1 00:00:19,000 --> 00:00:26,070 Let us continue our discussion on the modes of a planar symmetric waveguide. In the last 2 00:00:26,070 --> 00:00:33,750 lecture we had obtained the modal fields and propagation constants of symmetric and antisymmetric 3 00:00:33,750 --> 00:00:43,000 modes of asymmetric dielectric planar waveguide. Let us now look more carefully at these modes 4 00:00:43,000 --> 00:00:50,290 and understand what do they actually represent. What are the modal fields actually and what 5 00:00:50,290 --> 00:00:59,489 do the propagation constants represent. So, for that let me write down the field of 6 00:00:59,489 --> 00:01:07,820 for example, TE0 mode or any symmetric mode. 7 00:01:07,820 --> 00:01:14,580 So, it is given in the core or in the guiding film it is given by Ey of x is equal to a 8 00:01:14,580 --> 00:01:24,970 cosine kappa x. If I write down the complete solution that is the z part and t part also, 9 00:01:24,970 --> 00:01:34,000 then it would be a cosine kappa x e to the power i omega t minus beta z. 10 00:01:34,000 --> 00:01:39,740 Let me expand this cosine kappa x it is in the same way as I had done for planar mirror 11 00:01:39,740 --> 00:01:45,700 waveguide. So, it is E to the power I kappa x plus E to the power minus I kappa x divided 12 00:01:45,700 --> 00:01:56,410 by 2, times E to the power i omega t minus beta z. So, let me arrange the terms in a 13 00:01:56,410 --> 00:02:03,570 particular passion. And then what I see that this term has E to the power i omega t minus 14 00:02:03,570 --> 00:02:10,240 beta z minus kappa x and this has E to the power i omega t minus beta z plus kappa x. 15 00:02:10,240 --> 00:02:18,250 So, this is nothing but a plane wave propagating in plus x-z direction, making certain angle 16 00:02:18,250 --> 00:02:25,450 with z axis and this is another plane wave which is propagating in minus x-z direction 17 00:02:25,450 --> 00:02:36,000 making certain angle from z axis. So, I have this field comes out to be the 18 00:02:36,000 --> 00:02:42,480 superposition of 2 plane waves one going in plus x-z direction another going in minus 19 00:02:42,480 --> 00:02:52,510 x-z direction. So, these are the 2 constituent plane waves. 20 00:02:52,510 --> 00:02:57,200 And what happens is if I do this kappa square plus beta square it comes out to be k naught 21 00:02:57,200 --> 00:03:04,690 n1 and k naught, n1 is nothing but the propagation constant with which a plane wave propagates 22 00:03:04,690 --> 00:03:09,099 in an infinitely extended medium of refractive index n1. 23 00:03:09,099 --> 00:03:17,060 So, kappa is nothing but, kappa is nothing but the component of this along x. So, if 24 00:03:17,060 --> 00:03:23,909 I resolve this in x and z direction then k naught n1 sin theta will give me kappa. So, 25 00:03:23,909 --> 00:03:32,879 it is the component of k naught n1 along x. And beta is the component of this k naught 26 00:03:32,879 --> 00:03:42,010 n1 along z. So, in x direction I will get from here and here I will get 2 counter propagating 27 00:03:42,010 --> 00:03:47,780 waves, and these 2 counter propagating waves gives me standing wave in x direction. So, 28 00:03:47,780 --> 00:03:51,810 the energy stands in x direction it does not flow out. 29 00:03:51,810 --> 00:04:00,590 However this standing wave pattern flows in z direction with propagation constant beta, 30 00:04:00,590 --> 00:04:11,330 exactly in the same way as planar mirror waveguide. So, these modal fields these modal fields 31 00:04:11,330 --> 00:04:18,030 are the standing wave patterns in x direction, because I have Ey of x Ey of x and this is 32 00:04:18,030 --> 00:04:20,150 standing wave pattern. 