3. BASICS
What is Light??
Light Wave Vs Sound Wave
What is velocity of Light and Is it a Constant?
Spectrum of Visible Light?
What factor decides the colour of Light?
Why is sky Blue??
3
4. HISTORY OF OPTICAL COMN
4
Fire signals - 800BC.
Transmitted the message of the final conquest of
Trojan.
Heliographs
Telescope extended the range of
optical relays - 1600AD
Photophone by Graham Bell – 1876
6. HISTORY OF OPTICAL COMN
6
LASER – 1960 AD.
CK Kao in 1966 suggested that fiber will be used for comn purpose.
1955- it was an Indian scientist by the name of Narinder Singh Kapany
who had demonstrated for the first time the transmission of images
over a bundle of optic fibres.
7. HISTORY OF OPTICAL COMN
7
Devp of low loss fibers (20 dB/Km) at 633nm(He Ne
wavelength) – 1970 AD.
Initially loss in the range of 105 dB/ Km.
Due to impurities.
Presently, 0.2 dB/Km at 1.55µm.
Initially semiconductor lasers had short life of
few Hr, then 1000 Hr, Now 7000 Hr.
Originally sources were from alloy of AlGaAs ,
which emitted in near IR 0.8 to 0.9 µm
subsequently extended to 1.1 to 1.6 µm .
8. OPT FIBER VS TXN IN OTHER MEDIA
12
• Restd to LOS.
• Low BW Comn Link.
• Affected by atmospheric disturbances
Rain
Snow
Fog
Dust
• COAXIAL Cable. High Attenuation at high
frequency
9. 13
• Info carrying capacity is related to BW ,
which is limited to fraction of carrier freq-
• Optical carrier freq are at 1014 to 1016 Hz.
• BW at optical freq is 10,000 times the
microwave freq.
• Initial attn in optical fiber was very high.
• Lasers reduced to 5 db/km.
OPT FIBER VS TXN IN OTHER MEDIA
14. Information source provides the signal to the
transmitter.
The electrical stage drives an optical source to give
modulation of the light wave carrier.
The optical source provides the electrical-optical
conversion.
Txn medium is an optical fiber cable.
The rxr consists of an optical detector .
Detector drives the electrical stage and hence
provides demodulation of the optical carrier. It
provides the optical – electrical conversion.18
OPTICAL FIBER COMN
SYS
15. 19
ADVANTAGES
The advantages of fiber optic comn as compared to
copper, co-axial, MW and Satl include:
• Small Size. A 3/8-inch (12 pair) fiber/cable operating
at 140 mb/s can handle as many voice channels as a 3-
inch diameter copper (900) twisted-pair cable.
• Light Weight. The same fiber-optic cable weighs
approximately 60 Kg per kilometer. The twisted pair
cable weighs approximately 700 Kg.
• High Bandwidth. Fiber optics has been band-width
tested at over 10-billion bits per second(10 Gbps- over a
100 km distance. Theoretical rates of 50-billion bps are
obtainable.
16. 20
• Low Loss. SM fibers have losses as low as 0.2
dB/km. Multimode - 1 dB/Km (at 850 or 1300 nm).
This facilitates longer distances without costly
repeaters.
• Noise Immunity. Unlike wire systems, which require
shielding to prevent EM radiation or pick-up, fiber-
optic cable is a dielectric and is not affected by EM or
RF interference. The potential for lower bit error rates
can increase circuit efficiency.
• Transmission Security. Because the fiber is a
dielectric it does not radiate EM pulses, radiation, or
other energy that can be detected. This makes the
fiber/cable difficult to find and methods to tap into
fiber create a substantial system signal loss.
17. 21
• No Short Circuits: Since the fiber is glass and does not
carry electrical current, radiate energy, or produce heat or
sparks, the data is kept within the fiber medium.
• Wide Temperature Range. Fibers and cables can be
manufactured to meet temperatures from -40°F to +200°F.
• No Spark or Fire Hazard. Fiber optics provides a path for
data without transmitting electrical current. For
applications in dangerous or explosive environments, fiber
provides a safe transmission medium.
• Fewer Repeaters. Few repeaters are required because
of increased performance of light sources and continuing
increases in fiber performance.
18. 22
• Stable Performance.
Fiber optics is affected less by moisture, thus
lesser corrosion and degradation. Therefore, no
scheduled maintenance is required. Fiber also has
greater temperature stability than copper systems.
• Topology Compatibility.
