1
00:00:08,710 --> 00:00:27,619
So, in this lecture, let me continue with
the previous lecture where I was doing the
2
00:00:27,619 --> 00:00:31,399
analysis of planar mirror waveguide.
3
00:00:31,399 --> 00:00:47,120
So, what I was doing was to find out the modes
of a planar mirror waveguide and what I did
4
00:00:47,120 --> 00:00:58,890
that I took the planar mirror waveguide which
has refractive index n and metal deposited
5
00:00:58,890 --> 00:01:06,860
at x is equal to 0 and x is equal to d, and
the modes of this waveguide I found out as
6
00:01:06,860 --> 00:01:15,049
Ey is equal to A sinm pi x over d and corresponding
beta are given by beta m square is equal to
7
00:01:15,049 --> 00:01:20,920
k naught square n square minus m pi over d
square.
8
00:01:20,920 --> 00:01:33,920
So, what I got that I got this kind of variation
of electric field Ey with respect to x or
9
00:01:33,920 --> 00:01:44,189
this or this and these fields propagate in
the waveguide and sustain their shape as they
10
00:01:44,189 --> 00:01:52,030
propagate. They also propagate with certain
propagation constants beta1, beta2 beta3 and
11
00:01:52,030 --> 00:01:59,780
so on. Let me understand what they are.
12
00:01:59,780 --> 00:02:14,500
So, the field is Am sin kappa mx where kappa
m is equal to m pi over d if I write the complete
13
00:02:14,500 --> 00:02:23,040
solution this is only Ey m of x, x part of
the solution remember that there is t and
14
00:02:23,040 --> 00:02:35,590
z parts also. So, if I write down the complete
solution there it is some Am sin kappa mx,
15
00:02:35,590 --> 00:02:45,960
e to the power i omega t minus beta mz. Let
me write this sin kappa mx in the form of
16
00:02:45,960 --> 00:02:56,320
e to the power plus minus I kappa mx.
So, if I do that I get this and then let me
17
00:02:56,320 --> 00:03:09,060
put it in this form. So, what I have got from
here, I have got it e to the power i omega
18
00:03:09,060 --> 00:03:17,560
t minus b time z minus kappa mx and then e
to the power i omega t minus b time z plus
19
00:03:17,560 --> 00:03:28,200
kappa mx what they are and remember you see
now outside I have a constant it is not a
20
00:03:28,200 --> 00:03:38,680
function of x, the x dependence I have included
in the exponential now, the x dependence which
21
00:03:38,680 --> 00:03:48,080
was earlier here I have included in the exponential
and when I see this now this is nothing, but
22
00:03:48,080 --> 00:03:57,390
the plane wave because outside is a constant.
So, and this is the plane wave which is moving
23
00:03:57,390 --> 00:04:09,710
in this direction, moving in x-z plane making
certain angle from z axis. And this is a plane
24
00:04:09,710 --> 00:04:22,340
wave which is moving in plus z minus x plus
z minus x direction k. So, what I have got
25
00:04:22,340 --> 00:04:32,259
that this mode, this mode is nothing but the
superposition of 2 plane waves, this plane
26
00:04:32,259 --> 00:04:39,389
wave and this plane wave with some phases.
So, it is the superposition of 2 plane waves
27
00:04:39,389 --> 00:04:50,060
1 propagating in this direction another propagating
in this direction. If you look at the waveguide
28
00:04:50,060 --> 00:04:58,770
if you look at the waveguide this is x this
is z. So, this is x-z plane this is x this
29
00:04:58,770 --> 00:05:07,469
is z. So, one wave is moving like this another
wave is moving like this and this is what
30
00:05:07,469 --> 00:05:16,430
you have, that if you launch wave like this
then it will get reflected then reflected
31
00:05:16,430 --> 00:05:22,919
from here, reflected from here.
So, all the time you will have 2 waves one
32
00:05:22,919 --> 00:05:31,460
going in this direction and another going
in this direction. Let me put them together
33
00:05:31,460 --> 00:05:33,669
here in x-z plane.
