1 00:00:08,710 --> 00:00:27,619 So, in this lecture, let me continue with the previous lecture where I was doing the 2 00:00:27,619 --> 00:00:31,399 analysis of planar mirror waveguide. 3 00:00:31,399 --> 00:00:47,120 So, what I was doing was to find out the modes of a planar mirror waveguide and what I did 4 00:00:47,120 --> 00:00:58,890 that I took the planar mirror waveguide which has refractive index n and metal deposited 5 00:00:58,890 --> 00:01:06,860 at x is equal to 0 and x is equal to d, and the modes of this waveguide I found out as 6 00:01:06,860 --> 00:01:15,049 Ey is equal to A sinm pi x over d and corresponding beta are given by beta m square is equal to 7 00:01:15,049 --> 00:01:20,920 k naught square n square minus m pi over d square. 8 00:01:20,920 --> 00:01:33,920 So, what I got that I got this kind of variation of electric field Ey with respect to x or 9 00:01:33,920 --> 00:01:44,189 this or this and these fields propagate in the waveguide and sustain their shape as they 10 00:01:44,189 --> 00:01:52,030 propagate. They also propagate with certain propagation constants beta1, beta2 beta3 and 11 00:01:52,030 --> 00:01:59,780 so on. Let me understand what they are. 12 00:01:59,780 --> 00:02:14,500 So, the field is Am sin kappa mx where kappa m is equal to m pi over d if I write the complete 13 00:02:14,500 --> 00:02:23,040 solution this is only Ey m of x, x part of the solution remember that there is t and 14 00:02:23,040 --> 00:02:35,590 z parts also. So, if I write down the complete solution there it is some Am sin kappa mx, 15 00:02:35,590 --> 00:02:45,960 e to the power i omega t minus beta mz. Let me write this sin kappa mx in the form of 16 00:02:45,960 --> 00:02:56,320 e to the power plus minus I kappa mx. So, if I do that I get this and then let me 17 00:02:56,320 --> 00:03:09,060 put it in this form. So, what I have got from here, I have got it e to the power i omega 18 00:03:09,060 --> 00:03:17,560 t minus b time z minus kappa mx and then e to the power i omega t minus b time z plus 19 00:03:17,560 --> 00:03:28,200 kappa mx what they are and remember you see now outside I have a constant it is not a 20 00:03:28,200 --> 00:03:38,680 function of x, the x dependence I have included in the exponential now, the x dependence which 21 00:03:38,680 --> 00:03:48,080 was earlier here I have included in the exponential and when I see this now this is nothing, but 22 00:03:48,080 --> 00:03:57,390 the plane wave because outside is a constant. So, and this is the plane wave which is moving 23 00:03:57,390 --> 00:04:09,710 in this direction, moving in x-z plane making certain angle from z axis. And this is a plane 24 00:04:09,710 --> 00:04:22,340 wave which is moving in plus z minus x plus z minus x direction k. So, what I have got 25 00:04:22,340 --> 00:04:32,259 that this mode, this mode is nothing but the superposition of 2 plane waves, this plane 26 00:04:32,259 --> 00:04:39,389 wave and this plane wave with some phases. So, it is the superposition of 2 plane waves 27 00:04:39,389 --> 00:04:50,060 1 propagating in this direction another propagating in this direction. If you look at the waveguide 28 00:04:50,060 --> 00:04:58,770 if you look at the waveguide this is x this is z. So, this is x-z plane this is x this 29 00:04:58,770 --> 00:05:07,469 is z. So, one wave is moving like this another wave is moving like this and this is what 30 00:05:07,469 --> 00:05:16,430 you have, that if you launch wave like this then it will get reflected then reflected 31 00:05:16,430 --> 00:05:22,919 from here, reflected from here. So, all the time you will have 2 waves one 32 00:05:22,919 --> 00:05:31,460 going in this direction and another going in this direction. Let me put them together 33 00:05:31,460 --> 00:05:33,669 here in x-z plane. 