1 00:00:18,520 --> 00:00:26,070 In the last lecture we had carried out the analysis of asymmetric planar waveguide for 2 00:00:26,070 --> 00:00:33,020 TE-modes. In this lecture we will look into the TM-modes of the waveguide. 3 00:00:33,020 --> 00:00:39,010 So this is the waveguide again, and we will confine our analysis only to step index waveguides. 4 00:00:39,010 --> 00:00:43,340 So, this is the step index asymmetric planar waveguide. 5 00:00:43,340 --> 00:00:51,810 So, for TM-modes the non-vanishing components of electric and magnetic fields for this refractive 6 00:00:51,810 --> 00:00:59,980 index profile and z propagation are Hy, Ex and Ez. 7 00:00:59,980 --> 00:01:08,160 Again for guided modes the beta over k naught should lie between ns and nf, and if the beta 8 00:01:08,160 --> 00:01:16,490 over k naught is below ns then field radiates out. So, they are radiation mode. So, I again 9 00:01:16,490 --> 00:01:25,490 write down the wave equation in three different regions: the cover, the film, and the substrate. 10 00:01:25,490 --> 00:01:31,380 And now the equations are in the form d2Hy over dx square minus beta square minus k naught 11 00:01:31,380 --> 00:01:39,880 square nc square Hy is equal to 0 for the cover. And similarly in the film this is the 12 00:01:39,880 --> 00:01:42,000 equation, and in the substrate this is the equation. 13 00:01:42,000 --> 00:01:48,300 What I see that these equations are exactly the same. These equations are exactly the 14 00:01:48,300 --> 00:01:57,270 same in TE case these equations were in Ey and now these equations are in Hy. Again the 15 00:01:57,270 --> 00:02:04,280 definitions of gamma c kappa f and gamma s are the same. 16 00:02:04,280 --> 00:02:12,170 And if I write the solutions the solutions in the cover region is Hy of x is equal to 17 00:02:12,170 --> 00:02:19,930 A e to the power minus gamma c x. In the film it is B e to the power i kappa f x plus C 18 00:02:19,930 --> 00:02:28,890 e to the power minus i kappa f x. In the substrate it is D e to the power gamma s x, where gamma 19 00:02:28,890 --> 00:02:33,920 c kappa f and gamma s are defined by these expressions. 20 00:02:33,920 --> 00:02:40,220 Again I will have to apply the boundary conditions at the interfaces x is equal to 0 and x is 21 00:02:40,220 --> 00:02:49,530 equal to minus d; to obtain the relationships between A B C and D, and to obtain the eigenvalue 22 00:02:49,530 --> 00:02:50,530 equation. 23 00:02:50,530 --> 00:02:57,560 So, if I do this I again write down the solutions in different regions. And now apply the boundary 24 00:02:57,560 --> 00:03:05,390 conditions, remember again the boundary conditions are the tangential components of E and H are 25 00:03:05,390 --> 00:03:14,350 continuous. So, these boundary conditions now become Hy and 1 over n square dHy over 26 00:03:14,350 --> 00:03:22,160 dx are continuous at the interfaces x is equal to 0 and at x is equal to minus d 27 00:03:22,160 --> 00:03:32,000 So, when I now apply these boundary conditions to these fields and do mathematical manipulations 28 00:03:32,000 --> 00:03:41,550 similar to what we had done in the case of TE-modes. I get the eigenvalue equation as 29 00:03:41,550 --> 00:03:51,900 this. The only difference is now these factors of nf square over ns square associated with 30 00:03:51,900 --> 00:03:58,860 the substrate term, nf square over nc square associated with the cover term. So, these 31 00:03:58,860 --> 00:04:02,990 extra factors are there otherwise the equation is similar. 32 00:04:02,990 --> 00:04:08,940 So, this is the eigenvalue equation, after solving this I can find out the modes, the 33 00:04:08,940 --> 00:04:13,110 propagation constants of the modes and later on the fields. 