33 00:04:20,150 --> 00:04:29,710 And this Ey of x is propagating with propagation constant beta. So, if you look at it again 34 00:04:29,710 --> 00:04:50,770 this E (x, z, t) is equal to Ey of x E to the power i omega t minus beta z. Beta is 35 00:04:50,770 --> 00:05:00,650 k naught n effective. Which basically tells with what velocity this particular field this 36 00:05:00,650 --> 00:05:09,720 is standing wave will propagate. If you remember that in one of the previous lectures I had 37 00:05:09,720 --> 00:05:17,000 said that intuitively I can I can understand the effective index of the mode as effective 38 00:05:17,000 --> 00:05:26,000 refractive index felt by a particular field distribution because the field is now distributed 39 00:05:26,000 --> 00:05:34,230 in n1 and n2 regions. So, the effective refractive index would be somewhere between n2 and n1. 40 00:05:34,230 --> 00:05:39,160 And that was intuitively correct and that was not rigorously correct. And I see that 41 00:05:39,160 --> 00:05:49,500 rigorously, rigorously n effective is nothing but n effective is nothing but the velocity 42 00:05:49,500 --> 00:05:54,690 with which it will it will define the velocity with which a particular field pattern will 43 00:05:54,690 --> 00:06:00,070 travel ok. Rigorously I cannot get I cannot get from 44 00:06:00,070 --> 00:06:09,610 the distribution of power in n1 and n2 regions. So, this n effective is nothing but n1 cos 45 00:06:09,610 --> 00:06:20,830 theta. So, it is the component of the plane wave k naught n1 in z direction. And this 46 00:06:20,830 --> 00:06:28,530 distribution the effective refractive index as the distribution of field into 2 regions 47 00:06:28,530 --> 00:06:34,199 is also not rigorously correct in the sense that that I cannot get the effective index 48 00:06:34,199 --> 00:06:39,860 from that field distribution in that way. And also if you look at planar mirror waveguide 49 00:06:39,860 --> 00:06:47,919 then you will say that the field is always in n1 region then why it should have different 50 00:06:47,919 --> 00:06:54,100 effective indices for different modes and why it should be different from n1 or refractive 51 00:06:54,100 --> 00:06:58,889 index n. Similarly, if you talk about radiation mode 52 00:06:58,889 --> 00:07:08,110 in radiation modes the energy is distributed in n2 region it goes up to infinity, and the 53 00:07:08,110 --> 00:07:16,900 effective index for radiation modes as we will see later it is it is less than n2. So, 54 00:07:16,900 --> 00:07:26,920 rigorously effective index of the mode is nothing but the component of this along z 55 00:07:26,920 --> 00:07:34,380 direction. And it gives me the velocity with which a particular standing wave pattern will 56 00:07:34,380 --> 00:07:44,080 travel in z direction. And since different modes are different in standing wave patterns 57 00:07:44,080 --> 00:07:50,691 and constitute different plane waves which are at different angles from z axis. So, n 58 00:07:50,691 --> 00:07:59,330 effective would be different. If I look at the condition for guided mode 59 00:07:59,330 --> 00:08:08,410 then it is beta over k naught lies between n2 and n1. And beta over k naught is equal 60 00:08:08,410 --> 00:08:16,789 to n2 gives me the cutoff of a particular guided mode. From here what I get beta over 61 00:08:16,789 --> 00:08:24,490 k naught is nothing but n1 cos theta. So, if I replace this beta over k naught by n1 62 00:08:24,490 --> 00:08:34,030 cos theta here. And if I look at it this cos theta is nothing but sin phi if phi is the 63 00:08:34,030 --> 00:08:40,300 angle which the plane wave bakes with the normal to the interface of n1 and n2 regions 64 00:08:40,300 --> 00:08:47,100 than this is n2 less than n1 sin phi is less than n1. 65 00:08:47,100 --> 00:08:55,089 Let me divide the whole thing by n1. So, this gives me n2 over n1 is smaller than sin phi 66 00:08:55,089 --> 00:09:02,850 is smaller than 1 or sin phi is greater than n2 over n1. This is nothing but the condition 67 00:09:02,850 --> 00:09:12,330 for total internal reflection. So, I automatically get that that the cutoff condition translates 68 00:09:12,330 --> 00:09:23,050 to the condition for total internal reflection. If this angle phi is smaller than if this 69 00:09:23,050 --> 00:09:31,320 angle phi is smaller than sin inverse of n2 over n1, then this wave will not undergo total 70 00:09:31,320 --> 00:09:37,290 internal reflection at this interface and it would be refracted. And the corresponding 71 00:09:37,290 --> 00:09:45,930 mode would be radiated out it would not be guided anymore. So, this is how I can understand 72 00:09:45,930 --> 00:09:55,350 the mode of a waveguide. Let us work out again some examples. Let me 73 00:09:55,350 --> 00:10:03,860 consider a waveguide with n1 is equal to 1.5 n2 is equal to 1.46 d is equal to 2 micrometer. 