Lecture 3
Light Propagation In
Optical Fiber
By:
Mr. Gaurav Verma
Asst. Prof.
ECE, NIEC
06/08/14
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Introduction
 An optical fiber 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. When light enters one end of the fiber it
travels (confined within the fiber) until it leaves the fiber at the other
end. Two critical factors stand out:


Very little light is lost in its journey along the fiber
Fiber can bend around corners and the light will stay within it and be
guided around the corners.
 An optical fiber consists of two parts: the core and the cladding. The
core is a narrow cylindrical strand of glass and the cladding is a tubular
jacket surrounding it. The core has a (slightly) higher refractive index
than the cladding. This means that the boundary (interface) between
the core and the cladding acts as a perfect mirror. Light traveling along
the core is confined by the mirror to stay within it - even when the fiber
bends around a corner.
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BASIC PRINCIPLE
When a light ray travelling in one material hits a different material and
reflects back into the original material without any loss of light, total internal
reflection is said to occur.
Since the core and cladding are constructed from different compositions of
glass, theoretically, light entering the core is confined to the boundaries of
the core because it reflects back whenever it hits the cladding.
For total internal reflection to occur, the index of refraction of the core must
be higher than that of the cladding, and the incidence angle is larger than
the critical angle.
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What Makes The Light Stay in Fiber
 Refraction
 The light waves spread out along its beam.
 Speed of light depend on the material used called refractive index.
 Speed of light in the material = speed of light in the free
space/refractive index
 Lower refractive index  higher speed
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The Light is Refracted
Lower Refractive index Region
This end travels
further than the
other hand
Higher Refractive index Region
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Refraction
 When a light ray encounters a boundary separating two
different media, part of the ray is reflected back into the
first medium and the remainder is bent (or refracted) as it
enters the second material. (Light entering an optical fiber
bends in towards the center of the fiber – refraction)
Refraction
LED or
LASER
Source
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Reflection
 Light inside an optical fiber bounces off the cladding reflection
Reflection
LED or
LASER
Source
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Critical Angle
 If light inside an optical fiber strikes the cladding too steeply,
the light refracts into the cladding - determined by the critical
angle. (There will come a time when, eventually, the angle of
refraction reaches 90o and the light is refracted along the
boundary between the two materials. The angle of incidence
which results in this effect is called the critical angle).
n1Sin X=n2Sin90o
Critical Angle
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Angle of Incidence
 Also incident angle
 Measured from perpendicular
 Exercise: Mark two more incident angles
Incident Angles
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Angle of Reflection
 Also reflection angle
 Measured from perpendicular
 Exercise: Mark the other reflection angle
Reflection Angle
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Reflection
Thus light is perfectly reflected at an interface between two
materials of different refractive index if:
 The light is incident on the interface from the side of
higher refractive index.
 The angle θ is greater than a specific value called the
“critical angle”.
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Angle of Refraction
 Also refraction angle
 Measured from perpendicular
 Exercise: Mark the other refraction angle
Refraction Angle
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Angle Summary
 Three important angles
 The reflection angle always equals the incident angle
Refraction Angle
Incident Angles
Reflection Angle
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Index of Refraction
 n = c/v
c = velocity of light in a vacuum
v = velocity of light in a specific
medium
 light bends as it passes from one medium to
another with a different index of refraction
air, n is about 1
glass, n is about 1.4
Light bends
away from
normal - higher
n to lower n
Light bends in towards normal lower n to higher n
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Snell’s Law
 The angles of the rays are measured with respect to the
normal.
