CEC 345 OPTICAL COMMUNICATION AND NETWORKS

UNIT I
INTRODUCTION TO OPTICAL FIBER COMMUNICATION
Introduction – The General Systems – Advantages of Optical Fiber Communication- Ray Theory
Transmission: Total Internal Reflection, Acceptance Angle, Numerical Aperture, Skew Rays -
Electromagnetic Mode Theory for Optical Propagation: Modes in a Planar Guide, Phase and group
velocity – Cylindrical Fiber: Step index fibers, Graded index fibers – Single mode fibers: Cutoff
wavelength.

INTRODUCTION

FIBER OPTICS:

• It is a technology that uses glass (silica) to transmit data in the form of light
• It consists of a bundle of glass threads, each of which is capable of transmitting
messages modulated onto light waves.

ADVANTAGES OF OPTICAL FIBER CABLE:

• Transmission of optical signals are faster and they possess greater bandwidth.
• Optical fiber cables are less susceptible to noise and electromagnetic interference
• Thinner and lighter
• Does not generate excess heat
• More secure. It is more difficult to tap fiber-optic cables
• Maintenance cost is less

DISADVANTAGES OF OPTICAL FIBER CABLE:

• High installation cost

• Not suitable for mobile communication
• Difficult to reconfigure
• Difficult to splice since they are more fragile than wire

BASIC STRUCTURE OF OPTICAL FIBER:

The basic structure of an optical fiber consists of three parts:
the core, the cladding, and the coating or buffer.

CORE:

• The core is a cylindrical rod of dielectric material

• It does not conduct electricity
• Light propagates mainly along the core of the fiber
• The core is generally made of glass.
• The core is surrounded by a layer of material called cladding.

CLADDING:

• The cladding layer is made of a dielectric material.

• Cladding is generally made of glass or plastic.
• Reduces loss of light from the core into the surrounding air
• It reduces scattering loss at the surface of the core Protects the fiber from absorbing
surface contaminants.
• It adds mechanical strength
• For extra protection, the cladding is enclosed in an additional layer called
the coating or buffer.

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 prevents the optical fiber from scattering losses caused by microbends

EVOLUTION OF OPTICAL (OR) LIGHT WAVE SYSTEM

❖ First Generation:
• Loss produced is about 2 to 3 dB/km.
• Gallium Arsenide (GaAs) light sources are
1) LED
2) LASER
• The optical communication system allows a bit rate of 45 Mbits/s.
• Repeaters are spaced at a distance of 10km.
• GaAs laser operates at a wavelength of 850nm.
❖ Second Generation :
• Loss obtained from the fiber is 0.4 dB/km.
• InGaAsP light source is used.
• The optical communication system allows a bit rate of 100Mb/s.
• Repeaters are spaced at a distance of 50km.
• GaAsP laser operates at a wavelength of 1330nm.
• Since, single mode fibers have lower dispersion than multi mode fibers. The bit
rate-distance product can be increased by using single mode fibers instead of
multi mode fiber.
❖ Third Generation:
• Silica fiber is used.
• Operates at a wavelength of 1550nm.
• The optical communication system allows a bit rate of 4 GB/s.
• Repeaters are spaced at a distance of 100km.
• Loss obtained is about 0.2 dB/km.
• Signals are regenerated by amplification before the signal is again transmitted to
an optical fiber which causes loss and delay.
❖ Fourth Generation :
• Introduces a optical network technology with combination of dense wavelength
division multiplexer (DWDM) and optical amplifiers.
• Bit rate of 10Tb/s doubles for every 6 months.
❖ Fifth Generation:
• InP light source is used.
• Operating wavelength is about 1.53µm to 1.57µm.
• Bit rate is of 40 to 160 Gb/s
• Repeater spacing is about 24000km to 35000km.
• Loss is enormously reduced.

BLOCK DIAGRAM OF OPTICAL COMMUNICATION SYSTEM:

❖ Information source
• Consists of Audio, video and computer data.
• Non electrical messages.
❖ Transducer
• Message signals are converted into electrical signals
❖ Optical Transmitter
1) Driver circuit
• Carry the electrical signals from transducer and transfer it to the light source.
• Used to boost up the weak electrical signals.
2) Light source
• Converts electrical signals into optical signals.
• Two types of light source: LED and LASER.
• LED is simple and cheap when compared to LASER.
• For long distance communication, LASER is a suitable light source.

❖ Transmission medium
• Optical fiber cable is used which is a solid cylindrical dielectric waveguide.
• It is fabricated from SiO2 (Silicon dioxide).
• SiO2 is naturally available in sand.
• Cable consists of core, cladding and buffer(or) plastic coating.
• Inner layer is the core layer which is used to guide the light waves.
• Refractive index of core medium is n1.
• Core is surrounded by cladding layer.
• Refractive index of cladding medium is n2.
Advantages of cladding :
• Provides mechanical support and strength to the cable .
• Prevents scattering loss on the core surfaces .
• Splices: Permanent joint between two fiber cables.
Joint can be made by applying an electric arc.
• Connectors: Temporary joint between two fiber cables.
It can be easily disconnected and Reconnected.
• Fly lead :used to connect the fiber optic cable and the device.
❖ Optical receiver
1) Photo Detector
• Detect and converts optical signal into electrical signal.
• PIN and Avalanche photo diodes are used.
2) Amplifier
• Amplifies the weak electrical signals using BJT or FET amplifier.
• Repeater is used to amplify the weak optical signal.
❖ Transducer
• Converts amplified electrical signal into original information .
❖ Information source
• Original message information (audio,video or data) is reached at the destination.

ADVANTAGES OF OPTICAL COMMUNICATION SYSTEM:

• Low cost
• Repeaters used is reduced
• High data rate
• Low loss
• Increased transmission speed
• More immunity to electromagnetic interference.

RAY THEORY TRANSMISSION:

In optics, a ray is an idealized narrow beam of light .

❖ Law of Reflection
It states that the angle of incidence (θ1) is equal to the angle of reflection (θ2).

angle θ1= angle θ2

❖ Law of Refraction

When light ray travels from medium 1(air) to medium 2(glass), bending of light ray may occur.

❖ SNELL’S LAW

Snell’s law states that the ratio of the sines of the angles of incidence and refraction is
equivalent to the reciprocal of the ratio of the indices of refraction.