34 00:05:33,669 --> 00:05:43,440 So, one is moving an angle theta m with z axis and another one is moving minus theta 35 00:05:43,440 --> 00:05:53,770 m angle with z axis and this has to be the propagation constant in the medium of a plane 36 00:05:53,770 --> 00:05:57,780 wave these are now plane waves plane waves we will have propagation constant k naught 37 00:05:57,780 --> 00:06:10,889 n. So, this is the component of propagation constant in x direction this is the component 38 00:06:10,889 --> 00:06:21,110 of propagation constant in z direction. So, if I now do kappa m square plus beta m square 39 00:06:21,110 --> 00:06:31,240 is square root then it comes out to be k naught n and it comes out to be k naught n because 40 00:06:31,240 --> 00:06:48,050 you remember that kappa m square is equal to k naught square n square minus beta m square 41 00:06:48,050 --> 00:06:55,129 which means that this is indeed k naught n this is indeed k naught n. 42 00:06:55,129 --> 00:07:07,300 So, the propagation constant in z direction is k naught n cos theta m. Let us understand 43 00:07:07,300 --> 00:07:15,319 it. If I take the components in x direction I have 1 wave going in positive x another 44 00:07:15,319 --> 00:07:22,500 wave going in negative x, which means that in x direction I have 2 counter propagating 45 00:07:22,500 --> 00:07:29,819 waves and when I have 2 counter propagating waves they give you a standing wave solution 46 00:07:29,819 --> 00:07:38,969 they give you standing wave. So, in x direction I have a standing wave. In z direction if 47 00:07:38,969 --> 00:07:45,400 I see this also gives me propagation in positive z. 48 00:07:45,400 --> 00:07:57,960 So, I have a standing wave in x which propagates in z. The energy does not flow out in x direction 49 00:07:57,960 --> 00:08:06,659 the wave stands in x direction, but that standing wave pattern flows in z direction this is 50 00:08:06,659 --> 00:08:18,860 the mode. So, these modes are nothing, but the standing wave patterns and they are made 51 00:08:18,860 --> 00:08:28,050 out of two waves one going in this direction another going in this direction. 52 00:08:28,050 --> 00:08:33,780 I can now find out what is the effective refractive index of a given mode or effective index of 53 00:08:33,780 --> 00:08:42,790 the mode by beta m over k naught which is nothing, but kn cos theta m. So, for every 54 00:08:42,790 --> 00:08:52,180 mode I have different beta m and therefore, the angles that the constituent plane waves 55 00:08:52,180 --> 00:09:04,550 make with z axis would be different and these angles can be found out from here. For example, 56 00:09:04,550 --> 00:09:16,570 for n is equal to 1.5 d is equal to 1 micrometer and lambda naught is equal to 0.633 micrometer 57 00:09:16,570 --> 00:09:28,700 helium neon wavelength. I find out that there are these modes four modes. Let me look at 58 00:09:28,700 --> 00:09:33,010 them closely. 59 00:09:33,010 --> 00:09:42,800 The first one has effective index 1.4662 and the angles of constituent plane waves with 60 00:09:42,800 --> 00:09:54,140 z axis is z axis r plus minus 12.18 degrees. So, there are 2 plane waves one making plus 61 00:09:54,140 --> 00:09:59,990 12 degrees and other making minus 12 degree with z axis and this is the a standing wave 62 00:09:59,990 --> 00:10:06,890 pattern in x direction corresponding to those waves. 63 00:10:06,890 --> 00:10:18,250 The second one is with n effective is equal to 1.3599 with angles plus minus approximately 64 00:10:18,250 --> 00:10:27,620 25 degrees. Yet another one is this with the angles plus minus 39 degrees and this one 65 00:10:27,620 --> 00:10:29,860 plus minus 57.565 degrees. 66 00:10:29,860 --> 00:10:45,940 So, these are the modes how many modes are there for a given waveguide. Now let me look 67 00:10:45,940 --> 00:10:57,660 at this picture again and from here I know that kappa m d is equal to m pi. So, this 68 00:10:57,660 --> 00:11:09,610 m would be kappa m d over pi and kappa m is nothing, but k naught n sin theta m and the 69 00:11:09,610 --> 00:11:16,140 maximum value of this would be k naught n because the maximum value of sin theta is 70 00:11:16,140 --> 00:11:22,640 1. So, the maximum value of k m is k naught n 71 00:11:22,640 --> 00:11:29,750 and this will give me what is the maximum value of m. So, I put maximum value of k m 72 00:11:29,750 --> 00:11:36,940 here which is k naught