Graded Refractive-Index

Graded Refractive-Index

Common Devices

Methodologies for Graded Refractive Index

Methodologies: • Ray Optics • WKB • Multilayer Modelling

Solution requires: • some knowledge of index profile 𝑛2 𝑥

Ray Optics for graded refractive index

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

WKB Technique for graded refractive index

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

General Solution

𝐸𝑦 𝑥 =𝑐1

𝑘𝑥 𝑥𝑒+𝑗 𝑘𝑥 𝑥 𝑑𝑥 +

𝑐2

𝑘𝑥 𝑥𝑒−𝑗 𝑘𝑥 𝑥 𝑑𝑥

𝑘𝑥 𝑥 =𝜔

𝑐 𝑛2 𝑥 − 𝑁2

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

3D Waveguides

Channel Waveguides Optical Fibers

𝒕𝒚𝒑𝒊𝒄𝒂𝒍𝒍𝒚: 𝒘𝒊𝒅𝒕𝒉(𝑾) > 𝒕𝒉𝒊𝒄𝒌𝒏𝒆𝒔𝒔 (𝑻)

I V

II

III

IV

y x

y

x

𝑾

𝑻 𝑛1

𝑛3

𝑛2

𝑛4 𝑛5

Channel Waveguides

a) Marcatili’s Method

Hy

Ex TM-like modes: Hy & Ex

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

b) Effective Index Method

𝒏𝒄

𝒏𝒔

𝒏𝒔

𝒏𝒔 𝒏𝒔

𝒏𝒇

𝒏𝒇

𝑵𝑰

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

Graphical Representation

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.

Optical Attenuation

1 𝑑𝐵 = −10 𝑙𝑜𝑔 𝑇

0.16 dB = (3.6 %)

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

Coupled Mode Theory

A few examples:

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

Counter-Directional Couplers

𝐴 𝑧 2 =1 + 𝐹 𝑠𝑖𝑛ℎ2 𝛼 𝑧 − 𝐿

1 + 𝐹 𝑠𝑖𝑛ℎ2 𝛼 𝐿

𝐵 𝑧 2 =𝐹 𝑠𝑖𝑛ℎ2 𝛼 𝑧 − 𝐿

1 + 𝐹 𝑠𝑖𝑛ℎ2 𝛼 𝐿

𝐹 ≡ Κ𝑐2

Κ𝑐2 − ∆2

> 1

∆ = 0 Κ𝑐= 0.2

𝐿 = 5 𝐴 𝑧 2

𝐵 𝑧 2

𝐴 𝑧 2

𝐵 𝑧 2

𝐿 = 9

𝐿 ≳π

Κ𝑐

when: ∆ = 0