34 00:04:13,110 --> 00:04:20,400 In normalized parameters if I define this equation then it becomes tan 2 V square of 35 00:04:20,400 --> 00:04:29,440 1 minus b nf square over ns square square root of b divided by 1 minus b plus nf square 36 00:04:29,440 --> 00:04:37,659 over nc square square root of b plus a over 1 minus b divided by 1 minus nf square over 37 00:04:37,659 --> 00:04:46,979 ns square times square root of b over 1 minus b and nf square over nc square times square 38 00:04:46,979 --> 00:04:53,400 root of b plus a over 1 minus b. What are the cut offs? The cut offs are again 39 00:04:53,400 --> 00:05:00,990 defined by b is equal to 0. So, if I put b is equal to 0 here the equation define the 40 00:05:00,990 --> 00:05:09,060 cut offs for TM-modes are now tan 2Vc is equal to nf square over nc square times square root 41 00:05:09,060 --> 00:05:16,810 of a. So, if I compare it with the cut offs of TE-modes then I have this extra factor 42 00:05:16,810 --> 00:05:26,930 of nf square over nc square here. So, my cut offs for m-th TM-modes are now given by m 43 00:05:26,930 --> 00:05:35,250 pi by 2 plus half tan inverse of nf square over nc square square root of a. 44 00:05:35,250 --> 00:05:44,430 I can see from here that since nf is larger than nf then these cut offs of TM-modes are 45 00:05:44,430 --> 00:05:48,419 larger than the cut offs of corresponding TE-modes. 46 00:05:48,419 --> 00:05:59,530 So, let me now plot b-V curves here by solving the transcendental equation for different 47 00:05:59,530 --> 00:06:10,650 values of V and for the given value of a. So here, I have plotted for two cases: one 48 00:06:10,650 --> 00:06:20,560 is for symmetric waveguide and another is for asymmetric waveguide. For comparison I 49 00:06:20,560 --> 00:06:28,089 have also plotted the b-V curves for TE-modes. So, the b-V curves for TE-modes are given 50 00:06:28,089 --> 00:06:35,900 by solid line, for TM-modes they are given by dashed line. If I look at these curves 51 00:06:35,900 --> 00:06:47,199 now, for a is equal to 0 this is TE0 mode, this is TM0 mode. Solid line is TE0 mode dashed 52 00:06:47,199 --> 00:06:53,660 line is TM0 mode, and I see that both the modes have the same cut off. 53 00:06:53,660 --> 00:07:01,210 But if I take a naught equal to 0 that is I consider the case of asymmetric waveguide 54 00:07:01,210 --> 00:07:14,499 then the TE0 mode has cut off somewhere here, and TM0 mode has cut off somewhere here. And 55 00:07:14,499 --> 00:07:20,680 I can see that in this range, so this cut off point is defined by half tan inverse of 56 00:07:20,680 --> 00:07:28,289 square root a and this cut off point is defined by half tan inverse nf square over nc square 57 00:07:28,289 --> 00:07:32,819 square root of a. So, in this region if the value of V lies 58 00:07:32,819 --> 00:07:41,659 in this region then only TE0 mode is guided and all the other modes included in TM0 are 59 00:07:41,659 --> 00:07:53,469 cut off. So, in this range I guide strictly one mode and that mode is TE0 mode. Or rigorously 60 00:07:53,469 --> 00:08:02,289 speaking in this range I guide only one mode and one polarization only TE polarization. 61 00:08:02,289 --> 00:08:10,719 So, this range is also called SPSM range- Single Polarization Single Mode range. 62 00:08:10,719 --> 00:08:13,649 Let us look at the modal fields. 63 00:08:13,649 --> 00:08:20,960 Modal fields are similar but there is one difference we should mark, and that difference 64 00:08:20,960 --> 00:08:31,919 is discontinuity at on the slope at the interfaces. Because Hy is continuous so there is no discontinuity 65 00:08:31,919 --> 00:08:39,419 in the field, but dHy over dx is not continuous at the interface. So, there is discontinuity 66 00:08:39,419 --> 00:08:47,610 in the slope at the interfaces. The number of modes: how many modes are guided? 