. 552 00:39:52,156 --> 00:39:57,888 That is as far as I can go regarding the dividing lines of 553 00:39:57,888 --> 00:40:03,089 course grades. Now I will continue the idea of 554 00:40:03,089 --> 00:40:08,372 interference which has very far-reaching consequences. 555 00:40:08,372 --> 00:40:13,754 And I will turn to what we call thin film interference. 556 00:40:13,754 --> 00:40:19,634 I will cover normal incidence. It really doesn't change very 557 00:40:19,634 --> 00:40:23,521 much if you change to a different angle. 558 00:40:23,521 --> 00:40:30,000 The concept is the same. I have here a thin film of oil. 559 00:40:30,000 --> 00:40:33,750 And we will shortly see what I mean by thin. 560 00:40:33,750 --> 00:40:37,238 This is air. We have plane waves of light 561 00:40:37,238 --> 00:40:41,162 coming in this way, so it is normal incidence. 562 00:40:41,162 --> 00:40:45,436 And this is n1 which is air, so that is about one. 563 00:40:45,436 --> 00:40:49,186 And this is n2, which for oil is about 1.45. 564 00:40:49,186 --> 00:40:52,848 Let's make it 1.5, just so we round it off. 565 00:40:52,848 --> 00:40:56,424 And this is n3, which is again air in this 566 00:40:56,424 --> 00:41:02,469 case, so let's make it one. It does not have to be air, 567 00:41:02,469 --> 00:41:07,225 but I will make it air. And let the separation or the 568 00:41:07,225 --> 00:41:10,884 thickness, I should say, of the oil be d. 569 00:41:10,884 --> 00:41:16,463 And I call this surface A and this surface between oil and air 570 00:41:16,463 --> 00:41:18,018 B. Light comes in, 571 00:41:18,018 --> 00:41:23,231 normal incidence comes in from above over a large surface. 572 00:41:23,231 --> 00:41:28,353 I just put in one arrow here. And let that intensity be i 573 00:41:28,353 --> 00:41:32,893 zero. And now I am going to calculate 574 00:41:32,893 --> 00:41:36,533 how much is reflected back into the air. 575 00:41:36,533 --> 00:41:41,386 And I put the arrow here, so I offset it just for the 576 00:41:41,386 --> 00:41:45,679 purpose of clarity. Some of it is reflected and 577 00:41:45,679 --> 00:41:48,666 some of it goes straight through. 578 00:41:48,666 --> 00:41:53,800 All of you are more than capable of calculating how much 579 00:41:53,800 --> 00:41:56,506 is reflected. At that point A, 580 00:41:56,506 --> 00:42:02,293 the r, the reflectivity of the E vector, is n1 minus n2 divided 581 00:42:02,293 --> 00:42:07,298 by n1 plus n2. And that is minus 0.2. 582 00:42:07,298 --> 00:42:13,358 The minus sign tells you that there is a flip here in the E 583 00:42:13,358 --> 00:42:18,895 vector by 180 degrees. The intensity of the light that 584 00:42:18,895 --> 00:42:24,119 is reflected is then 4%, the square of this number. 585 00:42:24,119 --> 00:42:29,029 And, if 4% is reflected, clearly 96% go through. 586 00:42:29,029 --> 00:42:35,052 If this is 100 then this 4.0. And then this is 96. 587 00:42:35,052 --> 00:42:38,850 This light here hits the surface at b, 588 00:42:38,850 --> 00:42:42,135 the boundary between oil and air. 589 00:42:42,135 --> 00:42:47,268 And some of that light goes through, which I am not 590 00:42:47,268 --> 00:42:51,374 interested in, and some of it comes back. 591 00:42:51,374 --> 00:42:57,225 And so, I can calculate again the r value for that surface 592 00:42:57,225 --> 00:43:04,000 near b, which is now n2 minus n3 divided by n2 plus n3. 593 00:43:04,000 --> 00:43:06,464 This is n2. This is n3. 594 00:43:06,464 --> 00:43:12,066 And so, that is plus 0.2. There is no flip in the E 595 00:43:12,066 --> 00:43:17,668 vector as it returns. And the fraction of the light 596 00:43:17,668 --> 00:43:21,141 that is reflected is, again, 4%. 597 00:43:21,141 --> 00:43:27,414 However, it is 4% of the 96% because only 96% go through. 598 00:43:27,414 --> 00:43:33,623 And 4% of 96% is 3.84. When it reaches this surface 599 00:43:33,623 --> 00:43:38,120 here, again, 4% will be reflected and 96% will go 600 00:43:38,120 --> 00:43:41,211 through. What comes out here then, 601 00:43:41,211 --> 00:43:45,615 ultimately, is 96% of this number, which is 3.7. 602 00:43:45,615 --> 00:43:50,861 I now want to ask the question, what happens when you are 603 00:43:50,861 --> 00:43:56,201 looking here from above and you see this light coming back 604 00:43:56,201 --> 00:44:00,792 straight from a and this light coming back at you, 605 00:44:00,792 --> 00:44:04,539 which makes this journey through the oil, 606 00:44:04,539 --> 00:44:10,254 whether or not the combination of those two can constructively 607 00:44:10,254 --> 00:44:17,000 interfere and whether they can destructively interfere. 608 00:44:17,000 --> 00:44:20,831 Now, as far as destructive interference is concerned, 609 00:44:20,831 --> 00:44:23,410 let me make a qualitative statement. 