n. So, m max would be k naught n by pi and k naught is equal 73 00:11:36,940 --> 00:11:44,180 to 2 pi over lambda naught. So, this would be 2 nd over lambda naught. So, the maximum 74 00:11:44,180 --> 00:11:52,030 number of modes that can propagate in a planar mirror waveguide would be a number and integer 75 00:11:52,030 --> 00:12:01,860 which is closest to, but less than this, it is closest to but less than this. For example, 76 00:12:01,860 --> 00:12:08,970 if I have n is equal to 1.5, d is equal to 1 micrometer and lambda is equal to 0.633 77 00:12:08,970 --> 00:12:16,980 micrometer there should be micrometer here then 2nd over lambda naught comes out to be 78 00:12:16,980 --> 00:12:28,100 4.739 and this gives me the maximum number of modes as 4. So, there would be 4 mode supported. 79 00:12:28,100 --> 00:12:38,210 So, this is the analysis of planar mirror waveguide which gives me insight into the 80 00:12:38,210 --> 00:12:48,660 modes of a waveguide which structure is quite intuitive and light guidance and data structure 81 00:12:48,660 --> 00:12:57,610 is also quite intuitive, but this is a structure planner mirror waveguide is of very little 82 00:12:57,610 --> 00:13:10,930 practical use for 1 it is not feasible to make this waveguide that you have one micrometer 83 00:13:10,930 --> 00:13:21,450 film of certain material and you deposit metal on top and bottom of that and use it. Second 84 00:13:21,450 --> 00:13:30,020 since it involves metal coatings and metal is highly absorbing material. So, as these 85 00:13:30,020 --> 00:13:39,610 modes propagate they will attenuate very quickly. So, they have very high loss. So, this kind 86 00:13:39,610 --> 00:13:50,170 of waveguide is of little practical use although it gives a good understanding of mode propagation. 87 00:13:50,170 --> 00:14:02,950 So, now what we are going to do is to look at more practical waveguide which is asymmetric 88 00:14:02,950 --> 00:14:09,880 step index planar waveguide dielectric waveguide planar dielectric waveguide which is obtained 89 00:14:09,880 --> 00:14:23,230 by sandwiching high index lab between 2 lower refractive index labs. So, I have n1 n2, n2 90 00:14:23,230 --> 00:14:32,700 here and n1 is greater than n2. Again if you look there is refractive index discontinuity 91 00:14:32,700 --> 00:14:43,130 only in x direction, y direction it is infinitely extended in z direction also there is no indexed 92 00:14:43,130 --> 00:14:50,620 discontinuity and it is infinitely extended. If you look at the refractive index profile 93 00:14:50,620 --> 00:15:05,970 of this waveguide then it looks like this and let me put my x is equal to 0 access in 94 00:15:05,970 --> 00:15:14,310 the middle of the waveguide so that I can make use of the symmetry of the problem let 95 00:15:14,310 --> 00:15:26,220 it have of the width d. So, this is x is equal to d by 2 this is x is equal to minus d by 96 00:15:26,220 --> 00:15:38,190 2. I can write down the refractive index variation with respect to x in this fashion. 97 00:15:38,190 --> 00:15:47,200 So, this region is mod x less than d by 2 while these 2 reasons can be represented by 98 00:15:47,200 --> 00:15:55,390 mod x greater than d by 2. So, in mod x less than the d by 2 I have refractive index n1, 99 00:15:55,390 --> 00:16:06,120 in mod x greater than d by 2 I have refractive index n2 and now my problem is to find out 100 00:16:06,120 --> 00:16:16,630 Ey and corresponding beta for this given n of x. So, how do I do? This again go back 101 00:16:16,630 --> 00:16:23,040 to the wave equation corresponding to t e votes. Let me first solve the problem for 102 00:16:23,040 --> 00:16:24,230 TE-modes. 103 00:16:24,230 --> 00:16:36,110 So, this is the wave equation for TE-modes and this is the n(x). So, how do I solve it? 104 00:16:36,110 --> 00:16:47,190 Well I find that in this region and in this region and in this region in all the 3 regions 105 00:16:47,190 --> 00:16:55,640 the refractive index is constant although when I go from here to here and here to here 106 00:16:55,640 --> 00:17:03,440 I encounter index discontinuity, but on this side and on that side the refractive index 107 00:17:03,440 --> 00:17:10,380 remains constant. So, I make use of this and write down the 108 00:17:10,380 --> 00:17:17,919 wave equation in this region and in this region that is in this region and in these 2 regions. 