67 00:08:47,610 --> 00:08:55,020 Well, now all the modes are shifted by this much amount half tan inverse of nf square 68 00:08:55,020 --> 00:09:02,700 over nc square square root of a. So, the number of modes would now be an integer closest to 69 00:09:02,700 --> 00:09:10,530 but greater than this number. Let us look at how the number of TM-modes 70 00:09:10,530 --> 00:09:16,990 vary when we change wavelength and compare it with the TE case also. 71 00:09:16,990 --> 00:09:24,600 So, here I have plotted the number of modes as a function of wavelength, ok. If I find 72 00:09:24,600 --> 00:09:34,020 out the cut off wavelength of TE0 mode then for these waveguide parameters, then it comes 73 00:09:34,020 --> 00:09:43,590 out to be 1.7187 micrometre. If I now find out the cut off wavelength for the same waveguide 74 00:09:43,590 --> 00:09:54,450 for TM0 mode then it comes out to be 1.5916. So now, if I again start from a wavelength 75 00:09:54,450 --> 00:10:10,100 two micron and start decreasing the wavelength, then as I cross 1.7187 as I cross 1.7187 TE0 76 00:10:10,100 --> 00:10:17,810 mode starts appearing this blue one; TE0 mode starts appearing but TM0 is still not there. 77 00:10:17,810 --> 00:10:26,940 And as soon as I go below 1.5916 then TM0 also starts appearing. 78 00:10:26,940 --> 00:10:37,120 So, in this range only TE0 mode is there and this is SPSM range, while if I go below this 79 00:10:37,120 --> 00:10:47,800 value then I have both TE0 and TM0. Similarly if I go below this then TM1 will also start 80 00:10:47,800 --> 00:10:57,970 appearing. So, this is the range of wavelength 1.5916 to 1.7187, in this wavelength range 81 00:10:57,970 --> 00:11:05,200 I have single polarization single mode operation of the waveguide. 82 00:11:05,200 --> 00:11:08,750 Let us workout few examples here. 83 00:11:08,750 --> 00:11:16,630 Let me consider a dielectric asymmetric planar waveguide with nf is equal to1.5 and ns is 84 00:11:16,630 --> 00:11:24,160 equal to 1.48 and nc is equal to 1. Now, I want to calculate the range of normalized 85 00:11:24,160 --> 00:11:35,250 frequency for SPSM operation. So, I know that SPSM operation involves asymmetry parameter 86 00:11:35,250 --> 00:11:44,120 a, so first I calculate a for this waveguide and a comes out to be about 20. Then SPSM 87 00:11:44,120 --> 00:11:52,180 range is given by half tan inverse square root a smaller than V smaller than half tan 88 00:11:52,180 --> 00:11:59,480 inverse nf square over nc square square root of a. So now, if I calculate these then this 89 00:11:59,480 --> 00:12:08,960 range comes out to be 0.6753 less than V less than 0.7358. So, this is the range of normalized 90 00:12:08,960 --> 00:12:14,930 frequency V for SPSM operation. 91 00:12:14,930 --> 00:12:27,100 And remember that the definition of V which I have used is 2 pi over lambda naught times 92 00:12:27,100 --> 00:12:38,180 d by 2 times square root of nf square minus ns square. And I would like to bring out that 93 00:12:38,180 --> 00:12:44,370 in the text books in several textbooks the definition of V is 2 pi over lambda naught 94 00:12:44,370 --> 00:12:51,920 times d times nf square minus ns square, while I use d by 2. So, there would be a factor 95 00:12:51,920 --> 00:12:53,000 of 2. 96 00:12:53,000 --> 00:13:02,030 Second is for d equal to 1 micrometre what is the wavelength range for SPSM operation? 97 00:13:02,030 --> 00:13:11,640 So, I know that from the previous problem I know that V should lie between 0.6753 to 98 00:13:11,640 --> 00:13:20,780 0.7358 for SPSM operation. Now for this range of V I can find out the corresponding range 99 00:13:20,780 --> 00:13:32,120 of lambda if d is given. So, lambda in terms of V and d and nf ns is given by 2 pi over 100 00:13:32,120 --> 00:13:38,500 V d by 2 square root of nf square minus nc square. So, I simply find out the value of 101 00:13:38,500 --> 00:13:43,180 lambda naught corresponding to these values of V. 102 00:13:43,180 --> 00:13:55,130 So, it comes out to be lambda lies between 1.042 micrometre and 1.136 micrometer. 