1
00:00:18,720 --> 00:00:29,699
So, let us continue our analysis on step index
optical fiber. So, we were doing the analysis
2
00:00:29,699 --> 00:00:36,730
of this step index optical fiber whose core
has a radius a and refractive index n1 and
3
00:00:36,730 --> 00:00:40,659
the cladding has refractive index n2.
4
00:00:40,659 --> 00:00:51,440
So, what we had done we had formed equations
in capital R in the region r less than a and
5
00:00:51,440 --> 00:00:59,739
r greater than a and these equations have
the solutions which are in the form of Bessel
6
00:00:59,739 --> 00:01:00,739
functions.
7
00:01:00,739 --> 00:01:12,610
So, as we were discussing that in case of
planar waveguide for mod x less than d by
8
00:01:12,610 --> 00:01:30,390
2 where nx was equal to n1, I had the solutions
cosine kappa x and sin kappa x. While in the
9
00:01:30,390 --> 00:01:41,950
region mod x greater than d by 2 where nx
is equal to n2. I had solutions e to the power
10
00:01:41,950 --> 00:01:51,560
minus gamma x and e to the power gamma x.
So, I had oscillatory solutions in the high
11
00:01:51,560 --> 00:01:57,030
index region because n1 is greater than n2.
So, in the high index region, I had oscillatory
12
00:01:57,030 --> 00:02:04,310
solutions and exponentially decaying solutions
in the low index region so that the energies
13
00:02:04,310 --> 00:02:14,670
confined in the high index region.
Here I have sets of equations; you see this
14
00:02:14,670 --> 00:02:22,470
l can take several values l is equal to 0,
1, 2, 3 and so on. So, for each of these equations
15
00:02:22,470 --> 00:02:29,660
I have each of these functions for l is equal
to 0. I will have J0, Y0 l is equal to J1
16
00:02:29,660 --> 00:02:37,900
Y1 and so on. So, these are a kind of series
of functions of family of functions Jl, Yl,
17
00:02:37,900 --> 00:02:44,530
Kl, Il; however, in this case in planar wave
guide I did not have this kind of situation
18
00:02:44,530 --> 00:02:52,360
because the confinement was only in 1 direction
here the confinement is on also in phi direction,
19
00:02:52,360 --> 00:03:02,030
so from there this l is coming.
If I look at this cosine function, it is simply
20
00:03:02,030 --> 00:03:12,900
if I have a function something like this then
I call it cosine function similarly a function
21
00:03:12,900 --> 00:03:32,629
like this I labelled as sin function. Similarly,
if there is a function like this then I defined
22
00:03:32,629 --> 00:03:46,200
it as some exponentially decaying and a function
which goes like this I define as exponentially
23
00:03:46,200 --> 00:03:55,290
amplifying function. So, let me treat them
just labels, if there is this kind of variation
24
00:03:55,290 --> 00:04:09,570
I label this as cosine x if it is let me put
it cosine x itself and if the variation is
25
00:04:09,570 --> 00:04:20,379
like this then let me label it as sin x. Similarly,
these Bessel functions I will see that what
26
00:04:20,379 --> 00:04:27,569
kind of variations they have. So, let me plot
these functions let me look at these functions
27
00:04:27,569 --> 00:04:32,000
and just treat them as some particular variation
of x.
28
00:04:32,000 --> 00:04:46,910
So, if now for different values of l. I plot
these J0, Y0 and J1, Y1 and so on. Then I
29
00:04:46,910 --> 00:04:56,130
find that I find that J0 has this kind of
variation. So, let me just label a function
30
00:04:56,130 --> 00:05:03,871
which has this kind of variation as J0 just
like the function which has this kind of variation
31
00:05:03,871 --> 00:05:11,380
I label as cosine x. J1 x has this kind of
variation.