34
00:05:33,669 --> 00:05:43,440
So, one is moving an angle theta m with z
axis and another one is moving minus theta
35
00:05:43,440 --> 00:05:53,770
m angle with z axis and this has to be the
propagation constant in the medium of a plane
36
00:05:53,770 --> 00:05:57,780
wave these are now plane waves plane waves
we will have propagation constant k naught
37
00:05:57,780 --> 00:06:10,889
n. So, this is the component of propagation
constant in x direction this is the component
38
00:06:10,889 --> 00:06:21,110
of propagation constant in z direction. So,
if I now do kappa m square plus beta m square
39
00:06:21,110 --> 00:06:31,240
is square root then it comes out to be k naught
n and it comes out to be k naught n because
40
00:06:31,240 --> 00:06:48,050
you remember that kappa m square is equal
to k naught square n square minus beta m square
41
00:06:48,050 --> 00:06:55,129
which means that this is indeed k naught n
this is indeed k naught n.
42
00:06:55,129 --> 00:07:07,300
So, the propagation constant in z direction
is k naught n cos theta m. Let us understand
43
00:07:07,300 --> 00:07:15,319
it. If I take the components in x direction
I have 1 wave going in positive x another
44
00:07:15,319 --> 00:07:22,500
wave going in negative x, which means that
in x direction I have 2 counter propagating
45
00:07:22,500 --> 00:07:29,819
waves and when I have 2 counter propagating
waves they give you a standing wave solution
46
00:07:29,819 --> 00:07:38,969
they give you standing wave. So, in x direction
I have a standing wave. In z direction if
47
00:07:38,969 --> 00:07:45,400
I see this also gives me propagation in positive
z.
48
00:07:45,400 --> 00:07:57,960
So, I have a standing wave in x which propagates
in z. The energy does not flow out in x direction
49
00:07:57,960 --> 00:08:06,659
the wave stands in x direction, but that standing
wave pattern flows in z direction this is
50
00:08:06,659 --> 00:08:18,860
the mode. So, these modes are nothing, but
the standing wave patterns and they are made
51
00:08:18,860 --> 00:08:28,050
out of two waves one going in this direction
another going in this direction.
52
00:08:28,050 --> 00:08:33,780
I can now find out what is the effective refractive
index of a given mode or effective index of
53
00:08:33,780 --> 00:08:42,790
the mode by beta m over k naught which is
nothing, but kn cos theta m. So, for every
54
00:08:42,790 --> 00:08:52,180
mode I have different beta m and therefore,
the angles that the constituent plane waves
55
00:08:52,180 --> 00:09:04,550
make with z axis would be different and these
angles can be found out from here. For example,
56
00:09:04,550 --> 00:09:16,570
for n is equal to 1.5 d is equal to 1 micrometer
and lambda naught is equal to 0.633 micrometer
57
00:09:16,570 --> 00:09:28,700
helium neon wavelength. I find out that there
are these modes four modes. Let me look at
58
00:09:28,700 --> 00:09:33,010
them closely.
59
00:09:33,010 --> 00:09:42,800
The first one has effective index 1.4662 and
the angles of constituent plane waves with
60
00:09:42,800 --> 00:09:54,140
z axis is z axis r plus minus 12.18 degrees.
So, there are 2 plane waves one making plus
61
00:09:54,140 --> 00:09:59,990
12 degrees and other making minus 12 degree
with z axis and this is the a standing wave
62
00:09:59,990 --> 00:10:06,890
pattern in x direction corresponding to those
waves.
63
00:10:06,890 --> 00:10:18,250
The second one is with n effective is equal
to 1.3599 with angles plus minus approximately
64
00:10:18,250 --> 00:10:27,620
25 degrees. Yet another one is this with the
angles plus minus 39 degrees and this one
65
00:10:27,620 --> 00:10:29,860
plus minus 57.565 degrees.
66
00:10:29,860 --> 00:10:45,940
So, these are the modes how many modes are
there for a given waveguide. Now let me look
67
00:10:45,940 --> 00:10:57,660
at this picture again and from here I know
that kappa m d is equal to m pi. So, this
68
00:10:57,660 --> 00:11:09,610
m would be kappa m d over pi and kappa m is
nothing, but k naught n sin theta m and the
69
00:11:09,610 --> 00:11:16,140
maximum value of this would be k naught n
because the maximum value of sin theta is
70
00:11:16,140 --> 00:11:22,640
1.
