1 00:00:18,800 --> 00:00:28,849 In the last lecture we had done the modal analysis of optical fiber and we had obtained 2 00:00:28,849 --> 00:00:36,290 the transcendental equation satisfied by the propagation constants of the modes, we had 3 00:00:36,290 --> 00:00:40,050 also seen the cut offs of various modes. 4 00:00:40,050 --> 00:00:48,550 Now in this lecture we will extend the analysis and see how the modal fields look like. 5 00:00:48,550 --> 00:00:55,500 So, this is the step index fiber which we are analyzing, which has a high index core 6 00:00:55,500 --> 00:01:02,800 of refractive index n1 of radius a and the cladding of refractive index n2. 7 00:01:02,800 --> 00:01:10,580 And we have done this that the radial part of the modal fields is given by A Jl Ur over 8 00:01:10,580 --> 00:01:18,590 a in the region r less than a which is the core, and B Kl Wr over a in the region r greater 9 00:01:18,590 --> 00:01:21,860 than a which is the cladding. 10 00:01:21,860 --> 00:01:30,910 We had also defined the normalized frequency V normalized propagation constant b and U 11 00:01:30,910 --> 00:01:36,690 and W are defined by these relations and you can also express them in terms of normalized 12 00:01:36,690 --> 00:01:38,870 parameters V and b. 13 00:01:38,870 --> 00:01:46,880 So, we had seen that the propagation constants of the linearly polarized modes of this fiber 14 00:01:46,880 --> 00:01:52,670 satisfy these transcendental equations or the Eigen value equations. 15 00:01:52,670 --> 00:01:59,759 So, after solving these equations for a given value of V, I can find out the propagation 16 00:01:59,759 --> 00:02:07,660 constants b of various modes of the fiber, and if I plot them as a function of V, so 17 00:02:07,660 --> 00:02:08,910 they look like this. 18 00:02:08,910 --> 00:02:10,950 So, this we had done. 19 00:02:10,950 --> 00:02:15,200 Now, let us look at how the modal fields look like. 20 00:02:15,200 --> 00:02:23,750 So, the total modal field would be given by the radial part and the angular part or azimuthal 21 00:02:23,750 --> 00:02:25,690 part. 22 00:02:25,690 --> 00:02:35,220 In phi direction I can have solutions cosine l phi and sin l phi and I label these 2 solutions 23 00:02:35,220 --> 00:02:44,240 by 2 different names I call cosine alpha solution as even mode and sin alpha solution as odd 24 00:02:44,240 --> 00:02:45,240 mode. 25 00:02:45,240 --> 00:02:53,819 Now, let us look at the modal fields of LP01 and LP11 modes of this fiber. 26 00:02:53,819 --> 00:03:02,260 So, since we are looking at LP01 and LP11, which means here l is equal to 0 here l is 27 00:03:02,260 --> 00:03:03,560 equal to 1. 28 00:03:03,560 --> 00:03:13,630 So, in the core I will have solution for LP01 mode as J0 Ur over a, and for LP11 mode J1 29 00:03:13,630 --> 00:03:15,210 Ur over a. 30 00:03:15,210 --> 00:03:21,350 So, let me plot J0 x and J1 x, they look like this. 31 00:03:21,350 --> 00:03:30,000 In the cladding I will have the solutions K0 Wr over a and K1 Wr over a. 32 00:03:30,000 --> 00:03:34,390 So, let me plot how K0 x and K1 x look like. 33 00:03:34,390 --> 00:03:38,260 So, when I combine this k. 34 00:03:38,260 --> 00:03:45,370 So, in the core I will have this solution and in the cladding I will have this solution, 35 00:03:45,370 --> 00:03:51,319 and since these are the first modes of l is equal to 0 and l is equal to 1 respectively. 36 00:03:51,319 --> 00:04:02,200 So, I will not have any 0 in this and I will not have any 0 in this except a 0 at r is 37 00:04:02,200 --> 00:04:03,849 equal to 0. 