Addis Ababa Science and Technology University
College of Electrical & Mechanical Engineering
Electrical & Computer Engineering Department
Optics and Optical Communication (EEEg-4302)
Chapter Two
Fundamentals of Optical Fiber Waveguides
Fundamentals of Optical Fiber Waveguides
Outline:
 Geometric Optics
 Total Internal Reflection
 Acceptance Angle
 Numerical Aperture
 Electromagnetic Mode Theory for Optical Propagation
 Optical Fiber Types and Modes
 Signal Degradation in Optical Fibers
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Chapter 2 –Fundamentals of Optical Fiber Waveguides
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Refractive Index
 In any dielectric medium, the speed of light becomes
c
v
n
 The factor n is the index of refraction (or refractive index) of
the medium.
e.g. For air and gases, v ~ c, so that n ~ 1.
 At optic frequencies, the refractive index of water is 1.33.
e.g. Glass has many compositions, each with a slightly
different n.
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Chapter 2 –Fundamentals of Optical Fiber Waveguides
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Interference – Effects
 Interference in the voice channels causes crosstalk
 A subscriber hears interference in the background due to an
undesired transmission
 Interference in the control channels causes error in digital
signaling which causes
 Missed calls
 Blocked calls
 Dropped calls
 Interference is more severe in urban areas, due to the
greater RF noise floor and the large number of base
stations and mobiles
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Basic optical laws and definitions
•A light ray is reflected and refracted when it encounters the
boundary between two different transparent mediums.
•Light reflects or refracts depending on the angle at which it
strikes a surface and the difference in refractive index of
the two materials.
Reflection is the change in direction of a light wave at an
interface between two dissimilar media so that the wave
front returns into the medium from which it originated.
•Refraction is the change in direction of a light wave due to
a change in its velocity. This is most commonly seen
when a wave passes from one medium to another.
•The phenomenon of total internal reflection, responsible for
guiding of light in optical fibers.
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Chapter 2 –Fundamentals of Optical Fiber Waveguides
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Properties of Light
Law of Reflection
The angle of Incidence = The angle of reflection
Law of Refraction •
Light beam is bent towards the normal when passing into a
medium of higher refractive index.
•
Light beam is bent away from the normal when passing into
a medium of lower refractive index.
Index of Refraction –
n = Speed of light in a vacuum / Speed of light in a medium
Inverse square law - Light intensity diminishes with square of
distance from source.
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Snell’s Law
 When a ray is incident on the interface between two
dielectrics of different refractive indices (e.g. glass-air),
reflection and refraction occur.
 The angle of incidence θ1 and the angle of refraction θ2 are
related to each other and to the refractive indices of the
dielectrics by Snell’s law of refraction as:
n1 sin 1  n2 sin  2
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Chapter 2 –Fundamentals of Optical Fiber Waveguides
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Critical Angle
 For n1 > n2, the angle of refraction θ2 is always greater than
the angle of incidence θ1.
 When the angle of refraction θ2 is 900, the refracted ray
emerges parallel to the interface between the media.
 This is the limiting case of refraction and the angle of
incidence is known as the critical angle θc .
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Total Internal Reflection
 At angles of incidence θ > θc , the light is totally reflected
back into the incidence higher refractive index medium.
 This is known as total internal reflection.
Example : n1  1.44, n2  1, then  c  sin 1 (1 / 1.44)  440
For total internal reflection : 1   c  1  440
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Chapter 2 –Fundamentals of Optical Fiber Waveguides
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Reflection and Refraction of Light
Medium 1
1
1
2
n1
Boundary
1
n2
Incident
ray
2
1
1 1
n1
Reflected
ray
Medium 2
n1 < n2
2
Refracted
n2 ray
n1 > n2
Using the Snell's law at the boundary we have:
n1 sin 1 = n2 sin 2
or
n1 cos 1 = n2 cos 2
1 = The angle of incident
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Total Internal Reflection
n1 > n2
• As 1 increases (or 1
decreases) then there is no reflection
n2
1
• The incident angle
1 =  c = Critical Angle
c
n1
• Beyond the critical angle, light ray
n1 > n2
becomes totally internally reflected
When 1 = 90o (or c = 0o)
n1 sin 1 = n2
1<c
n2
1
1>
1  n2