So, the maximum value of k m is k naught n
71
00:11:22,640 --> 00:11:29,750
and this will give me what is the maximum
value of m. So, I put maximum value of k m
72
00:11:29,750 --> 00:11:36,940
here which is k naught n. So, m max would
be k naught n by pi and k naught is equal
73
00:11:36,940 --> 00:11:44,180
to 2 pi over lambda naught. So, this would
be 2 nd over lambda naught. So, the maximum
74
00:11:44,180 --> 00:11:52,030
number of modes that can propagate in a planar
mirror waveguide would be a number and integer
75
00:11:52,030 --> 00:12:01,860
which is closest to, but less than this, it
is closest to but less than this. For example,
76
00:12:01,860 --> 00:12:08,970
if I have n is equal to 1.5, d is equal to
1 micrometer and lambda is equal to 0.633
77
00:12:08,970 --> 00:12:16,980
micrometer there should be micrometer here
then 2nd over lambda naught comes out to be
78
00:12:16,980 --> 00:12:28,100
4.739 and this gives me the maximum number
of modes as 4. So, there would be 4 mode supported.
79
00:12:28,100 --> 00:12:38,210
So, this is the analysis of planar mirror
waveguide which gives me insight into the
80
00:12:38,210 --> 00:12:48,660
modes of a waveguide which structure is quite
intuitive and light guidance and data structure
81
00:12:48,660 --> 00:12:57,610
is also quite intuitive, but this is a structure
planner mirror waveguide is of very little
82
00:12:57,610 --> 00:13:10,930
practical use for 1 it is not feasible to
make this waveguide that you have one micrometer
83
00:13:10,930 --> 00:13:21,450
film of certain material and you deposit metal
on top and bottom of that and use it. Second
84
00:13:21,450 --> 00:13:30,020
since it involves metal coatings and metal
is highly absorbing material. So, as these
85
00:13:30,020 --> 00:13:39,610
modes propagate they will attenuate very quickly.
So, they have very high loss. So, this kind
86
00:13:39,610 --> 00:13:50,170
of waveguide is of little practical use although
it gives a good understanding of mode propagation.
87
00:13:50,170 --> 00:14:02,950
So, now what we are going to do is to look
at more practical waveguide which is asymmetric
88
00:14:02,950 --> 00:14:09,880
step index planar waveguide dielectric waveguide
planar dielectric waveguide which is obtained
89
00:14:09,880 --> 00:14:23,230
by sandwiching high index lab between 2 lower
refractive index labs. So, I have n1 n2, n2
90
00:14:23,230 --> 00:14:32,700
here and n1 is greater than n2. Again if you
look there is refractive index discontinuity
91
00:14:32,700 --> 00:14:43,130
only in x direction, y direction it is infinitely
extended in z direction also there is no indexed
92
00:14:43,130 --> 00:14:50,620
discontinuity and it is infinitely extended.
If you look at the refractive index profile
93
00:14:50,620 --> 00:15:05,970
of this waveguide then it looks like this
and let me put my x is equal to 0 access in
94
00:15:05,970 --> 00:15:14,310
the middle of the waveguide so that I can
make use of the symmetry of the problem let
95
00:15:14,310 --> 00:15:26,220
it have of the width d. So, this is x is equal
to d by 2 this is x is equal to minus d by
96
00:15:26,220 --> 00:15:38,190
2. I can write down the refractive index variation
with respect to x in this fashion.
97
00:15:38,190 --> 00:15:47,200
So, this region is mod x less than d by 2
while these 2 reasons can be represented by
98
00:15:47,200 --> 00:15:55,390
mod x greater than d by 2. So, in mod x less
than the d by 2 I have refractive index n1,
99
00:15:55,390 --> 00:16:06,120
in mod x greater than d by 2 I have refractive
index n2 and now my problem is to find out
100
00:16:06,120 --> 00:16:16,630
Ey and corresponding beta for this given n
of x. So, how do I do? This again go back
101
00:16:16,630 --> 00:16:23,040
to the wave equation corresponding to t e
votes. Let me first solve the problem for
102
00:16:23,040 --> 00:16:24,230
TE-modes.
103
00:16:24,230 --> 00:16:36,110
So, this is the wave equation for TE-modes
and this is the n(x). So, how do I solve it?
104
00:16:36,110 --> 00:16:47,190
Well I find that in this region and in this
region and in this region in all the 3 regions
105
00:16:47,190 --> 00:16:55,640
the refractive index is constant although
when I go from here to here and here to here
106
00:16:55,640 --> 00:17:03,440
I encounter index discontinuity, but on this
side and on that side the refractive index
107
00:17:03,440 --> 00:17:10,380
remains constant.
