graded refractive-index
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Methodologies for Graded Refractive Index
Methodologies: • Ray Optics • WKB • Multilayer Modelling
Solution requires: • some knowledge of index profile 𝑛2 𝑥
Phase-change due to propagation
𝑘𝑥 𝑇 = 𝑡𝑎𝑛−1𝛾𝑠𝑘𝑥+ 𝑡𝑎𝑛−1
𝛾𝑐𝑘𝑥+𝑚 𝜋
=𝜔
𝑐 𝑛 𝑥 𝑐𝑜𝑠 𝜃 𝑥 =
=𝜔
𝑐 𝑛2 𝑥 − 𝑛 𝑥 𝑠𝑖𝑛 𝜃 𝑥 2 =
=𝜔
𝑐 𝑛2 𝑥 − 𝑁2
𝑘𝑥 𝑥
𝜃 𝑥𝑖
𝜃 𝑥𝑖+1
𝑘𝑥 𝑥 𝑇
𝑘𝑥 𝑥𝑖 Δ𝑥𝑖𝑖
𝜔
𝑐 𝑛2 𝑥 − 𝑁2 𝑑𝑥𝑥𝑡
0
𝑛 𝑥𝑖+1
𝑛 𝑥𝑖
𝑁
Cladding-Film Interface
𝑘𝑥 𝑇 = 𝑡𝑎𝑛−1𝛾𝑠𝑘𝑥+ 𝑡𝑎𝑛−1
𝛾𝑐𝑘𝑥+𝑚 𝜋
𝑎𝑡 𝑥 = 0 𝛾𝑐 =𝜔
𝑐𝑁2 − 𝑛𝑐
2
𝑘𝑥 =𝜔
𝑐𝑛2 𝑥 = 0 −𝑁2=
𝜔
𝑐𝑛𝑓2 − 𝑁2
𝑡𝑎𝑛−1𝛾𝑐
𝑘𝑥= 𝑡𝑎𝑛−1
𝑁2−𝑛𝑐2
𝑛𝑓2−𝑁2
≅𝜋
2
𝑁
Turning Point “Interface”
𝑘𝑥 𝑇 = 𝑡𝑎𝑛−1𝛾𝑠𝑘𝑥+ 𝑡𝑎𝑛−1
𝛾𝑐𝑘𝑥+𝑚 𝜋
𝑎𝑡 𝑥 = 𝑥𝑡
𝛾𝑠 =𝜔
𝑐𝑁2 − 𝑛2 𝑥 = 𝑥𝑡 − ∆𝑥
𝑘𝑥 =𝜔
𝑐𝑛2 𝑥 = 𝑥𝑡 + ∆𝑥 −𝑁
2
𝑁
𝑥𝑡
𝑡𝑎𝑛−1𝛾𝑐
𝑘𝑥= 𝑡𝑎𝑛−1
𝑁2−𝑛2 𝑥=𝑥𝑡−∆𝑥
𝑛2 𝑥=𝑥𝑡+∆𝑥 −𝑁2≅ 𝑡𝑎𝑛−1 1 =
𝜋
4
𝑛 𝑥 = 𝑥𝑡 = 𝑁
Bringing all the pieces together:
𝑘𝑥 𝑇 = 𝑡𝑎𝑛−1𝛾𝑠𝑘𝑥+ 𝑡𝑎𝑛−1
𝛾𝑐𝑘𝑥+𝑚 𝜋
𝜔
𝑐 𝑛2 𝑥 − 𝑁2 𝑑𝑥𝑥𝑡
0
=3
4+𝑚 𝜋
𝜔
𝑐 𝑛2 𝑥 − 𝑁2 𝑑𝑥𝑥𝑡
0
=𝜋
4+𝜋
2+ 𝑚 𝜋
𝑁
𝑥𝑡
dispersion relation for a graded-refractive index waveguide
Solving for TE modes
𝑑2𝐸𝑦 𝑥
𝑑𝑥2+𝜔2
𝑐2𝑛2 𝑥 − 𝑁2 𝐸𝑦 𝑥 = 0
𝑘𝑥 𝑥 =𝜔
𝑐 𝑛2 𝑥 − 𝑁2
𝑛 𝑥 = 𝑛𝑓 𝑘𝑥 𝑥 = 𝑘𝑥 𝐸𝑦 𝑥 = 𝐴 𝑒𝑗 𝑘𝑥 𝑥 If:
When: Δ 𝑛2 𝑥 − 𝑁2
Δ𝑥/𝜆 𝑥≪ 1