74 00:10:03,860 --> 00:10:14,910 And it supports 2 modes TE0 and TE1 mode at lambda is equal to 1 micrometer. And if I 75 00:10:14,910 --> 00:10:24,390 find out the effective indices of these 2 modes they come out to be 1.4905 and 1.4665 76 00:10:24,390 --> 00:10:29,890 respectively. Now, let me calculate the penetration depths 77 00:10:29,890 --> 00:10:39,250 of these modes. And the angles of constituent plane waves from the waveguide axis. I know 78 00:10:39,250 --> 00:10:44,149 that penetration depth is given by 1 over gamma, where gamma is equal to beta square 79 00:10:44,149 --> 00:10:50,350 minus k naught square n2 square. Or k naught times square root of n effective square minus 80 00:10:50,350 --> 00:10:57,240 n2 square. So, k naught is 2 pi over lambda naught lambda naught is given n effective 81 00:10:57,240 --> 00:11:04,850 and n2 are given. So, if I now calculate these gamma for both the modes, then for TE0 mode 82 00:11:04,850 --> 00:11:13,580 gamma is equal to 1.884 micrometer inverse and correspondingly penetration depth is 0.53 83 00:11:13,580 --> 00:11:19,790 micrometer. For TE1 mode gamma comes out to be 0.867 micrometer 84 00:11:19,790 --> 00:11:28,340 inverse. And correspondingly the penetration depth comes out to be 1.15 micrometer. What 85 00:11:28,340 --> 00:11:35,140 are the angles of constituent plane waves from the waveguide axis? Well I know that 86 00:11:35,140 --> 00:11:43,190 these angles are given by n effective is equal to n1 cos theta. So, for these values of n 87 00:11:43,190 --> 00:11:50,410 effective I can immediately find out the values of theta, because I know the value of n1 also. 88 00:11:50,410 --> 00:11:58,860 So, for TE0 mode this theta would be cos inverse n effective over n1. So, it would be 6.45 89 00:11:58,860 --> 00:12:09,420 degrees and for TE1 it would be 12.13 degrees. So, TE0 mode is nothing but is nothing but 90 00:12:09,420 --> 00:12:17,820 the superposition of 2 plane waves, making angles making angles plus minus 6.45 degrees 91 00:12:17,820 --> 00:12:28,529 from the waveguide axis. And TE1 mode is the super position of 2 plane waves which make 92 00:12:28,529 --> 00:12:34,149 angles plus minus 12.13 degrees from the waveguide axis. 93 00:12:34,149 --> 00:12:47,060 So, after doing the analysis of the waveguide for TE modes, now in the same manner we can 94 00:12:47,060 --> 00:13:02,389 do the analysis for TM modes transfers magnetic modes. If I go back and see that for a waveguide 95 00:13:02,389 --> 00:13:10,720 which has confinement in x direction, that is refractive index variation in x direction 96 00:13:10,720 --> 00:13:31,389 and propagation in z direction, then the non-vanishing components of E and H are Hy, Ex and Ez. 97 00:13:31,389 --> 00:13:41,800 And these 3 components are related by these 3 relations. The procedure to obtain the modes 98 00:13:41,800 --> 00:13:50,090 and modal fields for TM modes is exactly the same as we had done in the case of TE polarization. 99 00:13:50,090 --> 00:13:59,880 So, what I need to do? I need to find out a differential equation for example, in Hy 100 00:13:59,880 --> 00:14:05,130 and solve it. So, to find out the differential equation 101 00:14:05,130 --> 00:14:14,399 in Hy, I substitute for Ex and Ez from these 2 equations into this equation, and rearrange 102 00:14:14,399 --> 00:14:23,860 the terms. Then I get a differential equation in Hy as d2Hy over dx square minus 1 over 103 00:14:23,860 --> 00:14:31,209 n square dn square over dx. dHy over dx plus k naught square n square minus beta square 104 00:14:31,209 --> 00:14:41,550 Hy is equal to 0. You may see that you may notice that this equation is different from 105 00:14:41,550 --> 00:14:49,589 the wave equation that