φ1= incident angle,

φ2= refracted angle
n1=refractive index of medium 1
n2=refractive index of medium 2 and n2< n1

Application:
Used to find refractive index of unknown material.
❖ CRITICAL ANGLE
• When we increase the incident angle with respect to normal, at some incident angle, the
refracted ray travels along the boundary or surface. Hence φ2 becomes 90.
• The angle of incidence for which angle of refraction becomes 90 degrees is called
critical angle.

The critical angles of different materials are as follows:

❖ Total Internal Reflection

• When the incident angle φ is greater than the critical angle φc , the light ray is reflected
back to medium 1.
• There will not be any light transmission (refraction) in medium 2.
• This is called Total Internal Reflection.

Conditions for Total Internal Reflection:

• The medium from which light is incident, must be optically denser than the medium to which it
is incident, n2 > n1.
• The angle of incidence in the denser medium must be greater than the critical angle of the
denser medium with respect to the rarer medium.

FIBER CLASSIFICATION:
Based on refractive index profile :
• Step index fiber
• Graded index fiber

Based on modes of propagation :

• Single mode fiber
• Multimode fiber

❖ STEP INDEX FIBER:

The step index (SI) fiber is a cylindrical waveguide core with central or inner core with
uniform refractive index n1.
 • The core is surrounded by outer cladding with uniform refractive index n2, n2 < n1.
• But there is an abrupt change in the refractive index at the core cladding interface
• Number of modes M= V2/2.
• Propagation of light takes the path of axial, meridional and skew rays.
• The refractive index profile is defined as
n(r) = {n1 when r < 𝑎 (𝑐𝑜𝑟𝑒)
n2 when a < 𝑟 < 𝑏 (cladding)

where r is the radial distance from fiber axis

• The core refractive index is uniform throughout the region and undergoes a sudden change at
the core- cladding interface. This is called step index fiber.

❖GRADED INDEX FIBER:

• In the graded index (GRIN) fiber the refractive index is not uniform within the core.
• It is high at the center and decreases smoothly and continuously with distance towards the
cladding.
• The refractive index profile across the core takes the parabolic nature.
• The light waves are bent by refraction towards the core axis .
• They follow a curved path because of change in refractive index as they move away from the
center of the core.
• A graded index fiber has high coupling efficiency and higher bandwidth than the step index
fiber.
• The core refractive index decreases radially (or) the refractive index is maximum at the fiber
axis and it radially decreases towards the boundary surface.
• The core refractive index is made to vary as a function of radial distance (wavelength) from the
center of the fiber .This is called Graded Index Fiber.
Advantages of Graded Index Fiber:

• Reduces the intermodal dispersion

• Facilitates higher bandwidth

❖ SINGLE MODE FIBER:

• Only one mode is allowed to propagate.

• Core diameter is too small in the range of 8µm to 12µm.
• In this mode, light takes axial path.
• The ray which passes only through the fiber axis is called axial ray.
• Fundamental mode always travels along the fiber axis or axial path.
• The mode which has zero cutoff frequency is called fundamental mode (LP 01) .
• For single mode operation, the normalized frequency number V<=2.405.
• Laser diode is used.

ADVANTAGES:

• Intermodal dispersion is absent.

• More information can be transmitted per unit of time.
• Higher bandwidth compared to multimode fiber.
• Used for long distance transmission .

DISADVANTAGES:

• LED is not a suitable light source, laser is suitable.

• Coupling light into core is difficult.
• Fabrication is difficult.

APPLICATION:

• Used in telecom networks.

• Used in CATV networks.
❖ MULTIMODE FIBER:

• Multimode fiber allows many modes to propagate along the fiber core.
• The light rays are propagated in the core in zigzag manner (meridonial ray).
• Numerous modes (light rays) are carried simultaneously through the waveguide.
•

• The core diameter is larger than single mode fiber.

• Core diameter in the range of 50µm to 200µm.
• Less optical power .
• LED is a suitable light source
• Easier to make.
• Low cost, longer lifetime.

ADVANTAGES:

• More than one mode can be transmitted.

• Coupling efficiency is large.
• Suitable for short distance..

DISADVANTAGES:

• Modal noise is present.

• Greater Intermodal dispersion.

APPLICATIONS:

• LAN.
• Security systems.
• General fiber networks.

Disadvantages of MultiMode Fiber are Intermodal Dispersion. When an optical pulse is

launched into the fiber, the optical power is distributed over most of the modes of the fiber .Each
of these modes travel at a slightly different velocity. This means that the modes in a given optical
pulse arrive at the fiber end at different times causing the pulse to spread out as it travels along
the fiber. This effect is known as intermodal dispersion.
COMPARISON OF SINGLE MODE AND MULTI MODE FIBERS:

S.NO SINGLE MODE FIBER MULTI MODE FIBER

1 Only a single ray passes through the More than one ray passes through the fiber at a
fiber time
2 Ray passes along the axis-axial ray MMSI-Meridional and Skew rays
MMGI- Paraxial Ray
3 Core diameter is small (8 to 12 micro Core diameter is large(50-200 micrometers)
meters)
4 Intermodal dispersion is not present Intermodal dispersion is present
5 Fabrication is difficult Fabrication is easy
6 Coupling efficiency is less Coupling efficiency is large
7 Led is not suitable Led is suitable

THREE CONFIGURATIONS BASED ON FIBER CLASSIFICATION:

❖ Single mode step index fiber

❖ Multimode step index fiber
❖ Multimode graded index fiber

SINGLE MODE STEP INDEX FIBER:(SMSI)

 • Single mode fiber permits only one mode to propagate and does not suffer from mode
delay differences.
• Core is very narrow so that only a single path exists through the cable core through which
light can travel.
• The single mode step index fiber has a central core with uniform refractive index with a
step change in core-cladding.
• The light rays are propagated in the fiber through reflection.
• Single mode fiber is also known as fundamental or monomode fiber.
• Numerical aperture is constant.
• Refractive index profile is step index profile.
• Used effectively with time division multiplex (TDM) and wavelength division multiplex
(WDM) systems operating in 1550nm wavelength region.
• Modes having highest cut-off frequency is called dominant mode.
• Modes having zero cut-off frequency is called fundamental mode.
Used in telecommunication and data networking industries
• Smaller core diameter size 8-12µm.
• Cladding diameter size is 125µm.
• Light ray which passes through fiber axis is called axial ray.
• For long distance communication this fiber cable is used.