610 00:44:23,410 --> 00:44:27,536 The intensity of this is 4.0 and the intensity of this is 611 00:44:27,536 --> 00:44:29,747 3.7. It can never be completely 612 00:44:29,747 --> 00:44:32,969 darkness. But you can never kill 3.7 613 00:44:32,969 --> 00:44:36,530 intensity with 4.0. But, if they are 180 degrees 614 00:44:36,530 --> 00:44:39,636 out of phase, which we will be able to do, 615 00:44:39,636 --> 00:44:41,984 there is very little light left. 616 00:44:41,984 --> 00:44:45,924 There will only be the difference, which is 0.3 left, 617 00:44:45,924 --> 00:44:49,787 so it will be quite dark. But I will still call that 618 00:44:49,787 --> 00:44:52,893 constructive and destructive interference. 619 00:44:52,893 --> 00:44:57,439 The question now is what is the phase difference between this 620 00:44:57,439 --> 00:45:02,380 light here and this light there? Phase difference, 621 00:45:02,380 --> 00:45:06,442 that is what matters. Well, let's first look at the 622 00:45:06,442 --> 00:45:10,016 difference in path. The difference in path is 623 00:45:10,016 --> 00:45:14,647 clearly 2d because this light has to go in this direction. 624 00:45:14,647 --> 00:45:18,058 That is it has to come back. So that is 2d. 625 00:45:18,058 --> 00:45:21,876 Now is the question, how many wavelengths fit on 626 00:45:21,876 --> 00:45:24,719 that separation on that distance 2d? 627 00:45:24,719 --> 00:45:27,075 Wavelengths in oil, of course. 628 00:45:27,075 --> 00:45:31,488 That is what matters. That is this many. 629 00:45:31,488 --> 00:45:34,134 What now is the phase difference? 630 00:45:34,134 --> 00:45:37,111 Now you have to multiply this by 2pi. 631 00:45:37,111 --> 00:45:41,741 Because, if this were three, then the phase difference is 632 00:45:41,741 --> 00:45:45,462 three times 2pi. For each wavelength you get a 633 00:45:45,462 --> 00:45:49,596 phase difference 2pi. Now you have to multiply this 634 00:45:49,596 --> 00:45:51,664 by 2pi. Now you may think, 635 00:45:51,664 --> 00:45:54,475 and that would not be unreasonable, 636 00:45:54,475 --> 00:45:58,196 that this is the phase difference between this 637 00:45:58,196 --> 00:46:04,434 radiation and this radiation. But what have you overlooked 638 00:46:04,434 --> 00:46:06,000 when you do that? 639 00:46:06,000 --> 00:46:12,000 640 00:46:12,000 --> 00:46:16,083 That is right. You have overlooked the fact 641 00:46:16,083 --> 00:46:21,916 that there is here a flip of 180 degrees which does not occur 642 00:46:21,916 --> 00:46:25,125 here. There is an additional pi in 643 00:46:25,125 --> 00:46:29,791 the phase difference, so I am going to write down 644 00:46:29,791 --> 00:46:37,020 here now the total delta. Delta is now 4pi times d. 645 00:46:37,020 --> 00:46:42,649 Now, I can write this as lambda oil. 646 00:46:42,649 --> 00:46:48,277 It is perfectly fine. I can do that. 647 00:46:48,277 --> 00:46:54,710 I would prefer to write it as lambda air. 648 00:46:54,710 --> 00:47:03,877 And then all I have to do, of course, is define the lambda 649 00:47:03,877 --> 00:47:09,827 air by the index of refraction of oil. 650 00:47:09,827 --> 00:47:16,742 And so, yes, I could have written lambda oil 651 00:47:16,742 --> 00:47:24,622 here, but I would prefer to write here lambda air. 652 00:47:24,622 --> 00:47:30,454 47:00 But, if someone wanted to be 653 00:47:30,454 --> 00:47:35,189 nasty to you on an exam, they could make n3 larger than 654 00:47:35,189 --> 00:47:37,643 n2. What would change in this 655 00:47:37,643 --> 00:47:40,449 equation? The pi would disappear. 656 00:47:40,449 --> 00:47:44,219 That is all. The rest would remain the same. 657 00:47:44,219 --> 00:47:47,989 Don't think of this as a universal equation. 658 00:47:47,989 --> 00:47:52,460 There is no such thing. This pi is the result of the 659 00:47:52,460 --> 00:47:57,983 fact that here n2 is larger than n1, but that here n3 is smaller 660 00:47:57,983 --> 00:48:03,652 than n2. That is why you get only pi. 661 00:48:03,652 --> 00:48:10,695 Now, we are in business. We now have here the criterion 662 00:48:10,695 --> 00:48:15,000 for constructive interference. 663 00:48:15,000 --> 00:48:27,000 664 00:48:27,000 --> 00:48:31,157 Thank you very much. How could I? 665 00:48:31,157 --> 00:48:33,885 It is d, right? Uh-oh. 666 00:48:33,885 --> 00:48:38,562 My Greek heritage somehow. All right. 667 00:48:38,562 --> 00:48:42,070 We agree that, in this case, 668 00:48:42,070 --> 00:48:47,917 it is the phase angle. Now we are going to ask 669 00:48:47,917 --> 00:48:55,192 ourselves the question what thickness are required to get 670 00:48:55,192 --> 00:49:03,174 minima and to get maxima? The first very interesting case 671 00:49:03,174 --> 00:49:09,010 is that if you make d equals zero that you kill this whole 672 00:49:09,010 --> 00:49:13,617 term and you only have pi. That is destructive 673 00:49:13,617 --> 00:49:17,098 interference. There is no question, 674 00:49:17,098 --> 00:49:21,092 destructive. Now, you may say zero film, 675 00:49:21,092 --> 00:49:26,109 what kind of nonsense is that? Is that meaningful? 676 00:49:26,109 --> 00:49:32,161 Yes, it is very meaningful. I will demonstrate it to you. 677 00:49:32,161 --> 00:49:36,334 I cannot make an oil film zero thickness, but that is not 678 00:49:36,334 --> 00:49:38,718 necessary. All that is necessary, 679 00:49:38,718 --> 00:49:42,966 that this d overlap the oil. It becomes exceedingly small. 