WEBVTT 00:00:00.085 --> 00:00:02.828 This is College Physics Answers with Shaun Dychko. 00:00:03.520 --> 00:00:05.720 An oil slick is on top of some water 00:00:06.062 --> 00:00:09.600 and the oil has a thickness of 120 nanometers we're told. 00:00:10.125 --> 00:00:13.948 It has an index of refraction of 1.4 and it's on top of water 00:00:13.948 --> 00:00:16.068 which has an index refraction of 1.3. 00:00:17.605 --> 00:00:23.074 The light ray of white light which means it consists of all the different colors 00:00:23.074 --> 00:00:28.200 mixed together comes in perpendicular to the oil slick 00:00:28.342 --> 00:00:33.268 but I've drawn it on an angle just so we can see the two reflected rays. 00:00:33.542 --> 00:00:36.611 And then I can distinguish between them in the drawing. 00:00:36.714 --> 00:00:39.325 If I were to draw it straight down then I'd have line straight down 00:00:39.325 --> 00:00:41.262 and another line straight up and they would all overlap. 00:00:41.640 --> 00:00:42.880 That would be confusing looking. 00:00:43.948 --> 00:00:47.377 So we imagine that these rays are actually coming straight down 00:00:47.577 --> 00:00:50.354 and then reflecting straight back up and mixing together. 00:00:52.045 --> 00:00:55.108 Now we need to know what is the phase change 00:00:55.108 --> 00:00:57.405 for each of these reflected rays here 00:00:57.857 --> 00:01:01.211 and the phase changes measured in number of wavelengths. 00:01:01.880 --> 00:01:05.942 So it's travelling through air initially which has an index refraction of one 00:01:06.337 --> 00:01:11.102 and when this ray number one bounces off the top of the oil surface 00:01:11.382 --> 00:01:15.062 it will have a phase change of half of a wavelength. 00:01:15.525 --> 00:01:17.702 And that's because it's going from a low index of 00:01:17.702 --> 00:01:22.000 1.0 to a material of high index of 1.4 . 00:01:22.394 --> 00:01:25.988 And so reflecting off of an interface from low to high, 00:01:26.485 --> 00:01:30.097 causes a phase change of half a wavelength. 00:01:32.222 --> 00:01:34.257 And then we have to figure out 00:01:34.257 --> 00:01:39.611 what is the phase shift of this second Ray number two. 00:01:40.954 --> 00:01:43.737 Now when there's a reflection off of this interface 00:01:43.737 --> 00:01:48.520 there will be no phase change due to reflection because it's starting. 00:01:48.520 --> 00:01:53.120 In this material with an index refraction of 1.4 00:01:53.514 --> 00:01:54.805 and bouncing off an interface 00:01:54.805 --> 00:01:57.537 with a material that has a lower index of refraction 00:01:57.794 --> 00:02:03.520 it's going from a high index and reflecting off of interface with a low index material 00:02:03.822 --> 00:02:09.337 produces no phase change as a result of that reflection. 00:02:10.742 --> 00:02:14.108 So we do not have a lambda over two term here. 00:02:15.748 --> 00:02:22.222 This, by the way, is the way I write phase change with this symbol Phi. 00:02:23.205 --> 00:02:27.582 And this is Phi 1 which is the phase change of ray one. 00:02:29.217 --> 00:02:31.034 So the phase change for Ray two is going to be just 00:02:31.034 --> 00:02:35.154 due to the additional path length of this ray 00:02:35.342 --> 00:02:38.251 through the thickness of the oil once and then twice. 00:02:39.382 --> 00:02:42.342 And how many wavelengths will it be shifted. 00:02:43.708 --> 00:02:46.045 Well it's going to be it's going through the thickness twice 00:02:46.045 --> 00:02:49.205 and so it's going to be travelling two times a thickness 00:02:49.828 --> 00:02:53.262 and we're going to divide that by the wavelength 00:02:53.325 --> 00:02:55.028 but I have a subscript n here 00:02:55.285 --> 00:02:59.600 because it's the wavelength in this oil material that we're concerned with. 00:03:00.457 --> 00:03:04.611 So how many wavelengths is this total thickness of 2t, 00:03:04.754 --> 00:03:06.908 this total path length additional 00:03:08.245 --> 00:03:11.388 and so that's 2t divided by the wavelength in the material. 