1 00:00:19,050 --> 00:00:28,220 Now, in this lecture we will start the analysis of optical waveguides using electromagnetic 2 00:00:28,220 --> 00:00:36,780 theory. Let us first look at types of waveguides, this is the bulk medium. 3 00:00:36,780 --> 00:00:43,989 So, this is not a waveguide, when the dimensions are very large as compared to wavelength. 4 00:00:43,989 --> 00:00:55,280 We can have a waveguide like this where we have a channel or a material in this geometry 5 00:00:55,280 --> 00:01:04,920 and the dimensions are comparable to the wavelength and it is surrounded by a lower index medium 6 00:01:04,920 --> 00:01:17,259 like this. Then in this way I create a channel for light to flow along this. This is a rectangular 7 00:01:17,259 --> 00:01:26,870 channel waveguide. I can have this channel in the form of a cylinder high index refractive 8 00:01:26,870 --> 00:01:36,160 index cylinder which is surrounded by lower refractive index medium. Again all are comparable 9 00:01:36,160 --> 00:01:43,220 to wavelength the dimensions are comparable to wavelength and this is optical fiber waveguide. 10 00:01:43,220 --> 00:01:52,110 And then I can also have a configuration like this we are this width of the channel can 11 00:01:52,110 --> 00:01:56,470 be infinitely extended. So, instead of having this kind of channel, 12 00:01:56,470 --> 00:02:06,709 I can have a slab thin slab of high refractive index. It is sandwiched between 2 other thick 13 00:02:06,709 --> 00:02:16,860 slaps of lower refractive indices. Then this is known as planner waveguide. In these 2, 14 00:02:16,860 --> 00:02:28,000 I have confinement in 2 dimension and propagation in this longitudinal direction. Here I have 15 00:02:28,000 --> 00:02:38,510 confinement only in one direction and propagation in this longitudinal direction. So, we are 16 00:02:38,510 --> 00:02:52,480 going to now do the electromagnetic wave analysis of such kind of media. We had seen that in 17 00:02:52,480 --> 00:02:59,750 infinitely extended medium where there were no such refractive index discontinuities, 18 00:02:59,750 --> 00:03:07,790 the solutions of the wave equations were plane waves. 19 00:03:07,790 --> 00:03:17,850 If you remember that the electric field associated with that wave in an infinitely extended medium 20 00:03:17,850 --> 00:03:35,430 if it is infinitely extended medium. I had seen that E was given by E0 e to the 21 00:03:35,430 --> 00:03:55,140 power i omega t minus kz and H was H0 e to the power i omega t minus kz and I had these 22 00:03:55,140 --> 00:03:56,840 vectors. 23 00:03:56,840 --> 00:04:19,940 So, they have components Ex, Ey, Ez, Hx, Hy, Hz, this is Hz. And the point to be notice 24 00:04:19,940 --> 00:04:39,289 here was that, these E0 and H0 they were constants. S0, E0, H0 purely constants; what we could 25 00:04:39,289 --> 00:04:50,639 see that in case of infinitely extended medium our direction of propagation was z. So, I 26 00:04:50,639 --> 00:04:59,610 had z part coming out like this and then time part coming out like this. Here also what 27 00:04:59,610 --> 00:05:10,090 I will do? I will take this direction which is propagation direction is z and z part would 28 00:05:10,090 --> 00:05:22,430 be like this itself. So, in order to now find out the propagation of light waves in such 29 00:05:22,430 --> 00:05:30,139 kind of medium where we have refractive index discontinuity refractive index is not uniform, 30 00:05:30,139 --> 00:05:38,849 when I consider Maxwell’s equations in homogeneous, linear isotropic charge free current free 31 00:05:38,849 --> 00:05:43,210 dielectric medium. In the previous case it was homogeneous medium 32 00:05:43,210 --> 00:05:49,529 now it is inhomogeneous medium. And again I write down all the Maxwell’s equations 33 00:05:49,529 --> 00:05:58,180 Maxwell’s equations go like this. But now if I look at constitutive relations, when 34 00:05:58,180 --> 00:06:07,999 D is equal to epsilon E this epsilon is now in general a function of x, y and z, which 35 00:06:07,999 --> 00:06:15,400 is nothing but epsilon naught times n square of x, y and z. So, this is not a constant 36 00:06:15,400 --> 00:06:24,809 scalar now, which was the case for infinitely extended medium. And again I am talking about 37 00:06:24,809 --> 00:06:32,500 dielectric. So, B is equal to mu H which is closely equal to mu naught times H because 38 00:06:32,500 --> 00:06:42,620 it is non-magnetic medium. So, what should I do? Know again I should form the wave equation. 