680 00:49:42,966 --> 00:49:44,158 For instant, 100. 681 00:49:44,158 --> 00:49:48,182 If I make d 100 of the wavelengths of light in oil then 682 00:49:48,182 --> 00:49:51,685 I am close enough. And you will indeed see then, 683 00:49:51,685 --> 00:49:55,858 I will do this with soap, that in reflection you will see 684 00:49:55,858 --> 00:49:58,764 no light. You get completely destructive 685 00:49:58,764 --> 00:50:02,997 interference. Apart from the fact which I 686 00:50:02,997 --> 00:50:07,065 mentioned that there is an imbalance between the 4.0 and 687 00:50:07,065 --> 00:50:09,580 the 3.7. That is a different issue. 688 00:50:09,580 --> 00:50:13,057 We will even be able to enjoy the d equals zero. 689 00:50:13,057 --> 00:50:15,498 But now I want to be more general. 690 00:50:15,498 --> 00:50:19,863 I want to give you a feeling quantitatively for what you are 691 00:50:19,863 --> 00:50:23,117 going to look at, and also a feeling for what 692 00:50:23,117 --> 00:50:28,000 thicknesses are really necessary to see these colors. 693 00:50:28,000 --> 00:50:35,966 And so I am going to address now only the case of destructive 694 00:50:35,966 --> 00:50:41,145 interference. And I am going to choose a 695 00:50:41,145 --> 00:50:47,784 lambda in air of 400 nanometers. That is my choice. 696 00:50:47,784 --> 00:50:54,954 That means that my lambda in oil is 400 divided by 1.5, 697 00:50:54,954 --> 00:51:02,866 which is about 267 nanometers. And now, I want to calculate 698 00:51:02,866 --> 00:51:08,293 for what thickness is d, I am going to get destructive 699 00:51:08,293 --> 00:51:12,491 inference. I am going to set this equal to 700 00:51:12,491 --> 00:51:16,484 2n minus one times pi. That is all I do. 701 00:51:16,484 --> 00:51:21,092 And I choose n equals one, and what do I find? 702 00:51:21,092 --> 00:51:24,266 I find d equals zero. Of course. 703 00:51:24,266 --> 00:51:30,000 We knew that already. We just discussed that. 704 00:51:30,000 --> 00:51:34,921 Because, if n equals one, then that requires that delta 705 00:51:34,921 --> 00:51:37,291 is pi. And, if delta is pi, 706 00:51:37,291 --> 00:51:41,484 then this term is zero. You see that d is zero. 707 00:51:41,484 --> 00:51:46,406 You can immediately see that. Now we take n equals two. 708 00:51:46,406 --> 00:51:51,510 And, if you substitute for n equals two in this equation, 709 00:51:51,510 --> 00:51:55,338 you get 3pi. And this is already pi so this 710 00:51:55,338 --> 00:51:59,166 has to be 2pi. And so, if now you calculate 711 00:51:59,166 --> 00:52:05,000 what d is, you will find that d is 133 nanometers. 712 00:52:05,000 --> 00:52:08,915 Well, that is obvious. It obviously must be half of 713 00:52:08,915 --> 00:52:11,578 the wavelength of the light in oil. 714 00:52:11,578 --> 00:52:15,415 Because, if the wavelength of light in oil is 267, 715 00:52:15,415 --> 00:52:18,391 you have to travel this distance twice. 716 00:52:18,391 --> 00:52:21,367 And, if you travel this distance twice, 717 00:52:21,367 --> 00:52:26,144 that is exactly one wavelength. And one wavelength corresponds 718 00:52:26,144 --> 00:52:31,000 to a phase angle of 2pi. And so, here you get 2pi. 719 00:52:31,000 --> 00:52:34,411 And it is this pi that kills you. 720 00:52:34,411 --> 00:52:39,955 That gives you the darkness. And then n equals three, 721 00:52:39,955 --> 00:52:45,500 you will find that d is approximately 267 nanometers. 722 00:52:45,500 --> 00:52:49,551 Now 2d translates into two wavelengths. 723 00:52:49,551 --> 00:52:53,709 Now I am going to choose this thickness. 724 00:52:53,709 --> 00:53:00,000 And, with that thickness now, I am going to work. 725 00:53:00,000 --> 00:53:07,280 And I am going to radiate white light onto depth thickness. 726 00:53:07,280 --> 00:53:14,811 I select that one for which I already know that in blue light 727 00:53:14,811 --> 00:53:19,205 I kill it, destructive interference. 728 00:53:19,205 --> 00:53:25,857 We now have white light. And we have d 133 nanometers, 729 00:53:25,857 --> 00:53:35,000 so that is nonnegotiable. D is 133 nanometers. 730 00:53:35,000 --> 00:53:47,250 I am going to write here lambda air in nanometers, 731 00:53:47,250 --> 00:54:02,000 I am going to write here delta, and I am going to write here 732 00:54:02,000 --> 00:54:11,000 the cosine squared of delta over two. 733 00:54:11,000 --> 00:54:21,250 That is a measure for the light intensity. 734 00:54:21,250 --> 00:54:32,000 It is proportional to that. 54:00 735 00:54:32,000 --> 00:54:37,244 We have green. Now I am going to also shine on 736 00:54:37,244 --> 00:54:42,721 it green light 500 nanometers. I know what d is. 737 00:54:42,721 --> 00:54:48,548 That is nonnegotiable. And now my lambda air is now 738 00:54:48,548 --> 00:54:53,909 500 nanometers. And so I can calculate now what 739 00:54:53,909 --> 00:54:57,872 delta is. Untouched by human hands, 740 00:54:57,872 --> 00:55:03,000 very simple. And I find now 2.6pi. 741 00:55:03,000 --> 00:55:10,758 And, if now I take cosine square of delta over two, 742 00:55:10,758 --> 00:55:16,965 I find 0.35. But now I go to red and I do 743 00:55:16,965 --> 00:55:23,948 650 nanometers. Now I calculate what delta is, 744 00:55:23,948 --> 00:55:30,000 and I find 2.2pi. Now I get 0.90. 