32
00:05:11,380 --> 00:05:18,340
So, what I see that these functions these
functions have oscillatory behavior just like
33
00:05:18,340 --> 00:05:29,080
the cosine function. However, as I go in x
direction the amplitude of oscillations it
34
00:05:29,080 --> 00:05:35,520
changes it is not constant as you have in
cosine and in cosine functions you know that
35
00:05:35,520 --> 00:05:41,990
this is pi by 2, this is pi, this is 3 pi
by 2, this is 2 pi and so on, but here it
36
00:05:41,990 --> 00:05:50,400
is not as simple as that. So, it is some complicated
function, but let me say that if this is oscillatory
37
00:05:50,400 --> 00:05:57,020
function of this kind then it is J0, this
is J1 if I look at y function then y function
38
00:05:57,020 --> 00:06:03,330
is also oscillatory it is also oscillatory,
but the only thing is that it blows up at
39
00:06:03,330 --> 00:06:14,259
x is equal to 0, Y0, Y1, Y2 and so on.
So, these are Bessel J and Bessel Y functions
40
00:06:14,259 --> 00:06:22,240
and if I now look at Bessel K and Bessel I
functions then they look like this. So, Bessel
41
00:06:22,240 --> 00:06:29,900
K functions they have asymptotically decaying
behavior and Bessel I functions they are exponent
42
00:06:29,900 --> 00:06:40,150
they are asymptotically amplifying behavior.
So, now, remember that I want to find out
43
00:06:40,150 --> 00:06:46,759
the guided modes, which functions to choose
in which region that I will have to take care
44
00:06:46,759 --> 00:06:51,850
of.
In the core I need oscillatory functions.
45
00:06:51,850 --> 00:06:59,550
So, of course, the solutions are oscillatory
here, but I should pay attention that in the
46
00:06:59,550 --> 00:07:05,590
core I have r is equal to 0 and if I have
r is equal to 0 then these functions will
47
00:07:05,590 --> 00:07:11,500
blow up at r is equal to 0. So, I cannot take
these functions. So, I cannot include Bessel
48
00:07:11,500 --> 00:07:21,780
Y functions in the core in the cladding because
r can go up to infinity. So, at infinity these
49
00:07:21,780 --> 00:07:28,110
2 functions these I functions will blow up.
So, I will have to discard I functions. So,
50
00:07:28,110 --> 00:07:33,560
in the core I will discard this function,
in the cladding I will discard this function.
51
00:07:33,560 --> 00:07:39,970
So, now, I have the solutions I have the solution
in the core I have the solution in the cladding.
52
00:07:39,970 --> 00:07:48,930
So, what are the solutions? So, if this is
the fiber, then the radial part has the solution
53
00:07:48,930 --> 00:07:57,770
some constant times Jl Ur over a in the core
and in the cladding some constant b times
54
00:07:57,770 --> 00:08:06,230
Kl Wr over a. So, these are the solutions
just like in case of planar wave guide in
55
00:08:06,230 --> 00:08:15,539
high index region I had A cosine kappa x plus
B sin kappa x and here I had C e to the power
56
00:08:15,539 --> 00:08:23,360
minus gamma x. So, similar are the solutions
here also.
57
00:08:23,360 --> 00:08:31,849
Next thing is now to find out the relationship
between B and A and to form a transcendental
58
00:08:31,849 --> 00:08:36,921
equation. So, for that I apply boundary conditions
what are the boundary conditions, boundary
59
00:08:36,921 --> 00:08:48,290
conditions are radial part R and dR over dr
are continuous at r is equal to a and since
60
00:08:48,290 --> 00:08:54,910
it is weekly guiding fiber. So, I can take
these as the boundary conditions. So, even
61
00:08:54,910 --> 00:09:03,850
though they are they are not exactly tangential
at r is equal to a, but these are approximate,
62
00:09:03,850 --> 00:09:11,550
but they are quite valid quite valid for weekly
guiding fiber.
63
00:09:11,550 --> 00:09:23,489
So if I put these here the continuity of capital
R at r is equal to a will give me A Jl U is
64
00:09:23,489 --> 00:09:31,689
equal to B Kl W and the continuity of its
derivative at r is equal to a will give me
65
00:09:31,689 --> 00:09:43,189
A U over A Jl prime U is equal to B W over
a Kl prime W. So, from here I can relate B
66
00:09:43,189 --> 00:09:52,060
to A and then I can eliminate A and B and
get an Eigen value equations in beta. So,
67
00:09:52,060 --> 00:09:58,370
this Eigen value equation comes out to be
if I divide this by this U Jl prime U divided
68
00:09:58,370 --> 00:10:04,850
by Jl U is equal to W Kl prime W divided by
Kl W.