00:03:12.720 --> 00:03:19.862 So the wavelength in the material is the wavelength in an air or a vacuum 00:03:20.205 --> 00:03:22.531 divided by the index of refraction of the material. 00:03:22.800 --> 00:03:27.108 And since we're dividing by this we're going to multiply by it's reciprocal. 00:03:27.108 --> 00:03:29.217 So we're multiplying by n over lambda 00:03:31.668 --> 00:03:35.874 so we have the phase change for Ray two is two times the thickness of the oil 00:03:35.874 --> 00:03:39.262 times Index of refraction of oil divided by the wavelength in air 00:03:39.554 --> 00:03:40.662 times the wavelength in air 00:03:40.960 --> 00:03:46.691 and these cancel giving us a phase shift of two times index refraction of the oil 00:03:47.017 --> 00:03:47.971 times the thickness. 00:03:50.434 --> 00:03:53.102 So constructive interference occurs 00:03:53.777 --> 00:03:59.160 when you have the total phase shift is some integer multiplied by the wavelength 00:04:01.451 --> 00:04:06.960 and so we can add these two phase shifts together. 00:04:07.611 --> 00:04:12.051 And by the way the question asks us what color will the oil b. 00:04:12.114 --> 00:04:14.977 Well it'll be whatever color constructively interference 00:04:16.765 --> 00:04:18.851 because constructive interference increases 00:04:18.851 --> 00:04:20.405 the amplitude of that particular wavelength 00:04:20.405 --> 00:04:22.617 and it will be the one that is more noticeable. 00:04:23.880 --> 00:04:27.120 So the total phase shift is gonna be the sum of the phase shifts 00:04:28.451 --> 00:04:31.011 from each of these rays one and two. 00:04:31.634 --> 00:04:34.850 And that has to equal m lambda for constructive interference. 00:04:35.577 --> 00:04:37.125 And then we substitute for each of these. 00:04:37.125 --> 00:04:40.371 So that's lambda over two phase shift from the reflection at the top of the oil 00:04:40.834 --> 00:04:44.240 plus two times index of refraction of oil times oil thickness, 00:04:44.342 --> 00:04:49.382 phase shift for the second Ray that bounces off the oil water interface. 00:04:49.691 --> 00:04:52.697 And that equals m lambda and then we get rid of fractions 00:04:52.697 --> 00:04:55.851 because fractions are a bit messy looking 00:04:55.851 --> 00:04:57.171 and we multiply everything by two 00:04:57.171 --> 00:05:00.720 and we get lambda plus 4nt equals to m lambda . 00:05:01.434 --> 00:05:03.988 And then subtract lambda from both sides 00:05:05.217 --> 00:05:09.234 and we get and then factor out the lambda as well on this right side now 00:05:09.451 --> 00:05:10.657 and then switch sides around. 00:05:10.982 --> 00:05:15.034 So we have lambda times bracket 2m minus one equals 4nt 00:05:15.691 --> 00:05:18.474 and then divide both sides by 2m minus one 00:05:20.274 --> 00:05:24.022 and so here is the wavelength that will experience constructive interference. 00:05:24.880 --> 00:05:27.685 It's 4nt over 2m minus one 00:05:28.120 --> 00:05:30.942 and because this number m is an integer that we can choose. 00:05:31.188 --> 00:05:33.325 You'll notice that there are many wavelengths 00:05:33.325 --> 00:05:34.948 that will have constructive interference 00:05:35.291 --> 00:05:37.800 but the one we're concerned with is the one that is visible. 00:05:38.348 --> 00:05:40.822 And so we're going to try different values of m . 00:05:40.822 --> 00:05:42.131 Let's try one first of all. 00:05:42.417 --> 00:05:44.348 So we have four times index of refraction of oil 00:05:44.348 --> 00:05:50.925 1.4 times the thickness of the oil slick 120 nanometres divided by two times m 00:05:50.925 --> 00:05:55.737 or to choosing to be one minus one which gives 672 nanometres 00:05:55.937 --> 00:05:58.262 and that is a lucky guess of m equals one 00:05:58.262 --> 00:06:01.011 because this is visible and this is the color red.