38 00:04:03,849 --> 00:04:12,180 So, for LP01 mode in the core the field will go like this, and in the cladding this function 39 00:04:12,180 --> 00:04:13,930 will take over. 40 00:04:13,930 --> 00:04:19,220 For LP11 mode the field in the core would be given by J1. 41 00:04:19,220 --> 00:04:28,011 So, it would go like this without any 0 except a0 at r is equal to 0 and then this K1 function 42 00:04:28,011 --> 00:04:29,540 will take over. 43 00:04:29,540 --> 00:04:35,900 So, this is how the radial part of the modal fields would look like, what about the angular 44 00:04:35,900 --> 00:04:36,900 part? 45 00:04:36,900 --> 00:04:43,499 Angular part is simple for l is equal to 0 you have cosine alpha is equal to 1 and there 46 00:04:43,499 --> 00:04:48,110 is no contribution from sin l phi because it would be 0 everywhere. 47 00:04:48,110 --> 00:04:51,719 So, I will have only this kind of solution. 48 00:04:51,719 --> 00:05:00,499 So, you take this in radial part and then you go in angular part rotate it in phi direction 49 00:05:00,499 --> 00:05:03,840 and you will get this kind of variation. 50 00:05:03,840 --> 00:05:13,449 However, in for LP11 mode, I can have cosine phi solution and sin phi solution. 51 00:05:13,449 --> 00:05:18,960 So, if I take cosine phi solution that is at phi is equal to 0. 52 00:05:18,960 --> 00:05:23,110 I have a maximum, and then phi is equal to pi by 2. 53 00:05:23,110 --> 00:05:26,550 I will have 0 and so on. 54 00:05:26,550 --> 00:05:34,610 So, if I rotate it like this in phi direction, then I will get this kind of variation this 55 00:05:34,610 --> 00:05:39,749 kind of density plot or intensity pattern. 56 00:05:39,749 --> 00:05:47,059 If I take sin phi solution, then sin phi would be 0 at phi is equal to 0 and then it would 57 00:05:47,059 --> 00:05:52,720 be maximum at phi is equal to pi by 2 and then if I now rotate it. 58 00:05:52,720 --> 00:05:56,699 So, it would be 0 at phi is equal to 0 it would be maximum at phi is equal to pi by 59 00:05:56,699 --> 00:05:57,699 2. 60 00:05:57,699 --> 00:06:00,819 So, it will give me this kind of solution. 61 00:06:00,819 --> 00:06:12,680 So, l is equal to 0 mode will only be of this kind that is that is I cannot have 2 fold 62 00:06:12,680 --> 00:06:19,449 degeneracy for l is equal to 0 mode as I can have for l is equal to 1 mode here or l is 63 00:06:19,449 --> 00:06:25,159 equal to nonzero mode l nonzero mode. 64 00:06:25,159 --> 00:06:27,659 I can look at the 3 D plot of these. 65 00:06:27,659 --> 00:06:30,580 So, if I look at the various modes. 66 00:06:30,580 --> 00:06:36,400 So, this is how the LP01 mode would look like and if you view it from the top it would look 67 00:06:36,400 --> 00:06:37,740 like this. 68 00:06:37,740 --> 00:06:46,660 LP02 mode LP02 mode will admit 10 because it is the second mode in l is equal to 0 series. 69 00:06:46,660 --> 00:06:50,309 So, it will admit 10 in the core. 70 00:06:50,309 --> 00:06:55,669 So, it will go down this is the intensity plot this is the intensity plot. 71 00:06:55,669 --> 00:07:00,240 So, if you look at look at it from the top it would look like this and there would be 72 00:07:00,240 --> 00:07:06,300 a 0 at r is equal to something here in the core itself. 