69
00:10:04,850 --> 00:10:13,879
So, by solving this equation I can find out
the modes of the fiber, but this equation
70
00:10:13,879 --> 00:10:22,720
has derivative in derivatives in Bessel functions.
So, so either I numerically solve them using
71
00:10:22,720 --> 00:10:30,679
the derivatives or I can represent them in
terms of Bessel functions itself non-prime
72
00:10:30,679 --> 00:10:39,319
functions for that I can make use of identities
of Bessel functions, the Bessel functions
73
00:10:39,319 --> 00:10:46,949
satisfy these identities these four identities
and we also make note that J minus 1 U is
74
00:10:46,949 --> 00:10:53,160
equal to minus J1 U and K minus 1 W is equal
to K1 W.
75
00:10:53,160 --> 00:11:03,009
So, if I use these then this transcendental
equation can now be written in 2 forms - one
76
00:11:03,009 --> 00:11:12,920
is U Jl plus 1 U over Jl U is equal to W Kl
plus 1 W over Kl W or U Jl minus 1 U over
77
00:11:12,920 --> 00:11:25,049
Jl U is equal to minus W Kl minus 1 W over
Kl W. So, I can get rid of these primed functions
78
00:11:25,049 --> 00:11:37,050
here. So, as we have been doing the representation
of all these equations in a normalized parameter.
79
00:11:37,050 --> 00:11:44,439
So, in fiber also it is worth representing
this in normalized parameters and do the analysis
80
00:11:44,439 --> 00:11:52,470
in normalized parameter so as to get rid of
any fiber parameters and wave length in depth.
81
00:11:52,470 --> 00:11:57,829
So, we can have independent of fiber parameters
and wave length.
82
00:11:57,829 --> 00:12:05,760
So, we have already the normalized frequency
which is 2 pi over lambda naught times a times
83
00:12:05,760 --> 00:12:10,889
the square root of n1 square minus n2 square,
you can see that this is square root of n1
84
00:12:10,889 --> 00:12:18,070
square minus n2 square is nothing, but n a
numerical aperture. We define the normalized
85
00:12:18,070 --> 00:12:23,759
propagation constant exactly in the same way
as we had done for planar wave guide.
86
00:12:23,759 --> 00:12:32,639
So, it is beta over k naught square minus
n2 square over n1 square minus n2 square.
87
00:12:32,639 --> 00:12:41,489
And now in order to in order to represent
these equations in terms of V and B. I will
88
00:12:41,489 --> 00:12:51,040
have to I will have to find out U and W in
terms of V and B. So, what is U? U square
89
00:12:51,040 --> 00:12:57,839
is a square times k naught square n1 square
minus beta square and W square is a square
90
00:12:57,839 --> 00:13:03,829
times beta square minus k naught square n2
square which are similar to kappa square and
91
00:13:03,829 --> 00:13:11,490
gamma square of planar wave guide. So, from
here I can immediately see that this B is
92
00:13:11,490 --> 00:13:17,709
nothing, but W square over V square if I take
k naught square downstairs and multiply the
93
00:13:17,709 --> 00:13:25,069
numerator and denominator by A square then
it becomes W square over V square.
94
00:13:25,069 --> 00:13:35,809
So, I can immediately get W is equal to V
times square root of B and I already have
95
00:13:35,809 --> 00:13:48,179
from previous slide that U square plus W square
is equal to V square since U square plus W
96
00:13:48,179 --> 00:13:54,769
square is equal to V square then I immediately
get that U is equal to V times square root
97
00:13:54,769 --> 00:13:57,089
of 1 minus B.