39 00:06:42,620 --> 00:06:51,330 So, that I can find out what are the electric and magnetic field solutions. So, the usual 40 00:06:51,330 --> 00:06:57,499 procedure to find out the wave equation is I take the curl of this equation and use other 41 00:06:57,499 --> 00:06:59,979 Maxwell’s equations into it. 42 00:06:59,979 --> 00:07:09,050 So, I get del of del dot E minus del square E is equal to minus del t of mu times del 43 00:07:09,050 --> 00:07:17,680 D over del t, thus in the same way as I had done in the previous case. But now here I 44 00:07:17,680 --> 00:07:26,889 should check that del dot E is not 0 I should be careful. In the previous case del dot E 45 00:07:26,889 --> 00:07:37,970 was 0, but now I have del dot D and D is equal to epsilon E epsilon depends upon x, y and 46 00:07:37,970 --> 00:07:38,970 z. 47 00:07:38,970 --> 00:07:44,990 So, I cannot take that epsilon out of this del operator. So, del dot E would not be 0 48 00:07:44,990 --> 00:07:56,919 if del dot E is not 0 then how this wave equation is going to change? So, let me know find out 49 00:07:56,919 --> 00:08:03,919 what is del dot E for that I take del dot D is equal to 0 and put D is equal to epsilon 50 00:08:03,919 --> 00:08:11,879 not n square En square is a function of x, y and z. So, I write it down as epsilon naught 51 00:08:11,879 --> 00:08:19,710 del of n square dot E plus n square del dot E and that should be equal to 0. This gives 52 00:08:19,710 --> 00:08:31,319 me del dot E is equal to minus gradient of n square over n square dot E. I put this back 53 00:08:31,319 --> 00:08:41,589 into this equation and rearrange the terms to get a wave equation in this particular 54 00:08:41,589 --> 00:08:48,779 form. What it is del square E plus del of 1 over 55 00:08:48,779 --> 00:08:55,760 n square del n square dot E minus mu naught epsilon naught n square del 2 E over del t 56 00:08:55,760 --> 00:09:01,470 square. So now, you can notice an extra term here, this is an extra term that we have got 57 00:09:01,470 --> 00:09:07,190 it as compared to the case of infinitely extended medium. 58 00:09:07,190 --> 00:09:20,060 So, this is a wave equation in an inhomogeneous medium. What is the implication of this term 59 00:09:20,060 --> 00:09:34,110 now? What complicates it can introduce in our analysis? Well if I expand this then I 60 00:09:34,110 --> 00:09:43,399 write down this x, y and z components and if expand this what I see that now it is not 61 00:09:43,399 --> 00:09:57,339 possible for me to separate out x, y and z. t can be separated out, that is not a problem. 62 00:09:57,339 --> 00:10:06,420 But I cannot separate out x, y and z. x, y and z solutions cannot be separated or which 63 00:10:06,420 --> 00:10:18,871 I could do in case of infinitely extended medium. So, what do I do now? Well similarly 64 00:10:18,871 --> 00:10:29,291 if I do it for the magnetic field I will obtain an equation in H something like this, again 65 00:10:29,291 --> 00:10:38,980 there would be a middle terms which makes it impossible to separate out x, y and z solutions. 66 00:10:38,980 --> 00:10:44,830 So now let me consider a case where refractive index where is only transfers direction which 67 00:10:44,830 --> 00:10:54,310 is the case of optical waveguides and optical fibers. So, so I take in general a case where 68 00:10:54,310 --> 00:11:01,690 n square is a function of x, y which can be a channel waveguide or optical fiber. So, 69 00:11:01,690 --> 00:11:10,940 n square does not vary with z, in this case I can separate out z and t parts. And if I 70 00:11:10,940 --> 00:11:17,829 can separate out z and t parts then the solution z and t solution can be written in the same 71 00:11:17,829 --> 00:11:30,740 way as this. So, I write down z and t solution like this. And x, y solution is still remains. 