we had got for the TE case. In the TE case this term was not 106 00:14:49,589 --> 00:14:57,589 there, excuse me in the TE case this term was not there, but here I have this term. 107 00:14:57,589 --> 00:15:04,110 So, let us see how do we take care of this now we need to solve this equation for given 108 00:15:04,110 --> 00:15:13,350 n square of x to find the modes. So, this is the waveguide which we are considering 109 00:15:13,350 --> 00:15:22,340 again symmetric dielectric planar waveguide where n of x is given like this. 110 00:15:22,340 --> 00:15:29,110 Procedure is again the same that I need to write this equation in this region and in 111 00:15:29,110 --> 00:15:40,930 this region. So, I do this for mode x less than d by 2 I get this equation. Why? Although 112 00:15:40,930 --> 00:15:48,720 I have this term in general for a variation n square of x, but when I write this equation 113 00:15:48,720 --> 00:15:54,100 in the individual regions in this kind of a step index waveguide the refractive index 114 00:15:54,100 --> 00:16:00,920 is uniform and when the effective index as uniform in this region. So, this term goes 115 00:16:00,920 --> 00:16:07,440 off. So, I have got this equation for mode x less than d by 2, and this equation for 116 00:16:07,440 --> 00:16:13,920 mode x greater than d by 2 and you can see these equations are exactly the same as we 117 00:16:13,920 --> 00:16:22,530 had obtained in the case of TE polarization. So, again I define this is kappa square and 118 00:16:22,530 --> 00:16:30,880 this is gamma square, and since I am interested in guided modes. So, kappa square and gamma 119 00:16:30,880 --> 00:16:33,440 square are positive. 120 00:16:33,440 --> 00:16:43,149 So, these are the equations and the solutions are again the same exactly the same, Hy is 121 00:16:43,149 --> 00:16:52,350 equal to a cosine kappa x plus b sin kappa x for mode x less than d by 2 and the decaying 122 00:16:52,350 --> 00:17:01,649 solutions for mode x greater than d by 2. And again A B C and D can be determined with 123 00:17:01,649 --> 00:17:05,130 the help of boundary conditions. 124 00:17:05,130 --> 00:17:11,120 In this case also I can make use of the symmetry of the problem and divide these solutions 125 00:17:11,120 --> 00:17:19,800 into symmetric and antisymmetric. So, for symmetric modes Hy would be a cosine kappa 126 00:17:19,800 --> 00:17:26,230 x in the guiding film, and Ce to the power minus gamma mode x for mode x greater than 127 00:17:26,230 --> 00:17:37,520 d by 2 that is in these regions. So, what next? What I should do? Now I should apply 128 00:17:37,520 --> 00:17:45,830 the boundary conditions, to this kind of modal field. And what are the boundary conditions 129 00:17:45,830 --> 00:17:51,250 the boundary conditions are again the same the tangential components of E and H should 130 00:17:51,250 --> 00:17:56,789 be continuous at x is equal to plus minus d by 2. 131 00:17:56,789 --> 00:18:03,370 What are the tangential components here? Well if I look at TM modes, the non-vanishing field 132 00:18:03,370 --> 00:18:10,640 components are Hy, Ex and Ez. So, the components which are tangential to x is equal to plus 133 00:18:10,640 --> 00:18:20,860 minus d by 2 planes are Hy and Ez. So, Hy and Ez should be continuous. How Ez is related 134 00:18:20,860 --> 00:18:30,240 to Hy? We will look at this. So, Ez is related to Hy with some constant times 1 over n square 135 00:18:30,240 --> 00:18:36,970 dHy over dx. So, this because this n square depends upon depends on x. So, I should take 136 00:18:36,970 --> 00:18:46,010 it to that side. So now, now the boundary conditions give me Hy and 1 over n square 137 00:18:46,010 --> 00:18:52,130 dHy over dx should be continuous. So, this is the difference with the TE case. 138 00:18:52,130 --> 00:19:00,490 In TE case I had the boundary conditions the boundary condition led to Ey and dEy over 139 00:19:00,490 --> 00:19:06,940 dx should be continuous at the boundaries, but here I have Hy and 1 over n square dHy 140 00:19:06,940 --> 00:19:16,830 over dx should be continuous at