103 00:13:55,130 --> 00:14:06,290 Third is if I choose a light source of wavelength lambda naught is equal to 1 micrometre then 104 00:14:06,290 --> 00:14:15,480 in what range of film thickness there would be SPSM operation. So, again I have the SPSM 105 00:14:15,480 --> 00:14:28,290 range of V from 0.6753 to 0.7358 and I simply now find out the corresponding values of d 106 00:14:28,290 --> 00:14:34,510 if lambda naught is given to me. So, this comes out to be between 0.88 micrometre 107 00:14:34,510 --> 00:14:45,550 and 0.96 micrometre. So, in this range in this very small range of d I will have SPSM 108 00:14:45,550 --> 00:14:48,070 operation. 109 00:14:48,070 --> 00:14:53,350 How much is the power associated with the modes. So, to find out the power associated 110 00:14:53,350 --> 00:15:00,790 with the mode I will have to calculate the pointing vector and then integrate it over 111 00:15:00,790 --> 00:15:09,250 the entire cross section. Since it is a planar waveguide, so I cannot integrate it over y, 112 00:15:09,250 --> 00:15:15,430 because in y direction it is infinitely extended. So, what I can have is power per unit length 113 00:15:15,430 --> 00:15:21,110 in y direction. So, for TE-mode I have already seen in case 114 00:15:21,110 --> 00:15:28,940 of symmetric waveguides that P is given by beta over 2 omega mu naught times integral 115 00:15:28,940 --> 00:15:38,150 minus infinity to plus infinity Ey square of xdx. So, if I now find out the integration 116 00:15:38,150 --> 00:15:47,471 Ey square of xdx for the modes; for the modal fields of asymmetric planar waveguide for 117 00:15:47,471 --> 00:15:56,100 TE-mode then I can find out the power associated with TE-modes and it is given by this. 118 00:15:56,100 --> 00:16:04,380 Similarly for TM-modes the power per unit length is given by beta over 2 omega epsilon 119 00:16:04,380 --> 00:16:15,070 naught integral minus infinity to plus infinity 1 over n square Hy square xdx. And if I do 120 00:16:15,070 --> 00:16:26,279 the same then the power corresponding to TM-modes comes out to be like this. 121 00:16:26,279 --> 00:16:35,029 The last thing that I would like to do in this lecture is- how do I excite a particular 122 00:16:35,029 --> 00:16:43,110 mode. Can I excite a particular mode? Selectively excite; I know if I have waveguide and if 123 00:16:43,110 --> 00:16:49,610 I launched the light from end then all the modes would be excited in different proportion 124 00:16:49,610 --> 00:16:56,250 depending upon what is the intensity profile or what is the amplitude profile of the incident 125 00:16:56,250 --> 00:17:01,820 light. But, can I selectively excite one particular 126 00:17:01,820 --> 00:17:11,819 mode? So for what is used is prism coupling technique. So, what you do? You have a waveguide 127 00:17:11,819 --> 00:17:18,429 this is an asymmetric planar waveguide whose cover is air. So, this can be glass and this 128 00:17:18,429 --> 00:17:26,010 can be polymer film for example, and this is air. Now what I do I put a prism on top 129 00:17:26,010 --> 00:17:39,190 of this and press it, clamp it, then what I have here even though it is pressed hard 130 00:17:39,190 --> 00:17:48,660 there is some air gap between the prism and the film. And what happens is now I launched 131 00:17:48,660 --> 00:17:58,270 light from here into the prism, this light beam gets reflected into the prism. 132 00:17:58,270 --> 00:18:06,430 And then depending upon this angle psi which can be translated, which can be related to 133 00:18:06,430 --> 00:18:14,430 this angle theta p with this beam makes with the normal to the prism base then depending 134 00:18:14,430 --> 00:18:21,260 upon this angle theta p this would get totally internally reflected. So, I can have total 135 00:18:21,260 --> 00:18:28,850 internal reflection at the prism base. I know that a total internal reflection is always 136 00:18:28,850 --> 00:18:36,280 associated with an evanescent tail. So, here what basically I have a standing wave here 137 00:18:36,280 --> 00:18:43,910 whose tail evanescent tail extends into the film. 138 00:18:43,910 --> 00:18:50,000 So, I have a standing wave here whose evanescent tail extends into the field and it is this 139 00:18:50,000 --> 00:19:01,400 evanescent field which excites the mode of the