Thus the critical angle c  sin  
 n1 
n1
c
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Light Ray Guiding Condition
 Light ray that satisfies total internal reflection at the interface
of the higher refractive index core and the lower refractive
index cladding can be guided along an optical fiber.
Example: Under what condition will light be trapped inside the
fiber core if n1 = 1.46 and n2 = 1.44?
 c  sin 1 (n2 / n1 )  sin 1 (1.44 / 1.46)  80.50 ,    c    80.50
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Optical Fiber Structures
 A typical bare fiber consists of a core, a cladding and a
polymer jacket (buffer coating).
 The polymer coating is the first line of mechanical
protection.
 The coating also reduces the internal reflection at the
cladding, so light is only guided by the core.
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Chapter 2 –Fundamentals of Optical Fiber Waveguides
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Optical Fiber Structures
An optical fibre is a very thin strand of silica glass in geometry
quite like a human hair. In reality it is a very narrow, very long
glass cylinder with special characteristics.
The basic structure of an optical fiber consists of three parts;
the core, the cladding, and the coating or buffer as it is shown
in figure.
Core
•It is a cylindrical rod of dielectric
material which conducts no electricity.
Light propagates mainly along the
core of the fiber.
•The core is mainly made of glass, in
low-and medium-loss fibers.
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Cladding
•Made of a dielectric material with an index of refraction less than that
of the core material.
•The cladding is generally made of glass or plastic. The cladding
performs the following functions:
 Reduces loss of light from the core into the surrounding air
Reduces scattering loss at the surface of the core
Protects the fiber from absorbing surface contaminants
 Adds mechanical strength
Coating or buffer
•The coating or buffer is a layer of material used to protect an optical
fiber from physical damage. The material used for a buffer is a type of
plastic.
•The buffer is elastic in nature and prevents abrasions.
•It also prevents the optical fiber from scattering losses caused by microbends.
•Micro-bends occur when an optical fiber is placed on a rough and
distorted surface.
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Silica Optical Fibers
 Both the core and the cladding are made from a type of
glass known as silica (SiO2) which is almost transparent in
the visible and near-IR.
 In the case that the refractive index changes in a “step”
between the core and the cladding, the fiber structure is
known as step-index fiber.
 The higher core refractive index (~ 0.3% higher) is typically
obtained by doping the silica core with germanium dioxide
(GeO2).
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Acceptance Angle
 The maximum angle to the axis of the core at which light
may enter the fiber in order to be propagated by total
internal reflection is referred to as acceptance angle.
 Ray A incident at the critical angle θc at the core-cladding
interface enters the fiber core at an angle θa to the fiber
axis and is refracted at the air-core interface.
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Acceptance Angle………
 Any rays which are incident into the fiber core at an angle
θ> θa have an incident angle less than θc at the core-cladding
interface.
 These rays will NOT be totally internal reflected, thus
eventually loss to radiation (at the cladding-jacket interface).
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Chapter 2 –Fundamentals of Optical Fiber Waveguides
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Numerical Aperture
 Fiber numerical aperture (NA) characterizes the fiber’s
ability to gather light from a source and guide the light.
 The numerical aperture (NA) is defined as:
NA  na sin  a
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Numerical Aperture………..
 We can relate the acceptance angle θa and the refractive
indices of the core n1, cladding n2 and air na as follows:
Using Snell' s law at the air - core interface, we have :
na sin  a  n1 sin  c
Launching the light from air (na  1), we get :
sin  a  n1 sin  c , but  c 

2
 c


 sin  a  n1 sin    c   n1 cos  c , but cos  c  1  sin 2  c

2
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Numerical Aperture………..
n2
 sin  a  n1 1  sin  c , but sin  c 
n1
2
2
n2
2
2
2
2
 sin  a  n1 1  2  n1  n2  NA  sin  a  n1  n2
n1
 The NA may also be given in terms of the relative refractive
index difference Δ between the core and the cladding
which is defined as:
n1  n2
n1 n2


, for   1
2
n1
2n1
2
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Chapter 2 –Fundamentals of Optical Fiber Waveguides
21
Numerical Aperture………..
 The NA is then given by:
NA  n1 2
 The numerical aperture is a very useful measure of the
light-collecting ability of a fiber.
 It is independent of the fiber core diameter and will hold for
diameters as small as 8μm.
 However, for smaller diameters , it is inapplicable as the
geometric optics approach is invalid.
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Chapter 2 –Fundamentals of Optical Fiber Waveguides
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Ray Propagation in Fibre - Bound Rays
 > c,  > max
5
2
c
3
2
2
1

a
4
Core n1
Air (no =1) Cladding n2
At high frequencies f , since a > light wavelength , lightt launched
From Snell’s Law:
into the fibre core would propagate as plane TEM wave with vp =
/1
n0 sin  = n1 sin (90 - )
At high frequencies f , since a < , light launched into the fibre
will propagate within the cladding, thus unbounded and
 = max when  = c
unguided plane TEM wave with vp = /2
Therefore, we have n2k0 = 2 <  < 1 = n1k0
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Ray Propagation in Fibre - contd.
n0 sin max = n1 (1 - sin2 c)0.5
Or
Since
1  n2