DISADVANTAGES:

• Interconnection of cables and interfacing with source is difficult because of extremely

small size.
• Light from an optical transmitter will have definite spectral width.
MULTIMODE STEP INDEX FIBER:(MMSI)

• Similar to SMSI except the core diameter is larger with multimode configuration.
• Multimode step index fiber is more widely used type due to large size.
• It is easy to manufacture and hence less expensive.
• Its core has large aperture and allows more light to enter the cable.
• Light rays take many paths and are propagated down the core in zigzag manner
(meridional rays)
• Core diameter is 50-200µm
• Cladding diameter is 125µm to 400µm
• Uses the principle of total internal reflection (TIR).
DISADVANTAGES:

• It causes the transmission bandwidth to be fairly small.

• Hence it is used in applications requiring distances of less than 1km.

MULTIMODE GRADED INDEX FIBER:(MMGI)

• The light rays no longer follow straight lines, they follow a parabolic path being
gradually bent back towards the center by the continuously declining refractive index.
• The light rays near the edge of the core takes a longer path but travel faster since the
index of refraction is lower.
• Non-uniform refractive index at the core.
• Due to this, light rays are bent smoothly and converged repeatedly at points along the
cable.
• Less modal dispersion due to the arrival of all light paths at one point simultaneously.
 • Its takes paraxial ray path.
• The light ray is propagated through refraction. The light ray enters the fiber at many
different angles.
• This cable is mostly used for long distance communication.
• Core diameter is 50 to 100µm.
• Cladding diameter is 125µm to 140µm.

ADVANTAGES:

• Comparatively cheaper.
• Intermodal dispersion is reduced.

COMPARISON OF STEP INDEX AND GRADED INDEX FIBER:

S.NO STEP INDEX FIBER GRADED INDEX FIBER

1 The core has uniform refractive index The core has high refractive index along the
with a step change in core-cladding axis which gradually decreases towards the
clad-core interface.
2 Axial ray-SMSI Paraxial rays-MMGI
Meridional rays and skew rays-MMSI
3 Intermodal dispersion is present in MMSI Intermodal dispersion is reduced in MMGI
4 Step index profile Graded index profile where alpha is the profile
factor
5 Numerical aperture is constant Numerical aperture is a function of radius
2
6 M=𝑉 /2 for step index fiber M=𝑉 2 /4 for graded index fiber
7 Fabrication is easy Fabrication is difficult
8 Coupling efficiency is higher Coupling efficiency is lower
9 Plastic or glass material is preferred Only glass is preferred
10 Typical light source is LED Light sources are LED , LASERs

RAY OPTICS:

RAYS

MERIDIONAL RAYS SKEW RAYS

BOUND RAYS UNBOUND RAYS

 ❖ MERIDIONAL RAYS:

• The meridional ray enters the core and passes through its axis.
• These rays are confined to the meridian planes of the fiber which are the planes
that contain the axis of symmetry of the fiber (the core axis).
• Lies in a single plane, its path is easy to track as it travel along the fiber.
• When the core surface is parallel, it will always be reflected to pass through the
center.
• Meridional ray takes comparatively lesser light ray path because of lesser
acceptance angle.
• These rays travel slower because of Total Internal Reflection (TIR).

Two classes of Meridional rays:

❖ Bound Rays:
• They are trapped inside the core
• Propagate along the fiber axis

❖ Unbound Rays:
• They are refracted out of the fiber core.

Skew Rays:
 • The skew ray does not pass through the center.
• The skew ray reflects off from the core cladding boundaries and again bounces
around the outside of the core.
• These rays are not confined to a single plane but follow a helical path along the
fiber.
• Since it takes helical path, it travels slower.

• The helical path traced through the fiber gives a change in direction of 2γ at
each reflection, where γ is the angle between the projection of the ray in two
dimensions and radius of the fiber core at the point of reflection.
• They are more difficult to track as they do not lie in a single plane.
• Skew rays take maximum possible light ray path because of greater acceptance
angle.
• Power loss is high because many of the skew rays to be trapped in the fiber are
leaky rays.
• Follows a helical path along the fiber.
• Rays are more difficult to track as they travel along the fiber.

Leaky Rays:

• Partially confined to the core and attenuate as the light travels along the optical wave
guide.
• Helical path traced through the fiber.

Skew rays meet the core-clad interface at many places due to helical path, lower order
mode in core couples with the higher order mode in the cladding. There is more power radiation.
Hence they are called leaky rays.

AXIAL RAYS:

• The axial ray travels along the axis of the fiber and stays at the axis all the time.
• Axial rays rarely take light ray path because the acceptance angle is zero.
• Axial rays travel faster
COMPARISON OF TYPES OF RAYS:
1.12 OPTICAL COMMUNICATION

Analysis of Meridional Rays

Meridional rays having the following parameters
i) Refractive index ii) Total internal Reflection
iii) Acceptance angle iv) Numerical Aperture
v) V-Number

1. Refractive Index

Low index n2
Exit ray
(air)
2

High index n1 Partial

(glass) internal
1 reflection

Incident ray

Figure 1.4: Light rays incident on high to low refractive index interface (e.g.. Glass-air)

Refractive index is defined as the ratio between the speed of light in vaccum to the speed of
light in material. It is denoted by “” or “n”.

speed of light in vaccum

 or n 
speed of light in material

When the ray is incident on the interface between two dielectrics of differing refractive indices,
part of the ray is reflected in to the first medium and the rest is refracted into the second medium. Let
us take n1 is the index profile of first medium and n2 is the index profile of second medium. The
condition for the refraction is n1 must be greater than n2. The refraction will occur at the interfaces
is due to the difference in the speed of light in two materials.
The incident angle and the reflected angle is related by using snell’s law of refraction.
Introduction to Optical Fibers 1.13

According to Snell’s law

n1sin 1 = n2sin 2 ... (1)

sin 1
 n 2 n1
sin  2

When the angle of refraction is 90° the refracted ray will become parallel to the interface
between two materials.
Therefore, when 2 = 90° the incident angle = Critical angle

sin 1 = n 2 n1

 sin c = n 2 n1

The critical angle c = sin–1 n 2 n1 ... (2)

2. Total Internal Reflection

Total internal reflection is defined as the complete reflection of light into the same medium
without any transmission of light .
Conditions for the Total Internal Reflection are
1) Light should travel from High refractive index material to lower refractive index
material.
2) Incident angle should be greater than the critical angle

Low index
medium

Incident
1
ray Totally
1 > C
reflected ray
High index
medium

Figure 1.5: Total internal Reflection

1.14 OPTICAL COMMUNICATION

Low index
cladding

   
High index   Core axis
core

Figure 1.6: Transmission of a light ray in a perfect optical fiber

In a fiber, when the light ray is incident at the core cladding interface, total internal reflection
will occur. Since the angle of incident at the core cladding boundary is greater than the critical angle
the light gets totally reflected back to the core.