98
00:13:57,089 --> 00:14:05,959
So, with the help of this I can get the transcendental
equation or Eigen value equation in terms
99
00:14:05,959 --> 00:14:14,100
of normalized parameters and this equation
I can put in 2 categories 1 is for l is equal
100
00:14:14,100 --> 00:14:22,029
to 0 and 1 for l greater than or equal to
1. So, for l is equal to 0 the equation becomes
101
00:14:22,029 --> 00:14:30,180
this V times square root of 1 minus b, J1
V square root of 1 minus b divided by J0 V
102
00:14:30,180 --> 00:14:36,689
square root of 1 minus b is equal to V square
root of b K1 V square root of b divided by
103
00:14:36,689 --> 00:14:43,970
K0 of V square root of b. So, you can see
on this side I will have K functions and with
104
00:14:43,970 --> 00:14:52,230
K functions I have associated V times square
root of b and with on this side I have J function
105
00:14:52,230 --> 00:14:59,689
and with that is associated V times square
root of 1 minus b because this is U this is
106
00:14:59,689 --> 00:15:07,809
W for l greater than or equal to 1 the equation
becomes like this.
107
00:15:07,809 --> 00:15:16,529
So, here I have Jl minus 1 upstairs and Jl
downstairs while I have J1 upstairs and J0
108
00:15:16,529 --> 00:15:23,179
downstairs. So, this is the only difference
we should take care of and also here I have
109
00:15:23,179 --> 00:15:34,889
positive sign here I have negative sign. So,
for any given value of l. I can have the equation
110
00:15:34,889 --> 00:15:44,810
the Eigen value equation and I can solve it
for different values of V and generate what
111
00:15:44,810 --> 00:15:50,779
are known as b-V curves which are universal
curves which are universal curves.
112
00:15:50,779 --> 00:16:00,129
So, now, for any given value of l if I start
solving this equation then depending upon
113
00:16:00,129 --> 00:16:10,809
the value of V. I can have various routes
and mth root of any equation for a given l
114
00:16:10,809 --> 00:16:22,040
will represent LPlm mode l represents the
solution in phi direction, while m represents
115
00:16:22,040 --> 00:16:31,139
the solution in radial direction. What are
the cutoffs of the modes what are the cut
116
00:16:31,139 --> 00:16:44,130
offs of the modes well if you look back our
b is defined as beta square over k naught
117
00:16:44,130 --> 00:16:55,110
square minus n2 square over n1 square minus
n2 square and for guided modes beta over k
118
00:16:55,110 --> 00:17:05,930
naught lies between n2 and n1 and beta over
k naught is equal to n2 defines the cut off.
119
00:17:05,930 --> 00:17:15,180
So, which means that b lies between 0 and
1 and b is equal to 0 corresponds to the cut
120
00:17:15,180 --> 00:17:21,449
off.
So, the cut offs of the modes are defined
121
00:17:21,449 --> 00:17:29,509
by b is equal to 0. So, if I put b is equal
to 0 here I will get the cut offs for different
122
00:17:29,509 --> 00:17:37,860
values of l. So, for l is equal to 0 this
is the equation and if I put b is equal to
123
00:17:37,860 --> 00:17:45,279
0 then the cut offs are defined by Vc J1 Vc
is equal to 0 where Vc are the cut offs, Vc
124
00:17:45,279 --> 00:17:52,889
J1 Vc is equal to 0. For l is all l greater
than or equal to 1 the cut offs will be defined
125
00:17:52,889 --> 00:18:00,580
from here which are Jl minus 1 Vc is equal
to 0. So, from here I can get the cut offs
126
00:18:00,580 --> 00:18:10,389
of various modes.
If you remember in case of planners symmetric
127
00:18:10,389 --> 00:18:18,381
wave guide the cut offs were very simple cut
offs for Vc is equal to m pi by 2. So, for
128
00:18:18,381 --> 00:18:27,340
mth mode you have the cut off m pi by 2. So,
TE0 mode no cut off TE1 mode pi by 2. TE2
129
00:18:27,340 --> 00:18:35,970
mode pi and so on, but here I will have to
find out the 0s of these Bessel functions
130
00:18:35,970 --> 00:18:43,360
to find the cut offs. There, they were the
0s of ten Vc and caught Vc, but here they
131
00:18:43,360 --> 00:18:51,360
are the 0s of J1 Vc and Jl minus 1 Vc.
132
00:18:51,360 --> 00:19:00,250
So, let us first find out the cut offs of
l is equal to 0 modes cut off condition is
133
00:19:00,250 --> 00:19:07,390
Vc J1 Vc is equal to 0. So, of course, the
first root is Vc is equal to 0 itself and
134
00:19:07,390 --> 00:19:17,429
then let us look at how J1 varies with x.