(constant) (constant) (constant)
𝐸𝑦 𝑥 = 𝐴 𝑥 𝑒𝑗 𝑘𝑥 𝑥 𝑥
𝑘𝑥 𝑥 𝑥 ≡ 𝜙 𝑥
𝐸𝑦 𝑥 = 𝐴 𝑥 𝑒𝑗 𝜙 𝑥
where:
Major steps in the derivation:
𝐸𝑦 𝑥 = 𝐴 𝑥 𝑒𝑗 𝜙 𝑥
𝑑2𝐸𝑦 𝑥
𝑑𝑥2+𝜔2
𝑐2𝑛2 𝑥 − 𝑁2 𝐸𝑦 𝑥 = 0
𝐴′′ + 2 𝑗 𝐴′𝜙′ + 𝑗 𝐴 𝜙′′ − 𝐴 𝜙′2= − 𝑘𝑥
2𝐴
𝐴′′ − 𝐴 𝜙′2= − 𝑘𝑥
2𝐴
2 𝐴′𝜙′ + 𝐴 𝜙′′ = 0
2 𝐴′𝜙′ + 𝐴 𝜙′′ = 0 𝐴2𝜙′ ′ = 0 𝐴 𝑥 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑘𝑥 𝑥
𝐴′′ − 𝐴 𝜙′2= − 𝑘𝑥
2𝐴 𝐴′′ = 𝜙′2− 𝑘𝑥
2 𝐴 𝜙′2− 𝑘𝑥
2 ≅ 0
𝜙 𝑥 = ± 𝑘𝑥 𝑥 𝑑𝑥
Cladding Region
𝑓𝑜𝑟 𝑥 < 0
𝑘𝑥 𝑥 =𝜔
𝑐𝑛2 𝑥 − 𝑁2 =
𝜔
𝑐𝑛𝑐2 − 𝑁2 = 𝑗
𝜔
𝑐𝑁2 − 𝑛𝑐
2= 𝑗 𝛾𝑐
𝐸𝑦 𝑥 =𝑐1
𝑗 𝛾𝑐𝑒− 𝛾𝑐
0𝑥 𝑑𝑥 +
𝑐2
𝑗 𝛾𝑐𝑒+ 𝛾𝑐
0𝑥 𝑑𝑥 𝐸𝑦 𝑥 =
𝐴
𝛾𝑐 𝑒 𝛾𝑐 𝑥
𝐸𝑦 𝑥 =𝑐1
𝑘𝑥 𝑥𝑒+𝑗 𝑘𝑥 𝑥 𝑑𝑥 +
𝑐2
𝑘𝑥 𝑥𝑒−𝑗 𝑘𝑥 𝑥 𝑑𝑥
𝑘𝑥 𝑥 =𝜔
𝑐 𝑛2 𝑥 − 𝑁2
𝑥
Guiding Region
𝑓𝑜𝑟 0 < 𝑥 < 𝑥𝑡
𝑘𝑥 𝑥 =𝜔
𝑐𝑛2 𝑥 − 𝑁2
𝐸𝑦 𝑥 =𝑐1
𝑘𝑥 𝑥𝑒+𝑗 𝑘𝑥 𝑥 𝑑𝑥 +
𝑐2
𝑘𝑥 𝑥𝑒−𝑗 𝑘𝑥 𝑥 𝑑𝑥
𝑘𝑥 𝑥 =𝜔
𝑐 𝑛2 𝑥 − 𝑁2
𝐸𝑦 𝑥 =𝐵
𝑘𝑥 𝑥𝑒+𝑗 𝑘𝑥 𝑥
𝑥𝑡𝑥 𝑑𝑥 +
𝐶
𝑘𝑥 𝑥𝑒−𝑗 𝑘𝑥 𝑥
𝑥𝑡𝑥 𝑑𝑥
𝑥
Beyond Turning Point
𝑓𝑜𝑟 𝑥 > 𝑥𝑡
𝐸𝑦 𝑥 =𝑐1
𝑘𝑥 𝑥𝑒+𝑗 𝑘𝑥 𝑥 𝑑𝑥 +
𝑐2
𝑘𝑥 𝑥𝑒−𝑗 𝑘𝑥 𝑥 𝑑𝑥
𝑘𝑥 𝑥 =𝜔
𝑐 𝑛2 𝑥 − 𝑁2
𝐸𝑦 𝑥 =𝐷
𝛾𝑠 𝑥𝑒− 𝛾𝑠 𝑥𝑥𝑥𝑡
𝑑𝑥
𝑥
𝑘𝑥 𝑥 =𝜔
𝑐𝑛2 𝑥 − 𝑁2 = 𝑗
𝜔
𝑐𝑁2 − 𝑛2 𝑥 = 𝑗 𝛾𝑠 𝑥
Transverse confinement along x-axis, tangential Hy
Region I: 𝐻𝑦 𝑥, 𝑦 = 𝐻1 𝑐𝑜𝑠 𝑘𝑥 𝑥 + 𝜙1
Region II :
Region III:
𝐻𝑦 𝑥, 𝑦 = 𝐻2 𝑒𝛾𝑥,2 𝑥+𝑇
𝐻𝑦 𝑥, 𝑦 = 𝐻3 𝑒−𝛾𝑥,3 𝑥
−𝑇 < 𝑥 < 0