109 00:17:17,919 --> 00:17:26,490 So, for mod x less than d by 2 I write it down as d2Ey over dx square plus k naught 110 00:17:26,490 --> 00:17:33,700 square n1 square because nx is equal to n1 in this region minus beta square times Ey 111 00:17:33,700 --> 00:17:46,090 is equal to 0. And I also keep in mind that my effective refractive index which is given 112 00:17:46,090 --> 00:17:58,679 by beta over k naught. Effective refractive index n effective which is given by beta over 113 00:17:58,679 --> 00:18:09,100 k naught this should lie between n2 and n1 for guided modes for guided modes n effective 114 00:18:09,100 --> 00:18:20,600 would lie between n2 and n1 which means that beta would lie between k naught n2 and k naught 115 00:18:20,600 --> 00:18:27,549 n1. So, beta would be greater than k naught n2 while it would be smaller than k naught 116 00:18:27,549 --> 00:18:29,220 n1. 117 00:18:29,220 --> 00:18:39,340 So, when I write this equation now in these 2 regions which are represented by mod x greater 118 00:18:39,340 --> 00:18:48,230 than d by 2 I write it as d2Ey over dx square minus beta square minus k naught square n2 119 00:18:48,230 --> 00:18:56,340 square times Ey because n x is equal n2 here. And why I have written it in this fashion. 120 00:18:56,340 --> 00:19:03,679 So, that this is positive here and this is also positive here because I am solving it 121 00:19:03,679 --> 00:19:12,649 going to solve it for this condition which is the condition for guided modes for which 122 00:19:12,649 --> 00:19:22,120 the energy would be confined into this region. So, I represent this as kappa square and I 123 00:19:22,120 --> 00:19:29,009 represent this as gamma square and this kappa square and gamma square are positive for guided 124 00:19:29,009 --> 00:19:38,730 modes as I have explained. So, let me write these equations down again here. 125 00:19:38,730 --> 00:19:45,990 So, for mod x less than d by 2 I have a got a got an equation d2E by over dx square plus 126 00:19:45,990 --> 00:19:54,379 kappa square Ey is equal to 0 and for mod x greater than d by 2 it is d2Ey over dx square 127 00:19:54,379 --> 00:20:02,419 minus gamma square Ey is equal to 0 where kappa square as this and gamma square is this. 128 00:20:02,419 --> 00:20:09,710 What are the solutions? I know the solutions of these differential equations very well, 129 00:20:09,710 --> 00:20:16,350 this equation gives me oscillatory solutions e to the power I kappa x or e to the power 130 00:20:16,350 --> 00:20:23,600 minus i kappa x or sin kappa x cosine kappa x while this equation gives me exponentially 131 00:20:23,600 --> 00:20:32,770 amplifying and decaying solutions. So, if I write the solutions in the regions mod x 132 00:20:32,770 --> 00:20:39,390 less than d by 2 and mod x greater than d by 2 I find that in this region it is oscillatory 133 00:20:39,390 --> 00:20:50,809 solution A cosine kappa x plus B sin kappa x and in the region mod x greater than d by 134 00:20:50,809 --> 00:20:59,960 2 I have exponential and decaying solutions, but out of these 2 solutions I retain only 135 00:20:59,960 --> 00:21:07,419 exponentially decaying solution while exponentially decaying solution and not exponentially amplifying 136 00:21:07,419 --> 00:21:16,870 solution because I want a solution which gives me guided modes and for guided modes the energy 137 00:21:16,870 --> 00:21:24,940 should be confined into the high index region and it should decay down as you go away from 138 00:21:24,940 --> 00:21:31,190 the high index region. As you go towards infinity x is equal to plus 139 00:21:31,190 --> 00:21:38,700 minus infinity the energy should decay down. So, I cannot take exponentially employee find 140 00:21:38,700 --> 00:21:44,880 solutions that is y for x greater than d by 2 I have chosen the form of solutions