72 00:11:30,740 --> 00:11:42,740 So, I put it with E0. So, the associated electric field I can write as E(x, y, t) is equal to 73 00:11:42,740 --> 00:11:50,889 E0 of x, y times e to the power i omega t minus beta z Similarly H. 74 00:11:50,889 --> 00:12:05,660 Now what I have got? I have got that the solutions here have this form. So, this is a function 75 00:12:05,660 --> 00:12:14,189 of x and y and this is the propagation in z direction. So, as if some function of x 76 00:12:14,189 --> 00:12:26,950 and y is propagating in z direction with some propagation constant beta, similarly for H. 77 00:12:26,950 --> 00:12:37,329 So, these are the modes of the system. And as I will find out that there can be only 78 00:12:37,329 --> 00:12:48,959 certain such functions possible which sustain their shape and propagate with certain propagation 79 00:12:48,959 --> 00:12:57,050 constant beta these are the modes of the system. So now, my problem reduces to find out these 80 00:12:57,050 --> 00:13:11,350 functions, E0 of x, y and H0 of x, y and their corresponding propagation constants beta. 81 00:13:11,350 --> 00:13:22,620 Let me start with doing a very simple problem where I remove any index discontinuity even 82 00:13:22,620 --> 00:13:31,000 in y direction. So, I take the simplest case here the variation of refractive index is 83 00:13:31,000 --> 00:13:37,230 only in x direction. So, I have n square as a function of x only 84 00:13:37,230 --> 00:13:48,660 and this is the case of say planner waveguide something like this, where you have this is 85 00:13:48,660 --> 00:13:59,350 x this is y and this is z. So, where y is infinitely extended, z is infinitely extended 86 00:13:59,350 --> 00:14:05,060 and you have index cn discontinuity only in x direction. So, here you have different refractive 87 00:14:05,060 --> 00:14:15,630 index here different and here different. So, n square is the function of x only. Now if 88 00:14:15,630 --> 00:14:24,870 it is a function of x only then I can separate out y part also. And the solution I can write 89 00:14:24,870 --> 00:14:34,269 as e to the power i omega t minus some gamma y minus beta z. And x part will be now associated 90 00:14:34,269 --> 00:14:42,319 with E0. So, E0 is not a constant is it is a function of x some function of x. Similarly, 91 00:14:42,319 --> 00:14:50,899 H is equal to H0x e to the power i omega t minus gamma y minus beta z. 92 00:14:50,899 --> 00:15:01,170 So, these are the form of solutions now. What I can do? I can always choose my direction 93 00:15:01,170 --> 00:15:13,440 of propagation if the light is propagating in this direction I can label it as z or I 94 00:15:13,440 --> 00:15:24,589 can label it as y it is up to me to choose my access. So, what I do? I choose z axis 95 00:15:24,589 --> 00:15:32,360 as the direction of propagation then without loss of any generality I put gamma is equal 96 00:15:32,360 --> 00:15:42,949 to 0. So, you can see that if this is infinitely extended, this is infinitely extended index 97 00:15:42,949 --> 00:15:52,459 discontinuity is only in x. So, you can launch light into this in this direction and let 98 00:15:52,459 --> 00:15:59,339 the light propagate along y or you can let the light propagate along z you can launch 99 00:15:59,339 --> 00:16:09,600 it from here. So, I choose to take the direction of propagation as z. So, I put gamma is equal 100 00:16:09,600 --> 00:16:17,001 to 0. Then I can write the solutions as E0 of x 101 00:16:17,001 --> 00:16:23,930 e to the power i omega t minus beta z and x 0 of x e to the power i omega t minus beta 102 00:16:23,930 --> 00:16:32,620 z. So, I have got similar form of solution, but it is not the same. Here E0 and H0 are 103 