Fiber is suitable to meet the changing
topologies and configurations necessary to meet
operation growth and expansions. Technologies such
as wavelength division multiplexing (WDM), optical
multiplexing, and drop and insert technologies are
available to upgrade and reconfigure system designs.
19. 23
• Decreasing Costs.
Costs are decreasing due to larger
manufacturing volumes, standardization of common
products, greater repeater spacing, and proven
effectiveness of older technologies such as
multimode.
• Non Obsolescence.
Expansion capabilities beyond current
technologies
• Material Availability.
Silica Glass, material required for the
production of fiber is readily available.
20. SUMMARY - ADVANTAGES
24
The advantages of fiber optic as compared to
copper, co-axial, MW and Satl, as media, include:
o Small Size.
o Light Wt.
o High BW.
o Low Loss.
o Noise Immunity.
o Txn Security.
o No Short Ccts.
o Wide Temp Rg.
o No Spark or Fire Hazard.
o Fewer Repeaters.
21. 25
o Stable Performance.
o Decreasing Costs.
o Non Obsolescence.
o Mtrl Availability.
o Immunity to Interference and x-talk.
o Ruggedness and flexibility.
o System Reliability and ease of Maint.
SUMMARY - ADVANTAGES
22. 26
THE OPTICAL FIBER CABLE IN THE FOREGROUND HAS
THE EQUIVALENT INFO-CARRYING CAPACITY OF THE
COPPER CABLE IN THE BACKGROUND.
23. DISADVANTAGES
27
• High initial cost.
• Prior Planning of drop points.
• Fragile.
• High deg of proficiency in laying and maint.
• Jointing and Splicing difficult.
24. Reflection
Light incident upon a surface will
in general be partially reflected
and partially transmitted as a
refracted ray.
The angle relationships for both
reflection and refraction can be
derived from Fermat's
principle.
The fact that the angle of
incidence is equal to the angle of
reflection is sometimes called the
“Law of Reflection"
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28
26. Refraction
Refraction is the
bending of a wave when
it enters a medium
where its speed is
different.
The refraction of light when
it passes from a fast medium
to a slow medium bends the
light ray toward the normal
to the boundary between the
two media.
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30
• The amount of bending depends on the indices of
refraction of the two media and is described quantitatively
by Snell's Law.
29. Refractive Index
The bending of refraction can be
visualized in terms of Huygens'
principle.
As the speed of light is
reduced in the slower medium,
the wavelength is shortened
proportionately. The
frequency is unchanged;
it is a characteristic of the source of
the light and unaffected by
medium changes.
The index of refraction is defined
as the speed of light in vacuum
divided by the speed of light in the
medium.
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33
Material n Material n
Vacuum 1.000
Ethyl
alcohol
1.362
Air 1.000277 Glycerine 1.473
Water 4/3 Ice 1.31
Carbon
disulfide
1.63
Polystyre
ne
1.59
Methylen
e iodide
1.74
Crown
glass
1.50-1.62
Diamond 2.417 Flint glass 1.57-1.75
30. Fiber Optic Transmission Bands
Near Infrared
Frequency
Wavelength
1.6
229
1.0 0.8 µm
0.6 0.4
1.8 1.4
UV
(vacuum) 1.2
THz
193 461
0.2
353
Longhaul Telecom
Regional Telecom
Local Area Networks
850 nm
1550 nm
1310 nm
CD Players
780 nm
HeNe Lasers
633 nm
•1000 nm wavelength signal is 300THz (~300,000,000,000,000 cycles/s).
•BW = 1-2% of Carrier Freq
•So at 1000 nm around 3 THz of bandwidth available
33. f = c λ/ n λ2
c
λ
f =?
λ = 1 nm, λc = 1550nm, n = 1.5
So 1nm spectral width at 1550 nm =______GHz BW
38
34. Plank’s Law- relates photon energy to freq
eV is the energy reqd to move an electron across a potential
of 1V. One electron-volt is equal to 1.6⋅10-19 joules
E = hν= hc/λ where hc (nm)= 1240 eV nm
H = planks constant = 6.63x 10 -34 J-sec or 6.58 × 10-16 eV -s
E(eV)= 1.24/λ(inμm)
39
35. The most common frequency grid used for optical
comn in Dense Wavelength Division Multiplexing
(DWDM) at wavelengths around 1550 nm and defined
by ITU-T G.694.1.
The grid is defined relative to 193.1 THz and extends from
191.7 THz to 196.1 THz with 100 GHz spacing. Which
covers the wavelength range of 1528.77 nm to
1563.86 nm with approximately a 0.9 nm channel spacing.
.