c  sin  
 n1 
  n 2 
n0 sin  max  n1 1   2  
  n1  
Then

 n12  n22

0.5
0.5


n12

2 0.5
 n2
 Numerical Aperture ( NA)
NA determines the light gathering
capabilities of the fibre
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Ray Propagation in Fibre - contd.
Therefore
n0 sin max = NA
1  NA 




sin
Fiber acceptance angle max
 n0 
Note n1  n2   Relative refractive index difference
n1
Thus
NA  n1 (2) 0.5
0.14< NA < 1
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Lens Coupling to Fiber End Faces
 By measuring the output couple ray cone angle, we can
measure the fiber acceptance angle.
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Chapter 2 –Fundamentals of Optical Fiber Waveguides
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Large NA Fibers
 Developing ways for fiber to collect light efficiently was an
important early step in developing practical fiber optic
communications.
 It seems logical to have optical fibers with NA as large as
possible with as large Δ as possible in order to couple
maximum amount of light into the fiber.
 Soon, we will find out that such large NA fibers tend to be
“multimode” and are unsuitable for high-speed
communications because of a limitation known as modal
dispersion.
 Relatively small NA fibers are therefore used for high-speed
optical communication systems.
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Chapter 2 –Fundamentals of Optical Fiber Waveguides
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Typical fiber NA
 Silica fibers for long range transmission are designed to
have numerical apertures from about 0.1 to 0.3.
 The low NA makes coupling efficiency tend to be poor, but
turns out to improve the fiber’s bandwidth.
 Plastic fibers are restricted to short lengths because of the
relatively high attenuation in plastic materials.
 Plastic optical fibers (POFs) are designed to have high
numerical apertures (typically, 0.4 – 0.5) to improve coupling
efficiency and so partially offset the high propagation losses.
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Chapter 2 –Fundamentals of Optical Fiber Waveguides
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Examples On Geometric Optics
Example-1:
What is the fiber acceptance angle when n1=1.46 and n2=1.44?
Solution:
 n2 
1  1.44 
0
0
0
 c  sin    sin 
  80.5   c  90   c  9.5
 1.46 
 n1 
1
sin  a  n1 sin  c , (taking na  1)
  a  sin 1 (n1 sin  c )  sin 1 (1.46 sin 9.50 )  140
 The acceptance angle  a  140
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Examples On Geometric Optics……...
Example-2:
A silica optical fiber with a core diameter large enough to
be considered by ray theory analysis has a core refractive
index of 1.50 and a cladding refractive index of 1.47.
Determine:
a. the critical angle at the core–cladding interface
b. the NA for the fiber
c. the acceptance angle in air for the fiber.
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Examples On Geometric Optics……...
Solution:
a. The critical angle  c at the core - cladding interface
is given by :
 n2 
1  1.47 
0
 c  sin    sin 
  78.5
 1 .5 
 n1 
1
b. The NA is :
NA  n1  n2  1.52  1.47 2  2.25  2.16  0.3
2
2
c. The acceptance angle in air  a is given by :
 a  sin 1 ( NA)  sin 1 (0.3)  17.40
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Examples On Geometric Optics……...
Example-3:
A typical relative refractive index difference for an optical
fiber designed for long distance transmission is 1%. When
the core index is 1.46, calculate:
a. the numerical aperture (NA) for the fiber
b. the critical angle at the core–cladding interface within
the fiber
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Examples On Geometric Optics……...
Solution:
a. With   0.01, the NA is given by :
NA  n1 2  1.46 0.02  0.21
b. The relative refractive index difference  is given by :

n1  n2
n
n
 1  1  1  1    1  0.01  0.99
n1
n2
n2
The critical angle  c at the core - cladding interface is then :
 n2 
 c  sin    sin 1 (0.99)  81.90
 n1 
1
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Chapter 2 –Fundamentals of Optical Fiber Waveguides
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Exercises On Geometric Optics
Exercise-1:
Consider a multimode silica fiber that has a core refractive
index n1=1.48 and a cladding refractive index n2=1.46.
Find:
a. the critical angle
b. the numerical aperture
c. the acceptance angel
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Chapter 2 –Fundamentals of Optical Fiber Waveguides
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Exercises On Geometric Optics…….
Exercise-2:
Consider a multimode fiber that has a core refractive index
of and a core-cladding index difference of 2%.
Find:
a. the numerical aperture (NA)
b. the acceptance angle
c. the critical angle
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Chapter 2 –Fundamentals of Optical Fiber Waveguides
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The End!
Any
Question?
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Chapter 2 –Fundamentals of Optical Fiber Waveguides
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