00:16:32,620 --> 00:16:42,630 constants; here E0 and H0 are the functions of x and these functions now I want to find 104 00:16:42,630 --> 00:16:43,879 out. 105 00:16:43,879 --> 00:16:53,720 I need to find out how E varies with x and how H varies with x, that is what I need to 106 00:16:53,720 --> 00:17:09,069 know and they will give me the modes. So, I have the solutions here and do not forget 107 00:17:09,069 --> 00:17:16,770 that I have vector sins here which means this E is nothing but Ex x cap plus Ey y cap plus 108 00:17:16,770 --> 00:17:29,940 Ez z cap and similarly H. So, basically I have 6 such equations 3 in E and 3 in H. So, 109 00:17:29,940 --> 00:17:37,100 if I write down the components of this then I can write them as Ej is equal to Ej x e 110 00:17:37,100 --> 00:17:45,820 to the power i omega t minus beta z and similarly Hz and similarly Hj where j can be x, y or 111 00:17:45,820 --> 00:17:58,470 z. Now, let me put these solutions into curl 112 00:17:58,470 --> 00:18:07,910 equations. Why I am doing this? Ultimately I want to find out how E varies with x and 113 00:18:07,910 --> 00:18:17,320 how H varies with x. So, I need to form a differential equation in E with respect to 114 00:18:17,320 --> 00:18:29,080 x. In order to do that and I know from here I will get del E over del x del H over del 115 00:18:29,080 --> 00:18:41,110 x terms. So, that is why I put these into these equations now. When I do this then this 116 00:18:41,110 --> 00:18:49,410 will give me 3 equations, one corresponding to Hx then Hy and Hz. And this will also give 117 00:18:49,410 --> 00:19:02,780 me 3 equations, Ex, Ey an Ez x, y and z components here, let me do it. The x component from here 118 00:19:02,780 --> 00:19:15,570 we will come out to be if you expand this, because you know that del cross E del cross 119 00:19:15,570 --> 00:19:37,150 E you can write as x cap y cap z cap del del x del del y del del z. And then you have Ex, 120 00:19:37,150 --> 00:19:49,750 Ey Ez is equal to So, this is del cross E minus mu naught del del t and then you have 121 00:19:49,750 --> 00:20:03,320 3 components here Hx, Hy and Hz. So, from here you can find out 6 equations, 122 00:20:03,320 --> 00:20:13,830 the first one would be i beta Ey is equal to minus i omega mu naught Hx. The x equation 123 00:20:13,830 --> 00:20:23,010 from here would be i beta Hy is equal to i omega epsilon naught n square of x times Ex. 124 00:20:23,010 --> 00:20:31,770 The second one from here would be minus i beta Ex minus de Ez over del x is equal to 125 00:20:31,770 --> 00:20:41,950 minus i omega mu naught Hy. And here it would be minus i beta Hx minus del Hz over del x 126 00:20:41,950 --> 00:20:46,120 minus i omega epsilon naught n square of x Ey. 127 00:20:46,120 --> 00:20:56,080 Third one would be del E by over del x minus i omega mu naught Hz and here it would be 128 00:20:56,080 --> 00:21:06,090 del Hy over del x, i omega epsilon naught n square of x Ez. So, I have got 6 equations 129 00:21:06,090 --> 00:21:18,810 which relate the electric and magnetic field components Ex, Ey and Ez and Hx, Hy and Hz. 130 00:21:18,810 --> 00:21:31,750 What do I do with these equations? Well, what I notice one thing that I can simplify the 131 00:21:31,750 --> 00:21:41,250 situation to certain extent. And how can I simplify the situation? Well if I have if 132 00:21:41,250 --> 00:21:50,530 I have a waveguide and I launch light into this when launching light into this I have 133 00:21:50,530 --> 00:22:04,020 some control on light and that is I can launch this polarization or this polarization. 