waveguide. So, this is the standing wave. 140 00:19:01,400 --> 00:19:08,190 And this is one typical guided mode. This is you can see that it is TE0 mode if launched 141 00:19:08,190 --> 00:19:20,090 polarization is TE. This guided mode is nothing but a superposition of two plane waves and 142 00:19:20,090 --> 00:19:26,830 these plane waves make angle plus minus theta f from the waveguide axis. So, if this angle 143 00:19:26,830 --> 00:19:36,800 is theta f then the propagation constant of this plane wave is k naught nf, then the propagation 144 00:19:36,800 --> 00:19:40,720 constant of the mode is k naught nf cos theta f. 145 00:19:40,720 --> 00:19:46,550 So, propagation constant of the mode is k naught nf cos theta f, which is the horizontal 146 00:19:46,550 --> 00:19:54,990 component of this. While if I look at this standing wave pattern then, what is the horizontal 147 00:19:54,990 --> 00:20:00,350 component of this plane wave. The horizontal component of this plane wave is k naught np 148 00:20:00,350 --> 00:20:08,090 sin theta p. So, if the horizontal component of this standing wave is the same as the horizontal 149 00:20:08,090 --> 00:20:16,750 component of this guided mode then well horizontal component of the plane wave corresponding 150 00:20:16,750 --> 00:20:24,370 to guided board then there would be phase matching, because these two horizontal components 151 00:20:24,370 --> 00:20:33,580 are now able to catch up each other. And that is how this would be able to resonantly couple 152 00:20:33,580 --> 00:20:40,600 energy into this mode. So, if this condition is satisfied I have 153 00:20:40,600 --> 00:20:49,179 excitation of that particular guided mode which corresponds to a particular angle theta 154 00:20:49,179 --> 00:20:55,360 f. And I know that this k naught nf cos theta f is nothing but k naught n effective. So, 155 00:20:55,360 --> 00:21:02,990 from here I can even find out the effective index of the mode if I know angle psi. How 156 00:21:02,990 --> 00:21:11,840 I have got this? Well, this angle psi can be related to this theta p. If I look at this 157 00:21:11,840 --> 00:21:22,950 triangle then A plus pi by 2 plus r where r is the angle of reflection plus pi by 2 158 00:21:22,950 --> 00:21:30,960 minus theta p should be equal to pi. Or I get r in terms of psi from Snellâ€™s law sin 159 00:21:30,960 --> 00:21:37,740 psi is equal to np sin r and if I put it here then I get theta p is equal to a plus sin 160 00:21:37,740 --> 00:21:45,380 inverse sin psi over np. And this gives me n effective in terms of 161 00:21:45,380 --> 00:21:54,370 the incident angle, the prism refractive index, prism angle A, and again prism reflective 162 00:21:54,370 --> 00:22:02,920 index. So, if I know this then by just measuring these psi I can find out the effective index 163 00:22:02,920 --> 00:22:11,820 of the mode. And you can see that this resonant excitation can take place when this condition 164 00:22:11,820 --> 00:22:19,200 is satisfied and theta f is discrete; theta f is discrete for different modes. For TE0 165 00:22:19,200 --> 00:22:24,820 mode it is different for TE1 mode it is different for TE2 mode it is different. 166 00:22:24,820 --> 00:22:33,670 So, for different values of psi I will have different values of theta p. And if they match 167 00:22:33,670 --> 00:22:40,980 to these discrete theta f then I will excite those particular modes. So, in this way I 168 00:22:40,980 --> 00:22:51,220 can selectively excite, I can selectively excite the modes of a planar waveguide. 169 00:22:51,220 --> 00:22:58,200 This is the experimental setup. So, you have a laser beam and this is a lens through which 170 00:22:58,200 --> 00:23:06,250 you focus this onto the prism coupling arrangement. This is a typical prism coupling arrangement, 171 00:23:06,250 --> 00:23:14,929 this is the waveguide, and this is the prism. And you can see that when I tune because this 172 00:23:14,929 --> 00:23:21,870 assembly can be rotated with respect to the beam, so I can change the angle psi. And I 173 00:23:21,870 --> 00:23:31,340 can see that for a particular value of psi the other input angle I see a mode is excited 174 00:23:31,340 --> 00:23:40,540 and I see a streak going down the length of the waveguide. 