Engineering Physics-II, As per RTU Syllabus

1. Coherence & Optical Fibres
Unit-III (B.Tech-II sem)

2. COHERENCE
Two wave sources are perfectly coherent if they have a constant phase
difference and the same frequency. Coherence is an ideal property
of waves that enables interference.
Coherent sources are those which emits light waves of same wave length
or frequency and have a constant phase difference.

4. Types of Coherence
Temporal Coherence
• It is measure of ability of a beam to interfere
of another portion of it self
Spatial Coherence
• It refer to ability of two separate portion of
wave to produce interference

5. Temporal Coherence Spatial Coherence
.
Temporal Coherence is concerned with the
phase correlation of waves at a given point in
space at two different instant of time.
Spatial coherence is concerned with the phase
correlation of two different points across a
wave front at a given instant of time
The type of coherence related with time The type of coherence related with position
It is known as longitudinal coherence. It is known as transverse coherence
The temporal coherence of light is related to
frequency bandwidth of the source.
monocromaticity
Spatial coherence is related to size of light
source
.
Temporal coherence measure in interferometer
such as Michelson Interferometer
Spatial coherence measure in interferometer
such as Young’s Double Slit Interferometer
.

6. Coherence time, Coherence Length and Spectral Purity Factor
The average time interval for which definite phase relationship exists in knowledge is known as
coherence time 
The distance L for which the wave field remains sinusoidal is given by coherence Length L
cL 
Coherence Length In Terms of Frequency






1
t
cL


c

----------------Eq 1
----------------Eq 2
----------------Eq 3
By Differentiate equation 3 we got




  22
cc
& Spectral Purity Factor is given by



Q
So
Q
c
cL 



 




2
QL 
c
Q
c
L 
 
---------Eq 4
---------Eq 5
---------Eq 6
---------Eq 7
---------Eq 8

7. Problems & Solutions
Q.1. Light of wavelength 4800Å has a length of 25 waves. What is the coherence length and
coherence time.
Sol:- Given Wavelength 4800Å so the coherence length L=25 X 4800 X 10-10 m
= 12 X 10-6m
So coherence Time s
sm
m
C
L 14
8
6
100.4
/103
1012 





Q.2. Imagine that we crop a continuous laser beam ( to be perfectly mochromatic
wavelength 6000Å) into 0.5ns pulse using some sort of shutter. Compute the coherence
length, band width and the line width.
Sol:- Given Wavelength 6000Å, time = 0.5 X 10 -9s,
msmscL 15.0/103105.0 89
 

zH
s
9
9
102
105.0
11


 


m
sm
Hm
c
z 13
8
9272
1016.28
/103
102)105.6( 










8. Interference Visibility
A quantitative measurement of the coherence of a light source is equal to the visibility V of
the fringes.
The fringe visibility is defined as
minmax
minmax
II
II
V


 ---------- eq (1)
A Value V ranging between 0 to 1 if V = 1 implies very high contrast fringes, the fringes
are washed away when V=0
A value greater than 0.88 indicates that light is highly coherent.

9. Interference Visibility as a measure of Coherence
Let us consider two interfering waves, each of intensity I0 . Both the waves consists of
coherent part Ic and Incoherent part Iinc .
If C degree of coherence then.
oc CII 
oinc ICI )1( 
------------- eq . 1
------------- eq . 2
When superimposed. The coherence part shall interfere adding there by their amplitudes
where as for the incoherent parts the intensities are simply added. Thus
incc III 24max 
incII 20min 
------------- eq . 3
------------- eq . 4
Putting the values from eq 1 and 2 to eq 3 and 4
oo ICCII )1(24max 
oICI )1(20min 
------------- eq . 5
------------- eq . 6
As we know interference visibility given as
C
ICICCI
ICICCI
II
II
V
oo
ooo







)1(2)1(24
)1(2)1(24
minmax
minmax
Therefore it is seen that degree of contrast
is a measure of degree of coherence
between waves of equal intensities000

10. Optical Fibre (Introduction)
• An optical fiber is essentially a waveguide for light
• It consists of a core and cladding that surrounds the core
• The index of refraction of the cladding is less than that of the core,
causing rays of light leaving the core to be refracted back into the core
• A light-emitting diode (LED) or laser diode (LD) can be used for the
source
• Advantages of optical fiber include:
– Greater bandwidth than copper
– Lower loss
– Immunity to crosstalk
– No electrical hazard

11. Optical Fiber Construction

12. Daniel Colladon first described
this “light fountain” or “light pipe”
in an 1842 article titled On the
reflections of a ray of light inside a
parabolic liquid stream. This
particular illustration comes from a
later article by Colladon, in 1884.
A wall-mount cabinet containing
optical fiber interconnects. The
yellow cables are single mode
fibers; the orange and aqua cables
are multi-mode fibers: 50/125 µm
OM2 and 50/125 µm OM3 fibers
respectively.