134 00:22:04,020 --> 00:22:15,180 This is x axis this is y axis. So, if I decide to launch this polarization that is Ey is 135 00:22:15,180 --> 00:22:26,960 non 0 and Ex is 0 then let me see which equations do I invoke. Do invoke all the 6 equations 136 00:22:26,960 --> 00:22:31,210 or I invoke only a few of them? 137 00:22:31,210 --> 00:22:45,330 And what I find that the equations which have E y non 0 and E x 0 are this one has E y non 138 00:22:45,330 --> 00:22:58,530 0 E x is equal to 0. This one has E y non 0 and this one has E y non 0. So, so if I 139 00:22:58,530 --> 00:23:07,120 if I launch y polarized wave then I invoke these 3 equations and when I launch exploitation 140 00:23:07,120 --> 00:23:16,050 I invoke these 3 equation. So, 3 equations can be involved at a time. So, this gives 141 00:23:16,050 --> 00:23:26,480 me this gives me a room to simplify the problem, because I need to now consider only 3 equations 142 00:23:26,480 --> 00:23:38,240 at a time. These 3 equations the blue ones are these and what I see there is they have 143 00:23:38,240 --> 00:23:47,200 only 3 non vanishing components of E and H and they are Ey, Hx and Hz. 144 00:23:47,200 --> 00:23:56,500 In these 3 I get that there is only one component of E and that is transverse. Then these modes 145 00:23:56,500 --> 00:24:06,090 are also known as TE modes or transverse electric modes or transverse electric polarization. 146 00:24:06,090 --> 00:24:16,611 While the other 3 have non vanishing components of E and H as Hy Ex and Ez and I see that 147 00:24:16,611 --> 00:24:24,460 there is only one component of magnetic field and that is transverse component then they 148 00:24:24,460 --> 00:24:34,140 are known as transverse magnetic modes or TM polarization and this will correspond to 149 00:24:34,140 --> 00:24:49,980 these 3 equations. So now, let me do the analysis of TE modes first. So, what I want to do? 150 00:24:49,980 --> 00:24:59,940 Again do not forget I want to find out how E and H vary with x and I need to find out 151 00:24:59,940 --> 00:25:11,100 a differential equation in E or H with respect to x. So, for TE modes I write down these 152 00:25:11,100 --> 00:25:13,820 3 equations. 153 00:25:13,820 --> 00:25:26,730 And what I can do now since these 3 equations relate Ey, Hx and Hz then if I know one of 154 00:25:26,730 --> 00:25:35,580 them then I can find out the others. So, for example, if I know Ey I can find out Hx from 155 00:25:35,580 --> 00:25:44,350 here and Hz from here ok. So, what I do let me find out Ey. So, I substitute 156 00:25:44,350 --> 00:25:55,290 for Hx and Hz from these 2 equations into the third equation. And when I do this I form 157 00:25:55,290 --> 00:26:06,350 a differential equation in Ey. And this comes out to be d2 Ey over dx square plus k naught 158 00:26:06,350 --> 00:26:15,030 square n square of x minus beta square times Ey is equal to 0. Where I have use the effect 159 00:26:15,030 --> 00:26:22,820 that k naught is omega be c, which is also 2 pi over lambda naught where lambda naught 160 00:26:22,820 --> 00:26:31,740 is free space wavelength. So now, I have got now I have got a differential equation in 161 00:26:31,740 --> 00:26:43,910 Ey for a given n square of x. So, if I know my planner waveguide that is I know n square 162 00:26:43,910 --> 00:26:56,350 of x then I can solve this equation for the given n square of x and obtain Ey. And that 163 00:26:56,350 --> 00:27:05,190 will give me the modes, that will tell me how electromagnetic wave propagates in that 164 00:27:05,190 --> 00:27:19,140 medium of n square x refractive index variation. So, let me apply it to a very simple waveguide, 165 00:27:19,140 --> 00:27:24,500 which I call planner mirror waveguide. What is a planner mirror waveguide? 166 00:27:24,500 --> 00:27:37,300 You take a very thin slab of refractive index and let us say glass. It has got a width d 167 00:27:37,300 --> 00:27:47,920 and refractive index n. And I polish and I sorry not polish and I deposit metal here 168 00:27:47,920 --> 00:27:56,450 and here. When I deposit metal on top and bottom and if I launch any light then that 169 00:27:56,450 --> 00:28:05,070 light will be reflected back and forth from this mirror and from this mirror and should 170 00:28:05,070 --> 00:28:08,690 be guided. So, this is the simplest waveguide I can think 171 00:28:08,690 --> 00:28:16,720 of let me do that. So, I deposit metal on