175 00:23:40,540 --> 00:23:49,850 So, this represents the propagation of mode. And I can see that as it goes then this becomes 176 00:23:49,850 --> 00:23:55,300 feeble and feeble it becomes week because of the losses. And in fact, you can see this 177 00:23:55,300 --> 00:23:59,270 because there are scattering losses. There are scattering losses that is how you can 178 00:23:59,270 --> 00:24:11,440 see. So, more strong the streak is bad is the quality of the waveguide. I can also do, 179 00:24:11,440 --> 00:24:19,110 I can put another prism here and I can decouple the modes in the same way as I have coupled 180 00:24:19,110 --> 00:24:25,560 them. So, I put another prism here. So, you can see that from this side I am coupling 181 00:24:25,560 --> 00:24:32,250 light this is the streak and then I decouple, then I decouple and when I put it here on 182 00:24:32,250 --> 00:24:39,290 the screen then I see these lines which correspond to the modes different modes. For example, 183 00:24:39,290 --> 00:24:46,720 in this I can see 1, 2, 3, 4, 5, 6 modes, and by changing the angle I can put light 184 00:24:46,720 --> 00:24:54,520 into one particular mode. For example, here it is coupled to third mode from the left. 185 00:24:54,520 --> 00:25:03,170 These are known as m lines. So, by measuring angles psi I can find out the effective index 186 00:25:03,170 --> 00:25:10,890 of the mode n effective. And once I have the values of n effective then using a technique 187 00:25:10,890 --> 00:25:18,390 called inverse WKB method I can do the refractive index profiling of the waveguide. 188 00:25:18,390 --> 00:25:25,500 So, by measuring the refractive indices of the modes I can find out the refractive index 189 00:25:25,500 --> 00:25:32,140 profile of the waveguide. The only thing is that I should have sufficient values of n 190 00:25:32,140 --> 00:25:38,710 effective, I should have sufficient values of n effective. For a single mode waveguide 191 00:25:38,710 --> 00:25:44,860 it would not work, if you have only one value of n effective it will not work. So, if it 192 00:25:44,860 --> 00:25:48,830 is a single mode waveguide at one particular wavelength then it will not work, you will 193 00:25:48,830 --> 00:25:56,860 have to use different wavelengths to have the values of n effective at different wavelengths 194 00:25:56,860 --> 00:26:04,100 and then you can do it; otherwise, at a single wavelength if it is highly multimode waveguide 195 00:26:04,100 --> 00:26:13,920 then it is more accurate. So, this is all in planar waveguides. And 196 00:26:13,920 --> 00:26:24,100 in the next lecture we will go into cylindrical geometry and find out how the modes are formed 197 00:26:24,100 --> 00:26:30,460 in an optical fiber, and how do we analyze the modes of an optical fiber, how do we find 198 00:26:30,460 --> 00:26:38,610 out the modes of an optical fiber. I have spent a lot of time in the analysis of planar 199 00:26:38,610 --> 00:26:47,500 waveguide. And in the optical fiber I would adopt all these results to cylindrical geometry, 200 00:26:47,500 --> 00:26:55,080 because the physics is now clear. And it was easier to understand all these 201 00:26:55,080 --> 00:27:02,430 in planar geometry, because it was one dimensional problem. And also the functions involved were 202 00:27:02,430 --> 00:27:09,790 very simple: sin cosine functions and exponentially amplifying and decaying functions. But in 203 00:27:09,790 --> 00:27:16,190 case of cylindrical geometry these functions would be different. However, the physics which 204 00:27:16,190 --> 00:27:23,480 we have understood from the analysis of planar waveguide would be applicable there also. 205 00:27:23,480 --> 00:27:25,000 Thank you.