Lecture Notes

Fiber Optics
• Light waves has large information carrying capacity.
• For long distance communication using light waves all we need is a medium for transmission.
• Optical Fiber are treated as medium for information carrying light waves.
• Light waves communication using glass fibers can transmit information with high speed tera bytes per
second.
• Which is 15 million time stronger than the simultaneous telephonic conversations.
• Optical fibers are being one of the extremely important technological achievements of 20th centaury.
• TIR (Total internal reflection that plays very important role in optical fibers).

• The light beam pulses are then fed into a fiber–optic cable where they are transmitted over long distances.
• At the receiving end, a light sensitive device known as a photocell or light detector is used to detect the
light pulses.
• This photo cell or photo detector converts the light pulses into an electrical signal.
• The electrical pulses are amplified and reshaped back into digital form.
 Fiber Optics
Classification of fibers
1. Based on the refractive index of core medium, optical fibers are classified into two
categories.
a) Step index fiber
b) Graded index fiber
2. Based on the number of modes of transmission, optical fibers are classified into two
categories
a) Single mode fiber
b) Multi mode fiber
3. Based on the material used, optical fibers are may broadly classified into four categories
a) All glass fibers
b) All plastic fibers
c) Glass core with plastic cladding fibers
d) Polymer clad silica fibers.
Optical Fiber Cable
Fiber Optic Cable consists of four parts.
 Core
 Cladding
 Buffer
 Jacket

Core Buffer
The core of a fiber cable is a cylinder of plastic/glass that runs all The main function of the buffer is to protect the
along the fiber cable’s length, and offers protection by cladding. fiber from damage and thousands of optical fibers
The diameter of the core depends on the application used. Due to arranged in hundreds of optical cables. These
internal reflection, the light travelling within the core reflects from bundles are protected by the cable’s outer covering
the core, the cladding boundary. The core cross section needs to be that is called jacket.
a circular one for most of the applications.

Cladding JACKET
Cladding is an outer optical material that protects the core. The Fiber optic cable’s jackets are available in different
main function of the cladding is that it reflects the light back into colors that can easily make us recognize the exact
the core. When light enters through the core (dense material) color of the cable we are dealing with. The color
into the cladding(less dense material), it changes its angle, and yellow clearly signifies a single mode cable, and
then reflects back to the core. orange color indicates multi mode.
Fiber Optics
Refraction of light
•As a light ray passes from one transparent medium to another, it changes direction; this phenomenon is called
refraction of light. How much that light ray changes its Direction depends on the Refractive index of the
Mediums.

Refractive Index
• Refractive index is the speed of light in a vacuum (abbreviated c,
c=299,792.458km/second) divided by the speed of light in a material
(abbreviated v). Refractive index measures how much a material refracts light.
Refractive index of a material, abbreviated as n, is defined as
• n=c/v
Snell’s Law
When light passes from one transparent material to another, it bends according to
Snell's law which is defined as:
n1sin(θ1)=n2sin(θ2)
where:n1 is the refractive index of the medium the light is leaving
θ1 is the incident angle between the light beam and the normal (normal is 90° to the
interface between two materials)
n2 is the refractive index of the material the light is entering
θ2 is the refractive angle between the light ray and the normal
Fiber Optics
Critical angle
• The critical angle can be calculated from Snell's law, putting in an angle of 90° for the angle of the refracted
ray θ2.
• This gives θ1: Since θ2 =90°
• So sin(θ2) =1
• Then θc = θ1 = sin-1(n2/n1)

n2 < n1
Fiber Optics
• Principle
• The basic principle of optical fiber in the transmission of optical signal is total internal reflection.

Total internal reflection

o When the light ray travels from denser medium to rarer medium
the refracted ray bends away from the normal.
o When the angle of incidence is greater than the critical angle, the
refracted ray again reflects into the same medium. This
phenomenon is called total internal reflection.
o The refracted ray bends towards the normal as the ray travels
from rarer medium to denser medium. The refracted ray bends
away from the normal as it travels from denser medium to rarer
medium.