73 00:07:06,300 --> 00:07:13,580 If you look at LP11 mode it would look like this, it would not have any 0 in r direction 74 00:07:13,580 --> 00:07:23,369 except a 0 at r is equal to 0 and these are the this is the plot of even mode even LP11 75 00:07:23,369 --> 00:07:25,839 mode. 76 00:07:25,839 --> 00:07:36,180 So, if I again look at these plots, now I have LP mode say or linearly polarized modes 77 00:07:36,180 --> 00:07:44,430 it means that I can have 2 independent or orthogonal polarization states, I can label 78 00:07:44,430 --> 00:07:49,830 them as x polarized and y polarized because that is the direction of propagation. 79 00:07:49,830 --> 00:07:53,589 So, I can label them as x polarized and y polarized. 80 00:07:53,589 --> 00:07:59,119 So, if I look at LP01 mode, then it can be x polarized or y polarized. 81 00:07:59,119 --> 00:08:11,219 So, this LP01 mode is twofold degenerate and it does not have any 0 in r direction. 82 00:08:11,219 --> 00:08:20,509 If you look at LP11 mode then I can have cosine phi solution, and in cosine phi solution itself 83 00:08:20,509 --> 00:08:27,649 I can have 2 polarizations y polarized and x polarized and I can have sin phi solution. 84 00:08:27,649 --> 00:08:31,419 So, there also I can have x polarized and y polarized. 85 00:08:31,419 --> 00:08:38,159 So, it would be 2 it would be fourfold degenerate, and again there would be no 0 in r direction 86 00:08:38,159 --> 00:08:43,860 except a0 at r is equal to 0. 87 00:08:43,860 --> 00:08:50,650 If I look at LP02 mode, then LP02 mode would again be 2 fold degenerate you will have x 88 00:08:50,650 --> 00:08:53,570 polarized and y polarized. 89 00:08:53,570 --> 00:09:02,720 If you look at the second mode in l is equal to one series that is LP12 mode, then it will 90 00:09:02,720 --> 00:09:08,180 have one 0 in r direction, except a 0 at r is equal to 0. 91 00:09:08,180 --> 00:09:16,740 So, the modal fields would look like this and again this would again be fourfold degenerate. 92 00:09:16,740 --> 00:09:22,980 This is LP21 mode; now in LP21 mode because l is equal to 2. 93 00:09:22,980 --> 00:09:26,410 So, l is equal to 2 though. 94 00:09:26,410 --> 00:09:35,060 So, the 5 solutions are cosine 2 phi and sin 2 phi and they will have 4 zeros in phi directions 95 00:09:35,060 --> 00:09:44,079 that you can see here and in r direction there would not be any 96 00:09:44,079 --> 00:09:46,220 0 at except at r is equal to 0. 97 00:09:46,220 --> 00:09:55,320 So, so this is how the modal fields would look like for even and odd modes. 98 00:09:55,320 --> 00:10:06,320 So, in general what I get for LPlm mode, the number of zeros in phi direction would be 99 00:10:06,320 --> 00:10:15,120 2 l and the number of zeros in r direction would be m minus 1 except any 0 at r is equal 100 00:10:15,120 --> 00:10:17,290 to 0. 101 00:10:17,290 --> 00:10:25,790 So, if I have various mode I can immediately find out the number of zeros in r and phi 102 00:10:25,790 --> 00:10:34,480 direction for example, in LP21 mode there would be 4 zeros in phi direction and no 0 103 00:10:34,480 --> 00:10:35,980 in r direction. 104 00:10:35,980 --> 00:10:45,839 I always exclude a0 at r is equal to 0, for LP31 mode, I will have 6 zeros in phi direction 105 00:10:45,839 --> 00:10:50,959 and no 0 in r direction. 106 00:10:50,959 --> 00:11:00,079 For LP54 mode I will have 10 zeros in phi direction, and 3 zeros 4 minus 1 3 zeros in 107 00:11:00,079 --> 00:11:02,180 r direction. 108 00:11:02,180 --> 00:11:11,110 So, if I know any mode if I am given any mode then I can immediately find out the number 109 00:11:11,110 --> 00:11:18,940 of zeros in phi and r direction, and I can also plot the intensity patterns of those 110 00:11:18,940 --> 00:11:21,860 modes. 