Graded Refractive-Index
Transcript of Graded Refractive-Index
Methodologies for Graded Refractive Index
Methodologies: β’ Ray Optics β’ WKB β’ Multilayer Modelling
Solution requires: β’ some knowledge of index profile π2 π₯
Phase-change due to propagation
ππ₯ π = π‘ππβ1πΎπ ππ₯+ π‘ππβ1
πΎπππ₯+π π
=π
π π π₯ πππ π π₯ =
=π
π π2 π₯ β π π₯ π ππ π π₯ 2 =
=π
π π2 π₯ β π2
ππ₯ π₯
π π₯π
π π₯π+1
ππ₯ π₯ π
ππ₯ π₯π Ξπ₯ππ
π
π π2 π₯ β π2 ππ₯π₯π‘
0
π π₯π+1
π π₯π
π
Cladding-Film Interface
ππ₯ π = π‘ππβ1πΎπ ππ₯+ π‘ππβ1
πΎπππ₯+π π
ππ‘ π₯ = 0 πΎπ =π
ππ2 β ππ
2
ππ₯ =π
ππ2 π₯ = 0 βπ2=
π
πππ2 β π2
π‘ππβ1πΎπ
ππ₯= π‘ππβ1
π2βππ2
ππ2βπ2
β π
2
π
Turning Point βInterfaceβ
ππ₯ π = π‘ππβ1πΎπ ππ₯+ π‘ππβ1
πΎπππ₯+π π
ππ‘ π₯ = π₯π‘
πΎπ =π
ππ2 β π2 π₯ = π₯π‘ β βπ₯
ππ₯ =π
ππ2 π₯ = π₯π‘ + βπ₯ βπ
2
π
π₯π‘
π‘ππβ1πΎπ
ππ₯= π‘ππβ1
π2βπ2 π₯=π₯π‘ββπ₯
π2 π₯=π₯π‘+βπ₯ βπ2β π‘ππβ1 1 =
π
4
π π₯ = π₯π‘ = π
Bringing all the pieces together:
ππ₯ π = π‘ππβ1πΎπ ππ₯+ π‘ππβ1
πΎπππ₯+π π
π
π π2 π₯ β π2 ππ₯π₯π‘
0
=3
4+π π
π
π π2 π₯ β π2 ππ₯π₯π‘
0
=π
4+π
2+ π π
π
π₯π‘
dispersion relation for a graded-refractive index waveguide
Solving for TE modes
π2πΈπ¦ π₯
ππ₯2+π2
π2π2 π₯ β π2 πΈπ¦ π₯ = 0
ππ₯ π₯ =π
π π2 π₯ β π2
π π₯ = ππ ππ₯ π₯ = ππ₯ πΈπ¦ π₯ = π΄ ππ ππ₯ π₯ If:
When: Ξ π2 π₯ β π2
Ξπ₯/π π₯βͺ 1
(constant) (constant) (constant)
πΈπ¦ π₯ = π΄ π₯ ππ ππ₯ π₯ π₯
ππ₯ π₯ π₯ β‘ π π₯
πΈπ¦ π₯ = π΄ π₯ ππ π π₯
where:
Major steps in the derivation:
πΈπ¦ π₯ = π΄ π₯ ππ π π₯
π2πΈπ¦ π₯
ππ₯2+π2
π2π2 π₯ β π2 πΈπ¦ π₯ = 0
π΄β²β² + 2 π π΄β²πβ² + π π΄ πβ²β² β π΄ πβ²2= β ππ₯
2π΄
π΄β²β² β π΄ πβ²2= β ππ₯
2π΄
2 π΄β²πβ² + π΄ πβ²β² = 0
2 π΄β²πβ² + π΄ πβ²β² = 0 π΄2πβ² β² = 0 π΄ π₯ =ππππ π‘πππ‘
ππ₯ π₯
π΄β²β² β π΄ πβ²2= β ππ₯
2π΄ π΄β²β² = πβ²2β ππ₯
2 π΄ πβ²2β ππ₯
2 β 0
π π₯ = Β± ππ₯ π₯ ππ₯
General Solution
πΈπ¦ π₯ =π1
ππ₯ π₯π+π ππ₯ π₯ ππ₯ +
π2
ππ₯ π₯πβπ ππ₯ π₯ ππ₯
ππ₯ π₯ =π
π π2 π₯ β π2
Cladding Region
πππ π₯ < 0
ππ₯ π₯ =π
ππ2 π₯ β π2 =
π
πππ2 β π2 = π
π
ππ2 β ππ
2= π πΎπ
πΈπ¦ π₯ =π1
π πΎππβ πΎπ
0π₯ ππ₯ +
π2
π πΎππ+ πΎπ
0π₯ ππ₯ πΈπ¦ π₯ =
π΄
πΎπ π πΎπ π₯
πΈπ¦ π₯ =π1
ππ₯ π₯π+π ππ₯ π₯ ππ₯ +
π2
ππ₯ π₯πβπ ππ₯ π₯ ππ₯
ππ₯ π₯ =π
π π2 π₯ β π2
π₯
Guiding Region