745 00:55:30,000 --> 00:55:33,171 Now I am very close to a maximum. 746 00:55:33,171 --> 00:55:39,216 And so this is very convincing that if you make your film this 747 00:55:39,216 --> 00:55:42,882 thin that it will look distinctly red. 748 00:55:42,882 --> 00:55:47,738 Maybe not perfect red, but it will look distinctly 749 00:55:47,738 --> 00:55:51,207 red. The red color dominates and the 750 00:55:51,207 --> 00:55:57,252 blue, practically absent at all, apart then from the imbalance 751 00:55:57,252 --> 00:56:03,000 that you may have between the 4.0 and the 3.7. 752 00:56:03,000 --> 00:56:07,217 The question that I have asked over the years at the final exam 753 00:56:07,217 --> 00:56:11,163 for 8.03 is whether thin film interference is the result of 754 00:56:11,163 --> 00:56:15,040 the difference between the index of refraction between the 755 00:56:15,040 --> 00:56:18,102 different colors? And the answer is absolutely 756 00:56:18,102 --> 00:56:20,006 not. The index of refraction, 757 00:56:20,006 --> 00:56:23,680 I have taken the same for all colors, for all practical 758 00:56:23,680 --> 00:56:25,925 purposes. That difference does not 759 00:56:25,925 --> 00:56:29,190 explain the colors. The color is explained by the 760 00:56:29,190 --> 00:56:33,000 different path length divided by lambda. 761 00:56:33,000 --> 00:56:38,023 That is why you see colors. It has nothing to do with the 762 00:56:38,023 --> 00:56:41,700 index of refraction. You can see thin film 763 00:56:41,700 --> 00:56:46,186 interference in soap, and I will try to demonstrate 764 00:56:46,186 --> 00:56:49,415 that. You can see it in oil spills on 765 00:56:49,415 --> 00:56:52,644 the road. I will show you an example. 766 00:56:52,644 --> 00:56:55,694 An only thin film gives you colors. 767 00:56:55,694 --> 00:57:00,000 Thick film does not give you colors. 768 00:57:00,000 --> 00:57:04,990 If you go through the exercise, which I want you to do, 769 00:57:04,990 --> 00:57:09,333 to make d 0.1 millimeter, which is, by practical 770 00:57:09,333 --> 00:57:14,415 standards, still very thin, you will not see any colors. 771 00:57:14,415 --> 00:57:19,590 And the reason is that if you go to this equation and you 772 00:57:19,590 --> 00:57:24,765 calculate for which colors, you will see constructive and 773 00:57:24,765 --> 00:57:30,230 destructive interference. There are so many colors in the 774 00:57:30,230 --> 00:57:34,546 visible spectrum for which you get constructive interference, 775 00:57:34,546 --> 00:57:38,287 so many, you will see, that your brains will tell you 776 00:57:38,287 --> 00:57:42,100 that you see white lights. There is not one color that 777 00:57:42,100 --> 00:57:44,690 dominates. The n values that you will 778 00:57:44,690 --> 00:57:47,784 need, by the way, are going to be very high. 779 00:57:47,784 --> 00:57:50,086 They are in the range 500 to 700. 780 00:57:50,086 --> 00:57:54,474 Huge values of n are necessary in order to get the destructive 781 00:57:54,474 --> 00:57:58,000 and the constructive interference. 782 00:57:58,000 --> 00:58:01,965 Go through this exercise on your own, and you will see very 783 00:58:01,965 --> 00:58:05,794 quickly that so many colors constructively interfere that 784 00:58:05,794 --> 00:58:09,555 the film will look white. The first thing that I want to 785 00:58:09,555 --> 00:58:12,153 do, I want to try, it is not that easy, 786 00:58:12,153 --> 00:58:16,461 is to make you see these colors with soap, which all of you must 787 00:58:16,461 --> 00:58:18,991 have seen. In fact, if you just take a 788 00:58:18,991 --> 00:58:22,683 shower and you soap yourself, the bubbles themselves in 789 00:58:22,683 --> 00:58:25,145 reflection already have these colors. 790 00:58:25,145 --> 00:58:30,000 I will try to make one that is slightly larger in size. 791 00:58:30,000 --> 00:58:34,333 But we do not always succeed. Let's try that. 792 00:58:34,333 --> 00:58:37,878 The idea is I want you to see colors. 793 00:58:37,878 --> 00:58:43,000 Well, maybe you did but that was a little quick. 794 00:58:43,000 --> 00:58:53,000 795 00:58:53,000 --> 00:58:55,177 Now you see colors. Did you? 796 00:58:55,177 --> 00:58:58,000 That was all I wanted you to see. 797 00:58:58,000 --> 00:59:13,000 798 00:59:13,000 --> 00:59:15,808 Did you see them? If you had said no, 799 00:59:15,808 --> 00:59:19,163 I would have done it again, but now I won't. 800 00:59:19,163 --> 00:59:21,737 Do you want me to do it once more? 801 00:59:21,737 --> 00:59:24,000 It is not easy, though. 802 00:59:24,000 --> 00:59:37,000 803 00:59:37,000 --> 00:59:38,450 Now you say it, right? 