40
43. PROPAGATION OF LIGHT IN A FIBER
49
Light travels more slowly in optically dense medium
than in less dense medium.
Refraction occurs when a light ray is incident on the
interface between two dielectrics of differing refractive
indices.
When light enters from a rarer to denser medium, it bends
towards the normal to the point of incidence.
45. 51
PROPAGATION OF LIGHT IN A FIBER
CRITICAL ANGLE AND TOTAL INTERNAL REFLECTION AT GLASS –AIR INTERFACE
• When light enters from a denser to rarer
medium (i.e. n1>n2), it bends away from the
normal to the point of incidence.
46. TIR: SNELL’S LAW
52
The angles of incidence 1 and refraction 2 are
related to each other and to the refractive indices of
the dielectrics by Snells’ law:
n1.sin 1 = n2.sin 2
or, sin 1 / sin 2 = n2 / n1
Angle of refraction is always greater than angle of
incidence.
47. 53
When angle of refraction is 90º, the ray emerges
parallel to the interface and the angle of incidence
must be less than 90º.
The angle of incidence for the condition when angle
of refraction is 90º, is called the Critical Angle c.
We have,
sin c = n2 / n1
At angles of incidence greater than the critical angle,
the light is reflected back into the originating
medium, the phenomenon is called Total Internal
Reflection.
PROPAGATION OF LIGHT IN A FIBER
49. 55
Those light rays entering the optical fiber
with angles of incidence greater than the
critical angle c may be considered to
propagate down the fiber
A ray that propagates through the fiber and
passes through the axis of the fiber core is
called a Meridional Ray.
A ray that propagates through the fiber but
does not passes through the fiber axis is
called a Skew Ray.
PROPAGATION OF LIGHT IN A FIBER
54. ANGLE OF ACCEPTANCE
60
Only rays with angle greater than c at the core-
cladding interface are transmitted by total internal
reflection.
All rays entering the fiber core will not continue to
be propagated down its length.
Consider a Meridional Ray A incident on the core
– cladding interface at critical angle c.
55. This ray enters the fiber core at an angle a to the fiber axis
and is refracted at the air-core interface before transmission
to the core-cladding interface at the critical angle.
Rays which are incident into the fiber core at an
angle greater than a would be transmitted to the
core – cladding interface at an angle less than c and
will not be totally internally reflected.
Ray B is lost by radiation.
61
56. Rays transmitted by total internal reflection within the fiber
core must be incident on the fiber core within an acceptance
cone defined by the conical half angle a.
a is the maximum angle to the axis at which light can enter
the fiber in order to be propagated and is called the
acceptance angle for the fiber.
If the fiber has a regular cross-section, an incident
Meridional Ray at greater than the critical angle would
continue to be reflected and will be transmitted through the
fiber.
62
57. 63
a
Core(n1)
Cladding(n2)
NA = n0 Sin a = n1²-n2²
2 a is the Cone of Acceptance ( Light will be guided with in this
angle only).
Guided Light
62. NUMERICAL APERTURE (NA)
68
Consider a Meridional Ray incident on the fiber core
at an angle 1 < a to the fiber axis.
The ray enters the fiber from a medium of lower RI
n0.
Core has a refractive index of n1, which is slightly
greater than the cladding RI n2.
Assume the fiber entrance face to be normal to the
axis.
63. By Snell’s law, at the air-core interface,
n0.sin 1 = n1.sin 2
Considering the right angled triangle ABC,
= /2 - 2,
where is greater than the critical angle at the core-
cladding interface.
n0.sin 1 = n1.cos = n1(1-sin2 )1/2
In the limiting case, becomes the critical angle for the
core-cladding interface and is given by sinc = n2 / n1
69
n0
n2
n1
Core
Cladding
θ1
θ2
A
B
C
64. n0.sin a = (n1²-n2² )1/2
The NA of a optical fiber is defined as
NA = n0.sin a = (n1²-n2² )1/2
for fiber in air, NA = sin a
Incident Meridional Rays over the range 0 a will
propagate through the fiber.
The NA is also given in terms of the relative refractive
index difference between the core and the cladding
which is defined as
≈ (n1 - n2 )/ n1 for <<1
So, NA = n1 (2 ) 1/2
70
65.
NA2 = (n1²-n2² )
= 2 n1 (n1 - n2)
(n1 - n2) = NA2/ 2 n1
(n1 - n2)/ n1 = NA2/ 2 n1²
= (n1 - n2)/ n1 = NA2/ 2 n1²
Refractive index difference between the core and the
cladding which is defined as
≈ (n1 - n2 )/ n1 for <<1
So, NA = n1 (2 ) 1/2
71
66. For a Step index fiber placed in air with
n1 = 1.5, = 0.015 and core radius = 25
μm find
n2= ?