3. Acceptance Angle
Acceptance angle is the maximum angle to the fiber axis at which light may enter the fiber axis
in order to be propagated.
Let us consider two rays are launched into the fiber. The first ray named as “A” which makes
an angle C within the core cladding interface. This ray making an angle a1 to the fiber axis and it
gets refracted at the air core interface and this refracted ray is getting propagated into the fiber. The
second ray “B” which is launched to fiber at an angle greater than a1. This ray B gets refracted into
the cladding and get lost by radiation. a must be less than the critical angle at the fiber core.

A Low  cladding
Cohical High
a1  cladding C C
half
angle C C
a2

Acceptance B
cone a1<C
a2>C

Figure 1.7: Illustrate the acceptance angle

According to Snell’s law of refraction at the core air interface

n0 sin a = n1 sin 2 ... (3)
1.16 OPTICAL COMMUNICATION

2 2
NA = sin a = n1  n 2 ... (6)

The relative refractive index difference is given by

2 2
n1  n 2
 = 2
2n1
n1  n 2
  ... (7)
n1

2 2
 NA = sina = n1  n 2

NA = sina =  n1  n 2   n1  n 2 

NA = sina = 2n1  n1  n 2    n1  n 2 

n1  n 2
NA = sina = 2n1  n1    
n1

2
= 2n1  = n1 2 ... (8)

Then the maximum acceptance angle is said to be a numerical aperture and it is

given by

NA = sin a = n1 2

a = sin-1  n1 2 

a = sin-1 (NA)

5. V Number
V number is used to find the number of modes that can support the fiber. It is a dimensionless
number and it is given by

2a 2 2
V = n1  n 2 ... (9)

Acceptance Angle for the Skew Rays

To calculate the acceptance angle for a skew ray, it is necessary to define the direction of the
ray in two perpendicular planes.

The skew ray incident on the fiber core at point “A”. The ray is refracted into the fiber core
and getting reflected at point “B”. The angle of incident and reflection at point B is φ to rescue this
ray path AB relative to the radius BR in two perpendicular plane requires multiplication by cos γ
and sin θ.

where,
φ - Angle between the core radius and the projection of ray.
γ - Angle between the ray and a line AT drawn parallel to the core axis.

Hence, the reflection at point B at an angle φ may be given by :

cos γ sin θ = cos φ

Using the trigonometrical relationship sin2 φ + cos2 φ = 1, the above equation becomes:
Introduction to Optical Fibers 1.19

2 2
n1  n 2
2 2
cos sin = 1 n n = 2
2 1 n1

1 2 2
cos sin = n n1  n 2
1

1 2 2
sin = n cos  n1  n 2
1

According to snell’s law at air-core interface

n0sina = n1sin.

n1
 sina = n sin 
0

n1 1 2 2
sina = n n cos  n1  n 2
0 1

1 2 2
sina = n1  n 2
n 0 cos 

1
sina =  NA  ... (2)
n 0 cos 

Therefore the Acceptance angle of skew ray is given by a = as

1 2 2
sin as = n1  n 2
n 0 cos 

1  1 2 2
as = sin  n1  n 2 
 n 0 cos  
far air n0 = 1

1  1 
as = sin   NA 
 cos  

NA = sinas . cos. ... (3)

1.20 OPTICAL COMMUNICATION

1.7 ELECTRO MAGNETIC MODE THEORY

To study the light propagation in detail, this should be described in terms of electro magnetic
waves. Generally electromagnetic waves are consist of two fields, they are
i) Electric field
ii) Magnetic field
The light which is travelling through the fiber is said to be guided modes of a fiber. These
guided mode is a pattern of electric and magnetic field distribution. By solving maxwell’s equations
in cylindrical co-ordinates we can find the modes of a fiber. A mode travelling along the fiber axis,
where the fiber axis referred to z axis of Cartesian coordinates system is given by

e
j wt z 

where w  Angular frequency of the field mode depends upon the time and direction
of propagation.
  Z component of propagation constant.
In the optical fibers, the core cladding boundary conditions lead to a coupling between the
electric and magnetic field components. This produces hybrid modes.
The hybrid modes are designated as HE or EH modes depending on whether the transverse
electric field or transverse magnetic field.

1.7.1 Review of Planar Wave Guide

Planar wave guide consists of a slab of dielectric with refractive index n1 and sandwiched
between two regions of refractive index n2. Plane wave is resolved into two components one is
propagating in the x direction and the other one is in the z direction. The components of phase
propagation constants in the z and x direction by
 z = n1 k cos.
 x = n1 k sin.

X X direction
P Q
x = n1 k cos 
n1>n2 2 = n 1 k sin  Core Z direction
n2 Cladding

Figure 1.10: Phase, Propagation constants of optical ray

 DERIVATIONS

1. Starting from Maxwell’s equation, derive an expression for wave equation of an

electromagnetic wave propagating through optical fiber.

Maxwell’s equation provides the basis for study of electromagnetic wave

propagation.
Consider Maxwell’s equations,
B
 E = −
t
----------- I
D
 H =
t
Where, E is the Electric field
B is the magnetic flux density
H is the Magnetic field
D is the electric flux density
Consider divergence condition,
.D = 0
----------- II
.B = 0
(no free charges and no free poles respectively)
Where,  is the vector operator
The four field vectors are related by,
D = E
----------- III
B = H
Where,  is the dielectric permittivity
 is the magnetic permeability of the medium.
Substituting III in I,
H
 E = −
t
----------- IV
E
 H =
t
Taking curl for equation IV,
H  (  H )
  (  E ) =   (−  ) = −
t t
 E
= − ( )
t t
 2E
   (  E ) = −  ---------- V
t 2
Similarly,
E (  E )  H
  (  H ) =   ( ) = =  (−  )
t t t t

2H
   (  H ) = −  ----------- VI
t 2
Vector identity equation is,
  (  E ) = (.E ) −  2 ( E )
From equation III, D
E=
Eε
D
   (  E ) = (. ) −  2 (E) = 0 −  2 (E)
ε
E
( .D = 0)
   (  E ) = −2 ( E ) ------------- VII
Equating V and VII, we get,
2E
−  2 = −2 ( E )
t
2E
  ( E ) =  2
2
--------------- VIII
t
Similarly,
2H
 2 H =  ---------------- IX
t 2
Equations VIII and IX are called as general wave equations for dielectric wave guide.
Let field  be E or H, then
Equation VIII can be replaced by phase velocity as
1
Vp =