804 00:59:38,450 --> 00:59:39,486 Yeah. Thank you. 805 00:59:39,486 --> 00:59:43,492 Soap bubbles give these colors because they are exceedingly 806 00:59:43,492 --> 00:59:45,909 thin. It also gives you a feeling of 807 00:59:45,909 --> 00:59:49,294 how thin they have to be. Really, they can be much 808 00:59:49,294 --> 00:59:52,678 thicker than a few times the wavelengths of light. 809 00:59:52,678 --> 00:59:55,993 If they become much thicker, you lose the colors, 810 00:59:55,993 --> 1:00:00,000 for the reason that I just mentioned to you. 811 1:00:00,000 --> 1:00:05,574 There are too many colors that constructively interfere, 812 1:00:05,574 --> 1:00:08,918 and then you don't see it anymore. 813 1:00:08,918 --> 1:00:13,682 Now I have a demonstration which is very tricky. 814 1:00:13,682 --> 1:00:18,344 It works most of the time but not all the time. 815 1:00:18,344 --> 1:00:23,108 I am going to make a soap film on a metal frame. 816 1:00:23,108 --> 1:00:26,554 It is going to be done in this box. 817 1:00:26,554 --> 1:00:32,041 We have a metal frame. And we dip it in soap, 818 1:00:32,041 --> 1:00:35,011 and so you get a soap film there. 819 1:00:35,011 --> 1:00:40,302 And then gravity will make the film thinner at the top and 820 1:00:40,302 --> 1:00:44,850 thicker at the bottom. As you make a cross-section 821 1:00:44,850 --> 1:00:49,676 through here then the film for this is then the soap. 822 1:00:49,676 --> 1:00:54,502 I can make this so thin that it is completely dark in 823 1:00:54,502 --> 1:00:58,029 reflection. That is what I promised you 824 1:00:58,029 --> 1:01:03,451 that d is not zero. But d divided by lambda is so 825 1:01:03,451 --> 1:01:07,679 small, maybe 120 or 130 of the wavelength of light, 826 1:01:07,679 --> 1:01:10,553 that it will turn completely black. 827 1:01:10,553 --> 1:01:14,865 And, as this proceeds, in the beginning the top will 828 1:01:14,865 --> 1:01:18,839 show you colors. It goes through these phases of 829 1:01:18,839 --> 1:01:24,166 colors, then it gets thinner and thinner and thinner and then it 830 1:01:24,166 --> 1:01:28,478 turns completely black. I will show it to you upside 831 1:01:28,478 --> 1:01:31,617 down. But don't get confused. 832 1:01:31,617 --> 1:01:34,712 It has to do with the way that we project it. 833 1:01:34,712 --> 1:01:38,580 The thinnest part of the film will be low on the screen, 834 1:01:38,580 --> 1:01:43,010 and then the part of the screen that is the thickest of the film 835 1:01:43,010 --> 1:01:45,894 will be the upper part. As you look at it, 836 1:01:45,894 --> 1:01:49,762 appreciate the fact that when we go into the dark phase, 837 1:01:49,762 --> 1:01:52,857 if we succeed, that you are talking about the 838 1:01:52,857 --> 1:01:56,303 thickness of the film, which is many times smaller 839 1:01:56,303 --> 1:02:00,312 than the wavelengths of light, which is typically a half a 840 1:02:00,312 --> 1:02:05,295 micron. Let me first get the film up. 841 1:02:05,295 --> 1:02:09,272 Actually, let's not get the film up. 842 1:02:09,272 --> 1:02:14,500 Let's just get the light up. Let's turn it off. 843 1:02:14,500 --> 1:02:20,977 It has to be completely dark. Didn't we agree it has to be 844 1:02:20,977 --> 1:02:27,000 completely dark? I am going to dip it in here. 845 1:02:27,000 --> 1:02:40,000 846 1:02:40,000 --> 1:02:44,405 The bottom is the thinnest. And you see the colors. 847 1:02:44,405 --> 1:02:48,193 You see them? And you see that the thickness 848 1:02:48,193 --> 1:02:52,863 changes over the film, and that is why there are areas 849 1:02:52,863 --> 1:02:57,885 where the red dominates and then other areas where the red 850 1:02:57,885 --> 1:03:02,717 dominates again. And so, the bottom gets thinner 851 1:03:02,717 --> 1:03:07,276 and thinner and thinner. The white light that you see 852 1:03:07,276 --> 1:03:10,695 now here is the result of Markos' trick. 853 1:03:10,695 --> 1:03:16,043 In order to avoid that the film bursts, he has to put glycerin 854 1:03:16,043 --> 1:03:19,375 in the soap. And what you see here is a 855 1:03:19,375 --> 1:03:24,197 reflection of the layer of glycerin that forms on top of 856 1:03:24,197 --> 1:03:27,528 the soap. We will have to wait a little 857 1:03:27,528 --> 1:03:32,000 bit for that glycerin to also go down. 858 1:03:32,000 --> 1:03:34,632 And then you will see the pure soap. 859 1:03:34,632 --> 1:03:38,167 And that is then the area. And, boy, look at it, 860 1:03:38,167 --> 1:03:40,874 black. That area that you are looking 861 1:03:40,874 --> 1:03:44,860 here now is much thinner than the wavelength of light. 862 1:03:44,860 --> 1:03:48,470 And so, the reflection is, well, when I say zero, 863 1:03:48,470 --> 1:03:52,908 of course, that depends again on the asymmetry that you have 864 1:03:52,908 --> 1:03:57,345 that we discussed here on the blackboard between the 4.0 and 865 1:03:57,345 --> 1:04:01,111 the 3.7. It is a little different for 866 1:04:01,111 --> 1:04:03,600 water than it is for oil. You see? 867 1:04:03,600 --> 1:04:06,240 Black. You can just look at it for a 868 1:04:06,240 --> 1:04:07,673 while. It may burst. 