So, if I plot J1 x then J1 x looks like this
135
00:19:17,429 --> 00:19:26,230
its first 0 is at x is equal to 0 itself and
that is coming from here also. So, first root
136
00:19:26,230 --> 00:19:34,970
is of course, Vc is equal to 0. Now let us
look at the second 0, second 0 appear somewhere
137
00:19:34,970 --> 00:19:44,279
here near 3.8 and then the third 0, four 0
and so on. So, if I list the 0s of J1 x then
138
00:19:44,279 --> 00:19:55,540
they are 0, 3.8317, then here it is 7.0156
and then here it is 10.1735.
139
00:19:55,540 --> 00:20:07,740
So, this is the cutoff of LP01 more this is
the first mode l is equal to 0 and first mode
140
00:20:07,740 --> 00:20:16,880
is m is equal to 1, so LP01. This is m is
equal to 2, so LP02; m is equal to 3 LP03
141
00:20:16,880 --> 00:20:25,409
and so on. So, I have the cut offs of various
l is equal to 0 modes here. Let us look at
142
00:20:25,409 --> 00:20:33,860
the cut offs of l is equal to 1 mode, l is
equal to 1 mode has cut off condition J0 Vc
143
00:20:33,860 --> 00:20:39,090
is equal to 0 because it is Jl minus 1 Vc
is equal to 0 and l is equal to 1. So, it
144
00:20:39,090 --> 00:20:47,580
becomes J0 Vc is equal to 0, you should remember
that Vc is equal to 0 can be the cutoff of
145
00:20:47,580 --> 00:20:53,830
only 1 mode and that is the fundamental mode.
So, this cut off cannot be shared with any
146
00:20:53,830 --> 00:20:58,970
other mode; however, non 0 cut offs can be
shared with other modes.
147
00:20:58,970 --> 00:21:10,590
So, what are the 0s of J0 now? Let us plot
J0 x and J0 x goes like this and we find that
148
00:21:10,590 --> 00:21:18,100
the first 0 is around 2.4 and then and so
on. Let us list the 0es of J1 x, J0 x, first
149
00:21:18,100 --> 00:21:35,710
0 is 2.4048, second is 5.5201, third is 8.6537
and 11.7915 and so on. So, you have l is equal
150
00:21:35,710 --> 00:21:43,841
to 1, m is equal to 1, LP11, 1 is equal to
1, m is equal to 2, LP12 and so on. So, you
151
00:21:43,841 --> 00:21:56,710
have the cutoff of LP11 more this LP12 mode,
LP13, LP14. So, all LP1m modes you have here.
152
00:21:56,710 --> 00:22:06,860
Now let us look at l is equal to 2 since the
cut offs are Jl minus 1. Vc is equal to 0.
153
00:22:06,860 --> 00:22:12,309
So, this becomes J1 Vc is equal to 0 for l
is equal to 2. So, it is the same as this
154
00:22:12,309 --> 00:22:21,860
one, but I will have to exclude this Vc is
equal to 0 here because this cut off cannot
155
00:22:21,860 --> 00:22:32,149
be shared with any other mode. So, now the
cut offs are 3.8317, 7.0156 and so on. For
156
00:22:32,149 --> 00:22:38,290
l is equal to 2 modes, so this will correspond
to m is equal to 1, so LP21 mode, LP22, LP23,
157
00:22:38,290 --> 00:22:50,559
LP24 and so on. l is equal to 3 will give
you J2 Vc is equal to 0 and these are the
158
00:22:50,559 --> 00:23:00,250
0s of J2. So, you will have these cut offs
for various l is equal to 3 modes. So, in
159
00:23:00,250 --> 00:23:07,460
this way for whatever value of l. I want then
I can have the cut offs.