the boundary. So, I apply these so first I apply that the 141 00:19:16,830 --> 00:19:22,340 field should be continuous Hy is continuous at x is equal to let us say plus d by 2. So, 142 00:19:22,340 --> 00:19:29,620 a cosine kappa d by 2 is equal to C e to the power minus gamma d by 2, and then I apply 143 00:19:29,620 --> 00:19:35,950 this 1 over n square dHy over dx should be continuous at x is equal to plus d by 2, then 144 00:19:35,950 --> 00:19:45,370 it is on this side I have 1 over n square minus a sin kappa d by 2 is equal to minus 145 00:19:45,370 --> 00:19:50,940 1 over n2 square gamma C e to the power minus gamma d by 2. 146 00:19:50,940 --> 00:19:57,530 So, this gives me the transcendental equation, if I divide this by this. Kappa tan kappa 147 00:19:57,530 --> 00:20:06,169 d by 2 is equal to n1 square over n2 square times gamma or by multiplying with d by 2 148 00:20:06,169 --> 00:20:13,080 on both the sides I get kappa d by 2 tan kappa d by 2 is equal to n1 square over n2 square 149 00:20:13,080 --> 00:20:19,750 times gamma d by 2. So, this is the transcendental equation the difference is this factor. I 150 00:20:19,750 --> 00:20:28,659 have this factor n1 square over n2 square extra in the case of TM polarization. 151 00:20:28,659 --> 00:20:41,220 So, for symmetric modes, if I now convert this equation into psi and V then I know psi 152 00:20:41,220 --> 00:20:49,010 is nothing but kappa d by 2, then in exactly the same way as I had done for TE case for 153 00:20:49,010 --> 00:20:56,669 symmetric modes I will have the equation as psi tan psi is equal to n1 square over n2 154 00:20:56,669 --> 00:21:02,990 square V square minus psi square and this is the field. Similarly for antisymmetric 155 00:21:02,990 --> 00:21:10,440 modes the equation would be transformed to minus psi cot psi is equal to n1 square over 156 00:21:10,440 --> 00:21:14,710 n2 square, square root of V square minus psi square and this is the field. 157 00:21:14,710 --> 00:21:23,510 So, the transcendental equation is this. So, if I use the graphical solution to solve this. 158 00:21:23,510 --> 00:21:28,640 So, I should equate this on the right hand side and left hand side both to eta in this 159 00:21:28,640 --> 00:21:34,850 case as well as in this case. So, for symmetric mode I will have eta is equal to psi tan psi 160 00:21:34,850 --> 00:21:40,990 and for antisymmetric is as well as eta is equal to minus psi cot psi and from the right 161 00:21:40,990 --> 00:21:48,260 hand side I will get psi square over V square plus eta square over V square times n1 square 162 00:21:48,260 --> 00:21:59,200 over n2 square whole square is equal to 1. In the case of TE modes this factor was not 163 00:21:59,200 --> 00:22:07,211 there and I had the circles here psi square plus eta square is equal to V square. But 164 00:22:07,211 --> 00:22:15,720 now, but now I no more have circles, but this is the equation of an ellipse. So, I have 165 00:22:15,720 --> 00:22:24,360 ellipse here, and what is the major axis of the ellipse it is along eta. So, what I will 166 00:22:24,360 --> 00:22:31,740 have to do now? To have graphical solutions I will have to plot these 2 together for symmetric 167 00:22:31,740 --> 00:22:38,030 modes and these 2 together for antisymmetric modes and look for points of intersections. 168 00:22:38,030 --> 00:22:52,830 So, I do this. So, again these red curves show eta is equal to n1 square over n2 square 169 00:22:52,830 --> 00:22:55,350 psi tan psi. 170 00:22:55,350 --> 00:23:03,460 And so, this is for symmetric modes this is for antisymmetric modes. So, I this is eta 171 00:23:03,460 --> 00:23:12,929 is equal to psi tan psi it is eta is equal to minus psi cot psi. Now the solid lines 172 00:23:12,929 --> 00:23:21,320 the solid lines represent the circles, psi square plus eta square is equal to V square. 