NA =?
Max acceptance angle θm =?
72
67. For MMF < 1 % Telecom fiber
For SMF < 0.3 % only
Why kept small ?
73
68. For typical fibers diff betwn n1 and n2 is small ie
around 1%
sin-1 (n2/n1) = about 82o = Critical Angle
This implies that 2 = about 8o
74
69. Q1 For a Fiber n1 = 1.5 and n2 = 1.47 Find the Fwg
Critical Angle c at core cladding interface
NA of the fiber
Acceptance angle in air of the Fiber a
75
70. Q1 For a Fiber n1 = 1.5 and n2 = 1.47 Find the Fwg
Critical Angle c at core cladding interface
NA of the fiber
Acceptance angle in air of the Fiber a
Ans 78.5o, 0.30 and 17.4o
76
71. Q2 For a Fiber with = 1% and n1 = 1.46 find NA,
Critical Angle c
77
72. Q2 For a Fiber with = 1% and n1 = 1.46 find NA,
Critical Angle c
Answer – 0.21 and 81.9o
78
74. TYPE OF FIBER
80
• Based on Refractive index.
• Step Index(SI).
• Graded Index(GI).
• Based on mode of wave txn.
• Single Mode (SM)fibers (mono mode).
• Multimode fibers(MM).
• Based on mtrl.
• Glass (silica) which includes SM SI, MM SI, and
MM GI fibers.
• Plastic clad silica (PCS).
• Plastic.- High Attenuation
75. 81
PLASTIC FIBERS
Plastic fibers are the least efficient but tend to
be larger, more economical, and more rugged
The most commonly used material is is Poly
Methyl Methyl Acrylate or PMMA. Doping PMMA
fibers change the transmission properties
Attenuation is very high compared with glass
fiber at all wavelengths; Tx windows using visible
light near 570 nm and 650 nm are feasible for
short dist (up to 100 meters). At 100m , step
index fiber has a bandwidth of about 125
MHz, while for graded index fibers
bandwidths of better than 500 MHz have been
reported .
Core diameter is 100 times bigger than single-
mode fiber
Used in applications not requiring long
transmission distances, such as medical
instrumentation, automobile and aircraft
control systems, and consumer electronics
76. OTW
Q List out major difference between Silica
fibers, Plastic clad silica (PCS) Fibers , and
Plastic fibers.
82
78. 84
Propagation of light in Multi mode fiber
Multi mode: Number of reflections from core cladding interface
Propagation of light in single mode fiber
Single mode : Ideally no reflections from cladding but
practically some reflections takes place, Single mode
operation depends on cutoff wavelength of fiber
81. SMF Vs MMF
The typical best bandwidth–length products for the three fibers are
25 MHz km - Multimode step index fibers
1 GHz km - Multimode graded index fibers
Upto 1000 GHz km - single-mode step index fibers@1310nm and 4000GHz Km
@1550nm
SMF can sp thousands of kilometers at 10 Gbit/s, and several hundred
kilometers at 40 Gbit/s.
Due to large core size MMF
Large NA, MMF has higher "light-gathering" capacity than SMF. – sensor
applications
Simplifies connections and also allows the use of lower-cost electronics
such as LEDs
Sp more than one propagation mode; which causes INTRAMODAL
DISPERSION
The LED used with MMF produce a range of wavelengths and each
propagate at different speeds. Causing Chromatic Dispersion limiting the
87
82. GRADED -INDEX (GI) FIBER
89
Refractive index (RI) of core material decreases
continuously with distance from the fiber
n
r
1.475
1.460
83. 90
n2
n1
nr
Core RI n(r) = n1[ 1- 2(r/a)²] 0< r < a and for r > a
nr = n2
Step index profile
Graded index profile
nr
84. 91
•The maximum refractive index of the core is at the axis and it decreases gradually towards
the periphery of the core and then in the cladding it is constant at n2.
•The velocities of the rays increase as their lateral displacement from the axis increase
because they encounter regions of lower refractive index. This causes them to travel
together without any delay between themselves and thus reduce the pulse broadening
to a considerably low value.
•The perfect profiling of the refractive index has not yet been possible practically. Hence the
delay between the rays is never practically zero though it may be very small.