𝑥 < −𝑇
𝑥 > 0
𝑘𝑥 𝑇 = 𝑡𝑎𝑛−1𝛾𝑥,2𝑛22
𝑛12
𝑘𝑥+ 𝑡𝑎𝑛−1
𝛾𝑥,3𝑛32
𝑛12
𝑘𝑥+ 𝑝 𝜋
Lateral confinement along y-axis, tangential Ex
Region I: 𝐸𝑥 𝑥, 𝑦 = 𝐸1 𝑐𝑜𝑠 𝑘𝑦 𝑦 + 𝜙2
−𝑊
2< 𝑥 <
𝑊
2
𝑘𝑦 𝑊 = 𝑡𝑎𝑛−1𝛾𝑦,4𝑘𝑦+ 𝑡𝑎𝑛−1
𝛾𝑦,5𝑘𝑦+ 𝑞 𝜋
Region IV : 𝐸𝑥 𝑥, 𝑦 = 𝐸4 𝑒
− 𝛾𝑦,4 𝑦−𝑊2
𝑦 < −𝑊
2
𝐸𝑥 𝑥, 𝑦 = 𝐸5 𝑒− 𝛾5,𝑦 𝑦+
𝑊2
Region V:
𝑦 >𝑊
2
Finding the propagation constants:
𝑘𝑥 𝑇 = 𝑡𝑎𝑛−1𝛾𝑥,2𝑛22
𝑛12
𝑘𝑥+ 𝑡𝑎𝑛−1
𝛾𝑥,3𝑛32
𝑛12
𝑘𝑥+ 𝑝 𝜋
𝑘𝑦 𝑊 = 𝑡𝑎𝑛−1𝛾𝑦,4
𝑘𝑦+ 𝑡𝑎𝑛−1
𝛾𝑦,5
𝑘𝑦+ 𝑞 𝜋
𝐹1 𝑘𝑥, 𝛾𝑥,2, 𝛾𝑥,3, 𝛽 = 0
𝑛12 𝜔2
𝑐2= 𝑘1
2 = 𝑘𝑥2 + 𝑘𝑦
2 + 𝛽2 𝐺1 𝑘𝑥, 𝑘𝑦 , 𝛽 = 0
𝐹2 𝑘𝑦, 𝛾𝑥,4, 𝛾𝑥,5, 𝛽 = 0
𝑛22 𝜔2
𝑐2= 𝑘2
2 = −𝛾𝑥,22 + 𝑘𝑦
2 + 𝛽2
𝑛32 𝜔2
𝑐2= 𝑘3
2 = −𝛾𝑥,32 + 𝑘𝑦
2 + 𝛽2
𝑛42 𝜔2
𝑐2= 𝑘4
2 = 𝑘𝑥2 − 𝛾𝑦,4
2+ 𝛽2
𝑛52 𝜔2
𝑐2= 𝑘5
2 = 𝑘𝑥2 − 𝛾𝑦,5
2+ 𝛽2
𝐺2 𝛾𝑥,2, 𝑘𝑦 , 𝛽 = 0
𝐺3 𝛾𝑥,3, 𝑘𝑦 , 𝛽 = 0
𝐺4 𝑘𝑥, 𝛾𝑦,4, 𝛽 = 0
𝐺5 𝑘𝑥, 𝛾𝑦,5, 𝛽 = 0
TM-like modes
𝑛𝑐 , 𝑛𝑓, 𝑇 , 𝑛𝑠
𝐹 𝑎𝑀, 𝑏𝑀 𝑁𝐼 , 𝑉𝐼 = 0
I)
𝑁𝐼
II) 𝑛𝑠, 𝑁𝐼,𝑊 , 𝑛𝑠
𝐹 𝑎𝐸 , 𝑏𝐸 𝑁𝐼𝐼 , 𝑉𝐼𝐼 = 0
𝑁𝐼𝐼
(TM)
(TE)
Criteria for Single-Mode Operation
𝑛𝑓 = 𝑛𝑠 + ∆𝑛 ∆𝑛 ≪ 𝑛𝑠 with
I) cut-off condition for mode 𝑝 :
𝑏𝐼,𝑝 = 0
𝑉𝐼,𝑝 = 𝑡𝑎𝑛−1 𝑎𝐼 + 𝑝 𝜋
Requirement for existence of only one mode in transverse direction:
Transverse confinement:
𝑡𝑎𝑛−1 𝑎𝐼 < 𝑉𝐼 < 𝑡𝑎𝑛−1 𝑎𝐼 + 𝜋
Transverse confinement:
