1 00:00:18,550 --> 00:00:25,260 Till now we have carried out the analysis of symmetric planar wave guides where the 2 00:00:25,260 --> 00:00:32,790 high index region is sandwiched between two lower index regions of same refractive index. 3 00:00:32,790 --> 00:00:40,050 Now in this lecture we will extend the analysis to asymmetric planar waveguide asymmetric 4 00:00:40,050 --> 00:00:46,089 planar waveguides are more practical wave guides. So, we will extend this analysis to 5 00:00:46,089 --> 00:00:54,480 asymmetric planar waveguide and see how the waveguide hence is done in these kind of waveguides. 6 00:00:54,480 --> 00:01:00,590 So, what is an asymmetric planar waveguide you for example, take a glass of substrate 7 00:01:00,590 --> 00:01:10,110 of refractive index ns, and you deposit a film of refractive index nf of thickness d, 8 00:01:10,110 --> 00:01:17,060 this film can be for example, of a polymer material, the refractive index nf is higher 9 00:01:17,060 --> 00:01:26,330 than the refractive index ns and then it is surrounded by air. So, air works as cover; 10 00:01:26,330 --> 00:01:36,640 also you can put a cover of polymer material itself of refractive index nc, where nc is 11 00:01:36,640 --> 00:01:44,130 smaller than ns and ns is smaller than nf. So, this kind of waveguide we can see that 12 00:01:44,130 --> 00:01:53,670 is has asymmetric structure, because ns is not equal to nc here. If I look at the refractive 13 00:01:53,670 --> 00:02:02,420 index profile of this waveguide, then this is the substrate an example is glass, this 14 00:02:02,420 --> 00:02:09,530 is the guiding film the example is polymer, and this is the cover region which can be 15 00:02:09,530 --> 00:02:19,840 air or another polymer of lower refractive index. Here at x = 0, I have made the interface 16 00:02:19,840 --> 00:02:29,830 between the film and the cover, and at x = -d represents the interface between the substrate 17 00:02:29,830 --> 00:02:36,670 and the film. Since the refractive indices of the individual 18 00:02:36,670 --> 00:02:44,400 layers are uniform. So, this waveguide is step index waveguide. I can also have a waveguide 19 00:02:44,400 --> 00:02:51,590 in which the refractive index in the guiding film is not uniform, but it varies with x. 20 00:02:51,590 --> 00:03:01,080 So, how I can make this waveguide well I take a substrate of material glass, and then I 21 00:03:01,080 --> 00:03:07,010 diffuse certain ions into this pay over a certain distance. So, I will have a diffusion 22 00:03:07,010 --> 00:03:12,920 profile for example, I can diffuse silver ions into glass and that will change the refractive 23 00:03:12,920 --> 00:03:22,380 index of the material near the surface. So, now this nf would be a function of x now, 24 00:03:22,380 --> 00:03:31,790 and then I can have cover as air or cover of a polymer layer. So, now, this nc is smaller 25 00:03:31,790 --> 00:03:41,540 than ns, and then the refractive index of the film at the surface of the film nf at 26 00:03:41,540 --> 00:03:50,150 f = 0. So, ns is smaller than that. If I look at the refractive index profile of this structure 27 00:03:50,150 --> 00:03:56,620 now, so, I have a substrate and this is the guiding film. This guiding film now has refractive 28 00:03:56,620 --> 00:04:05,120 index with varies which varies with x and ultimately it merges into the refractive index 29 00:04:05,120 --> 00:04:10,390 of glass. So, this is the surface index nf(0), and this 30 00:04:10,390 --> 00:04:15,950 nf(0)is larger than ns and which is larger than nc. 31 00:04:15,950 --> 00:04:24,349 So, this is a typical graded index ion exchange waveguide. Let us do the modal analysis of 32 00:04:24,349 --> 00:04:30,420 this step index waveguide, the step index planar asymmetric waveguide. So, this is the 33 00:04:30,420 --> 00:04:35,539 refractive index profile and I can write down the refractive index in various regions as 34 00:04:35,539 --> 00:04:44,580 nx = nc, which is defined by the region x > 0 which is the cover, then in the region 35 00:04:44,580 --> 00:04:55,009 minus –d < x <0 the refractive index is nf which is the film region, and for x < -d 36 00:04:55,009 --> 00:04:58,500 which represents the substrate region, the refractive index is ns. 37 00:04:58,500 --> 00:05:05,010 Now, the modes of this waveguide are here guided modes, whose refractive index is lie 38 00:05:05,010 --> 00:05:16,479 between ns and nf. So, these are some representative guided modes, and for, the field would radiate 39 00:05:16,479 --> 00:05:27,120 out in the substrate region and forthe field would radiate out in the cover region as well. 40 00:05:27,120 --> 00:05:30,789 So, in this region you will have radiation modes. 41 00:05:30,789 --> 00:05:38,689 We are interested in the guided modes of the structure, which are defined in this range, 42 00:05:38,689 --> 00:05:46,650 and let us first do the TE mode analysis of the waveguide, and as I can see that the refractive 43 00:05:46,650 --> 00:05:53,419 index variation is in x-direction, and I am considering propagation in z-direction. So, 44 00:05:53,419 --> 00:06:02,930 the non- vanishing components corresponding to TE modes are Ey, Hx and Hz. 45 00:06:02,930 --> 00:06:11,629 I know from my previous analysis that the wave equation satisfied by TE modes of such 46 00:06:11,629 --> 00:06:25,979 a waveguide is given by . So, the procedure is again the same I write down this equation 47 00:06:25,979 --> 00:06:33,020 in all the three regions, I write down the solutions of these three equations and then 48 00:06:33,020 --> 00:06:41,229 apply boundary conditions to match the solutions. So, this is the equation in cover region x 49 00:06:41,229 --> 00:06:54,389 = 0, which is , because in this region nx = nc, and I 50 00:06:54,389 --> 00:07:05,939 have taken minus sign outside and represented it in this form because , so that this quantity 51 00:07:05,939 --> 00:07:12,000 is positive for guided modes. In the film region which is defined by minus 52 00:07:12,000 --> 00:07:28,830 -d < x < 0, the equation becomes, again for guided modes this quantity in the square brackets 53 00:07:28,830 --> 00:07:42,550 is positive. In the substrate region defined by x < -d, the equation is 54 00:07:42,550 --> 00:07:48,499 and again for guided modes this quantity is positive. 55 00:07:48,499 --> 00:08:02,400 So, now I can represent this as , this as and this as and then find out what are the 56 00:08:02,400 --> 00:08:06,929 solutions of these three equations. 57 00:08:06,929 --> 00:08:22,639 So, these are the three equations now, where , , and . 58 00:08:22,639 --> 00:08:42,380 So, what are the solutions? The solution of this differential equation is. And as you 59 00:08:42,380 --> 00:08:50,240 know that we are interested in guided modes. So, we cannot have exponentially amplifying 60 00:08:50,240 --> 00:08:57,130 solution. So, I am only considering the exponentially decaying solutions in cover regions and substrate 61 00:08:57,130 --> 00:09:04,019 region. Now, in the film this equation will give you 62 00:09:04,019 --> 00:09:13,740 the oscillatory solutions, which I write as , 63 00:09:13,740 --> 00:09:25,060 and in the substrate region it is because x < -d. So, it is negative. So, this again 64 00:09:25,060 --> 00:09:34,250 gives you exponentially decaying solutions. Now a b c and d are constants and these constants 65 00:09:34,250 --> 00:09:44,089 are determined by the boundary conditions. So, let us apply boundary conditions and obtain 66 00:09:44,089 --> 00:09:50,250 relationships between A, B, C and D obtain the Eigen value equation or characteristic 67 00:09:50,250 --> 00:09:57,260 equation which is satisfied by the propagation constants of the guided modes. 