1  2
   = 2 2 ------------- X
2

V p t
General wave equation is,
  =  0 e j ( t − z )

Where,  0 is amplitude of electric field

EM wave travels with frequency  , propagation constant  and along Z-direction.
Laplacian operation of planar waveguides described by rectangular Cartesian co-ordinates
(x, y, z) is
 2  2  2
2 = + + ------------ XI
x 2 y 2 z 2
Laplacian operation of circular fibers described by cylindrical polar co-ordinates (r,  , z) is
 2 1  1  2  2
2 = + + + -------------- XII
r 2 r r r 2  2 z 2
Equating X and XII,
1  2  2 1  1  2  2
= 2 + + + ------------- XII*
V p2 t 2 r r r r 2  2 z 2

But  =  0e j ( t −  z ) --------------- XIII

Where,  0 is the amplitude of electric field
Differentiating above equation w.r.t. z,

Ψ
=  0 e j (t − z ) .(− j )
z
 2
=  0 e j (t − z ) (− j )( − j )
z 2

= − 0 e j (t − z ) . 2

 2
= −  2 ------------- XIV
z 2

( 0e j (t − z ) =  )
Differentiating equation XIII w.r.t ‘t’,

=  0 e j ( t −  z ) . j 
t
 2
=  0 ( j  ) 2 e j ( t −  z )
t 2
  2
= − 2 ------------- XV
t 2

Substituting equation XIV and XV in XII*, we get

 2 1  1  2 1
+ + 2 + (−  2 ) = 2 (− 2 )
r 2
r r r  2
Vp
Refractive index is given by
c
n=
Vp
(Since, core refractive index is n1 )
c
Vp =
n1
c
− n12 (2 ) 2
1 1 − n (2f ) 
2 2

 2
(− 2 ) = 2
(− 2 ) = 1
=
Vp (c n1 ) c2 c 2

2
= −n12 ( ) 2 .


(− 2 ) = −n12k 2 ------------ XVI ( propagation constant k= 2 )

1

2
Vp
 2 1  1  2
Rewriting equation X11, + + −  2 = −n12 k 2
r 2
r r r 
2 2

 2 1  1  2
+ + +  [n12 k 2 −  2 ] = 0 ------------- XVII
r 2 r r r 2  2
Equation XVII is the scalar wave equation for cylindrical optical fiber.

2. Derive an expression for linearly polarized modes in optical fibers and obtain
the equation for V- number.
(or)
Mode theory of circular waveguide or cylindrical fibers.
Wave equation for cylindrical waveguide is
  2 1  1  2
+ + 2 2 +  (n12k 2 −  2 ) = 0 --------------- I
r r r r 
2

Where, k is the propagation constant

a. In the core region, the electromagnetic field  is real, that is,
n12 k 2   2
--------------- II
   n1k
b. In cladding region,
n12 k 2 −  2  n22 k 2 −  2
( n1  n2 )
 should fall rapidly with distance

 n12 k 2 −  2 = 0
n22 k 2   2
 n2 k  
  lies between n2 k and n1k (i.e.), n2k    n1k ------------- III
If equation III is satisfied, then the mode is said to be bounded or guided or trapped
mode
If equation III is not satisfied, then the mode is said to be unbounded or radiation mode.

According to separation of variable solution for equation I

 = E(r ) F1 ( ) F2 ( ) F3 ( z) -------------- IV
Let time and z dependent factors be
F1 ( ) = e jl --------------- V
F2 ( ) F3 ( z ) = e j (t − z ) --------------VI
Substituting V and VI in IV, we get
 = E (r )e jl e j (t − z ) ---------------- VII
Differentiating equation VII w.r.t. ‘r’, we get,
 E ( r ) jl j (t − z )
= e e --------------- VIII
r r
 2  2 E (r ) jl j (t − z )
= e e --------------- IX
r 2 r 2
Differentiating equation VII w.r.t ‘  ’, we get
 
= E (r )e jl ( jl )e j (t − z ) --------------- X

 2
= E (r )e jl ( jl ) 2 e j (t − z )
 2

 2
= −l 2 E (r )e jl e j (t − z ) --------------- XI
 2

Substituting equation VIII, IX and XI in equation I,

 2 1  Ψ 1  2
+ + 2 2 +  (n12 k 2 −  2 ) = 0
r 2
r r r 
 2 E (r ) jl j (t − z ) 1 E (r ) jl j (t − z ) 1 2
e e + [ e e ] + 2 [−l E (r )e jl e j (t − z ) +  (n12 k 2 −  2 ) = 0
r 2
r r r

Substituting expression for  from equation VII in the above equation,

 2 E jl j (t − z ) 1 E (r ) jl j (t − z ) l 2

.e e + [ e e ] − 2 E (r )e jl e j (t − z ) + E (r )e jl e j (t − z ) (n12k 2 −  2 ) = 0
r 2
r r r

jl j (t − z )  2 E (r ) 1 E (r ) l2
e e [ + + (n1 k −  − 2 ) E (r )] = 0 ------------- XII
2 2 2

r 2 r r r
Bessel’s differential equation is arrived as follows,
 2 E 1 E l2
+ + (n1 k −  − 2 ) E ] = 0
2 2 2

r 2 r r r
Solution of Bessel differential equation is called Bessel function. Bessel differential
equation has two solutions
a. Bessel’s function of first kind J l (U r ) : In core region: ’E’ must be finite.
b. Bessel’s function of second kind kl ( wr ) : in cladding region, ‘E’ must decay
from ‘r’ tends to infinity.
Er = {GJ l (UR), R  1, (core )
k r (WR )
GJ l (U ) , R  1, (cladding )}
kl (W )
U and W are Eigen values
R is the normalized radius
G is the amplitude constant
r
R= , Where, ‘a’ is the radius of the core
a
U and W are defined as
MODE

Mode is defined as the distribution of electromagnetic field pattern in the core.

Possible modes in optical cylindrical fiber:

TE (Transverse Electric) Mode:

• Electric field is perpendicular to the direction of propagation.

• There is no electric field component along the direction of propagation.
EZ = 0 and
Hz exists

TM (Transverse Magnetic) Mode:

• Magnetic field is perpendicular to the direction of propagation.