𝒏𝟏𝒔𝒊𝒏𝜽𝟏 = 𝒏𝟐𝒔𝒊𝒏𝜽𝟐 𝜽𝟐 = 𝟗𝟎𝟎 𝒏𝟏𝒔𝒊𝒏𝜽𝟏𝒄 = 𝒏𝟐

Fiber Optics
 An optical fiber is thin transparent, flexible strands that consist of core surrounded by cladding.
 The core and cladding of optical fiber made of same material, a type of glass called silica and they are differ
only in there refractive index.
 Refractive index is number showing the optical properties of material i.e. how strong material resist to
transmission of light.
 Refractive index n is given by 𝑛 = , v is velocity of light in medium, c velocity of light in vacuum.
 Core has refractive index n1 and cladding has different refractive index n2.

 There is abruptly the refractive

index changes across the fiber.
 That is why this called step index
fiber.

 The difference in the refractive

index is achieved by doping silica
with different dopants.
 There is abruptly the refractive
index changes across the fiber.
 That is why this called step index
fiber.
Fiber Optics
 Third layer of optical fiber is applied to
protect the entire structure.
 Coating is made up of different material than
core and cladding, this act as first line of
defense for fragile core-cladding structure,
without it installer and user can not work
with optical fiber.
 To achieve the total internal reflection in the
optical fiber, this is possible only when core
refractive index n1 is greater than the
cladding refractive index n2.
 Under this condition light can travel inside
the core not along its central pathway, but
also at various angle to centerline, without
leaving the core.
 Optical fiber can store light even if its bent.
Critical incident angle and critical propagation angle
 Critical propagation angle 𝛼 , is the angle beam makes with center line of optical fiber.
 The critical incident angle 𝜃 , is angle the beam makes with line perpendicular to optical boundary
between core and cladding.
 Consider the triangle ABC, 𝛼 = 90 − 𝜃 .
 Why is the critical propagation angle, 𝛼 is important?

o The refractive index of core and cladding of silica fiber

are 1.48 and 1.46 respectively, under what condition
will light trapped inside the core? (𝜃 =80.57 )

 Suppose beam travel in optical fiber at, 𝛼 = 10 > 𝛼 , Means 𝜃 = 80 < 𝜃

 Condition for total internal reflection is violated.
 Incident beam will divide into two: a reflected beam, which will be saved, refractive beam which will be lost.
 What happened in this condition?
 𝑆𝑖𝑛𝜃 = ⁄ , 𝑠𝑖𝑛𝑐𝑒 𝛼 = 90 − 𝜃 , 𝑆𝑖𝑛𝜃 = 𝐶𝑜𝑠𝛼
 ⁄ = 𝐶𝑜𝑠𝛼 , Sin𝛼 = 1 − 𝐶𝑜𝑠𝛼
𝑛2
𝛼 = 𝑆𝑖𝑛 1−
𝑛1
Acceptance angle
 How we can direct the beam so that it does
indeed at or below critical propagation angle?
 The beam at angle 𝜃 is incident beam and the
beam at angel 𝛼 is the launched one, which is
the refracted beam with respect to the gap-core
interface.
 The relation between 𝜃 and 𝛼 can be derived
using Snell’s law.
 𝑛 𝑠𝑖𝑛𝜃 = 𝑛 𝑠𝑖𝑛𝛼
 Where 𝜃 is spatial angle.
 If the gap between the source and fiber is air
 Light will be saved inside the fiber if it comes from a
than 𝑛 is very close to 1.
light source bounded by the cone 2𝜃 .
 𝑠𝑖𝑛𝜃 = 𝑛 𝑠𝑖𝑛𝛼
 That is why we call angle 2𝜃 as acceptance angle.
 To save all light inside the a fiber all rays must
 If rays coming at an angle exceeding the acceptance
propagate at critical angle 𝛼 or less. In order to
angle 𝜃 outside the fiber.
maintain the light inside the fiber at this angle
 The ray will travel in the fiber at an angle exceeding the
we have to direct it from outside at angle 𝜃 or
critical propagation angle 𝛼 .
less.
 Ray is not with in the acceptance cone, it will be lost
while traveling in the fiber.
What is the acceptance angle for the fiber with n1= 1.48 and n2=1.46?
Numerical Aperture
Numerical aperture NA is
𝑁𝐴 = 𝑠𝑖𝑛𝜃
What is the numerical aperture for the fiber with
n1= 1.48 and n2=1.46?