So, I make use of this and write down the
108
00:17:10,380 --> 00:17:17,919
wave equation in this region and in this region
that is in this region and in these 2 regions.
109
00:17:17,919 --> 00:17:26,490
So, for mod x less than d by 2 I write it
down as d2Ey over dx square plus k naught
110
00:17:26,490 --> 00:17:33,700
square n1 square because nx is equal to n1
in this region minus beta square times Ey
111
00:17:33,700 --> 00:17:46,090
is equal to 0. And I also keep in mind that
my effective refractive index which is given
112
00:17:46,090 --> 00:17:58,679
by beta over k naught. Effective refractive
index n effective which is given by beta over
113
00:17:58,679 --> 00:18:09,100
k naught this should lie between n2 and n1
for guided modes for guided modes n effective
114
00:18:09,100 --> 00:18:20,600
would lie between n2 and n1 which means that
beta would lie between k naught n2 and k naught
115
00:18:20,600 --> 00:18:27,549
n1. So, beta would be greater than k naught
n2 while it would be smaller than k naught
116
00:18:27,549 --> 00:18:29,220
n1.
117
00:18:29,220 --> 00:18:39,340
So, when I write this equation now in these
2 regions which are represented by mod x greater
118
00:18:39,340 --> 00:18:48,230
than d by 2 I write it as d2Ey over dx square
minus beta square minus k naught square n2
119
00:18:48,230 --> 00:18:56,340
square times Ey because n x is equal n2 here.
And why I have written it in this fashion.
120
00:18:56,340 --> 00:19:03,679
So, that this is positive here and this is
also positive here because I am solving it
121
00:19:03,679 --> 00:19:12,649
going to solve it for this condition which
is the condition for guided modes for which
122
00:19:12,649 --> 00:19:22,120
the energy would be confined into this region.
So, I represent this as kappa square and I
123
00:19:22,120 --> 00:19:29,009
represent this as gamma square and this kappa
square and gamma square are positive for guided
124
00:19:29,009 --> 00:19:38,730
modes as I have explained. So, let me write
these equations down again here.
125
00:19:38,730 --> 00:19:45,990
So, for mod x less than d by 2 I have a got
a got an equation d2E by over dx square plus
126
00:19:45,990 --> 00:19:54,379
kappa square Ey is equal to 0 and for mod
x greater than d by 2 it is d2Ey over dx square
127
00:19:54,379 --> 00:20:02,419
minus gamma square Ey is equal to 0 where
kappa square as this and gamma square is this.
128
00:20:02,419 --> 00:20:09,710
What are the solutions? I know the solutions
of these differential equations very well,
129
00:20:09,710 --> 00:20:16,350
this equation gives me oscillatory solutions
e to the power I kappa x or e to the power
130
00:20:16,350 --> 00:20:23,600
minus i kappa x or sin kappa x cosine kappa
x while this equation gives me exponentially
131
00:20:23,600 --> 00:20:32,770
amplifying and decaying solutions. So, if
I write the solutions in the regions mod x
132
00:20:32,770 --> 00:20:39,390
less than d by 2 and mod x greater than d
by 2 I find that in this region it is oscillatory
133
00:20:39,390 --> 00:20:50,809
solution A cosine kappa x plus B sin kappa
x and in the region mod x greater than d by
134
00:20:50,809 --> 00:20:59,960
2 I have exponential and decaying solutions,
but out of these 2 solutions I retain only
135
00:20:59,960 --> 00:21:07,419
exponentially decaying solution while exponentially
decaying solution and not exponentially amplifying
136
00:21:07,419 --> 00:21:16,870
solution because I want a solution which gives
me guided modes and for guided modes the energy
137
00:21:16,870 --> 00:21:24,940
should be confined into the high index region
and it should decay down as you go away from
138
00:21:24,940 --> 00:21:31,190
the high index region.
As you go towards infinity x is equal to plus
139
00:21:31,190 --> 00:21:38,700
minus infinity the energy should decay down.
So, I cannot take exponentially employee find
140
00:21:38,700 --> 00:21:44,880
solutions that is y for x greater than d by
2 I have chosen the form of solutions C e
141
00:21:44,880 --> 00:21:53,779
to the power minus gamma x and for x less
than minus d by 2 I have chosen the form D
142
00:21:53,779 --> 00:22:02,070
e to the power gamma x this A, B, C, D are
some constants which can be determined by
143
00:22:02,070 --> 00:22:09,429
the boundary conditions. What are the boundaries?