869 1:04:07,673 --> 1:04:10,917 You also see some very interesting thin film 870 1:04:10,917 --> 1:04:15,291 interference in the glycerin itself, which is the oil layer 871 1:04:15,291 --> 1:04:18,761 that is on top of the soap. It is not uncommon. 872 1:04:18,761 --> 1:04:23,060 You may remember that when you see oil spills on the road, 873 1:04:23,060 --> 1:04:26,756 that you sometimes see these very nice structures, 874 1:04:26,756 --> 1:04:30,000 brownish, reddish, bluish. 875 1:04:30,000 --> 1:04:34,551 But look at this here. Isn't that incredible? 876 1:04:34,551 --> 1:04:39,517 Isn't that fabulous? Boy, the film is holding up, 877 1:04:39,517 --> 1:04:42,206 Markos. We cannot complain. 878 1:04:42,206 --> 1:04:47,793 You see these nice oil structures here in the glycerin? 879 1:04:47,793 --> 1:04:53,793 The nightmare that we have always is that it pops too fast, 880 1:04:53,793 --> 1:05:00,000 and then you have to do again and again and again. 881 1:05:00,000 --> 1:05:04,344 And our experience is if somehow the soap is not right 882 1:05:04,344 --> 1:05:08,278 then, in general, no matter how often you try it, 883 1:05:08,278 --> 1:05:11,229 it keeps popping. Look how beautiful, 884 1:05:11,229 --> 1:05:16,311 look at how incredibly nice the [CNT?] film interference in the 885 1:05:16,311 --> 1:05:18,278 glycerin. And then, here, 886 1:05:18,278 --> 1:05:22,540 the total darkness because the thickness of the film, 887 1:05:22,540 --> 1:05:27,049 of the soap is many times smaller than the wavelength of 888 1:05:27,049 --> 1:05:30,000 the light in the soap. 889 1:05:30,000 --> 1:05:38,000 890 1:05:38,000 --> 1:05:40,641 I never get bored when I look at it. 891 1:05:40,641 --> 1:05:44,339 Isn't it gorgeous? Think about the physics that is 892 1:05:44,339 --> 1:05:46,000 going on in there. 893 1:05:46,000 --> 1:05:52,000 894 1:05:52,000 --> 1:05:55,008 All right. Let's look at some other 895 1:05:55,008 --> 1:05:59,256 phenomenon of interference. Try to remember this. 896 1:05:59,256 --> 1:06:02,000 This is a gorgeous picture. 897 1:06:02,000 --> 1:06:12,000 898 1:06:12,000 --> 1:06:18,721 There is a whole family of ways that you can see [CNT?] film 899 1:06:18,721 --> 1:06:23,164 interference. And, in your problem sets, 900 1:06:23,164 --> 1:06:30,000 we give you two problems to work on that quantitatively. 901 1:06:30,000 --> 1:06:32,977 And there is also this take-home experiment that 902 1:06:32,977 --> 1:06:36,081 supports the problem. And I want to concentrate on 903 1:06:36,081 --> 1:06:39,438 those, too, because they are very nice to demonstrate. 904 1:06:39,438 --> 1:06:42,733 I can demonstrate them to you in monochromatic light. 905 1:06:42,733 --> 1:06:46,470 Whenever you have monochromatic light, that means you do not 906 1:06:46,470 --> 1:06:49,004 have white light. Of course, the fringes, 907 1:06:49,004 --> 1:06:52,298 we call these fringes, the dark and the bright areas, 908 1:06:52,298 --> 1:06:55,339 we call them fringes, show up so much better than 909 1:06:55,339 --> 1:06:59,718 when you have white light. Because then you get the 910 1:06:59,718 --> 1:07:04,084 superposition of all the colors. And the maxima and the minima 911 1:07:04,084 --> 1:07:07,735 depend on the color. Remember, from one and the same 912 1:07:07,735 --> 1:07:11,672 thickness, you may have destructive interference for one 913 1:07:11,672 --> 1:07:15,895 color, but you have constructive interference for the other. 914 1:07:15,895 --> 1:07:19,976 If you do it in monochromatic light, you can see this very 915 1:07:19,976 --> 1:07:22,839 dramatically. And then one of the classic 916 1:07:22,839 --> 1:07:27,420 examples is, and you can do that at home, and you should do it at 917 1:07:27,420 --> 1:07:31,000 home, is the two microscope slides. 918 1:07:31,000 --> 1:07:34,782 You have a microscope slide which is way too thick to give 919 1:07:34,782 --> 1:07:38,631 you thin film interference. Don't even think about the fact 920 1:07:38,631 --> 1:07:41,617 that there is glass there. It will do nothing. 921 1:07:41,617 --> 1:07:44,537 Way too thick. Probably tens of a millimeter. 922 1:07:44,537 --> 1:07:47,324 It is ridiculous. And then there is another 923 1:07:47,324 --> 1:07:50,643 microscope glass here. And I exaggerate highly this 924 1:07:50,643 --> 1:07:54,890 area in between because they are lying flat on top of each other, 925 1:07:54,890 --> 1:07:57,279 but there is an air gap between them. 926 1:07:57,279 --> 1:08:02,492 And this is that air gap. And it is this air gap now that 927 1:08:02,492 --> 1:08:05,869 acts like a thin film. You are going to get 928 1:08:05,869 --> 1:08:10,371 interference now of light that bounces off this layer and 929 1:08:10,371 --> 1:08:14,633 bounces off this layer. And so, it is this thickness d 930 1:08:14,633 --> 1:08:19,135 now of the air layer that determines the darkness and the 931 1:08:19,135 --> 1:08:23,718 brightness, destructive and the constructive interference. 