 Numerical aperture describe the ability of

an optical fiber to gather light from a source
and than the ability to preserve or save this  Fiber optics technology operates not with the refractive
light inside the fiber because of the total indexes of the core and the cladding, but with there
internal reflection. difference ∆𝑛.
 It represent the condition of total internal  ∆𝑛 = 𝑛1 − 𝑛2
reflection inside the optical fiber, a  This value is always positive, its very common to use the
condition that is absolutely necessary if we relative difference of the refractive indexes
want to use the optical fiber for ∆= ⁄ , where n is the average refractive index given
communication. by

𝟐 𝟐
∆𝒏 𝟐
𝑵𝑨 = 𝒏𝟏 − 𝒏𝟐 = 𝒏𝟏 − 𝒏𝟐 𝒏𝟏 + 𝒏𝟐 = (∆𝒏)(𝟐𝒏) = 𝟐 𝒏 𝑵𝑨 = 𝒏 𝟐∆
𝒏
This shows that n1 and n2 are not important in themselves but only there average and relative difference.
By varying these two parameters we able to change NA over a wide range(0.1 to 0.3 for silica fiber).
Modes
 Numerical aperture NA is number that characterize the ability of specific optical fiber to gather light.
 Larger the NA easier is to direct the light into the optical fiber.
 Greater the NA larger the amount of light that can be directed into the and saved into the optical fiber.
 This means that we need numerical aperture as large as possible.
 Plastic fiber with NA-0.5192,Silica fiber with 0.2425. Which one is better?
 But this is not true always, now we need to understand the modes of the optical fiber.
 Light can propagate inside the optical fiber only as set of separate beams or ray.
 Looking inside the optical fiber we found that beam traveling at distinct propagation angle 𝛼 ranging from 0 to
critical propagation angle 𝛼 .
 These different beams are called modes, The smaller the modes propagating angle the lower the order of the
mode.
 The mode traveling precisely along the fiber’s central axis called zero order mode and at critical angle is called
the highest order mode possible in fiber.
 The zero order mode is called fundamental mode.
 As many modes exist in fiber, so fiber having many modes
is called multimode fiber.
Modes
 How many modes an optical fiber can carry?
 Depends on the optical and geometrical characteristic of the fiber.
 Larger the core diameter, more light core can accommodate so there will be greater number of modes.
 The shorter the wavelength of light more will be number of modes.
 As more is the numerical aperture more is the ability of fiber to gather the light and hence more will be the
number of modes.
 Hence the number of modes inside the optical fiber are directly proportional to diameter d and numerical
aperture NA, inversely proportional to the wavelength of light 𝜆.
𝝅𝒅 𝟐 𝟐
𝝅𝒅
𝑽= 𝒏𝟏 − 𝒏𝟐 𝑽= 𝑵𝑨 𝑵𝑨 = 𝒏 𝟐∆
𝝀 𝝀
𝝅𝒅 ∆= ⁄ is relative refractive index.
𝑽= 𝒏 𝟐∆
𝝀 n= average refractive index.

 For larger V number, the formula for step index fiber is Claclulate the number of modes for graded-
𝑁= index optical fiber if its core diameter
 For graded index fiber d=62.5𝜇𝑚, its numerical aperture is NA=0.275
𝑉 and its operating wavelength is 1300 nm.
𝑁=
4
Importance of Modes
 Light emerging from a light source into fiber break down in set of modes inside the fiber
 Within the fiber total light power is carried by individual modes so that at the fiber output, these small portion
combine, producing a output beam with its power.

o Light is made up of electromagnetic waves.

o The phase at which specific wave meets the core
cladding interface are different and depends upon
distance travel by the waves.
o Distance inside the optical fiber is measured in terms of
propagation angle.
o This different waves travelling within the core at o The waves experience the phase shift when it is
different propagation angle will strike the core cladding reflected: this depends upon the propagation
phase at different angle. angle.

o Only those waves will survive at the output, whose

phase remain the same before and after the reflection
at the core cladding interface.
o Those waves which change phase during reflection from
the core cladding interface, will not produce any effect
at the output of optical fiber
How input pulse is delivered with in a fiber?
 We are considering the fiber optic communication, therefore we are looking to use the light to carry a
communication signal.
 Most popular communication in digital transmission, a light pulse represent the logic 1 and no light pulse
represent the logic 0.
 Such light pulse radiated by the light source entered the optical fiber, where each pulse is breakdown in set of
small pulses carried by the each mode.
 At the output, the individual pulse recombine and since they are overlapping, receiver sees one long light pulse
whose rising edge is from fundamental mode and falling edge from the critical propagation mode.