 n1sin 1=n2sin 2
 Where
 n1 and n2 are refractive index of two materials
 1and 2 the angle of incident and refraction respectively
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Snell’s Law
 The amount light is bent by refraction is given by
Snell’s Law:
n1sin1 = n2sin2
 Light is always refracted into a fiber (although there
will be a certain amount of Fresnel reflection)
 Light can either bounce off the cladding (TIR) or refract
into the cladding
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Snell’s Law
Normal
Refraction
Angle(2)
Lower Refractive index(n2)
Ray of light
Higher Refractive index(n1)
Incidence
Angle(1)
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Snell’s Law (Example 1)
 Calculate the angle of refraction at the air/core interface
 Solution - use Snell’s law: n1sin1 = n2sin2
1sin(30°) = 1.47sin(refraction)
 refraction = sin-1(sin(30°)/1.47)
 refraction = 19.89°
nair = 1
ncore = 1.47
ncladding = 1.45
incident = 30°
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Snell’s Law (Example 2)
 Calculate the angle of refraction at the core/cladding interface
 Solution - use Snell’s law and the refraction angle from Example 3.1
 1.47sin(90° - 19.89°) = 1.45sin(refraction)
 refraction = sin-1(1.47sin(70.11°)/1.45)
 refraction = 72.42°
nair = 1
ncore = 1.47
ncladding = 1.45
incident = 30°
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Snell’s Law (Example 3)
 Calculate the angle of refraction at the core/cladding interface for
the new data below
 Solution: 1sin(10°) = 1.45sin(refraction(core))
 refraction(core) = sin-1(sin(10°)/1.45) = 6.88°
 1.47sin(90°-6.88°) = 1.45sin(refraction(cladding))
 refraction(cladding) = sin-1(1.47sin(83.12°)/1.45)
= sin-1(1.0065) = can’t do
light does not refract into
cladding, it reflects back
into the core (TIR)
nair = 1
ncore = 1.47
ncladding = 1.45
incident = 10°
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Critical Angle Calculation
 The angle of incidence that produces an angle of refraction
of 90° is the critical angle
 n1sin(c) = n2sin(°)
n1 = Refractive index of the core
 n1sin(c) = n2
n2 = Refractive index of the cladding
-1
 c = sin (n2 /n1)
 Light at incident angles
greater than the critical
angle will reflect back
into the core
Critical Angle, c
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NA Derivation
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Acceptance Angle and NA
 The angle of light entering
a fiber which follows the
critical angle is called the
acceptance angle, 
 = sin-1[(n12-n22)1/2]
 Numerical Aperature (NA)
describes the lightgathering ability of a fiber
n1 = Refractive index of the core
n2 = Refractive index of the cladding
Acceptance Angle, 
NA = sin
Critical Angle, c
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Numerical Aperture
 The Numerical Aperture is the sine of the largest angle contained
within the cone of acceptance.
 NA is related to a number of important fiber characteristics.
 It is a measure of the ability of the fiber to gather light at the input
end.
 The higher the NA the tighter (smaller radius) we can have bends
in the fiber before loss of light becomes a problem.
 The higher the NA the more modes we have, Rays can bounce at
greater angles and therefore there are more of them. This means
that the higher the NA the greater will be the dispersion of this
fiber (in the case of MM fiber).
 Thus higher the NA of SM fiber the higher will be the attenuation
of the fiber
Typical NA for single-mode fiber is 0.1. For multimode, NA is between 0.2
and 0.3 (usually closer to 0.2).
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Acceptance Cone
 There is an imaginary cone of acceptance with an angle 
 The light that enters the fiber at angles within the
acceptance cone are guided down the fiber core
Acceptance Angle, 
Acceptance Cone
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Acceptance Cone
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Formula Summary
 Index of Refraction
Snell’s Law
c
n=
v
n 1 sin θ 1 =n2 sin θ 2
θ c = sin
Critical Angle
Acceptance Angle
Numerical Aperture
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−1
n2
n1
()
α= sin− 1 (√
n12− n 22)
NA=sin α= √
n12 − n22
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Practice Problems
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Practice Problems (1)
 What happens to the light which approaches the
fiber outside of the cone of acceptance? The angle
of incidence is 30o as in Fig.1 (calculate the angle of
refraction at the air/core interface, r/ critical angle,
c/ incident angle at the core/cladding interface, i/)
does the TIR will occur?
Practice Problems (2)
Calculate:
angle of refraction at the
air/core interface, r
critical angle , c
incident angle at the
core/cladding interface , i
Will this light ray propagate
down the fiber?
core/cladding interface
air/core interface
nair = 1
ncore = 1.46
ncladding = 1.43
incident = 12°
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Answers:
r = 8.2°
c = 78.4°
i = 81.8°
light will propagate
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Refractive Indices and Propagation Times
6
5
4
3
2
1
0
Vacuum
Air
Water
Fused Silica
Belden Cable (RG59/U)
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Refractive
Index
Propagation
Time (ns/m)
1
1.003
1.333
1.458
N/A
3.336
3.346
4.446
4.863
5.551
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Propagation Time Formula
 Metallic cable propagation delay
 cable dimensions
 frequency
 Optical fiber propagation delay
 related to the fiber material
formula
t = Ln/c
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t = propagation delay in seconds
L = fiber length in meters
n = refractive index of the fiber core
c = speed of light (2.998 x 108 meters/second)
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Temperature and Wavelength
 Considerations for detailed analysis
 Fiber length is slightly dependent on temperature
 Refractive index is dependent on wavelength
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