πππ 0 < π₯ < π₯π‘
ππ₯ π₯ =π
ππ2 π₯ β π2
πΈπ¦ π₯ =π1
ππ₯ π₯π+π ππ₯ π₯ ππ₯ +
π2
ππ₯ π₯πβπ ππ₯ π₯ ππ₯
ππ₯ π₯ =π
π π2 π₯ β π2
πΈπ¦ π₯ =π΅
ππ₯ π₯π+π ππ₯ π₯
π₯π‘π₯ ππ₯ +
πΆ
ππ₯ π₯πβπ ππ₯ π₯
π₯π‘π₯ ππ₯
π₯
Beyond Turning Point
πππ π₯ > π₯π‘
πΈπ¦ π₯ =π1
ππ₯ π₯π+π ππ₯ π₯ ππ₯ +
π2
ππ₯ π₯πβπ ππ₯ π₯ ππ₯
ππ₯ π₯ =π
π π2 π₯ β π2
πΈπ¦ π₯ =π·
πΎπ π₯πβ πΎπ π₯π₯π₯π‘
ππ₯
π₯
ππ₯ π₯ =π
ππ2 π₯ β π2 = π
π
ππ2 β π2 π₯ = π πΎπ π₯
πππππππππ: πππ ππ(πΎ) > πππππππππ (π»)
I V
II
III
IV
y x
y
x
πΎ
π» π1
π3
π2
π4 π5
Channel Waveguides
Transverse confinement along x-axis, tangential Hy
Region I: π»π¦ π₯, π¦ = π»1 πππ ππ₯ π₯ + π1
Region II :
Region III:
π»π¦ π₯, π¦ = π»2 ππΎπ₯,2 π₯+π
π»π¦ π₯, π¦ = π»3 πβπΎπ₯,3 π₯
βπ < π₯ < 0
π₯ < βπ
π₯ > 0
ππ₯ π = π‘ππβ1πΎπ₯,2π22
π12
ππ₯+ π‘ππβ1
πΎπ₯,3π32
π12
ππ₯+ π π
Lateral confinement along y-axis, tangential Ex
Region I: πΈπ₯ π₯, π¦ = πΈ1 πππ ππ¦ π¦ + π2
βπ
2< π₯ <
π
2
ππ¦ π = π‘ππβ1πΎπ¦,4ππ¦+ π‘ππβ1
πΎπ¦,5ππ¦+ π π
Region IV : πΈπ₯ π₯, π¦ = πΈ4 π
β πΎπ¦,4 π¦βπ2
π¦ < βπ
2
πΈπ₯ π₯, π¦ = πΈ5 πβ πΎ5,π¦ π¦+
π2
Region V:
π¦ >π
2
Finding the propagation constants:
ππ₯ π = π‘ππβ1πΎπ₯,2π22
π12
ππ₯+ π‘ππβ1
πΎπ₯,3π32
π12
ππ₯+ π π
ππ¦ π = π‘ππβ1πΎπ¦,4
ππ¦+ π‘ππβ1
πΎπ¦,5
ππ¦+ π π
πΉ1 ππ₯, πΎπ₯,2, πΎπ₯,3, π½ = 0
π12 π2
π2= π1
2 = ππ₯2 + ππ¦
2 + π½2 πΊ1 ππ₯, ππ¦ , π½ = 0
πΉ2 ππ¦, πΎπ₯,4, πΎπ₯,5, π½ = 0
π22 π2
π2= π2
2 = βπΎπ₯,22 + ππ¦
2 + π½2
π32 π2
π2= π3
2 = βπΎπ₯,32 + ππ¦
2 + π½2
π42 π2
π2= π4
2 = ππ₯2 β πΎπ¦,4
2+ π½2
π52 π2
π2= π5
2 = ππ₯2 β πΎπ¦,5
2+ π½2
πΊ2 πΎπ₯,2, ππ¦ , π½ = 0
πΊ3 πΎπ₯,3, ππ¦ , π½ = 0
πΊ4 ππ₯, πΎπ¦,4, π½ = 0
πΊ5 ππ₯, πΎπ¦,5, π½ = 0
TM-like modes
ππ , ππ, π , ππ
πΉ ππ, ππ ππΌ , ππΌ = 0
I)
ππΌ
II) ππ , ππΌ,π , ππ
πΉ ππΈ , ππΈ ππΌπΌ , ππΌπΌ = 0
ππΌπΌ
(TM)
(TE)
Criteria for Single-Mode Operation
ππ = ππ + βπ βπ βͺ ππ with
I) cut-off condition for mode π :
ππΌ,π = 0
ππΌ,π = π‘ππβ1 ππΌ + π π
Requirement for existence of only one mode in transverse direction:
Transverse confinement:
π‘ππβ1 ππΌ < ππΌ < π‘ππβ1 ππΌ + π
Transverse confinement:
Lateral confinement:
II) cut-off condition for mode π :
ππΌπΌ,π = 0
ππΌπΌ,π = π‘ππβ1 ππΌπΌ + π π