 Let consider length of optical

fiber L.
 For zero mode or fundamental
mode travelling along the central
axis.
 Time taken is given by 𝑡 = ⁄
 L is length of optical fiber 𝑣 =
⁄ .

Time taken highest order mode at critical angle-the critical mode- needs time.
How input pulse is delivered with in a fiber?
 Let consider length of optical fiber L.
 For zero mode or fundamental mode travelling
along the central axis.
 Time taken is given by 𝑡 = ⁄
 L is length of optical fiber 𝑣 = ⁄ .

Time taken highest order mode at critical angle-the critical mode-

needs time. 𝑡 = ⁄
𝑛2
𝑛1 = 𝐶𝑜𝑠𝛼
𝑡 = 𝑛1𝐿 𝑐 𝑡 = 𝑛1𝐿 𝐿𝑛1 𝑛1 − 𝑛2
𝑐 𝑛2 𝑛1
Δ𝑡 = 𝑡 − 𝑡 =
𝑐 𝑛2
𝑛1 − 𝑛2
Δ= Δ𝑡 = 𝑡 − 𝑡 = Δ 𝐿
𝑛 Δ𝑡 = 𝑡 − 𝑡 = 𝑁𝐴
2𝑐𝑛2

How much is will light pulse spread after travelling along 5 km of

step index fiber whose NA=0.275 and n1=1.487?

Δ𝑡 = 423.8𝑛𝑠 Δ𝑡
𝐿 = 84.76𝑛𝑠
 How dispersion affect the transmission
How much is will light pulse spread after travelling
along 5 km of step index fiber whose NA=0.275 and
n1=1.487?

Δ𝑡 = 423.8𝑛𝑠 Δ Δ𝑡
𝐿 = 84.76𝑛𝑠
 Suppose we want to transmit information at 10Mbits/second.
 Means we want to transmit 10*106 pulses every second. What will be the
duration between each pulse?

 100 ns is the duration of the each pulse.

 The duration of the input pulse is negligible short.
 We have seen each pulse will be spreaded up 84.76ns every kilometer.
 Dispersion of first pulse will be 84.76 ns after first kilometer and 169.52 after
the second kilometer.
 Maximum bit rate can not be more than 12Mbits/sec.
 What is the benefit of optical fiber than?
 Same data speed can be achieved by coaxial cable.
How dispersion affect the transmission
Δ𝑡 Δ𝑡 = 423.8𝑛𝑠 Δ
𝐿 = 84.76𝑛𝑠
 Basic idea and structure of the a graded-index fiber
 Zero-order mode travels along the central axis and the higher order mode travel at or less than the critical
propagation angle.
 Beam travels over same velocity but over different distances and they arrive at the receiver end at different
time.
 If we could arrange it so that they arrive simultaneously, we would solve the problem.
 Recall the relation 𝑣 = ⁄ .
 We can design the core with different refractive index so that
the beam travelling the highest distance does so at highest
velocity and beam travelling the shortest distance does so with
lowest velocity.
 Such fibers are called the graded-index fiber.
 This shows that how refractive index value varies gradually from n1 at the
core center to n2 at core cladding boundary.
 The higher-order mode moves from higher to the lower refractive index
at each point along their path.
 Result in change of direction of propagation.
 The core of graded-index fiber can also be seen as a set of thin layers
whose refractive index changes slightly, in such a way that layer at the
center has refractive index n1 and layer at the boundary has refractive
index n2.
 How does a graded-index fiber reduce the modal dispersion?
 Input pulse is delivered within the fiber core in fractions and each of these fractions is carried by a different
mode.
 The mode propagating along the centerline of a graded index fiber, shortest distance traveled at lowest speed
because it meets the highest refractive index.
 The mode travelling closer to the fiber cladding, longest distance traveled at high speed because it meets the
lower refractive index.
 Hence the fraction of the input pulse delivered by the different modes arrive at the receiever end more or less
simultaneously.
 Therefore the intermodal dispersion is reduced and the bit rate will be increased.
 For the graded index fiber the pulse spreading is given by
(𝐿𝑁 ∆ )
∆𝑡 =
8𝑐
 A graded index fiber has N1=1.487 and ∆=1.71%. For 5km long optical
fiber, compute the pulse spreading due to the modal dispersion and
determine the maximum bit rate possible. (∆𝑡 =0.18ns/KM)