• There is no magnetic field component along direction of propagation.
HZ = 0 and
Ez exists

TEM (Transverse Electromagnetic Mode):

When total field lies in the transverse plane, transverse electromagnetic (TEM) wave exists
where both EZ and Hz=0
Introduction to Optical Fibers 1.31

Applying these condition to eliminate the arbitrary constants A, B, C, D.

 J V  ua  0 K V  a  0 
 
  v J V  ua  j
J v '  ua 
v
K V  a 
j
K V '  a  
 au 2 u a 2  
 0 J V  ua  0  K v  a  = 0 ... (8)
 
  j1 v  j 2 v 
 u J V '  ua  au 2
J V  ua 

K V ' a 
a 2
K V  a  

Simplifying this we get the equation

2 2
   1 1 
 I v  H1   K 2
I  K2 Hv
1 v
2
  1  2  2  ... (9)
 a  u v 

I'v  ua 
where Iv =
u J v  ua 

K 'v  a 
Hv =
 K v  a 

1.9 LINEARLY POLARIZED MODES

The characteristic of TE0l, TM0l and HE2l are similar i.e these modes are having same electric
field pattern and the propagation constant and the modes are said to be degenerate. The degenerate
modes are called linearly polarised modes.
Linear polarized mode classification depends upon the light intensity distribution rather than
the electric field configuration. In LPm modes, m and l denote the azimuthal variation of Ez or Hz and
propagation constant of the mode respectively.
Modes which are characterised by a common set of m and l and hence having a common
phase constant are called LPm modes regardless of their TE, TM, EH or HE field configurations.
i) Each LP0 mode is derived from an HE1 mode.
ii) Each LP1 mode comes from TE0, TM0 and HE2 modes.
iii) Each LPn mode (n  2) comes from an HMn+1 mode and EHn–1 mode and so on.
1.32 OPTICAL COMMUNICATION

Core Core Core

HE21 + TE01 LP11

Core Core Core

HE21 + TM01 LP11
Figure 1.13: Configuration of LP modes

The propagation constants of the guided mode  lies in the range

n2 k  n, K
When  < < 1, then K12 = K22 = 2. This approximation is used in the eigen value equation for .
2 2
 V   1 1 
2 2
(JV + KV) (K JV + K KV) = 
1 2   2 2
 a  u w 
2
V 1 1 
It becomes, JV + KV =   2 2
a u w 
The eigen values in the core and cladding are represented by u and w.
u = a (n12 K2 –  2) 1/2
w = a ( 2 – n22 K22) 1/2

The normalized frequency

V2 = (u2 + w2) 1/2
V2 = Ka (n12 – n22)1/2
The linearly polarized modes are formed from the traditionalexact modes. Here are some examples.
Introduction to Optical Fibers 1.33

Sl.No Linearly Polarized Mode Traditional Mode

1. LP01 HE11

2. LP11 TE01, TMO1, HE21

3. LP21 EH11, HE31

4. LP02 HE12

5. LP31 EH21, HE41

6. LP12 TE02, HE22

7. LP03 HE13

Advantages
i) LP mode concept is very useful in understanding and analysing the transmission characteristics
of optical fibers.
ii) The polarity of field changes also their will be an equivalent solutions.
iii) Visualization of a mode is quick and easy.
iv) Four discrete mode patterns can be obtained from a single LPm by taking two possible
directions of polarizations.
v) Number of exact modes can be converted into a single LP mode.
Demerit
 LP mode concept is valid only for weakly guiding approximation.

1.10 MODES OF A STEP INDEX FIBER

The solution of the above equation will give the characteristics of a mode.
The figure 1.14 shows about the Bessel function Jv (x) for the orders v = 0, 1, 2. Each order
Bessel function have “m” number of roots.
When v = 0, the equation get reduced to
J0 + K0 = 0
1.34 OPTICAL COMMUNICATION

y
1.0 V=0
0.8 V=1
0.6 V=2
0.4
0.2
0
2 4 6 8 10 x

Figure 1.14: Bessel function Jv (x) for the orders v = 0, 1, 2

When v  0 the equation is solved by using numerical methods and these modes are hybrid
modes.

1.10.1 Power Flow in Step Index Fiber

Lowest order mode

HE11

First set of
higher order modes

HE21
Core Core
TE01 TM01

Figure 1.15: Cross section of the electric field vectors in a step index fiber

In step index fiber electro magnetic energy is carried by both core and cladding. After cutoff
frequency, the field is penetrating more into the cladding and more amount of energy is propagates
in the cladding. But other than cut off were amount of field is transmitting through the core. At cutoff
frequency the field is fully transmitted through the core and the mode is acting as a radiating mode.
Introduction to Optical Fibers 1.35

Power flowing through the core and cladding is obtained by integrating the pointing vector in
axial direction.
a 2
1
Pcore =
2   r E
0 0
x H*y  E y H *x  d dr

 2
1 *
Pcladd =   r  E H
x y  E y H *x  d dr
2 0 0

where a is a core radius of a fiber.

Hy* and Hx* are the complex conjugate fields.

Pcore P
 1  clad
P P
where P is the total power in the particular mode “v”.

 Pcladd  4 1
 P   m 2
  total 3

where m – number of modes

1.11 GRADED INDEX FIBER STRUCTURE

The index profile variation of a graded index fiber is given by,
1
  2 
 
n1 1  2 r
n  r    
 
a 
for 0  r  a 

 1 
n1 1  2  2  n1 1     n 2 for n  a 

The index profile will obey the power law relationship in graded index fiber. The refractive
index is not constant in graded index fiber but it decreases with radial distance from a maximum
value of n1 at the axis to a constant value n2 in the cladding.
when  = 1, the profile is triangular
 = 2, the profile is parabolic
 = , the profile is constant and it will become a step index fiber.
1.36 OPTICAL COMMUNICATION

n1


=2

=1
n2 n2

Figure 1.16: Index profile variation for graded index fiber

The inder difference  for the graded index fiber is

n12 – n 22n1 – n 2
 = 
2n12 n1
The numerical aperture for graded index fiber is given by

 1 
 n 2  r  – n 22  2  NA  0  1–  r  ra
 a 
NA (r) = 

 0 ra

The arial numerical aperture is given by,

1
NA(0) =  n 2  0 – n 2  2   n 2  0  – n 2   n 2
 2 1 2 1

1.11.1 Modal Equation for Graded Index Fiber

Finding the modes in a graded index fiber, some of the assumption are made using WKBJ
(Wentzel, Kramer, Brillouin, Jeffries) approximation.