160
00:23:07,460 --> 00:23:16,909
How to arrange these modes? I have so many
modes 01, 02, 03, 11, 12, 13, 21, 22, 23,
161
00:23:16,909 --> 00:23:26,509
31, 32, 33 and so on. In what order their
propagation constants are arranged increase
162
00:23:26,509 --> 00:23:33,440
or decrease in case of planar wave guide it
was very simple TE0 has the highest propagation
163
00:23:33,440 --> 00:23:41,610
constant then TE1 then TE2 then TE3 and so
on. So, the propagation constant they decrease
164
00:23:41,610 --> 00:23:51,500
with mode number, but here it is not obvious
and also the cut offs you can see the numbers
165
00:23:51,500 --> 00:23:58,570
the numbers are weird. So, I cannot fit any
pattern there in planar wave guide it was
166
00:23:58,570 --> 00:24:08,530
very simple TE0 mode 0 TE1 mode pi by 2, TE2
mod pi. So, m pi by 2 as more number increases.
167
00:24:08,530 --> 00:24:15,799
So, the cut offs are every pi by 2 there is
a cut off of the higher order mode the next
168
00:24:15,799 --> 00:24:24,179
mode, but here I will have to really look
into these numbers and then put them together.
169
00:24:24,179 --> 00:24:30,789
So, I will have to arrange them in order of
their cut offs.
170
00:24:30,789 --> 00:24:41,749
Then we can generate we can generate b-V curves.
So, so I choose l is equal to 0 and solve
171
00:24:41,749 --> 00:24:48,909
the transcendental equation corresponding
to l is equal to 0 and list the highest root
172
00:24:48,909 --> 00:24:56,289
always list the value of highest root and
plot the value of highest root as a function
173
00:24:56,289 --> 00:25:04,539
of V then it will go like this, this is LP01
mode the highest root. The first one corresponding
174
00:25:04,539 --> 00:25:11,669
to l is equal to 0. So, this is LP01 mode
and this is the fundamental mode this is the
175
00:25:11,669 --> 00:25:20,879
fundamental mode. It is cut off is 0, so this
is always guided this is always guided and
176
00:25:20,879 --> 00:25:28,890
remember that we are doing the scalar analysis
under weekly guiding approximation. So, there
177
00:25:28,890 --> 00:25:35,560
they are LP modes in terms of vector mode
they are different in terms of vector mode
178
00:25:35,560 --> 00:25:40,881
they are different. So, we are not going into
the details of vector modes we will confine
179
00:25:40,881 --> 00:25:51,500
ourselves to the scalar modes in this course.
Now what I see that the cut off the next cut
180
00:25:51,500 --> 00:26:00,289
off appears for LP11 mode. So, this is 0 the
next number is 2.4048 which corresponds to
181
00:26:00,289 --> 00:26:12,159
LP11 mode. So, now I plot the propagation
constant of LP11 mode that is the highest
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root corresponding to l is equal to 1 which
is LP11. So, I plot the propagation constant
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of that with respect to V and it goes like
this. So, this is LP11 mode and it is cut
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00:26:26,929 --> 00:26:35,309
off at 2.4048.
If I look for the next number I find the next
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number is 3.8317 and it appears twice here
for LP02 and LP21. So, both will start from
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here at 3.8317 and then if I solve the transcendental
equation corresponding to these modes then
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the propagation constant vary like this. So,
LP21 mode goes like this and LP02 mode go
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goes like this similarly the next 1 is LP31
and the next one is LP12. So, I can see that
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a fiber would never guide 3 modes either there
will there would be 1 2 or 4, because these
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2 modes share the cut off, these 2 modes share
the cut off. So, these are b versus V curves
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and the cut offs of various modes.
Now if I want to find out how many modes are
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guided then it is not as easy as it was in
case of planar wave guide that I just calculate
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the value of V divided by pi by 2 get the
number and find the closest, but greater than
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00:28:00,559 --> 00:28:06,809
the integer for corresponding to that number,
but it is not that easy here. Here what I
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will have to do? If I have a fiber and wavelength
I calculate the value of V and then find out
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what modes are guided and what modes are cut
off looking in to these numbers, looking in
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to these numbers and then only I shall be
able to find out the find out the number of
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modes.
So, in the next lecture I will look into the
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modal fields and I will work out some examples
also on number of modes and what happens if
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the value of V is very large then what are
the approximate number of modes.
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00:28:50,340 --> 00:28:51,429
Thank you.