0 < ππΌπΌ < π
Lateral confinement:
ππΌπΌ = 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:
ππΌ =π
ππ ππ
2 β ππ 2
ππΌπΌ =π
ππ ππΌ
2 β ππ 2 =π
ππ ππΌ ππ
2 β ππ 2 =
π
π ππΌ ππΌ
ππΌ β ππΌ2 β ππ
2
ππ2 β ππ
2
0 < ππΌπΌ < π 0 <π
π<π
ππΌ ππΌ
π‘ππβ1 ππΌ < ππΌ < π‘ππβ1 ππΌ + π
and
ππΌ β β
single-mode region
Optical Fibers st
ep-
index
multimod
e
step-
index
sing
lemod
e
GRIN
a cylindrical dielectric waveguide
Modes in Optical Fibers
π2πΈ π₯, π¦
ππ₯2+π2πΈ π₯, π¦
ππ¦2+π2π2
π2 β π½2 πΈ π₯, π¦ = 0
π2πΈ π, π
ππ2+1
π
ππΈ π, π
ππ+1
π2π2πΈ π, π
ππ2+π2π2
π2 β π½2 πΈ π, π = 0
Cartesian coordinates
Cylindrical coordinates
π¬ π₯, π¦, π§, π‘ = πΈ π₯, π¦ ππ π π‘ β π½ π§
π¬ π, π, π§, π‘ = πΈ π, π ππ π π‘ β π½ π§
Solutions
π2π’
ππ2+1
π
ππ’
ππ+π2π2
π2 β π½2 β
π2
π2π’ = 0
πΈ π, π = π’ π π π π π π
ππ2 β‘πππ2π2
π2 β π½2
πΎ2 β‘ π½2 βπππ2π2
π2
Boundary conditions for
πΈπ§, π»π§, πΈπ, π»π
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
π = 4
π2π2
π = 2π π
π πππ
2 β πππ2 = 2π
π
πππ΄
V-number
ππ΄ = π0 sin π0 = πππ2 β πππ
2
Numerical Aperture
π < 2.405
single-mode operation
πΌπ‘ π₯, π¦, π§ = ππΌ π§
πΌ
π¬π‘,πΌ π₯, π¦ πβπ π½πΌ π§ + radiation modes
βπ‘ π₯, π¦, π§ = ππΌ π§
πΌ
π―π‘,πΌ π₯, π¦ πβπ π½πΌ π§ + ππππππ‘πππ πππππ
Decomposition into the eigenmodes of the original structure
After a Careful (& Long) Analysis, Final Result:
Β± πππππ§= βπ ππΌ π§ π
βπ π½πΌβπ½π π§
πΌ
Ξπ,πΌ
4 Ξπ‘π,πΌ β‘ π βπ π₯, π¦, π§ π¬π‘,πββ π¬π‘,πΌ
β
ββ
ππ₯ ππ¦
4Ξπ§π,πΌ β‘ π π βπ
π + βππ¬π§,π
β
β π¬π§,πΌ
β
ββ
ππ₯ ππ¦
Ξπ,πΌ β‘ Ξπ‘π,πΌ+ Ξπ§π,πΌ
Ξπ,πΌ = ΞπΌ,πβ whenever π is a real number, then
Co-Directional Couplers:
π΄ π§ 2
π΅ π§ 2
πΉ β‘Ξ2
π½π2 =
Ξ2
Ξ 2 + β2
1 β πΉ π ππ2 π½π π§ πππ 2 π½π π§ +β2
π½π2π ππ2 π½π π§ = =
πΉ π ππ2 π½π π§ = = Ξ2
π½π2 π ππ
2 π½π π§
πΉ = 0.2 πΉ = 0.8
πΉ = 1
π΄ π§ 2
π΅ π§ 2
β β‘ π½π βπ½π2= 0
when: π½π = π½π
π
2
π§π½π
π§π½π
π½π β‘ Ξ2 + β2= Ξ πΏ =
π
2 π½π=π
2 Ξ
Counter-Directional Couplers
π΄ π§ 2 =1 + πΉ π ππβ2 πΌ π§ β πΏ
1 + πΉ π ππβ2 πΌ πΏ
π΅ π§ 2 =πΉ π ππβ2 πΌ π§ β πΏ
1 + πΉ π ππβ2 πΌ πΏ
πΉ β‘ Ξπ2
Ξπ2 β β2
> 1
β = 0 Ξπ= 0.2
πΏ = 5 π΄ π§ 2
π΅ π§ 2
π΄ π§ 2
π΅ π§ 2
πΏ = 9
πΏ β³Ο
Ξπ
when: β = 0