C e 141 00:21:44,880 --> 00:21:53,779 to the power minus gamma x and for x less than minus d by 2 I have chosen the form D 142 00:21:53,779 --> 00:22:02,070 e to the power gamma x this A, B, C, D are some constants which can be determined by 143 00:22:02,070 --> 00:22:09,429 the boundary conditions. What are the boundaries? Boundaries are x is equal to plus d by 2 and 144 00:22:09,429 --> 00:22:22,899 x is equal to minus d by 2, but here I can make some more simplification what simplification 145 00:22:22,899 --> 00:22:32,860 I can make is to utilize the symmetry of the problem how can I utilize the symmetry of 146 00:22:32,860 --> 00:22:36,080 the problem let us see. 147 00:22:36,080 --> 00:22:50,059 I have the wave equation for TE-modes like this and this is the n of x. I see that this 148 00:22:50,059 --> 00:22:56,419 is symmetric about x is equal to 0 which means that n square of minus x is equal to n square 149 00:22:56,419 --> 00:22:57,419 of x. 150 00:22:57,419 --> 00:23:09,090 Now, in this equation if you replace x by minus x if you replace x by minus x, then 151 00:23:09,090 --> 00:23:30,330 you get d2Ey of minus x divided by d of minus x square that is x square plus k naught square 152 00:23:30,330 --> 00:23:45,779 n square of minus x minus beta square Ey of minus x is equal to 0 and this is nothing, 153 00:23:45,779 --> 00:23:57,470 but n square of x because of the symmetry of the profile which means that which means 154 00:23:57,470 --> 00:24:08,149 that I now have 2 possibilities which are Ey of minus x is equal to Ey of x if I put 155 00:24:08,149 --> 00:24:18,539 Ey of minus x is equal to Ey of x I get back the same equation or if I put Ey of minus 156 00:24:18,539 --> 00:24:24,809 x is equal to minus Ey of x then also I get back the same equation. 157 00:24:24,809 --> 00:24:37,340 So, there are now 2 possibilities one is this another is this and they are known as symmetric 158 00:24:37,340 --> 00:24:45,890 modes because this is the property of a symmetric function and this is the property of antisymmetric 159 00:24:45,890 --> 00:24:54,500 function. So, these modes are known as symmetric modes and these modes are known as antisymmetric 160 00:24:54,500 --> 00:25:05,739 modes. The simplification it introduces is that out of those 2 functions cosine and sin 161 00:25:05,739 --> 00:25:14,269 I can pick one here and one here since cosine is the symmetric function and sin is the antisymmetric 162 00:25:14,269 --> 00:25:18,539 function. So, symmetric modes would now be represented 163 00:25:18,539 --> 00:25:28,590 by Ey of x is equal to a cosine kappa x for mod x less than d by 2 and C e to the power 164 00:25:28,590 --> 00:25:39,830 minus gamma mod x for mod x greater than d by 2 for antisymmetric modes I will have Ey 165 00:25:39,830 --> 00:25:48,470 of x is equal to b sin kappa x for x less than for mod x less than d by 2 and then I 166 00:25:48,470 --> 00:25:54,220 have D e to the power minus gamma mod x for mod x less than d by 2, but I shall have to 167 00:25:54,220 --> 00:26:04,799 take care of sin when I go from left side or right side. What do I mean to say is this 168 00:26:04,799 --> 00:26:20,710 if this is x and this is d by 2 this is minus d by 2 this is x is equal to 0. 169 00:26:20,710 --> 00:26:32,600 Now, if you plot a sin function in the region mod x less than d by 2. So, it goes something 170 00:26:32,600 --> 00:26:49,059 like this something like this and now in the n2 region for positive x values you will for 171 00:26:49,059 --> 00:26:56,450 x greater than d by 2 you will approach from here and then it decays down while for x less 172 00:26:56,450 --> 00:27:01,530 than minus d by 2 it you will approach from here from negative side. So, here you start 173 00:27:01,530 --> 00:27:09,019 from positive side and decay down here you start from negative side and then decay down. 174 00:27:09,019 --> 00:27:17,119 So, to take care of that sin I will have to put x over mod x here. 175 00:27:17,119 --> 00:27:29,690 So, these are the symmetric modes and remember that I am doing the analysis of TE-modes have 176 00:27:29,690 --> 00:27:42,399 non vanishing components of E and H has Ey, Hx and Hz. What is left now? I need to find 177 00:27:42,399 --> 00:27:52,340 out the relationship between A and C I need to find out how these solutions are connected 178 00:27:52,340 --> 00:28:00,269 at the boundaries for that I will use boundary conditions. And I have learnt that the boundary 179 00:28:00,269 --> 00:28:08,090 conditions are when you encounter an interface between 2 dielectric media the boundary conditions 180 00:28:08,090 --> 00:28:15,990 are tangential components of E and H should be continuous. 