Lateral confinement:
II) cut-off condition for mode 𝑞 :
𝑏𝐼𝐼,𝑞 = 0
𝑉𝐼𝐼,𝑞 = 𝑡𝑎𝑛−1 𝑎𝐼𝐼 + 𝑞 𝜋
0 < 𝑉𝐼𝐼 < 𝜋
Lateral confinement:
𝑎𝐼𝐼 = 0 it is a symmetric waveguide, so we have:
Requirement for existence of only one mode in lateral direction:
A little bit of algebra leads to:
𝑉𝐼 =𝜔
𝑐𝑇 𝑛𝑓
2 − 𝑛𝑠2
𝑉𝐼𝐼 =𝜔
𝑐𝑊 𝑁𝐼
2 − 𝑛𝑠2 =𝜔
𝑐𝑊 𝑏𝐼 𝑛𝑓
2 − 𝑛𝑠2 =
𝑊
𝑇 𝑉𝐼 𝑏𝐼
𝑏𝐼 ≅𝑁𝐼2 − 𝑛𝑠
2
𝑛𝑓2 − 𝑛𝑠
2
0 < 𝑉𝐼𝐼 < 𝜋 0 <𝑊
𝑇<𝜋
𝑉𝐼 𝑏𝐼
𝑡𝑎𝑛−1 𝑎𝐼 < 𝑉𝐼 < 𝑡𝑎𝑛−1 𝑎𝐼 + 𝜋
and
𝑎𝐼 ≅ ∞
single-mode region
Optical Fibers st
ep-
index
multimod
e
step-
index
sing
lemod
e
GRIN
a cylindrical dielectric waveguide
Modes in Optical Fibers
𝜕2𝐸 𝑥, 𝑦
𝜕𝑥2+𝜕2𝐸 𝑥, 𝑦
𝜕𝑦2+𝑛2𝜔2
𝑐2 − 𝛽2 𝐸 𝑥, 𝑦 = 0
𝜕2𝐸 𝑟, 𝜙
𝜕𝑟2+1
𝑟
𝜕𝐸 𝑟, 𝜙
𝜕𝑟+1
𝑟2𝜕2𝐸 𝑟, 𝜙
𝜕𝜙2+𝑛2𝜔2
𝑐2 − 𝛽2 𝐸 𝑟, 𝜙 = 0
Cartesian coordinates
Cylindrical coordinates
𝑬 𝑥, 𝑦, 𝑧, 𝑡 = 𝐸 𝑥, 𝑦 𝑒𝑗 𝜔 𝑡 − 𝛽 𝑧
𝑬 𝑟, 𝜙, 𝑧, 𝑡 = 𝐸 𝑟, 𝜙 𝑒𝑗 𝜔 𝑡 − 𝛽 𝑧
Solutions
𝑑2𝑢
𝑑𝑟2+1
𝑟
𝑑𝑢
𝑑𝑟+𝑛2𝜔2
𝑐2 − 𝛽2 −
𝑙2
𝑟2𝑢 = 0
𝐸 𝑟, 𝜙 = 𝑢 𝑟 𝑒 𝑗 𝑙 𝜙 𝑒
𝑘𝑇2 ≡𝑛𝑐𝑜2𝜔2
𝑐2 − 𝛽2
𝛾2 ≡ 𝛽2 −𝑛𝑐𝑙2𝜔2
𝑐2
Boundary conditions for
𝐸𝑧, 𝐻𝑧, 𝐸𝜙, 𝐻𝜙
Power Confinement
Right above the cut-off, very little power is inside the core. As the core diameter increases, the power of the mode becomes confined inside the core.