173 00:23:21,320 --> 00:23:29,120 And correspond to TE modes while these ellipses which are represented by dotted lines they 174 00:23:29,120 --> 00:23:41,660 are the ellipse corresponding to the TM modes. So, these are the ellipse corresponding to 175 00:23:41,660 --> 00:23:47,470 TM modes. What I see here if I take a particular value 176 00:23:47,470 --> 00:23:56,400 of V, if I take a particular value of V and see and see the points of intersection of 177 00:23:56,400 --> 00:24:02,870 these TE and TM modes, then it should not be V naught it should be V. So, for example, 178 00:24:02,870 --> 00:24:10,289 for V is equal to 6, for V is equal to 6 the point of intersection corresponding to TE 179 00:24:10,289 --> 00:24:20,120 mode is somewhere here. And for TM mode it is at slightly larger value of psi. So, what 180 00:24:20,120 --> 00:24:28,740 I have for a given for a given mode whether it is m is equal to 0 or m is equal to 1 or 181 00:24:28,740 --> 00:24:38,539 m is equal to 2, psi for TM mode is always greater than psi for TE mode. And since psi 182 00:24:38,539 --> 00:24:44,679 is equal to kappa d by 2 which is d by 2 square root of k naught square n1 square minus beta 183 00:24:44,679 --> 00:24:54,900 square, then beta for TM is always less than beta of TE. 184 00:24:54,900 --> 00:25:02,299 In terms of normalized parameters in the same way as I had done for TE case I can also write 185 00:25:02,299 --> 00:25:09,730 down the transcendental equation for TM modes, symmetric and antisymmetric. 186 00:25:09,730 --> 00:25:17,650 So, equations are similar, only thing is this extra factor of n1 square over n2 square in 187 00:25:17,650 --> 00:25:23,650 both the cases symmetric and antisymmetric cases. 188 00:25:23,650 --> 00:25:29,679 Cut off conditions. What are the cut off conditions? Well beta over k naught should be equal to 189 00:25:29,679 --> 00:25:36,010 n2, which means b should be equal to 0, if I put b is equal to 0 here and here I will 190 00:25:36,010 --> 00:25:43,830 get the cut off conditions for symmetric and antisymmetric modes. And you can see that 191 00:25:43,830 --> 00:25:51,080 since b has to be 0 if I put b is equal to 0 then this factor is absorbed here in 0. 192 00:25:51,080 --> 00:25:56,520 So, it would not have any effect it would not have any effect on the cut offs. 193 00:25:56,520 --> 00:26:05,110 So, for symmetric modes the cut off condition remains Vc tan Vc is equal to 0, and for antisymmetric 194 00:26:05,110 --> 00:26:15,370 mode it remains Vc cot Vc is equal to 0. So, in general I have the cut offs for TMm modes 195 00:26:15,370 --> 00:26:25,809 as Vcm is equal to m pi by 2, which is the same as in the case of TE modes ok. 196 00:26:25,809 --> 00:26:34,330 Let us now plot b-V curves. So, so that I will have to plot for a given value of n1 197 00:26:34,330 --> 00:26:41,270 over n2, let me see how does it look like schematically for n1 over n2 is equal to 1.5. 198 00:26:41,270 --> 00:26:47,940 So, the first thing is the cut offs cut offs are the same. So, the plots will start from 199 00:26:47,940 --> 00:27:00,200 the same value for TE and TM. So, if I plot it for TE0 mode then for TE0 mode it both 200 00:27:00,200 --> 00:27:07,620 will start from 0 because they have the same cut off. So, TE0 mode goes like this and TM0 201 00:27:07,620 --> 00:27:14,871 will also start from here, but whether this curve would be below this or above this. And 202 00:27:14,871 --> 00:27:22,140 I see that since the propagation constants of TM modes are smaller than the propagation 203 00:27:22,140 --> 00:27:30,190 constants of TE modes. So, the curve corresponding to TM mode would lie below the curve corresponding 204 00:27:30,190 --> 00:27:38,300 to TE modes. So, it will go like this, for m is equal to 1, So this would be TE1 and 205 00:27:38,300 --> 00:27:47,460 this is TM1 similarly TE2 and TM2. Again these curves are universal for given 206 00:27:47,460 --> 00:27:57,419 value of n1 over n2 now. So, here I should also take care what is this ratio n1 over 207 00:27:57,419 --> 00:28:05,950 n2. Modal fields, if you look at modal fields then again in the same way the modal field 208 00:28:05,950 --> 00:28:17,150 for TM0 would look like this. This is for low index contrast waveguide and for TM1 mode 209 00:28:17,150 --> 00:28:23,909 it looks like this. What I should take care here is and why it I should pay attention 210 00:28:23,909 --> 00:28:33,440 to is that in case of TE polarization Ey and dEy over dx was continuous. 