932 1:08:23,718 --> 1:08:27,175 And you're being asked, in your problem set, 933 1:08:27,175 --> 1:08:32,000 to calculate where those maxima and minima are. 934 1:08:32,000 --> 1:08:36,419 And you can also do that then with the take- home experiment. 935 1:08:36,419 --> 1:08:39,218 And I will demonstrate it very shortly. 936 1:08:39,218 --> 1:08:42,680 There is another one, a classic that I will also 937 1:08:42,680 --> 1:08:46,805 demonstration and which is also part of your problem set. 938 1:08:46,805 --> 1:08:50,046 This is a glass lens. The curvature is highly 939 1:08:50,046 --> 1:08:53,214 exaggerated. And then here is a glass plate. 940 1:08:53,214 --> 1:08:56,970 And so here is now an air gap. And now, if you shine 941 1:08:56,970 --> 1:09:00,138 monochromatic light on this then, of course, 942 1:09:00,138 --> 1:09:05,000 seen from above there is symmetry about this line. 943 1:09:05,000 --> 1:09:08,708 You will see rings of darkness and rings of brightness. 944 1:09:08,708 --> 1:09:12,692 And these rings of darkness and brightness will be very far 945 1:09:12,692 --> 1:09:16,469 apart here and they will be very closely together there. 946 1:09:16,469 --> 1:09:20,109 This is part of your homework assignment to show that. 947 1:09:20,109 --> 1:09:24,368 It has to do with the fact that the change here in thickness is 948 1:09:24,368 --> 1:09:28,351 very little because of the curvature and the change here is 949 1:09:28,351 --> 1:09:32,005 much faster. This is called the Newton 950 1:09:32,005 --> 1:09:34,732 rings. And then, when you just walk 951 1:09:34,732 --> 1:09:38,582 out on the street, preferably on a day when it is 952 1:09:38,582 --> 1:09:42,914 not sunny but is cloudy, you look on the ground and you 953 1:09:42,914 --> 1:09:46,042 see oil spills. And they have phenomenal 954 1:09:46,042 --> 1:09:49,090 colors. I would like to show you one of 955 1:09:49,090 --> 1:09:53,663 my own pictures that I took some time ago of an oil spill. 956 1:09:53,663 --> 1:09:57,673 Let me first turn it on, and then we will make it a 957 1:09:57,673 --> 1:10:00,000 little darker. 958 1:10:00,000 --> 1:10:10,000 959 1:10:10,000 --> 1:10:15,279 I will make it a little darker so that you can see it better. 960 1:10:15,279 --> 1:10:18,184 I photograph countless oil spills. 961 1:10:18,184 --> 1:10:22,055 And, actually, the reason why I photograph so 962 1:10:22,055 --> 1:10:27,336 many is that always I try to reconstruct thickness of the oil 963 1:10:27,336 --> 1:10:30,152 on the road. And I was never very 964 1:10:30,152 --> 1:10:33,037 successful. For one thing, 965 1:10:33,037 --> 1:10:37,259 you can only get a spot like this if a car loses some oil. 966 1:10:37,259 --> 1:10:41,481 So there is one drop of oil and then it starts to run out. 967 1:10:41,481 --> 1:10:45,555 You would always think that it is thickest here and then 968 1:10:45,555 --> 1:10:48,148 gradually tapers off near the edges. 969 1:10:48,148 --> 1:10:52,666 And I have always wondered why then here you see so much white 970 1:10:52,666 --> 1:10:55,259 light. You would expect that perhaps 971 1:10:55,259 --> 1:10:59,972 it would be much darker. And so, many attempts that I 972 1:10:59,972 --> 1:11:04,059 have made to convince myself that it is really thicker here 973 1:11:04,059 --> 1:11:08,145 than there, many have failed. But there is no question what 974 1:11:08,145 --> 1:11:11,175 you are looking at, of course, is a striking 975 1:11:11,175 --> 1:11:13,500 example of thin film interference. 976 1:11:13,500 --> 1:11:17,022 And, not only that, but you know that this film has 977 1:11:17,022 --> 1:11:20,334 to be extremely thin, clearly very close in that 978 1:11:20,334 --> 1:11:24,490 range of 100 to 200 nanometers. Otherwise, you would not see 979 1:11:24,490 --> 1:11:29,000 over the entire area such beautifully equal blue. 980 1:11:29,000 --> 1:11:32,506 And here you see this reddish that is not pure, 981 1:11:32,506 --> 1:11:35,098 it never is, and very well-defined. 982 1:11:35,098 --> 1:11:39,291 Clearly, it has a spherical symmetry so it is probably a 983 1:11:39,291 --> 1:11:42,874 drop that then spread out. And so, now I want to 984 1:11:42,874 --> 1:11:46,000 demonstrate then your microscope. 985 1:11:46,000 --> 1:11:54,000 986 1:11:54,000 --> 1:11:58,842 The microscopes we have here, the microscope glass. 987 1:11:58,842 --> 1:12:03,104 Turn on the CD-TV there, if everything works. 988 1:12:03,104 --> 1:12:07,366 And we do it with [neo?] monochromatic light, 989 1:12:07,366 --> 1:12:13,178 this light from a mercury lamp which is largely in the green. 990 1:12:13,178 --> 1:12:17,246 So you see very well-established maxima and 991 1:12:17,246 --> 1:12:19,764 minima. That is the beauty, 992 1:12:19,764 --> 1:12:23,348 if you do it with monochromatic light. 