Core
Index

Cladding

Radius
Figure 1.17: Index profile variation with respect to radius in graded index fiber
Introduction to Optical Fibers 1.37

1. The Refractive index profile is radially symmetry as shown in figure 1.17.

2. The core diameter is 50 times greater than operating wavelength.
3. Index change withing the core region is small.
4. Variation of the refractive index is small over a distance of the order of wavelength.
The general equation of the fiber in cylindrical co-ordinate system is given by

d 21  r  1 d1  r   2 2 m2 
  k n  r   2  2  1  r   0 ... (1)
dr 2 r dr  r 

where k  propagation constant also known as wave number

  z dependent propagation constant
m  order of mode
r  radius of a fiber
The trival of form 1 (r) = Aejks(r) is assumed then

d 21 ds  r 
2
 jk 1  jks '  r  1 ... (2)
dr dr

d1 d 
  A.jk.s '  r  e jks r   ... (3)
dr dr
Now (1) can be written as

2 jk  m2 
jks ''  ks '  s '  k 2n 2  r   2  2   0 ... (4)
r  r 
s can be expanded in a power series of 1/K.
1 1
s  s0  s1  2 s2  .....
k k
1 1
s '  s '0  s1'  2 s 2'  .....
k k
1 1
s ''  s ''0  s1''  2 s2''  .....
k k
1.38 OPTICAL COMMUNICATION

By neglecting higher order terms and (4) can be written as

2
 1 1   1 1 
jk s ''0  s1''  2 s''2   k 2 s'0  s1'  2 s ''2 
 k k   k k 
jk  ' 1 ' 1 ''   2 2 2 m2 
 s  s  s  k n  r      0
r  k 2   r 2 
0 1 2
k
By expanding the above equation in terms of equal power of k, the result will be

2  m2 
k 2  s '0   k 2n 2  r   2  2   0 ... (5)
 r 

jk '
jks''0  2ks1' s '0  s0  0 ... (6)
r
2 j
js1''   s1'   s1'  2 s2' s0'  0 ... (7)
r
By integrating the equation (54) and it must be real for the bound mode to exist in the fiber

m2
k 2 n 2  r   2   0
r2
m2
at the interface k 2 n 2  r   2   0.
r2
The maximum number of bound mode is found by m = m (k n1) = kn (1 – )
Total number of modes
 2 2 2
M = m (k n1) = a kn 
2
V2   
M=  
2   2 
For the parabolic profile  = 2.
V2
 Total number modes =
4
For step index fiber  = 
V2
M = 2 k2 n2  =
2
Introduction to Optical Fibers 1.39

1.12 SINGLE MODE FIBER (SM)

 Only one mode is propagation over the fiber. All the remaining modes are attenuated by
absorption or loss.
 Small core size of 8-12 m, and small index difference of 0.2 to 1.0 percent.
 The fundamental mode LP01, can exist. SM fibers are having greater transmission bandwidths
and the lowest losses.
 They are compatible with the developing optical technology.
 They have long lifetime. Hence they are recently used for optical communication medium
and telecommunications.
 The normalized frequency valus is V = 2.405
Cutoff Wavelength ( c)
1
2 a n1   2
c = 2
Vc

Divide both sides by V.

1
2
c 2 a n1  2  
=  1
V Vc 2
2 a n1  2 

c 
=
V Vc

V
c =
Vc

V
c 
2.405

For single mode fiber c is in range from 1.1 to 1.28 m.

1.40 OPTICAL COMMUNICATION

Propagation Modes in Single Mode Fibers

Two independent degenerate modes propagating in the fiber. One mode is in horizontal (H)
polarization and the another one is in vertical polarization (V)

y y

x
Core x

ny

Horizontal mode Vertical mode

Figure 1.18: Mode Propagation in SM fiber

The electric field of the light depends on there two modes.

Birefringence
These polarization modes propagate with different phase velocities. The Birefriengence is the
difference between their effective refractive indices.
B f  ny – nx

Fiber Beat Length

When light is injected into the fiber, both the modes are excited. When they propagate, due
the variation in phase velocity one will be delayed in phase relative to the other.
When this phase difference is an integral multiple of 2 radians, the two modes will beat at this
point and reproduce the input polarization. The length overwhich this beating occurs in known as
fiber beat length.
2
LP =

2
=
K0  n y – n x 
Introduction to Optical Fibers 1.41

2
K0 = = Free space propagation constant

2
LP = 2
 y
 n – nx 

=
n y – nx 

LP 
Bf
 = Optical signal wavelength
Large Birefringence
The modal birefringence is maximized by reducing the beat length LP to around 1mm or less.
Small Birefringence
The modal birefringence is minimized by increasing the beat length LP to around 50 m or more.

1.12.1 Definitions used in Single Mode Fibers

1. Mode Field Diameter (MFD)
MFD is the basic parameter of a single mode fiber. The characteristic of single mode fiber is
described by a Gaussian distribution and the distribution is given by

 r2 
E (r) = E 0 exp  2  ... (1)
 w0 
where
r  radius of fiber core.
E0  field strength at the axis or zero radius.
w0  is the width of the electric field distribution.
1.42 OPTICAL COMMUNICATION

Eo

–1/e 1/e

–10 –8 –6 –4 –2 0 2 4 6 8 10
MFD
2wo

Figure 1.19: Illustration of electric field distribution to find MFD

MFD is generally taken as the distance between the opposite 1e field amplitude points. The
MFD of LP01 mode can be defined

1
  3 2  2

 2  r E  r  dr 
2w 0   0 
 2

  r E  r  dr 
 0 

2. Mode Field Radius Spot Size

The spot size is denoted as “w0” and “w0” is the nominal half width of the input excitation.

3. Fiber Beat Length

Single mode fiber allow the propagation of two nearly degenerate mode with orthogonal
polarizations. Fiber behaves as a birefringent medium due to effective refractive index difference.
So two modes are excited in the fiber, one mode is delayed in phase relative to the other as they
propagate. When the phase difference between two modes is integral multiple of 2, the two
modes will beat at this point and the input polarization is reproduced. The length over which the
beating occurs is known as fiber beat length.

2
Lp =

Phase and Group velocity:

Phase velocity:
It is defined as the ratio of angular frequency of wave to propagation constant.

Phase velocity Vp = ω / β

Where, ω is the angular frequency of wave

𝛽 is the propagation constant

Group velocity:

Group of waves having similar frequencies does not travel at phase velocity of individual
waves but travel with a group velocity V g.