Boundaries are x is equal to plus d by 2 and
144
00:22:09,429 --> 00:22:22,899
x is equal to minus d by 2, but here I can
make some more simplification what simplification
145
00:22:22,899 --> 00:22:32,860
I can make is to utilize the symmetry of the
problem how can I utilize the symmetry of
146
00:22:32,860 --> 00:22:36,080
the problem let us see.
147
00:22:36,080 --> 00:22:50,059
I have the wave equation for TE-modes like
this and this is the n of x. I see that this
148
00:22:50,059 --> 00:22:56,419
is symmetric about x is equal to 0 which means
that n square of minus x is equal to n square
149
00:22:56,419 --> 00:22:57,419
of x.
150
00:22:57,419 --> 00:23:09,090
Now, in this equation if you replace x by
minus x if you replace x by minus x, then
151
00:23:09,090 --> 00:23:30,330
you get d2Ey of minus x divided by d of minus
x square that is x square plus k naught square
152
00:23:30,330 --> 00:23:45,779
n square of minus x minus beta square Ey of
minus x is equal to 0 and this is nothing,
153
00:23:45,779 --> 00:23:57,470
but n square of x because of the symmetry
of the profile which means that which means
154
00:23:57,470 --> 00:24:08,149
that I now have 2 possibilities which are
Ey of minus x is equal to Ey of x if I put
155
00:24:08,149 --> 00:24:18,539
Ey of minus x is equal to Ey of x I get back
the same equation or if I put Ey of minus
156
00:24:18,539 --> 00:24:24,809
x is equal to minus Ey of x then also I get
back the same equation.
157
00:24:24,809 --> 00:24:37,340
So, there are now 2 possibilities one is this
another is this and they are known as symmetric
158
00:24:37,340 --> 00:24:45,890
modes because this is the property of a symmetric
function and this is the property of antisymmetric
159
00:24:45,890 --> 00:24:54,500
function. So, these modes are known as symmetric
modes and these modes are known as antisymmetric
160
00:24:54,500 --> 00:25:05,739
modes. The simplification it introduces is
that out of those 2 functions cosine and sin
161
00:25:05,739 --> 00:25:14,269
I can pick one here and one here since cosine
is the symmetric function and sin is the antisymmetric
162
00:25:14,269 --> 00:25:18,539
function.
So, symmetric modes would now be represented
163
00:25:18,539 --> 00:25:28,590
by Ey of x is equal to a cosine kappa x for
mod x less than d by 2 and C e to the power
164
00:25:28,590 --> 00:25:39,830
minus gamma mod x for mod x greater than d
by 2 for antisymmetric modes I will have Ey
165
00:25:39,830 --> 00:25:48,470
of x is equal to b sin kappa x for x less
than for mod x less than d by 2 and then I
166
00:25:48,470 --> 00:25:54,220
have D e to the power minus gamma mod x for
mod x less than d by 2, but I shall have to
167
00:25:54,220 --> 00:26:04,799
take care of sin when I go from left side
or right side. What do I mean to say is this
168
00:26:04,799 --> 00:26:20,710
if this is x and this is d by 2 this is minus
d by 2 this is x is equal to 0.
169
00:26:20,710 --> 00:26:32,600
Now, if you plot a sin function in the region
mod x less than d by 2. So, it goes something
170
00:26:32,600 --> 00:26:49,059
like this something like this and now in the
n2 region for positive x values you will for
171
00:26:49,059 --> 00:26:56,450
x greater than d by 2 you will approach from
here and then it decays down while for x less
172
00:26:56,450 --> 00:27:01,530
than minus d by 2 it you will approach from
here from negative side. So, here you start
173
00:27:01,530 --> 00:27:09,019
from positive side and decay down here you
start from negative side and then decay down.
174
00:27:09,019 --> 00:27:17,119
So, to take care of that sin I will have to
put x over mod x here.