993 1:12:23,348 --> 1:12:28,869 And so these two microscope slides are just on top of each 994 1:12:28,869 --> 1:12:32,913 other. But, yeah, there is always a 995 1:12:32,913 --> 1:12:37,013 little bit of air in between. And so, you can sort of make up 996 1:12:37,013 --> 1:12:40,430 your own mind that they may be touching each other, 997 1:12:40,430 --> 1:12:43,847 really touching where they have practically no air, 998 1:12:43,847 --> 1:12:47,742 for reasons that are not so clear, which must be somewhere 999 1:12:47,742 --> 1:12:49,519 here. And that is where the 1000 1:12:49,519 --> 1:12:52,799 separation between the dark lines is the largest. 1001 1:12:52,799 --> 1:12:55,874 And then, somehow here, you see these fringes, 1002 1:12:55,874 --> 1:12:58,949 the areas of darkness and maximum constructive 1003 1:12:58,949 --> 1:13:02,944 interference. When I push on one side of 1004 1:13:02,944 --> 1:13:06,385 these slides and, of course, I change this very 1005 1:13:06,385 --> 1:13:10,425 subtle structure of light in between these microscopes, 1006 1:13:10,425 --> 1:13:14,763 you will then see immediate change in the separation of the 1007 1:13:14,763 --> 1:13:17,307 lines. I am just pushing on it now, 1008 1:13:17,307 --> 1:13:20,000 changing the configuration. 1009 1:13:20,000 --> 1:13:28,000 1010 1:13:28,000 --> 1:13:31,106 Now this is the point of contact, more or less. 1011 1:13:31,106 --> 1:13:35,091 And you can really see that. What I am looking at right now, 1012 1:13:35,091 --> 1:13:38,130 I see it just as well as you can see it there. 1013 1:13:38,130 --> 1:13:42,115 And so, you should really do that experiment at home because 1014 1:13:42,115 --> 1:13:44,749 it is part of your take-home experiment. 1015 1:13:44,749 --> 1:13:48,531 I mentioned in the write-up that it is not necessary that 1016 1:13:48,531 --> 1:13:51,773 you turn in a written summary of your experiment, 1017 1:13:51,773 --> 1:13:54,880 so it won't count for your problem set as such, 1018 1:13:54,880 --> 1:13:59,000 but I will hold you responsible on the final. 1019 1:13:59,000 --> 1:14:03,544 So I advise you to do the experiments because I may ask a 1020 1:14:03,544 --> 1:14:08,657 few questions which you can only answer if you actually did this 1021 1:14:08,657 --> 1:14:11,335 experiment. Here I have two flats, 1022 1:14:11,335 --> 1:14:15,068 like the microscopes, but thicker so they are a 1023 1:14:15,068 --> 1:14:18,719 little sturdier. And this is the way opticians 1024 1:14:18,719 --> 1:14:23,264 actually test their optics. There is here an air gap that 1025 1:14:23,264 --> 1:14:26,510 grows in size. And when I push now on it, 1026 1:14:26,510 --> 1:14:33,677 you see now I make them flat. Now I see almost no fringes. 1027 1:14:33,677 --> 1:14:40,377 Now I push so hard that I open them again like this. 1028 1:14:40,377 --> 1:14:47,733 I make one go up like this. And now you see many fringes. 1029 1:14:47,733 --> 1:14:54,957 You've got a huge angle now. And then I want to show you 1030 1:14:54,957 --> 1:15:01,000 this, which is called the Newton rings. 1031 1:15:01,000 --> 1:15:05,566 We have it arranged in such a way that with some set screws we 1032 1:15:05,566 --> 1:15:10,208 can actually push down on this, so we can also make the air gap 1033 1:15:10,208 --> 1:15:13,802 larger and smaller. And that we are going to show 1034 1:15:13,802 --> 1:15:15,000 you there. 1035 1:15:15,000 --> 1:15:27,000 1036 1:15:27,000 --> 1:15:29,615 I can actually make it even darker. 1037 1:15:29,615 --> 1:15:33,615 That may be nicer for you. You already see the center 1038 1:15:33,615 --> 1:15:37,384 portion of the contact. Where the contact is here, 1039 1:15:37,384 --> 1:15:40,153 the air gap doesn't change very much. 1040 1:15:40,153 --> 1:15:43,538 And, therefore, you will see that the spacing 1041 1:15:43,538 --> 1:15:47,461 of the rings is large. But, when you go further out, 1042 1:15:47,461 --> 1:15:52,076 then the spacing gets smaller and it becomes difficult to see 1043 1:15:52,076 --> 1:15:54,769 perhaps. But, if you are close here, 1044 1:15:54,769 --> 1:15:59,230 I can really see the rings here, much closer than the rings 1045 1:15:59,230 --> 1:16:03,608 here. And I can put some pressure on 1046 1:16:03,608 --> 1:16:08,304 there, even though it has changed the geometry a little 1047 1:16:08,304 --> 1:16:11,260 bit. Now I squeeze a little harder. 1048 1:16:11,260 --> 1:16:15,695 I hope I don't break it. And when you squeeze harder 1049 1:16:15,695 --> 1:16:20,652 they touch each other over a larger surface in the center. 1050 1:16:20,652 --> 1:16:23,347 You see? Therefore, it opens up. 1051 1:16:23,347 --> 1:16:27,608 Over this very large area now the gap is the same. 1052 1:16:27,608 --> 1:16:32,391 And, when I loosen it up, then you expect those rings to 1053 1:16:32,391 --> 1:16:36,899 become smaller. And this is something, 1054 1:16:36,899 --> 1:16:41,820 I believe, that is also part of your take-home experiment. 1055 1:16:41,820 --> 1:16:45,965 I cannot wait to see the result of the mini-quiz. 1056 1:16:45,965 --> 1:16:50,714 I will be proud of you if you did well, let me tell you. 1057 1:16:50,714 --> 1:16:52,009 And I see then, I hope, on Thursday.