111 00:11:21,860 --> 00:11:31,070 For example, these are the intensity plots of various higher order modes, this is LP32 112 00:11:31,070 --> 00:11:35,759 mode LP32 mode will have 6 zeros in phi direction. 113 00:11:35,759 --> 00:11:45,029 So, I have 6 zeros 1, 2, 3, 4, 5, 6 zeroes in phi direction, and 2 minus 1 that is one 114 00:11:45,029 --> 00:11:50,019 0 in r direction except a 0 at r is equal to 0. 115 00:11:50,019 --> 00:11:59,290 LP43 mode will have 8 zeros in phi direction and 2 zeros in r direction. 116 00:11:59,290 --> 00:12:10,390 LP54 mode 10 zeros in phi direction and 3 zeros in r direction 1 2 3. 117 00:12:10,390 --> 00:12:21,430 LP63 mode 12 zeroes in phi direction 2 zeroes 3 minus 1 2 zeroes in r direction and so on. 118 00:12:21,430 --> 00:12:30,949 So, if I am given any mode then I can immediately draw the intensity pattern corresponding to 119 00:12:30,949 --> 00:12:32,980 that mode. 120 00:12:32,980 --> 00:12:43,019 Now, the question is how do I estimate the number of modes here, in planar wave guide 121 00:12:43,019 --> 00:12:51,720 it was easy I just calculate the value of V divide that value of V by pi by 2. 122 00:12:51,720 --> 00:12:59,970 And find out the closest, but greater integer and that will give me the number of modes; 123 00:12:59,970 --> 00:13:10,319 here the cut offs are not evenly spaced and there are several series corresponding to 124 00:13:10,319 --> 00:13:11,319 l. 125 00:13:11,319 --> 00:13:15,439 So, where do they fit I do not know immediately. 126 00:13:15,439 --> 00:13:20,350 So, what is the procedure of estimating the number of modes here? 127 00:13:20,350 --> 00:13:25,509 So, the first step is the same you first calculate the value of V. 128 00:13:25,509 --> 00:13:33,770 So, if you are given a fiber and the wave length, first you calculate the value of V, 129 00:13:33,770 --> 00:13:39,269 then you find out the cut offs of various modes using the cut off conditions and the 130 00:13:39,269 --> 00:13:40,839 zeros of Bessel functions. 131 00:13:40,839 --> 00:13:47,150 So, if you have the zeros of Bessel functions and cut off conditions for various modes you 132 00:13:47,150 --> 00:13:54,769 know for l is equal to 0 mode, the cut off condition for l not equal to 0 cut off conditions 133 00:13:54,769 --> 00:13:58,139 which are in terms of the zeros of Bessel functions. 134 00:13:58,139 --> 00:14:06,430 So, you find the cut offs and then arrange the modes in the increasing order of their 135 00:14:06,430 --> 00:14:07,569 cut offs. 136 00:14:07,569 --> 00:14:15,699 So, for example, the fundamental mode LP01 mode has 0 cutoff, LP11 mode has 2.4048 3.8317 137 00:14:15,699 --> 00:14:27,490 is for LP02 and LP21, 5.1356 is LP31, 5.5201 is LP12 and so on. 138 00:14:27,490 --> 00:14:35,689 So, you arrange them arrange the modes in increasing order of their cut offs, and then 139 00:14:35,689 --> 00:14:46,170 locate the value of V here and find out how many modes are above that value of V and those 140 00:14:46,170 --> 00:14:51,759 would be the number of modes supported by the fiber. 141 00:14:51,759 --> 00:15:00,959 Let me work out an example if for a given fiber and given wavelength I calculate the 142 00:15:00,959 --> 00:15:11,819 value of V and it comes out to be 5.3 then I locate this 5.3 in this table which is here 143 00:15:11,819 --> 00:15:17,029 and I count the modes above this, 1, 2, 3, 4, 5. 