Graded-index fiber was the first solution to the modal dispersion problem,
but at price cost.
Graded-index is popular for transmission medium for short and intermediate
distance network.
 The structure of single mode optical fiber
 This modal dispersion is arising due to the present of multimode in the optical fiber.
 Why not allow the light to travel in the optical fiber in single mode?
 The number of modes inside the optical fiber are directly proportional to diameter d and numerical aperture NA, inversely
proportional to the wavelength of light 𝜆. 𝑉 = 𝑛 2∆
 These number modes can be reduced if one restrict the core diameter and relative refractive indexes.
 For a single mode optical fiber d and ∆ are small as 8.3μm and 0.37%
respectively.
 Compare to the number 62.5μm and 2% of the graded-index fiber.
 This is how one can restrict the fiber to carry one single mode.
For a real single mode fiber 𝑉 ≤ 2.405

1. Single mode optical fiber affords the best solution to modal dispersion
problem.
2. Drawback is that it is most expensive fiber to manufacture and most
difficult to maintain. Because of difficulty of maintaining the an accurate
core size.
3. Single mode optical fiber is more prone to macro and micro core bending
losses and many other problem during the installation and operations.
 Attenuation
 We measure the light power before directed into the optical fiber and measure it again as it emerges from the
fiber. Would you expect to get the same number?
 In optical fiber technology the attenuation is the decrease in the light power during light propagation along an
optical fiber.
 When light coupled to an optical fiber for the purpose of communication, attenuation in optical fiber means a
power losses other than failure to achieve towards the total internal reflection.
Bending Losses
Macrobending Mode
 The beam forms a critical propagation angle with fiber’s central axis
at the straightened end, or flat part of the optical fiber.
 But the same beam forms the propagation angle which is more than
the critical angle when it strike the boundary of the optical fiber.
 Hence bending of the optical fiber introduces the losses in the light
power or attenuation..
 This is one of the major cause of losses the optical fiber experiences
while propagation through the optical fiber.
 Attenuation
 We measure the light power before directed into the optical fiber and measure it again as it emerges from the
fiber. Would you expect to get the same number?
 In optical fiber technology the attenuation is the decrease in the light power during light propagation along an
optical fiber.
 When light coupled to an optical fiber for the purpose of communication, attenuation in optical fiber means a
power losses other than failure to achieve towards the total internal reflection.
Bending Losses
Microbending Mode
 The beam forms a critical propagation angle with fiber’s central axis
at the straightened end, or flat part of the optical fiber.
 Microbending losses are caused by the microdeformations of the
fiber structure.
 Fiber-optics user can do nothing to overcome microbending loss
except ask manufacturer to improve the quality of the fiber.
 Attenuation
 We measure the light power before directed into the optical fiber and measure it again as it emerges from the
fiber. Would you expect to get the same number?
 In optical fiber technology the attenuation is the decrease in the light power during light propagation along an
optical fiber.
 When light coupled to an optical fiber for the purpose of communication, attenuation in optical fiber means a
power losses other than failure to achieve towards the total internal reflection.
Scattering
 The beam forms a critical propagation angle or less will change the
direction after it meets the obstacle in the path, light will be
scattered.
 Even a small change in the value of the core’s refractive index will be
seen by travelling beams as an optical obstacle and this obstacle will
change the direction of the original beam.

 Bending and scattering losses are caused by violation of the condition of total internal reflection.
 This means light which initially satisfy the condition of total internal reflection might violate the condition when the fiber is
bent or its core refractive index varies.
Attenuation

Absorption
 If an incoming photon has such a frequency that its energy is equal to
energy gap of material ∆𝐸, this photon will be absorbed by the
material. ∆𝐸 is the energy difference between two levels.
 We can not do anything to change the energy gap between the
energy levels.

Transparent windows
 Since we can not solved the problem by eliminating the OH- molecules.
 There are three major regions called the transparent window.
 The first located at 850nm, the second at near 1300nm and third near 1550nm.
 Typically we can expect the attenuation about 4db/km near 850 nm, about 0.5 db/km near 1300nm and about 0.3db/km
near 1550nm.
 Later is the most widely used wavelength today in long distance communications.