•GIOFs are not as better in bandwidth as SMOF but do have higher N.A. than SMOFs. This is
why, where light gathering is more a concern over bandwidth, GIOFs becomes the
appropriate choice. GIOFs are better than MMOFs in terms of bandwidth.
85. MODE
92
• A mode is simply a path that a light ray can
follow in traveling down a fiber.
• The number of modes supported by a fiber
ranges from one to over 100,000.
• Thus a fiber provides a path of travels for
one or thousands of light rays, depending
on its size and properties.
89. Light energy launched into the fiber may be considered to
travel in the form of numerous rays in accordance to the
Ray-Model
These rays travel different paths inside the core of an
optical fiber because different light rays are incident on the
tip of the optical fiber at different angles within the
acceptance cone
This causes different light rays in the acceptance cone to
travel along different paths in the core and accordingly take
different time intervals to travel a given distance, which
leads to a phenomenon of pulse broadening inside the core
of the optical fiber
96
PULSE BROADENING- MMSI FIBER
90. Thus the pulse of light which might originally be of
width T seconds now might be of T+ΔT seconds
inside the fiber core.
The amount of broadening is measured in terms of
the increase in the pulse time width and is denoted
by ΔT. the value of ΔT is given by
97
PULSE BROADENING- MMSI FIBER
92. Refer fig at prev slide
Time taken to traverse the distance ‘AB=L’,
PULSE DISPERSION IN MM SI OPTICAL FIBER
99
• If we assume all rays between =0 to = c is
present,
93. Time interval ‘T ’ at the output will be:
100
• For a typical fiber, assuming :
95. 102
For non overlapping light pulses in
OFC Link The data rate must be
less than reciprocal of broadened
pulse duration
Bt ≤ 1/2 Δ t
For Higher bandwidth of transmission the pulse broadening, Δt
should be as low as possible.
Δ t α L and Δ t dictates max tx BW
Keeping other parameters const ie
Δ and n1
BW x L = Constant
96. For no overlapping of light pulses down on an optical fiber link
the digital bit rate BT must be less than the reciprocal of the
broadened (through dispersion) pulse duration (2τ).
ie BT ≤1/ 2 Δ t for NRZ where 2Δ t is the broadened pulse
duration.
Δ t
This assumes that the pulse broadening due to dispersion on the
channel is Δ t which dictates the input pulse duration which is
also Δ t
103
97. The conversion of bit rate to bandwidth in hertz depends on
the digital coding format used.
A non return-to-zero code is employed, there are two bit
periods in one wavelength(i.e. 2 bits per second per hertz)
Maximum bandwidth B is one-half the maximum data rate or:
BT(max) Bits /sec = 2B hertz
For return-to-zero code data rate is equal to the bandwidth
in hertz BT(max) Bits /sec = B hertz
With same BW in Hz NRZ will achieve
higher datarate
104
NRZ
RZ
98. Q2 For a MMSI Fiber n1 = 1.5, n2 = 1(Air)
and Length = 10 Km Find the Tx BW it will
support
Ans = _____Kbps
105
So air cannot be used as cladding
99. 106
Q For a MMSI Fiber n1 = 1.5 and Δ= 0.01 and
L= 1Km Find the Tx BW
100. Time interval ‘T ’ or ΔT or Δt at the output will be:
107
101. ANALYSIS OF RESULT
The value of ΔT is dependent on the value of L, the
difference (n1 – n2) and n1/n2. But reducing the value of L
would signify the reduction in the length of the optical fiber,
which is not desirable.
As 1<n2<n1, the ratio, n1 / n2 is very close to 1. Thus for
low ΔT values, the only option available with us is to
decrease the value (n1–n2) or in other words, to increase
the refractive index of the cladding n2.
One can now notice that a contradictory situation has been
generated as to whether the cladding should be removed
for high NA or to use a cladding of large refractive
index value for higher bandwidth?
108
102. The answer to this query is purely application specific. That means if an
optical fiber is used as a sensor (say), where lowest possible light has
to be accepted, we use fiber with low n2 values.
When the optical fiber is used for data communication, fibers with high
values of n2 are used. For practical communication purposes the value
of (n1 – n2) is made of the order of about 10-3 to 10-4.
If the cladding is removed, the value of n2 becomes 1 and the value of
the above difference becomes about 0.5. The bandwidth corresponding
to this value of n1-n2 is of the order of few Kilohertz, which is even
worse than that of a normal twisted pair of wires. Thus cladding is an
extremely important requirement for optical fiber when the
bandwidth is the prime concern of the application and its refractive
index is made as close to that of the core as the available technology
permits, but not made equal.