Fraction of the power propagating inside the core against the V-number.
Number of Guided Modes in an Optical Fiber
𝑀 = 4
𝜋2𝑉2
𝑉 = 2𝜋 𝑎
𝜆 𝑛𝑐𝑜
2 − 𝑛𝑐𝑙2 = 2𝜋
𝑎
𝜆𝑁𝐴
V-number
𝑁𝐴 = 𝑛0 sin 𝜃0 = 𝑛𝑐𝑜2 − 𝑛𝑐𝑙
2
Numerical Aperture
𝑉 < 2.405
single-mode operation
𝔼𝑡 𝑥, 𝑦, 𝑧 = 𝑎𝛼 𝑧
𝛼
𝑬𝑡,𝛼 𝑥, 𝑦 𝑒−𝑗 𝛽𝛼 𝑧 + radiation modes
ℍ𝑡 𝑥, 𝑦, 𝑧 = 𝑎𝛼 𝑧
𝛼
𝑯𝑡,𝛼 𝑥, 𝑦 𝑒−𝑗 𝛽𝛼 𝑧 + 𝑟𝑎𝑑𝑖𝑎𝑡𝑖𝑜𝑛 𝑚𝑜𝑑𝑒𝑠
Decomposition into the eigenmodes of the original structure
After a Careful (& Long) Analysis, Final Result:
± 𝑑𝑎𝜇𝑑𝑧= −𝑗 𝑎𝛼 𝑧 𝑒
−𝑗 𝛽𝛼−𝛽𝜇 𝑧
𝛼
Κ𝜇,𝛼
4 Κ𝑡𝜇,𝛼 ≡ 𝜔 ∆𝜖 𝑥, 𝑦, 𝑧 𝑬𝑡,𝜇∗∙ 𝑬𝑡,𝛼
∞
−∞
𝑑𝑥 𝑑𝑦
4Κ𝑧𝜇,𝛼 ≡ 𝜔 𝜖 ∆𝜖
𝜖 + ∆𝜖𝑬𝑧,𝜇
∗
∙ 𝑬𝑧,𝛼
∞
−∞
𝑑𝑥 𝑑𝑦
Κ𝜇,𝛼 ≡ Κ𝑡𝜇,𝛼+ Κ𝑧𝜇,𝛼
Κ𝜇,𝛼 = Κ𝛼,𝜇∗ whenever 𝜖 is a real number, then
Co-Directional Couplers:
𝐴 𝑧 2
𝐵 𝑧 2
𝐹 ≡Κ2
𝛽𝑐2 =
Κ2
Κ 2 + ∆2
1 − 𝐹 𝑠𝑖𝑛2 𝛽𝑐 𝑧 𝑐𝑜𝑠2 𝛽𝑐 𝑧 +∆2
𝛽𝑐2𝑠𝑖𝑛2 𝛽𝑐 𝑧 = =
𝐹 𝑠𝑖𝑛2 𝛽𝑐 𝑧 = = Κ2
𝛽𝑐2 𝑠𝑖𝑛
2 𝛽𝑐 𝑧
𝐹 = 0.2 𝐹 = 0.8
𝐹 = 1
𝐴 𝑧 2
𝐵 𝑧 2
∆ ≡ 𝛽𝑏 −𝛽𝑎2= 0
when: 𝛽𝑏 = 𝛽𝑎
𝜋
2
𝑧𝛽𝑐
𝑧𝛽𝑐
𝛽𝑐 ≡ Κ2 + ∆2= Κ 𝐿 =
𝜋
2 𝛽𝑐=𝜋
2 Κ
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