TỔNG ÔN TẬP THI VÀO LỚP 10 MÔN TIẾNG ANH NĂM HỌC 2023 - 2024 CÓ ĐÁP ÁN (NGỮ Â...
Laser lecture 09 (applications, fiber optics)
1. 27/02/1437
1
LASER APPLICATIONS
Lecture -9
Dr. Mohamed Fadhali
1Dr. Fadhali - Laser Applications,
2014
Optical communication
Physics of Optical fibers
Dr. Fadhali - Laser Applications,
2014
2
Tyndall (UK), 1870
Light guiding in a
thin water jet
First experiment for Total
internal reflection
4. 27/02/1437
4
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.
7
Structure of fiber optics
8
1. According to the difference in refractive index profile
Optical Fibers and Wave Propagation
Dr. Fadhali - Laser Applications,
2014
5. 27/02/1437
5
9
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
Dr. Fadhali - Laser Applications,
2014
10
• 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)
6. 27/02/1437
6
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
11
From Snell’s Law
Dr. Fadhali - Laser Applications, 2014 12
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
7. 27/02/1437
7
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
Dr. Fadhali - Laser Applications,
2014
13
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 < 2.1m,4.2)(
2 2/12
2
2
1 nn
a
V
Solution:
Solution:
Dr. Fadhali - Laser Applications,
2014
14
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 1.3m
8. 27/02/1437
8
Dr. Fadhali - Laser Applications,
2014
15
,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 1.3m
Dr. Fadhali - Laser Applications,
2014
16
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