144 00:15:17,029 --> 00:15:25,459 So, these modes LP01, 11, 02, 21 and 31 these modes are guided. 145 00:15:25,459 --> 00:15:32,430 I can also find out the total number of modes by including their degeneracies, I find that 146 00:15:32,430 --> 00:15:36,019 there are 2 modes which have l is equal to 0. 147 00:15:36,019 --> 00:15:39,560 So, there will be only 2 fold degeneracy. 148 00:15:39,560 --> 00:15:50,949 So, they comprise of 4 modes, and then I have 1 2 3 modes which are l is equal to non L 149 00:15:50,949 --> 00:15:54,649 as L is equal to not 0. 150 00:15:54,649 --> 00:16:02,410 So, so they will have 4-fold degeneracy and they will comprise of 12 modes. 151 00:16:02,410 --> 00:16:04,870 So, I will have 12 plus 4. 152 00:16:04,870 --> 00:16:07,329 16 total modes including the degeneracies. 153 00:16:07,329 --> 00:16:14,720 So, this is how I can calculate the number of modes. 154 00:16:14,720 --> 00:16:22,510 Of course, if V is less than 2.4048, then there would only be one mode supported of 155 00:16:22,510 --> 00:16:25,800 course, there would be it is twofold degenerate. 156 00:16:25,800 --> 00:16:33,059 So, rigorously it supports 2 modes corresponding 2 polarizations, but as a convention I call 157 00:16:33,059 --> 00:16:41,459 it single mode fiber because for a given polarization it has only mode. 158 00:16:41,459 --> 00:16:51,749 So, for V less than 2.4048 I have a single mode fiber and this is how the cut off condition 159 00:16:51,749 --> 00:16:58,490 or for single mode operation is coming; because 2.4048 is the cutoff of first higher order 160 00:16:58,490 --> 00:17:01,690 mode which is the LP11 mode. 161 00:17:01,690 --> 00:17:11,360 If V is much much larger than one typically more than 10 or around 10, then the number 162 00:17:11,360 --> 00:17:21,310 of modes can be estimated using the approximate formula which are given as for a step index 163 00:17:21,310 --> 00:17:29,550 fiber, the approximate number of modes are V square by 2 and for a graded index fiber 164 00:17:29,550 --> 00:17:36,920 which power law profile which is characterized by profile parameter q, the number of modes 165 00:17:36,920 --> 00:17:42,890 can be estimated by half of q over q plus 2 times V square. 166 00:17:42,890 --> 00:17:51,580 So, if it is parabolic index fiber then q is equal to 2 then the number of modes would 167 00:17:51,580 --> 00:17:56,640 be V square by 4. 168 00:17:56,640 --> 00:18:05,490 Let me work out some examples, let me consider a step index optical fiber with co refractive 169 00:18:05,490 --> 00:18:13,900 index n1 is equal to 1.45, cladding refractive index n2 is equal to 1.44 and core radius 170 00:18:13,900 --> 00:18:22,890 8 micron, and I want to find out the total number of modes including degeneracies supported 171 00:18:22,890 --> 00:18:29,620 by the fiber at lambda naught is equal to 1.5 micrometer. 172 00:18:29,620 --> 00:18:38,070 So, the procedure is very simple you have to first find out the value of V, if you find 173 00:18:38,070 --> 00:18:47,340 out the value of V for these parameters then it comes out to be 5.697. 174 00:18:47,340 --> 00:18:55,490 If you locate this value of V in the table which you have created by arranging the modes 175 00:18:55,490 --> 00:19:03,510 in increasing order of their cut offs, then you find that these modes are supported LP01, 176 00:19:03,510 --> 00:19:09,180 11, 02, 21, 31 and 12. 