181 00:28:15,990 --> 00:28:22,379 Tangential components of E and H should be continuous what are the tangential components 182 00:28:22,379 --> 00:28:31,700 to the boundaries. This is the waveguide this is x this is x is equal to plus d by 2 this 183 00:28:31,700 --> 00:28:43,710 is x is equal to minus d by 2, this is y this is z. So, to this interface which is x is 184 00:28:43,710 --> 00:28:49,330 equal to plus d by 2 or x is equal to minus d by 2 here the tangential components are 185 00:28:49,330 --> 00:29:01,629 y and z y and z, x is a normal component and here the nonvanishing components are Ey, Hx 186 00:29:01,629 --> 00:29:13,690 and Hz. So, the tangential components are Ey and Hz. So, these Ey and Hz should be continuous 187 00:29:13,690 --> 00:29:23,169 and if you look back to the 3 equations which relate these 3 components of E and H then 188 00:29:23,169 --> 00:29:35,460 I find that Hz is nothing, but some constant times dEy over dx. So, Ey and dEy over dx 189 00:29:35,460 --> 00:29:43,899 should be continuous at x is equal to plus minus d by 2. So, let me apply these boundary 190 00:29:43,899 --> 00:29:50,100 conditions, let me do it for x is equal to plus d by 2 the same would be obtained for 191 00:29:50,100 --> 00:29:56,799 x is equal to minus d by 2. So, I put Ey should be continuous at x is 192 00:29:56,799 --> 00:30:03,799 equal to plus d by 2 which means a cosine kappa d by 2 should be equal to C e to the 193 00:30:03,799 --> 00:30:14,409 power minus gamma d by 2 and then for derivative dEy over dx. So, I will get minus a kappa 194 00:30:14,409 --> 00:30:25,359 sin kappa d by 2 is equal to minus gamma C e to the power minus gamma d by 2. If I divide 195 00:30:25,359 --> 00:30:34,249 this by this I get kappa tan kappa d by 2 is equal to gamma and remember that these 196 00:30:34,249 --> 00:30:41,999 kappa have dimensions of 1 over meter. So, I can make them dimensionless, so I can multiply 197 00:30:41,999 --> 00:30:50,299 both sides by d by 2 and get an equation like this which is kappa d by 2 tan kappa d by 198 00:30:50,299 --> 00:30:52,299 2 is equal to gamma d by 2. 199 00:30:52,299 --> 00:31:01,629 What is kappa and what is gamma? So, you remember that kappa a square is equal to k naught square 200 00:31:01,629 --> 00:31:14,549 m1 square minus beta square and gamma square is beta square minus k naught square n2 square 201 00:31:14,549 --> 00:31:22,789 k naught is equal to 2 pi over lambda naught. So, for a given waveguide and wavelength for 202 00:31:22,789 --> 00:31:29,479 a given waveguide and wavelength the only unknown in this in these kappa and gamma is 203 00:31:29,479 --> 00:31:38,870 beta. So, this is nothing, but a transcendental equation in beta. So, beta satisfy this equation. 204 00:31:38,870 --> 00:31:44,669 So, there are only certain values of beta which are possible and those values of theta 205 00:31:44,669 --> 00:31:52,350 satisfied this equation only those values of beta are possible. So, this makes this 206 00:31:52,350 --> 00:31:56,850 makes the modes discreet. So, from here I find out what are the possible 207 00:31:56,850 --> 00:32:06,039 values of beta and for those beta I can find out the fields. So, that is how I can get 208 00:32:06,039 --> 00:32:13,539 the modes of a symmetric planar waveguides and these are symmetric modes. 209 00:32:13,539 --> 00:32:24,100 In the next lecture I will continue with this to solve to find the solutions for antisymmetric 210 00:32:24,100 --> 00:32:25,770 modes. Thank you.