109
ANALYSIS OF RESULT
103. LIMITATIONS OF RAY THEORY
During TIR Energy confined to core only, however in
optical fibers energy travels in cladding also.
Ray model does not speak about discrete fd patterns
for propagation inside the fiber
Valid only for large core radius ie a/λ ratio high-
MMF
Ray theory does not explain why light launched only
at certain discrete angles will propagate in fiber
WAVE MODEL gives complete description of light
propagation inside optical fiber.
110
105. LINEARLY POLARISED WAVE
112
• The behavior of the electric field as a function of time is called the polarization of
light
• It is a quantity which illustrates the vector nature of light unlike other quantities like
intensity, wavelength and spectral width which show scalar nature of light
• The locus of the tip of the electric field vector with respect to time, gives the
polarization of the wave
108. WAVE MODEL OF PROPOGATION INSIDE
FIBER CORE
115
•The red and green coloured dotted lines represent the wave fronts of
the light rays which are perpendicular to their direction of propagation
•The distance between a red and a green wave-front corresponds to a
phase diff of 1800
•The similar coloured wave-fronts have either 00 or 3600 phase
difference between them
•When two similar coloured wave-fronts meet, they interfere
constructively and dissimilar coloured wave-fronts interfere destructively
109. 116
•In the core, the interference between the incident and the reflected
wave-fronts constitutes a standing wave pattern of varying light intensity
with discrete maxima and minima in a direction normal to the core-
cladding interface
•Electromagnetic wave theory of light, shows that at total internal
reflection, the light intensity inside the cladding is not completely
zero and, there exist some decaying fields in the cladding, which do
not carry any power but support the total internal reflection phenomenon
WAVE MODEL OF PROPOGATION INSIDE
FIBER CORE
110. 117
•These fields are called as Evanescent fields.
•The Ray-model of light does not offer any explanation about the
evanescent fields, which are as equally important as the fields in the
core for total internal reflection to occur.
•The slightest disturbance to these fields in the cladding could lead to
the failure of the TIR at the core-cladding boundary accompanied by
leakage of optical power to the cladding.
• Thus ray-model is inadequate in explaining the phenomena
exhibited by light
WAVE MODEL OF PROPOGATION INSIDE
FIBER CORE
111. TIR INSIDE FIBER CORE
118
• Evanescent fields are decaying fields, they never become zero, atleast
theoretically. In other words, they remain present upto infinite distance from the
core-cladding boundary.
•But in practice, these fields decay down to a negligibly small value as we move
away from the core-cladding boundary deeper into the cladding.
•Larger the value of the angle of incidence of the incident ray at the core-
cladding boundary, sharper is the decay of the evanescent fields.
•Thus there must me a sufficient thickness of cladding provided for these
evanescent fields to be accommodated so that they decay to a negligibly small
value in the cladding and cannot be disturbed by external sources.
112. Single Mode Optical Fiber
The light basically consists of wave fronts. A line perpendicular to a wave
front is called the ray. Light is an electromagnetic wave and when we say
it travels like a ray it is a collection of wavefronts which move.
Let us take an optical fiber with light rays propagating in it. The rays
and the wave fronts which are perpendicular to the rays, are as shown in
figure 7:
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119
Figure (7): Core of optical fiber, rays with wave
fronts
113. Propagation of Wave in Optical Fiber
Let us consider a phase front corresponding to the ray AB and passing through
the point B. This phase front also meets the ray CD at point E.
In other words, the phase of the ray at B (just before the reflection) is same as that
of the ray at point E.
That is to say that the phase change corresponding to the distance BCE added
with the phase (δ) of the reflection coefficient at points B and C should be a
multiple of 2π.
This is what is called the condition for the constructive interference.
From simple geometric considerations we have
θ + 2φ = = π/2
BC = d sec φ
CE=BC sin θ
= d sec φ sin ((π/2) - 2φ)
= d sec φ cos 2φ
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120
114. Condition for Constructive Interference
Phase change from B to E is
Φ = (2π/λ)n1 (BC + CE) + 2δ
= (2π/λ)n1 {d sec φ + d sec φ cos 2φ} + 2δ
For constructive interference the phase change should be
multiple of 2π
Therefore, Φ = 2mπ
Simplifying equations we get a condition for sustained
propagation of light rays inside the core as
(2πn1d cos φ)/λ + δ = mπ
It can be noted that for φ = π/2 (i.e. the ray along the axis of the
fiber), δ = 0 and the condition is satisfied with m = 0 for any
value of n1, d and λ.