177 00:19:09,180 --> 00:19:18,140 Now if you include the degeneracies of all these modes and calculate the total number 178 00:19:18,140 --> 00:19:22,520 of modes then they come out to be 20. 179 00:19:22,520 --> 00:19:34,920 So, with this value of V the number of modes would be around 20, you can also see that 180 00:19:34,920 --> 00:19:46,360 if you calculate using V square by 2, V square by 2 would roughly be because it is 6 it is 181 00:19:46,360 --> 00:19:47,560 close to 6. 182 00:19:47,560 --> 00:19:52,850 So, 6 squares 36, 36 divided by 2 is approximately 18. 183 00:19:52,850 --> 00:19:56,180 So, little more than 18 and here you are getting20. 184 00:19:56,180 --> 00:20:00,890 So, they are close. 185 00:20:00,890 --> 00:20:08,290 Second is the wave length at which the fiber is single moded, if I want to find out the 186 00:20:08,290 --> 00:20:14,020 wave length at which the fiber is single moded or the wavelength range in which the fiber 187 00:20:14,020 --> 00:20:21,910 is single moded then I know the single mode condition for a fiber is V should be less 188 00:20:21,910 --> 00:20:29,180 than 2.4048 and I have the expression for V. 189 00:20:29,180 --> 00:20:39,920 Which goes as V is equal to 2 pi over lambda naught times a times square root of n1 square 190 00:20:39,920 --> 00:20:42,810 minus n2 square. 191 00:20:42,810 --> 00:20:50,620 So, if I find lambda naught from here, then I will see that for lambda naught greater 192 00:20:50,620 --> 00:20:58,110 than 3.55 micrometer, V would be less than this and therefore, in this range of wave 193 00:20:58,110 --> 00:21:01,670 length the fiber would be single moded. 194 00:21:01,670 --> 00:21:10,840 Third part is approximate number of modes at lambda naught is equal to 0.5 micrometer. 195 00:21:10,840 --> 00:21:20,050 So, I again calculate the value of V at 0.5 micrometer, and it comes out to be about 17 196 00:21:20,050 --> 00:21:28,770 which is much larger than one, then I can find out the approximate number of modes by 197 00:21:28,770 --> 00:21:34,280 V square by 2 and it comes out to be about 146. 198 00:21:34,280 --> 00:21:39,480 So, this fiber will support 146 modes. 199 00:21:39,480 --> 00:21:52,500 Now, let me work out some examples on identification of modes from they are given intensity patterns, 200 00:21:52,500 --> 00:21:58,820 how do I identify various modes from their given intensity patterns. 201 00:21:58,820 --> 00:22:05,610 So, let us look at this, this one is easy. 202 00:22:05,610 --> 00:22:11,150 If I go in phi direction, I take any value of r and go in phi direction I do not encounter 203 00:22:11,150 --> 00:22:12,710 any 0. 204 00:22:12,710 --> 00:22:20,310 So, the number of zeros in phi direction is 0 which means 12 is 0, which means l is equal 205 00:22:20,310 --> 00:22:23,140 to 0. 206 00:22:23,140 --> 00:22:28,120 Second thing I do is I count the number of zeros in r direction. 207 00:22:28,120 --> 00:22:35,760 So, I take any value of phi and for that value of phi I move in r direction and an end I 208 00:22:35,760 --> 00:22:41,540 encounter 1 and 2 zeros, the number of zeros in r direction is 2. 209 00:22:41,540 --> 00:22:46,940 So, m is equal to 3 because m minus 1 is equal to 2. 210 00:22:46,940 --> 00:22:58,450 So, this mode is LP03 mode, similarly for this mode the number of zeros in phi direction 211 00:22:58,450 --> 00:23:01,210 the number of zeros in phi direction is 4. 