68 00:09:57,260 --> 00:10:04,750 So, these are the solutions in different regions in substrate in the film and in the cover 69 00:10:04,750 --> 00:10:10,680 and what are the boundary conditions? Boundary conditions are again the tangential components 70 00:10:10,680 --> 00:10:20,600 of E and H are continuous, and the tangential components here are Ey and Hz. So, they give 71 00:10:20,600 --> 00:10:31,209 me that Ey andbecause Hz is related to Ey as. So, they are continuous at the interfaces 72 00:10:31,209 --> 00:10:38,579 x = 0 and at x = -t. So, let us apply these boundary conditions. 73 00:10:38,579 --> 00:10:48,190 So, if I look at the continuity ofat. So, I apply it here. So, if I approach from the 74 00:10:48,190 --> 00:10:58,210 film and if I approach from the cover towards x = 0 is equal to 0. So, I will get B + C 75 00:10:58,210 --> 00:11:08,369 = A. Now continuity of at x = -d which is the substrate film interface, this gives me 76 00:11:08,369 --> 00:11:25,069 . Now, let 77 00:11:25,069 --> 00:11:44,670 us see the continuity of the derivatives at x = 0 and at x = d. So, at x = 0 gives meand 78 00:11:44,670 --> 00:11:49,191 continuity ofat x = -d interface gives me . 79 00:11:49,191 --> 00:12:09,769 So, I have now four equations relating these four 80 00:12:09,769 --> 00:12:19,199 constants A, B, C and D and by mathematical manipulation of these four equations. I can 81 00:12:19,199 --> 00:12:30,480 do two things one is obtain B, C and D in terms of A, and then eliminating A, B, C, 82 00:12:30,480 --> 00:12:38,560 D all together and form a transcendental equation inwhich is also known as Eigen value equation 83 00:12:38,560 --> 00:12:52,579 or characteristic equation. So, that equation comes out to be:. 84 00:12:52,579 --> 00:13:05,939 So, this is the equation which is satisfied bybecause the only unknown here is, appears 85 00:13:05,939 --> 00:13:17,930 in and . So, if I solve this equation and find out the roots of this equation then I 86 00:13:17,930 --> 00:13:23,540 can know what are the propagation constants of the modes, what are the modes, which are 87 00:13:23,540 --> 00:13:32,290 supported by this asymmetric step index planar waveguide. 88 00:13:32,290 --> 00:13:40,160 As we have done earlier also that it is always a good idea to obtain everything in normalized 89 00:13:40,160 --> 00:13:48,379 parameters. So, that we are more or less independent of waveguide parameters, here we cannot have 90 00:13:48,379 --> 00:13:53,949 complete independence from waveguide parameters, but to certain extent it is still we can draw 91 00:13:53,949 --> 00:14:03,360 some universal curve. So, how do we define normalized parameters here? First is normalized 92 00:14:03,360 --> 00:14:05,310 frequency V. 93 00:14:05,310 --> 00:14:14,029 If you remember that in symmetric planar waveguide we had only 2 refractive indicesand , . So, 94 00:14:14,029 --> 00:14:26,269 there we had defined it in terms of . Now, now in place of, I have, but low refractive 95 00:14:26,269 --> 00:14:33,459 indices are now up toand which one we should take in order to define V. So, it is very 96 00:14:33,459 --> 00:14:42,959 simple to choose becausedefines the cut-off of guided modes. So, I should choosebecause 97 00:14:42,959 --> 00:14:50,730 as soon as the effective index of a mode falls belowit is no more guided. So, I chooseto 98 00:14:50,730 --> 00:14:58,009 define the normalized frequency v here, and similarly for defining the normalized propagation 99 00:14:58,009 --> 00:15:09,939 constant b I choose. So, b is defined as . There is another parameter 100 00:15:09,939 --> 00:15:19,009 here in this structure, which tells you how asymmetric this waveguide is. The asymmetry 101 00:15:19,009 --> 00:15:25,249 in the structure is introduced by the different values ofand. So, what is the difference between 102 00:15:25,249 --> 00:15:34,810 and? So, let us define by asymmetry parameter a, which is given as. 103 00:15:34,810 --> 00:15:47,000 So, now I have here three normalized parameters in place, and now I can represent my transcendental 104 00:15:47,000 --> 00:15:53,430 equation or Eigen value equation in terms of these normalized parameters. If you remember 105 00:15:53,430 --> 00:16:11,110 that in the transcendental equation the three terms appear which are, and , which contain. 