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121
115. Condition for Propagation of Modes
in Optical Fiber
As (n1d /λ) increases (either due to increase of the diameter of the core
or refractive index of the core, or decrease in wavelength) more values of
m satisfy the condition and therefore have sustained propagation inside
the fiber.
The above phase condition can be satisfied only by discrete rays entering
the structure i.e. rays at finite number of angles are accepted by the
optical fiber.
The ensemble of rays entering at a specific angle from the axis of the
fiber gives discrete optical intensity distributions. These are called
the modes of an optical fiber.
From the expression of the phase matching condition we find that as d
increases, the number of rays accepted by the optical fiber
increases and as d decreases the number of rays decreases.
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116. Single Mode Propgn in Fibers
Since the dispersion is due to presence of multiple rays (modes), if only
one ray is made to propagate inside the fiber, there is no dispersion.
So if we take a value of d small enough such that it satisfies the
phase condition only the lowest value of m, then only one mode will
propagate inside the fiber.
The lowest value of m corresponds to the ray traveling along
the axis of the fiber.
In fact this ray does not have any constraint on the size of the
fiber etc, as it does not really go through the total internal
reflection at the core cladding boundary. This ray therefore
always propagates.
The optical fiber in which only one ray travels along the axis
of fiber is called the single mode optical fiber.
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117. 124
• Ray 1 and ray 2 are parallel rays. The phase-front AE is common to both Ray
1 and Ray 2. The phase-front DB is common to Ray 2 and Ray2 (BF).
•The Ray 2 is thus common to both the phase-fronts. Hence for a sustained
constructive interference, the distance between these two phase-fronts must be
multiples of 2π ie, the the phase change undergone by Ray 1 in travelling
distance s1 and the Ray 2 in travelling s2 must be 0 or integral multiples
of 2π.
•The total phase change undergone by Ray 1 in travelling s1 and Ray 2 in
travelling s2 is given by ϕ1 and ϕ2 respectively
•For a sustained constructive interference, both ϕ1 and ϕ2 must have a phase
difference of either 0 or integral multiples of 2π. That is, for an integer m
(=0,1,2,3,…) the following condition must be satisfied
118. 125
•If δ is the phase change undergone in each TIR of Ray 1, then the total phase
change undergone by Ray 1 in travelling s1 is given by
Φ 1 = 2πn1S1 +2δ
λ
Where n1 = refractive index of core; λ = Wavelength of the light in the core.
The phase change undergone by Ray 2 in travelling s2 is given by
Φ 2= 2πn1S2
λ
119. 126
•For a sustained constructive interference, both ϕ1 and ϕ2 must have a phase
difference of either 0 or integral multiples of 2π. That is, for an integer m
(=0,1,2,3,…) the following condition must be satisfied:
Φ 1 - Φ 2 = 2mπ
2πn1 (S1 - S2 ) + 2δ = 2mπ
λ
2πn1 d sinθ+ δ = mπ
λ
120. 127
The significance of the equation is that only those rays, which are incident
on the tip of the fiber at angles such that their angle of refraction in the
core satisfies equation can successfully travel along the fiber.
Since ‘m’ can take only discrete integral values, the value of angle θ is
also discrete.
This suggests that there are only some discrete launching angles within
the acceptance cone (N.A. cone) for which the rays can propagate inside
the fiber core .
121. 128
•The acceptance cone can no longer be assumed as a solid cone of rays, launched
at all possible angles (smaller than acceptance angle), but is composed of discrete
annular conical rings of rays which are launched at the tip of the fiber core at angles
which satisfy equation.
•Thus light can only be launched at certain discrete angles within the N.A. cone
leading to a further decrease in the light gathering efficiency of the optical fiber. Any
ray that is not launched at these discrete angles will not propagate inside the optical
fiber.
• This discretization in the values of launching angles lead to formation modes in an
optical fiber, which are different patterns of light intensity distribution around the
axis of the core
122. 129
•Only the fundamental zero-order mode is transmitted in a single-mode fiber.
•The SMF propagates only the fundamental mode, modal dispersion (the
primary cause of pulse overlap) is eliminated and the bandwidth is much higher
than that of a MMF.
•This simply means that pulses can be transmitted much closer together in time
without overlap. Because of this higher bandwidth, single-mode fibers are used
in all modern long-range communication systems.
•Typical core diameters are between 5 and 10 µm.
SMF
123. 130
•Single-mode fiber is characterized by the wavelength cutoff value, which is
dependent on core diameter, NA and wavelength of operation. Below the
cutoff wavelength, higher-order modes may also propagate, which
changes the fiber’s characteristics.