13. Optical Fiber
• Optical fiber is made from thin strands of either glass or plastic
• It has little mechanical strength, so it must be enclosed in a
protective jacket
• Often, two or more fibers are enclosed in the same cable for
increased bandwidth and redundancy in case one of the fibers
breaks
• It is also easier to build a full-duplex system using two fibers, one
for transmission in each direction

14. Optical Fibre Working Principle
• Optical fibers work on the principle of total internal reflection
• With light, the refractive index is listed
• The angle of refraction at the interface between two media is governed by
Snell’s law:
n1 sin1  n2 sin2

15. Types of Fiber on the basis of modes of propagation
In digital multimode fiber systems, a light pulse separates into multiple spatial paths or
modes. Each component reaches the receiver at a slightly different time, broadening the
received pulse. Single-mode fiber solves the differential mode delay problem, allowing
data rates to be increased until chromatic dispersion.
Single mode fiber has a lower power loss
characteristic than multimode fiber, which
means light can travel longer distances
through it than it can through multimode fiber.
Within a data center, it's typical to use
multimode which can get you 300-400 meters.
If you have very long runs or are connecting
over longer distance, single mode can get you
10km, 40km, 80km, and even farther.
The multi-mode fiber has has much larger core
diameter than single mode fiber.
The core diameter of multimode fiber is
typically 50–100 micrometers, while that of
single mode fiber is between 8 and 10.5
micrometers.

16. Modes and Materials
• Since optical fiber is a waveguide, light can propagate in a number of
modes
• If a fiber is of large diameter, light entering at different angles will
excite different modes while narrow fiber may only excite one mode
• Multimode propagation will cause dispersion, which results in the
spreading of pulses and limits the usable bandwidth
• Single-mode fiber has much less dispersion but is more expensive to
produce. Its small size, together with the fact that its numerical
aperture is smaller than that of multimode fiber, makes it more
difficult to couple to light sources

17. Types of Fiber on the basis on Index
• In step-index fibers the index of refraction changes radically between the core
and the cladding.
• Graded-index fiber is a compromise multimode fiber, but the index of
refraction gradually decreases away from the center of the core
• Graded-index fiber has less dispersion than a multimode step-index fiber

18. Numerical Aperture and Angle of Acceptance
• The numerical aperture of the fiber is
closely related to the critical angle and
is often used in the specification for
optical fiber and the components that
work with it
• The numerical aperture is given by the
formula:
• The angle of acceptance is twice that
given by the numerical aperture
22
sin.. cladcoreoiAN  

19. i
r
P R
Q
Incident ray
Launching
end
Total Internally
reflected ray Axis of
Fibre
µ core
Cladding µclad
Normal
Reflected ray
µo
θ
Numerical Aperture and Angle of Acceptance
Consider a cylindrical fibre wire which consists of an inner core of refractive index µcore
and an outer cladding of refractive index µclad.
Let µo. be the refractive index of the medium from which the light ray enters the fibre. This
end is known as launching end. Let a ray of light enters the fibre at an incidence angle of i.
to the axis of fibre as shown in figure. This ray refracted at an angle r and strikes the core-
cladding interface at an angle θ. Let θ is greater than critical angle θc. As long as the angle θ
is greater than critical angle θc, the light will stay within the optical fibre.

20. Now we shall calculate the angle of incidence i for which θ is greater than and equal to θc So
that the light remains within the core.
Applying the snell’s law of refraction at the
point of entry P.
ri coreo sinsin   ----------- eq 1
From triangle PQR it is seen that
o
r 90 ,  o
r 90
Or
 cos)90sin(sin  o
r ----------- eq 2
Substituting the value of sin r from equation
2 to eq 1
 cossin coreo i  ----------- eq 3



cossin
o
core
i 
If Incidence angle i is increased beyond a
limit , θ will drop below the critical value θc
and the ray will escape from the side walls of
the fibre. The largest value of i (imax) occures
when θ = θc.
----------- eq 4
So the eq 4 can be written as
c
o
core
i 


cossin max  ----------- eq 5
Applying the snell’s law of refraction
at the core cladding interface
o
cladccore 90sinsin  
core
clad
c


 sin
2
2
1cos
core
clad
c


  ----------- eq 6
So the eq 5 can be written as
2
22
2
2
max 1sin
o
cladcore
core
clad
o
core
i





 

NAi cladcore  22
maxsin 
221
max sin cladcorei   
Here imax is the angle of acceptance

21. So the angle of acceptance is defined as the maximum angle that a light ray can have relative
to the axis of the fibre and propagate down the fibre.
Numerical Aperture:- It is also known as figure of merit for optical fibre. It is defined as sine
of acceptance angle. NA= io = imax
Propagation Condition
NA in terms of Fractional Refractive index Δ
If i is the angle of incidence of an incident ray, then the ray will be able to propagate only if
i < io or sin i < sin io
or 22
sin cladcorei  
core
cladcore

 

The fractional refractive index change Δ is defined as the ratio of refractive index difference
between core and cladding to the refractive index of core. It is expressed as
core
cladcore
core
cladcore
cladcorecladcorecladcoreNA 



 2
2
)()(
))((22 

core
cladcore
NA 

2
2
)( 

Or if
core
cladcore




2
)(

22.  22 2
corecoreNA 
than
V- Number
This is an important parameter of optical fibre given by the relation
222
cladcore
a
V 


 Where a is the radius of the core
and λ is free space wave length.
The maximum number of modes (Nm) supported by a single mode step index fibre is
determined by.
2
2
1
VNm 
If V< 2.405, the fibre will support only one mode and known as single mode optical fibre
If V>2.405, the fibre will support many modes simultaneously. This is known as multi-
mode fibre
The wavelength corresponding to the value V=2.0405 known as cutoff wavelength this
is expressed as
405.2
V
c

 

23. Problems and Solutions
Q.1. Calculate the refractive indices of the core and cladding material of a fibre from the
following data NA=0.22 and Δ=0.012
Sol:- Given NA=0.22 and Δ=0.012
We know the relation
 22 2
corecoreNA 
42.1
012.02
22.0
2





NA
core
core
cladcore

 

42.1
42.1
012.0 clad
so
40.1clad

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