106 00:16:11,110 --> 00:16:24,279 So, now I will have to represent these , and in terms of V, b and a. And we can see that 107 00:16:24,279 --> 00:16:30,110 . and . 108 00:16:30,110 --> 00:17:03,430 So, keeping these in mind and looking at the expressions here I would now try to obtain 109 00:17:03,430 --> 00:17:20,240 , and in terms of V, b and a. So, let us first try for, and I can see that if I do 1- b then 110 00:17:20,240 --> 00:17:33,500 I can obtain this in the numerator of this. So, b is given by this. So, if I do 1- b it 111 00:17:33,500 --> 00:17:34,720 becomes, and . So, if I multiple here by d /2 and then I 112 00:17:34,720 --> 00:17:38,990 can havehere or I havedirectly here and this is nothing, but V2d/2. So, this becomes . So, 113 00:17:38,990 --> 00:17:43,540 this gives me. If I look atgamma s square, . So, it is directly 114 00:17:43,540 --> 00:17:47,420 appearing herethis thing. So, this will immediately give me . What about now. So, I can see from 115 00:17:47,420 --> 00:17:51,240 here that if I do b + a then I can only obtainin the numerator. So, I do b + a. So, it becomesdivided 116 00:17:51,240 --> 00:17:52,610 by this. So, which isdivided byand this gives me . 117 00:17:52,610 --> 00:17:58,100 So, in this way I have now obtained, and in terms of V, b and a. So, I can now represent 118 00:17:58,100 --> 00:18:00,050 my eigen value equation in normalized parameters. So, this equation simply becomes: . 119 00:18:00,050 --> 00:18:01,820 And of course, since and for cut offguided mode,lies between ns and nf. So, b would lie 120 00:18:01,820 --> 00:18:08,750 as usual between 0 and 1. So, I solve this equation for a given value of V, V is if you 121 00:18:08,750 --> 00:18:14,130 remember. So, this V contains all the waveguide parameters and the wavelength. So, for a given 122 00:18:14,130 --> 00:18:19,290 waveguide and wavelength I have V and for that value of V I can solve this equation 123 00:18:19,290 --> 00:18:20,290 to obtain the normalized propagation constant b. 124 00:18:20,290 --> 00:18:26,360 What are the cut offs? Cut offs are defined byor b = 0. So, if I put it there then the 125 00:18:26,360 --> 00:18:31,790 cut off equation becomes or cut off for TE mode mth TE mode is now given by . You can 126 00:18:31,790 --> 00:18:35,160 see here that the cut off of even TE0 mode is finite. 127 00:18:35,160 --> 00:18:42,530 So, in a particular range of V which is defined by V even the TE0 mode is not guided. 128 00:18:42,530 --> 00:18:46,580 This is the difference as compared to the symmetric planar waveguide. In symmetric planar 129 00:18:46,580 --> 00:18:52,040 waveguide TE0 mode has 0 cut off. So, TE0 mode was always guided, but here it is not 130 00:18:52,040 --> 00:18:57,640 the case. Now let us solve this equation the characteristic equation or Eigen value equation 131 00:18:57,640 --> 00:19:03,040 for different values of V, and plot the roots as obtained as a function of V. So, here I 132 00:19:03,040 --> 00:19:07,550 have plotted the roots for both the cases a = 0 which represents the symmetric planar 133 00:19:07,550 --> 00:19:11,830 waveguide, and then for a very high value of a which is 40 for asymmetric planar waveguide. 