211 00:28:33,440 --> 00:28:40,400 So, the field as well as it is slope was continuous at the boundaries x is equal to plus minus 212 00:28:40,400 --> 00:28:51,090 d by 2, but you remember that in TM polarization Hy and 1 over n square dHy over dx are continuous 213 00:28:51,090 --> 00:28:52,110 at the boundaries. 214 00:28:52,110 --> 00:29:00,929 So, the slope is not continuous. Although here I see that the slope is also continuous 215 00:29:00,929 --> 00:29:10,630 at the boundary, but rigorously it is not continuous. It is the discontinuity in the 216 00:29:10,630 --> 00:29:16,320 slope is. So, small that it does not show up because if you look at n1 and n2. n1 and 217 00:29:16,320 --> 00:29:23,899 n2 are very close. So, if it is weakly guiding if the waveguide is weekly guiding the n1 218 00:29:23,899 --> 00:29:31,510 the values of n1 and n2 are very close to each other than that would not show up here. 219 00:29:31,510 --> 00:29:37,860 If on the other hand if I plot this for high index contrast waveguide you now look at n 220 00:29:37,860 --> 00:29:43,299 the values of n1 and n2, the factor n1 over n2 is now 3.5. 221 00:29:43,299 --> 00:29:50,570 Now, if I plot the modal fields the modal fields for TE0 mode are like this there is 222 00:29:50,570 --> 00:29:57,610 no discontinuity in the slope while here this discontinuity shows up, because the factor 223 00:29:57,610 --> 00:30:06,179 n1 over n2 is very large. So, this is the first thing I will see, similarly 224 00:30:06,179 --> 00:30:19,190 in the case of TM1 mode there would be huge discontinuity. Again just as in the case of 225 00:30:19,190 --> 00:30:29,520 TE modes in TM modes also mth TM mode will have m number of 0s, and since the propagation 226 00:30:29,520 --> 00:30:40,059 constant of TM mode this is wrong it should be it should be beta TM is less than beta 227 00:30:40,059 --> 00:30:49,710 TE please make correction. So, the penetration depth of TM modes are larger, you can see 228 00:30:49,710 --> 00:30:56,649 the penetration depth of TM modes are larger than the penetration depths of TE modes. 229 00:30:56,649 --> 00:31:03,169 I can look at some examples if I consider this waveguide with n1 is equal to 1.5 n2 230 00:31:03,169 --> 00:31:06,179 is equal to 1.46 d is equal to 2 micrometer. 231 00:31:06,179 --> 00:31:14,220 At lambda is equal to 1 micrometer n effective of TE0 mode comes out to be this and of TM0 232 00:31:14,220 --> 00:31:21,110 mode comes out to be this. So, you can see if the index difference is very small, TE0 233 00:31:21,110 --> 00:31:29,670 and TM0 modes have nearly the same propagation constants, very close to each other. And if 234 00:31:29,670 --> 00:31:35,399 I now calculate the penetration depths for the 2 modes the penetration depths are also 235 00:31:35,399 --> 00:31:44,450 very close to each other 0.5305, 0.5329 differences only about 0.45 percent, but if I take high 236 00:31:44,450 --> 00:31:54,080 index contrast waveguide where n1 is 3.5 n2 is 1. Then n effective of TE0 mode is 3.1916 237 00:31:54,080 --> 00:32:02,690 and that of TM0 mode is 2.92. So, which is very different there is huge difference between 238 00:32:02,690 --> 00:32:08,140 the propagation constants. Now, if I calculate the penetration depths then the difference 239 00:32:08,140 --> 00:32:15,220 is about 10 and half percent. With this I have done the analysis of dielectric 240 00:32:15,220 --> 00:32:22,059 planar waveguide for symmetric modes as well as antisymmetric modes. I have calculated 241 00:32:22,059 --> 00:32:28,789 the propagation constants, I have found out the modal fields, their behaviors, penetration 242 00:32:28,789 --> 00:32:34,669 depth. Now only thing that remains is how much energy do they carry, how much energy 243 00:32:34,669 --> 00:32:40,080 how much power these modes carry. We will go into that in the next lecture. 244 00:32:40,080 --> 00:32:40,700 Thank you.