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.