175
00:27:17,119 --> 00:27:29,690
So, these are the symmetric modes and remember
that I am doing the analysis of TE-modes have
176
00:27:29,690 --> 00:27:42,399
non vanishing components of E and H has Ey,
Hx and Hz. What is left now? I need to find
177
00:27:42,399 --> 00:27:52,340
out the relationship between A and C I need
to find out how these solutions are connected
178
00:27:52,340 --> 00:28:00,269
at the boundaries for that I will use boundary
conditions. And I have learnt that the boundary
179
00:28:00,269 --> 00:28:08,090
conditions are when you encounter an interface
between 2 dielectric media the boundary conditions
180
00:28:08,090 --> 00:28:15,990
are tangential components of E and H should
be continuous.
181
00:28:15,990 --> 00:28:22,379
Tangential components of E and H should be
continuous what are the tangential components
182
00:28:22,379 --> 00:28:31,700
to the boundaries. This is the waveguide this
is x this is x is equal to plus d by 2 this
183
00:28:31,700 --> 00:28:43,710
is x is equal to minus d by 2, this is y this
is z. So, to this interface which is x is
184
00:28:43,710 --> 00:28:49,330
equal to plus d by 2 or x is equal to minus
d by 2 here the tangential components are
185
00:28:49,330 --> 00:29:01,629
y and z y and z, x is a normal component and
here the nonvanishing components are Ey, Hx
186
00:29:01,629 --> 00:29:13,690
and Hz. So, the tangential components are
Ey and Hz. So, these Ey and Hz should be continuous
187
00:29:13,690 --> 00:29:23,169
and if you look back to the 3 equations which
relate these 3 components of E and H then
188
00:29:23,169 --> 00:29:35,460
I find that Hz is nothing, but some constant
times dEy over dx. So, Ey and dEy over dx
189
00:29:35,460 --> 00:29:43,899
should be continuous at x is equal to plus
minus d by 2. So, let me apply these boundary
190
00:29:43,899 --> 00:29:50,100
conditions, let me do it for x is equal to
plus d by 2 the same would be obtained for
191
00:29:50,100 --> 00:29:56,799
x is equal to minus d by 2.
So, I put Ey should be continuous at x is
192
00:29:56,799 --> 00:30:03,799
equal to plus d by 2 which means a cosine
kappa d by 2 should be equal to C e to the
193
00:30:03,799 --> 00:30:14,409
power minus gamma d by 2 and then for derivative
dEy over dx. So, I will get minus a kappa
194
00:30:14,409 --> 00:30:25,359
sin kappa d by 2 is equal to minus gamma C
e to the power minus gamma d by 2. If I divide
195
00:30:25,359 --> 00:30:34,249
this by this I get kappa tan kappa d by 2
is equal to gamma and remember that these
196
00:30:34,249 --> 00:30:41,999
kappa have dimensions of 1 over meter. So,
I can make them dimensionless, so I can multiply
197
00:30:41,999 --> 00:30:50,299
both sides by d by 2 and get an equation like
this which is kappa d by 2 tan kappa d by
198
00:30:50,299 --> 00:30:52,299
2 is equal to gamma d by 2.
199
00:30:52,299 --> 00:31:01,629
What is kappa and what is gamma? So, you remember
that kappa a square is equal to k naught square
200
00:31:01,629 --> 00:31:14,549
m1 square minus beta square and gamma square
is beta square minus k naught square n2 square
201
00:31:14,549 --> 00:31:22,789
k naught is equal to 2 pi over lambda naught.
So, for a given waveguide and wavelength for
202
00:31:22,789 --> 00:31:29,479
a given waveguide and wavelength the only
unknown in this in these kappa and gamma is
203
00:31:29,479 --> 00:31:38,870
beta. So, this is nothing, but a transcendental
equation in beta. So, beta satisfy this equation.
204
00:31:38,870 --> 00:31:44,669
So, there are only certain values of beta
which are possible and those values of theta
205
00:31:44,669 --> 00:31:52,350
satisfied this equation only those values
of beta are possible. So, this makes this
206
00:31:52,350 --> 00:31:56,850
makes the modes discreet.
So, from here I find out what are the possible
207
00:31:56,850 --> 00:32:06,039
values of beta and for those beta I can find
out the fields. So, that is how I can get
208
00:32:06,039 --> 00:32:13,539
the modes of a symmetric planar waveguides
and these are symmetric modes.
209
00:32:13,539 --> 00:32:24,100
In the next lecture I will continue with this
to solve to find the solutions for antisymmetric
210
00:32:24,100 --> 00:32:25,770
modes.
Thank you.