212 00:23:01,210 --> 00:23:09,110 So, 2 l is equal to 4 l is equal to 2 and number of zeros in r direction you exclude 213 00:23:09,110 --> 00:23:10,110 0 here. 214 00:23:10,110 --> 00:23:11,110 So, it is only one. 215 00:23:11,110 --> 00:23:15,560 So, m minus 1 is equal to 1 which means m is equal to 2. 216 00:23:15,560 --> 00:23:19,650 So, this is LP22 mode. 217 00:23:19,650 --> 00:23:27,440 Similarly here it is LP51 mode and this is LP32 mode. 218 00:23:27,440 --> 00:23:32,940 So, in this way I can identify any mode. 219 00:23:32,940 --> 00:23:40,390 Next example is schematically draw the contour intensity plots or density plots of these 220 00:23:40,390 --> 00:23:46,140 modes, and I just want to draw the patterns of even mode. 221 00:23:46,140 --> 00:23:49,740 So, cosine alpha solution. 222 00:23:49,740 --> 00:23:52,970 So, let me do it for LP21 mode. 223 00:23:52,970 --> 00:23:59,960 So, for LP21 mode what I have l is equal to two. 224 00:23:59,960 --> 00:24:03,370 So, number of zeros in phi direction are 2 l. 225 00:24:03,370 --> 00:24:05,180 So, they are 4. 226 00:24:05,180 --> 00:24:10,290 So, now, where these 4 zeros are located? 227 00:24:10,290 --> 00:24:30,280 I know that if you go in phi direction then phi ranges from 0 to 2 pi, now you have solution 228 00:24:30,280 --> 00:24:35,210 which has cosine 2 phi. 229 00:24:35,210 --> 00:24:50,090 So, you now find out the zeros of this cosine 2 phi they will occur at pi by 2 then 3 pi 230 00:24:50,090 --> 00:24:58,730 by 2 and where do you stop you stop because phi goals phi goes up to 2 pi then 2 phi will 231 00:24:58,730 --> 00:25:01,720 go up to 4 phi. 232 00:25:01,720 --> 00:25:06,070 So, until you cross or reach 4 pi you do not stop. 233 00:25:06,070 --> 00:25:13,641 So, 3 pi by 2, phi pi by 2 and then 7 pi by 2, after that it would be nine pi by 2 which 234 00:25:13,641 --> 00:25:15,330 is more than 2 pi. 235 00:25:15,330 --> 00:25:16,750 So, you stop here. 236 00:25:16,750 --> 00:25:26,060 So, here these 4 zeros would be located which means phi is equal to pi by 4, 3 pi by 4, 237 00:25:26,060 --> 00:25:31,850 5 pi by 4 and 7 pi by 4. 238 00:25:31,850 --> 00:25:41,910 So, here you have 0 this is pi by 4, this is 2 pi by 4, this is 3 pi by 4, 4 pi by 4, 239 00:25:41,910 --> 00:25:45,340 5 pi by 4, 6 and this is seven pi by 4. 240 00:25:45,340 --> 00:25:55,200 So, here you have 0 and in between you will have the intensity and m is equal to 1. 241 00:25:55,200 --> 00:26:00,210 So, there would not be any 0 in r direction except a 0 at r is equal to 0. 242 00:26:00,210 --> 00:26:06,020 So, you will have intensity something like this. 243 00:26:06,020 --> 00:26:11,520 So, this is how you will plot the intensity. 244 00:26:11,520 --> 00:26:20,440 So, this is how it would look like similarly for LP32 mode you will have 6 zeros in phi 245 00:26:20,440 --> 00:26:24,580 direction and you can in the same way you can calculate the values of phi where the 246 00:26:24,580 --> 00:26:27,540 zeros will occur and since m is equal to 2. 247 00:26:27,540 --> 00:26:33,780 So, there would be 1 0 in r direction, in the same way you can plot for LP41 mode and 248 00:26:33,780 --> 00:26:34,940 LP13 mode. 249 00:26:34,940 --> 00:26:47,480 So, today we have learned about the mode modal fields, in the next class we would look in 250 00:26:47,480 --> 00:27:00,560 to what fraction of power of these modes is confined in the core and of and about various 251 00:27:00,560 --> 00:27:03,630 parameters of a single mode fiber. 252 00:27:03,630 --> 00:27:04,610 Thank you.