134 00:19:11,830 --> 00:19:16,720 These solid lines are corresponding to symmetric planar waveguide a = 0, and these dash lines 135 00:19:16,720 --> 00:19:25,090 are corresponding to asymmetric planar waveguide And what I can see that TE0 mode has cut off 136 00:19:25,090 --> 00:19:32,892 here, TE1 mode has cut off here, TE2 mode has cut off here. In case of symmetric planar 137 00:19:32,892 --> 00:19:37,710 waveguide this was the cut off was the corresponding cut off. Now all the cut offs have been shown 138 00:19:37,710 --> 00:19:41,570 shifted by this much amount which is . So, this is one thing another thing that I see 139 00:19:41,570 --> 00:19:44,840 here is that if I take a particular value of V then the propagation constant of the 140 00:19:44,840 --> 00:19:48,150 mode is now smaller than the propagation constant of the corresponding symmetric waveguide mode. 141 00:19:48,150 --> 00:19:51,770 And it is it is understandable that because asymmetric is introduced by introducing the 142 00:19:51,770 --> 00:19:56,560 cover region. So, if this is the symmetric waveguide, this is the symmetric waveguide. 143 00:19:56,560 --> 00:20:02,860 So, this is ns this is nf this is again the level ns, but now in asymmetric planar waveguide 144 00:20:02,860 --> 00:20:09,670 I have nc. So, I am reducing the refractive index in the cover region. So, the effect 145 00:20:09,670 --> 00:20:20,470 is to pull down the effective indices of the modes towards lower side so that we can see 146 00:20:20,470 --> 00:20:26,570 from here itself that the propagation constants of asymmetric planar waveguides are smaller 147 00:20:26,570 --> 00:20:29,610 than those corresponding to symmetric planar waveguide. 148 00:20:29,610 --> 00:20:34,990 Let us look at modal fields. So, these are the modal fields of TE0, TE1 and TE2 modes 149 00:20:34,990 --> 00:20:41,820 of a typical asymmetric planar waveguide. So, I can see that the modal fields are asymmetric 150 00:20:41,820 --> 00:20:57,660 and you can see that the penetration gap is more in the substrate and less in the cover 151 00:20:57,660 --> 00:21:05,930 which is understandable, because here at this interface the index contrast is smaller as 152 00:21:05,930 --> 00:21:09,240 compared to the index contrast at this interface ok. 153 00:21:09,240 --> 00:21:18,190 So, the field extends more into the substrate region as compared to in the cover region. 154 00:21:18,190 --> 00:21:26,590 How many modes are supported? If you go back to symmetric planar waveguide the number of 155 00:21:26,590 --> 00:21:33,810 modes are the integer which is closest to, but greater than , but now all the cut offs 156 00:21:33,810 --> 00:21:41,860 are shifted by this much amount. So, the number of modes would now be. So, you obtain this 157 00:21:41,860 --> 00:21:46,090 number and find out the integer closest to, but greater than this number. 158 00:21:46,090 --> 00:21:50,620 What are the cut off wavelengths of various modes and if I change the wavelength how the 159 00:21:50,620 --> 00:21:55,480 number of modes would change. So, I can see from here that the cut off wavelength for 160 00:21:55,480 --> 00:22:00,430 mth TE mode would be given bywhich comes directly from the definition of normalized frequency 161 00:22:00,430 --> 00:22:06,170 V. So, I can write it down also which we will quite often use. 162 00:22:06,170 --> 00:22:15,030 . So, from here I get these cut off wavelengths, now if I find out the cut off wavelengths 163 00:22:15,030 --> 00:22:24,780 corresponding to various modes then I see that for TE0 mode the cut off wavelength is 164 00:22:24,780 --> 00:22:27,270 1.7187 for these parameters of waveguide, and for TE1 mode it is 0.4914 and for TE2 165 00:22:27,270 --> 00:22:32,860 mode it is 0.2867. So, if I start from a longer wavelength let 166 00:22:32,860 --> 00:22:39,900 us say 2, then until I cross this TE0 mode is not guided. So, from here to here there 167 00:22:39,900 --> 00:22:46,450 is no mode guided by the structure, and as soon as I cross this go below this wavelength, 168 00:22:46,450 --> 00:22:53,330 then TE0 mode starts appearing and TE0 mode would be guided now for all the wavelengths 169 00:22:53,330 --> 00:23:00,970 is smaller than this. If I further reduce the wavelength and as 170 00:23:00,970 --> 00:23:19,970 soon as I go below 0.49TE1 mode starts appearing and below these this wavelength I will have 171 00:23:19,970 --> 00:23:30,890 both TE0 and TE1 and so on. So, this is how the waveguide would guide