Group velocity is defined as the transmission velocity of wave packet, which is made of many
photons with different frequencies and phase velocities.
RELATIONSHIP BETWEEN ACCEPTANCE ANGLE ∅𝒂 AND NUMERICAL
APERTURE:

a = sin −1 ( NA)
When the acceptance angle is made to rotate about the fiber axis, we get the
acceptance cone.
When the fiber has larger acceptance cone, then larger amount of light rays enters
into the core.
The Normalized frequency number for step index fiber,
2𝜋𝑎
V= (NA)
λ
Where a= core radius
λ=wavelength
NA=Numerical Aperture
2𝜋𝑎
V= √𝑛12 − 𝑛22
λ

NA=𝑛1 √2∆

𝑛1 −𝑛2
Where index difference ∆=
𝑛1

Number of modes guided by step index fiber is given by

𝑉2
M=
2

ANNA UNIVERSITY FRE QUENTLY ASKED QUESTIONS

Part – A

1. State Snell’s Law. (Apr-May 2015 , 2017 ) (R)

2. What is total internal reflection? (Nov-Dec 2015) (R)
3. Define Numerical Aperture and Acceptance angle. (Nov-Dec 2014, 2016) (R)
4. Define Phase and group velocity. (Nov-Dec 2015), (May-June 2016) (R)
5. What is meant by linearly Polarized mode? (May-June 2013) (R)
 6. For n1 = 1.55 and n2 =1.52 , Calculate the critical angle and numerical aperture. (May-
June 2013) (U)
7. List any two advantages of single mode fibers. (Nov-Dec 2014) (U)
8. Calculate the critical angle of incidence between two substances with different refractive
indices, where n1 =1.5 and n2 =1.46. (Apr-May 2015, 2017) (U)
9. Calculate the cutoff wavelength of a single mode fibre with core radius of 4 m and
 = 0.003 (Nov-Dec 2012) (U)
10. For a Fibre with core refractive index of 1.54 and fractional refractive index difference
of 0.01 calculate its numerical aperture. (Nov-Dec 2012) (U)
11. The refractive indexes of the core and cladding of a silica fiber are 1.48 and 1.46
respectively. Find the acceptance angle for the fiber. (Nov-Dec 2013) (U)
12. Determine the normalized frequency at 820nm for a step index fiber having a 25 m
radius. The refractive indexes of the cladding and the core are 1.45 and 1.47
respectively. How many propagate in this fiber at 820nm? (Nov-Dec 2013) (U)
13. State the reasons to opt for optical fiber communication. (Apr-May 2018) (U)
14. Sketch the cross sectional view of the transverse electric filed vectors for the four lowest
order modes in a step index fiber. (Apr-May 2018) (U)
15. Why partial reflection does not suffice the propagation of light? (Nov-Dec 2017) (U)
16. A graded index optical fiber has a core with a parabolic index profile which has a
diameter of 50 µm. The fiber has a numerical aperture off 0.2. Calculate the total number
of guided modes in the fiber when it is operating at a wavelength of 1µm. (Nov-Dec
2017) (U)
17. What are the conditions for light to be propagated inside a fiber? (Nov-Dec 2016) (U)
18. What are the conditions for single mode propagation? (May-June 2016) (R)

Part – B

1. i) With the help of neat block diagram explain the different

components of an optical fiber link. (Nov-Dec 2013) (U)
ii)Compare the optical fiber link with a satellite link. (Nov-Dec 2013) (AZ)
2. Derive an expression for Acceptance angle and Numerical Aperture of a fiber with the help
of neat figure showing all the details. (Nov-Dec 2013), (May - Jun 2016) (U)
3. i) Explain the differences between meridional and skew rays. (Nov-Dec 2013) (U)
ii) Bring out the differences between phase and group velocities. (Nov-Dec 2013) (U)
4. i) Derive the mode equations for a circular fiber using maxwell’s equations.(May-June
2013) (A)
ii) Calculate the NA of a fiber having n1 = 1.6 and n2 = 1.49 and another fiber having n1 =
1.448 and n2 = 1.405. Which fiber has greater acceptance angle? (May-June 2013) (AZ)
5. i) Explain the ray theory of a fiber with a special mention about TIR, Acceptance angle and
 NA. (May-June 2013) (U)
ii) Describe single mode fibers and their mode field diameter. What are the propagation
modes in them. (May-June 2013) (U)
6. i) Starting from maxwell’s equation, derive an expression for wave equation of an
electromagnetic wave propagating through optical fiber.(Nov-Dec 2012), (May - Jun 2016)
(A)
ii) Describe the ray theory behind the optical fiber communication by total internal
reflection. State the application of snell’s law in it. (Nov-Dec 2012) (U)
7. i) A SI fiber with silica-core refractive index of 1.458, V=75 and NA=0.3 is to be operated
at 820 nm, what should be its core size and cladding refractive index? Calculate the total
number of modes entering this fiber. (Nov-Dec 2012) (AZ)
ii) Derive the expression of linearly polarized modes in optical fibers and obtain the equation
for V-number. (Nov-Dec 2012) (A)
8. For multi-mode step-index fibre with glass core (n1 =1.5) and a fused quartz cladding (n2

=1.46), determine the acceptance angle (𝜽𝒊𝒏 ) and numerical aperture. The source to fibre
medium is air. (Apr-May 2015) (A)
9. Explain the ray propagation into and down an optical fibre cable.
Also derive the expression for acceptance angle. (Apr-May 2015) (U)

10. Describe a step index and graded index cable. (Apr-May 2015) (U)

11.Contrast the advantages and disadvantages of step-index, graded-index, single-mode

propagation and multi-mode propagation. (Apr-May 2015) (U)

12.Classify fibers and explain them. (Nov-Dec 2015) (U)

13.Describe and derive the modes in planar guide. (Nov-Dec 2015), (Apr-May 2017) (AZ)

14.Define the normalized frequency for an optical fiber and explain its use. (Nov-Dec 2014) (U)

15.Discuss on the transmission of light through graded index fiber. (Nov-Dec 2014) (U)

16.Explain the features of multimode and single mode step index fiber and compare them.
(Nov-Dec 2014) (U)
17.A Single mode step index fiber has a core diameter of 7µm and a core refractive index of
1.49. Estimate the shortest wavelength of light which allows single mode operation when the
relative refractive index difference for fiber is 1%. (Nov-Dec 2014) (AZ)

18. Explain phase shift with total internal reflection and evanescent field. (Nov-Dec 2017) (U)

19. Discuss whether TEM waves exist in an optical fiber. If not what type of mode will
propagate in a practical optical fiber? (Nov-Dec 2017) (U)