The actual number of modes that can be propagated through a fiber depends
on the core diameter, the numerical aperture and the wavelength of the
light being transmitted. These may be combined into the normalized frequency
parameter or V number
•V - Parameter = (2πa/) n1
2-n2
2 = 2aπ NA/ .
124. 131
•. The condition for single-mode operation is that:
V ≤ V Cutoff =2.405
•cutoff wavelength. This is the wavelength below which the fiber will allow
propagation of multiple modes and can be expressed as:
≥ Cutoff
A fiber is typically chosen with a cutoff wavelength slightly below the desired
operating wavelength. For lasers typically used as sources (with output
wavelengths between 850 and 1550 nm), the core diameter of a single-mode
fiber is in the range of 3 to 10 µm.
132. FIBRE PARAMETERS.
140
Numerical Aperture (NA) = Sin a
= n1
2-n2
2
V - Parameter = (2πa/) n1
2-n2
2 = 2πa NA/ .
Number of Modes, N =V2/2 for SI MM
Number of Modes, N =V2/4 FOR GI MM
At Cut Off for Single Mode Fibre,
V = Vc = 2.405. c = 2aNA/2.405.
133. The V number determines the fraction of the optical
power in a certain mode which is confined to the fiber
core. For single-mode fibers, that fraction is low for
low V values (e.g. below 1), and reaches ≈ 90% near the
single-mode cut-off at V ≈ 2.405.
There is also the so-called Marcuse equation for estimating
the mode radius of a step-index fiber from the V number;
A low V number makes a fiber sensitive to micro-bend
losses and to absorption losses in the cladding.
However, a high V number may increase scattering losses
in the core or at the core–cladding interface.
141
134. Determine the cutoff wavelength for a step index
fiber to exhibit single-mode operation when the core
refractive index and radius are 1.46 and 4.5 μm,
respectively, with the relative index difference being
0.25%
Ans. 1214 nm
142
135. Q A multimode step index fiber with a core diameter of 80
μm and a relative index difference of 1.5% is operating at
a wavelength of 0.85 μm. If the core refractive index is
1.48, estimate
(a) the normalized frequency for the fiber
(b) the number of guided modes.
Ans V = 75.8
Ms = 2873
V - Parameter = (2πa/) n1
2-n2
2 = 2πa NA/ .
NA = n1 (2 ) 1/2
NA = n0.sin a = (n1²-n2² )1/2
143
136. Q Estimate the maximum core diameter for single-
mode operation. Given optical fiber with the relative
RI difference 1.5% and core refractive index 1.48
operating at a wavelength of 0.85 μm. Further,
estimate the new maximum core diameter for single-
mode operation when the relative refractive index
difference is reduced by a factor of 10.
Ans= 2.6 μm and 8 μm
144
138. It is clear from last Example that in order to obtain single-mode
operation with a maximum V number of 2.4, the single-mode fiber
must have a much smaller core diameter than the equivalent
multimode step index fiber
However, it is possible to achieve single-mode operation with a
slightly larger core diameter, albeit still much less than the
diameter of multimode step index fiber, by reducing the relative
refractive index difference of the fiber.*
Both these factors create difficulties with single-mode fibers. The
small core diameters pose problems with launching light into the
fiber and with field jointing, and the reduced relative refractive
index difference presents difficulties in the fiber fabrication
process
146
139. 147
Q.When the mean optical power launched into an 8 km length
of fiber is 120 μW, the mean optical power at the fiber output
is 3 μW.
Determine:
(a) The overall signal attenuation or loss in decibels
through the fiber assuming there are no connectors or
splices.
(b) The signal attenuation per kilometer for the fiber.
(c) The overall signal attenuation for a 10 km optical link
using the same fiber with splices at 1 km intervals, each
giving an attenuation of 1 dB;
(d) The numerical input/output power ratio in (c).
143. FIRST TASK
Explore the internet for laser applns
and answer these questions.
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144. ASSIGNMENT - 1
Therefore, 06 gps.
Write Mil paper of at least 1000 words on each of the statements
below.
ASSIGNMENT QUESTIONS
1. How will laser affect the way future wars will be fought?
2. Why is laser considered a solution looking for a problem to be solved?
3. Why laser can be considered as force multiplier in advancement of
Science and technology?
4. How Laser has increased our day to day comfort and luxury directly as
well as indirectly?
5. Countries which are leaders in Laser and Nanotechnology will always be
economically ahead of other countries.
6. How can lasers be used as a potent weapon in Star Wars?
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