different modes 172 00:23:30,890 --> 00:23:36,560 if I change the wavelength. How the thickness affects the guided modes. So, from here I 173 00:23:36,560 --> 00:23:48,130 can find out the cut off thickness of mth TE mode from here itself. So, if I find out 174 00:23:48,130 --> 00:24:00,300 d in terms of V now and put the cut off of various modes in terms of V, then I can find 175 00:24:00,300 --> 00:24:10,440 out the cut off of thickness ok. So, if I have 0 thickness it means no waveguide 176 00:24:10,440 --> 00:24:23,210 no mode. If I start increasing the thickness then up to 0.58 there is no mode guided because 177 00:24:23,210 --> 00:24:31,160 TE0 mode has finite cut off and as soon as I cross this then TE0 mode starts appearing, 178 00:24:31,160 --> 00:24:37,990 and when I cross 2.03then TE1 mode starts appearing and so on. So, as I increase the 179 00:24:37,990 --> 00:24:52,730 waveguide thickness the number of modes start increasing I should have the label here which 180 00:24:52,730 --> 00:25:15,490 is d in. So, the label here is t in micron. Let us work out few examples. So, I take a 181 00:25:15,490 --> 00:25:18,850 dielectric step index asymmetric planar waveguide. 182 00:25:18,850 --> 00:25:28,340 Defined by nf =1.5 and ns =1.48 and nc = 1, and d = 4. Now the first thing is to calculate 183 00:25:28,340 --> 00:25:36,230 the number of modes at lambda naught is equal to 0.5. So, I first calculate the value of 184 00:25:36,230 --> 00:25:44,890 asymmetric parameter a which comes out to be about 20, then I find out what is the value 185 00:25:44,890 --> 00:25:50,390 of V at = 0.5, and this comes out to be about 6.13. 186 00:25:50,390 --> 00:26:00,620 Then I find out this number. So, this comes out to be 3.4. So, the number of modes are 187 00:26:00,620 --> 00:26:05,660 4. 188 00:26:05,660 --> 00:26:14,910 The second is wavelength range in which the waveguide does not support anymore. So, I 189 00:26:14,910 --> 00:26:22,670 know there would not be any more support it if , which if I put the value V here the expression 190 00:26:22,670 --> 00:26:30,220 for V here. Then this gives me the condition in terms ofasshould be greater than this, 191 00:26:30,220 --> 00:26:41,380 and if I plug in all these numbers I find out for > 4.5427 the waveguide would not support 192 00:26:41,380 --> 00:26:42,380 anymore. 193 00:26:42,380 --> 00:27:00,960 Third is what 194 00:27:00,960 --> 00:27:38,100 is the cut off wavelength of TE2 mode? So, I find out what is the V value for cut off 195 00:27:38,100 --> 00:27:56,160 of TE2 mode. So, which is given by this with m/2. So, if I put m /2 then Vc is equal to 196 00:27:56,160 --> 00:28:34,910 this, and the cut off Vc for cut off V for TE2 mode is this if I translate it to the 197 00:28:34,910 --> 00:28:42,520 wavelength it comes out to be 0.8038. 198 00:28:42,520 --> 00:29:02,060 What is the range of d so, that only TE0 and TE1 modes are guided 199 00:29:02,060 --> 00:29:13,860 at. So, I know that for mth mode the cut off 200 00:29:13,860 --> 00:29:39,820 thickness is this, and I want TE0 and TE1 both the modes guided. So, I find out this 201 00:29:39,820 --> 00:30:06,890 dc for both the modes, and I also know thisfor mth mode is given by this. So, I find outfor 202 00:30:06,890 --> 00:30:23,060 d0 mode, Vc is this for TE1 mode Vc is 203 00:30:23,060 --> 00:30:41,480 this correspondingly if I now find out dc for these 2 modes, then for TE0 mode cut off 204 00:30:41,480 --> 00:31:08,700 thickness is 0.88, and for this is 2.93. So, if TE0 and TE1 modes are 205 00:31:08,700 --> 00:31:23,610 to be guided 206 00:31:23,610 --> 00:33:56,990 then thickness should be 207 00:33:56,990 --> 00:34:15,490 between 208 00:34:15,490 --> 00:34:28,350 this 209 00:34:28,350 --> 00:35:57,440 and this. So, this is all in this lecture in the next 210 00:35:57,440 --> 00:36:01,580 lecture, we will extend the analysis to TM modes. 211 00:36:01,580 --> 00:36:02,650 Thank you.