Laser Physics- lecture notes

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LASER APPLICATIONS
Lecture -9
Dr. Mohamed Fadhali
1Dr. Fadhali - Laser Applications,
2014
Optical communication
Physics of Optical fibers
Dr. Fadhali - Laser Applications,
2014
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Tyndall (UK), 1870
Light guiding in a
thin water jet
First experiment for Total
internal reflection

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Dr. Fadhali - Laser Applications,
2014
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Optical Communication System
Optical
Transmitter
Comm.
Channel
Optical
Receiver
OutputElectrical
Signal Input
• Wavelength: 0.85, 1.3, 1.55
• SM & MM
• DFB, VCSEL, FP, DBR
• Modulation: Direct, external, integrated
modulator
Considerations:
• SM & MM
• Dispersion
• Loss
Optical fiber communication
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2014
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Advantages of fiber optics

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Core – Thin glass center of the
fiber where light travels.
Cladding – outer optical material
surrounding the core
Buffer Coating – plastic coating that
protects the fiber.
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Structure of fiber optics
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1. According to the difference in refractive index profile
Optical Fibers and Wave Propagation
Dr. Fadhali - Laser Applications,
2014

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2. According to number of propagated modes:
1) Single mode fiber(SMF)
A single mode(fundamental mode) can propagate in the fiber.
2) Multi-mode fiber(MMF)
Multi-modes can propagate in the fiber.
Typical dimensions of different fibers
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• Rays bend at interfaces between surfaces of different refractive
indices
– Snell’s law
• Light remains confined in the core if the propagation angle is
greater than the critical angle





 
2
1
21
1
sinsin  n
n
1 1 2 2sin sinn n 
1 2
1
sinc
n
n
      
 
1
2
sin
n
n
c 221 90sinsin nnn c 
Guiding light using fibers (rays analysis)

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Dr. Fadhali - Laser Applications, 2014
Acceptance angle: (the max angle for coupling light into the fiber)
Let’s consider the meridional rays
α c
n2
n2
α0
θc n1
n0
0 maximum coupling angle of the ray into the fiber
c is the critical angle for total internal reflection.
c is the critical propagation angle
  ccc
nnnn  cos2/sinsinsin 11100

n0  1 for air
 
2
2 2 22
0 0 1 1 1 1 2
1
sin cos 1 sin 1c c
n
n n n n n n
n
  
             
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From Snell’s Law
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Numerical Aperture (NA) of the fiber
NA = sine of the half of the maximum acceptance angle,
typically, NA=0.15 for single mode fiber and 0.3 for MMF
   2 2 2 1 2
1 2 1 2 1 2 1
1
2
n n
NA n n n n n n n
n

     
2 2 1/2 1/2
1 2 1
2 2
1 2 1 2
2
11
( ) (2 )
where
2
NA n n n
n n n n
nn
   
 
  
Assume that ∆ is the partial
variation of refractive index

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V-number (normalized frequency)
,)/996.01428.1( 2
Vb 
,)(
2 2/12
2
2
1 nn
a
V 


,41.2)(
2 2/12
2
2
1  nn
a
V
c
cutoff


Number of modes when V>>2.41
,
2
2
V
M 
Normalized propagation constant
for V between 1.5 – 2.5.
Mode field diameter (MFD)
),
1
1(22
V
aw 
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Optical fiber parameters
a: fiber radius
Examples --- single mode and multi-mode fibers
1. Calculate the number of allowed modes in a multimode
optical fiber of radius a = 100 m. The refractive indices of Its core and
cladding are 1.468 and 1,447 respectively for the wavelength of 850 nm
,44.91)(
2 2/12
2
2
1  nn
a
V


,4181
2
2

V
M
Solution:
a < 2.1m,4.2)(
2 2/12
2
2
1  nn
a
V


Solution:
Solution:
Dr. Fadhali - Laser Applications,
2014
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2. What should be the value of the fiber radius of a single mode fiber
with refractive indices of core and cladding of 1.468 and1.447
respectively for the wavelength of 1.3m

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2014
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,1.10)/11(22 0 mVaw 
3. Calculate the mode field diameter of a single mode fiber
with a core refractive index of 1.458 and cladding refractive
index of 1.452 for the wavelength of 1.3m
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HW1
A step index fiber of core radius and index 5um and 1.46 respectively. If the
refractive index fractional difference is 0.01 , then find the numerical aperture
and the maximum operating frequency for single mode propagation
1 2 2
2 1
1 1
1 0.01 0.99 0.99(1.46) 1.45
n n n
n n
n n

        
1 2 1.46 2(0.01) 0.206NA n   
max
max 1 1
min
6
max8
14
max
22
2.405 2.405 2 2
2
(5 10 )(0.2064)
3 10
1.112 10
a
V V n a n
c
Hz






      
  

  