top and bottom if I look at the refractive 172 00:28:16,720 --> 00:28:26,800 index profile. Then I find that in this region between 0 and d. I have refractive index n 173 00:28:26,800 --> 00:28:34,690 and here at the boundaries I have metal. When it boundaries I have metal then the electric 174 00:28:34,690 --> 00:28:42,470 field at the metal boundary should be 0 that is what I know. So, what I do know I write 175 00:28:42,470 --> 00:28:53,090 down the wave equation the equation which I obtained in the previous slide. And I put 176 00:28:53,090 --> 00:29:00,300 n square of x as n square and I write it down in the region between 0 and d. 177 00:29:00,300 --> 00:29:08,230 So, this would be the equation. So, I from here I can find out how Ey varies in this 178 00:29:08,230 --> 00:29:19,040 layer. And I know that the fields has to be 0 here. Now let me defined this k naught square 179 00:29:19,040 --> 00:29:31,290 n square minus beta square as some kappa square since I know that beta which is the propagation 180 00:29:31,290 --> 00:29:37,030 constant of the wave in this region has to be less than k naught n it cannot be greater 181 00:29:37,030 --> 00:29:46,270 than k naught n, because propagation constant cannot be greater than the propagation constant 182 00:29:46,270 --> 00:29:50,470 of the medium itself k of infinitely extended medium. 183 00:29:50,470 --> 00:29:59,720 So, beta is less than k naught n. So, kappa square is always greater than 0 which means 184 00:29:59,720 --> 00:30:08,710 that the solution of this equation would be Ey is equal to A sin kappa x plus B cosine 185 00:30:08,710 --> 00:30:17,410 kappa x. Now my feel has to be 0 here and here. So, I apply these boundary conditions 186 00:30:17,410 --> 00:30:26,080 Ey is equal to 0 at x is equal to 0 and at x is equal to d and this gives me B is equal 187 00:30:26,080 --> 00:30:40,730 to 0 and kappa d is equal to m pi. So, since B is equal to 0. So, this term goes off and 188 00:30:40,730 --> 00:30:43,120 kappa is equal to m pi over d. 189 00:30:43,120 --> 00:30:49,820 So, I put kappa is equal to m pi over d. So, my solution now becomes E y is equal to a 190 00:30:49,820 --> 00:31:01,640 sig m pi x over d where m can take integer values now what are. So, I have got E y of 191 00:31:01,640 --> 00:31:09,890 x what is left corresponding beta from where beta are coming from here because I know kappa 192 00:31:09,890 --> 00:31:18,020 d is equal to m pi and kappa square is equal to k naught square n square minus beta square. 193 00:31:18,020 --> 00:31:27,150 So, this gives me that there would be only certain discrete values of beta defined by 194 00:31:27,150 --> 00:31:34,250 beta m and given by beta m square is equal to k naught square n square minus m pi over 195 00:31:34,250 --> 00:31:43,930 d whole square. So, I have got for a planer mirror waveguide 196 00:31:43,930 --> 00:31:53,520 only certain functions only certain functions which has certain propagation constants they 197 00:31:53,520 --> 00:32:03,410 will be sustained. If I plot them then for m is equal to one it will look like this for 198 00:32:03,410 --> 00:32:11,570 m is equal to 2 it would look like this m is equal to 3 like this. They are nothing 199 00:32:11,570 --> 00:32:21,390 but if you if you look carefully they look like as the modes of vibrations of a stretched 200 00:32:21,390 --> 00:32:28,340 string modes of vibrations of a stretched string. So, it is similar to that. So, in 201 00:32:28,340 --> 00:32:39,620 the next lecture I am going to understand what do they exactly represent. I know that 202 00:32:39,620 --> 00:32:48,860 in a waveguide in a waveguide if I launch if I launch ray like this then it will we 203 00:32:48,860 --> 00:32:54,210 reflected back and forth or if I launcher wave then wave would be reflected back and 204 00:32:54,210 --> 00:33:07,240 forth, but how do these represent the guidance in an a waveguide? So, let us understand it 205 00:33:07,240 --> 00:33:09,370 in the next lecture. Thank you.