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Lecture 7: Transcendental Equations

Boundary Condition: Phase Matching at an Interface

As the spatial rate of change of phase at the boundary on the core side (n1 side) must match with that on the cladding side (n2 side), we have β1= β2. This condition is known as “phase-matching”. Phase-matching allows coupling of oscillating field between two media.

β1= β2 n1k sinϕ1 = n2k sinϕ2 (Snell’s Law)

Phase Matching at TIR

ϕ1 > ϕc (i. e. sinϕ1 >n2

n1)

β1(= β2) = n1k sinϕ1 > n1kn2

n1

β1 > n2k

n1k > β1 > n2k

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k-Vector Triangle in Cladding (n2 Medium)

kx2 = [(n2k)2 − (𝛽2)2]1

2⁄

kx2 = j[(𝛽2)2 − (n2k)2]1

2⁄

kx2 = jα [ here 𝛼2 = 𝛽2 − 𝑘2𝑛22 ]

The General Solution of Wave Equation (for TE polarization):

Let’s assume E = �̂�𝐸𝑦, and no y dependence, i.e. 𝜕

𝜕𝑦= 0.

The solution of the wave equation for electric field has the (assuming the TE

polarization):

𝐸 = �̂�𝐸𝑦𝑒𝑗(𝛽𝑧−𝜔𝑡) (1)

𝜕2𝐸

𝜕𝑧2 = 𝑗2𝛽2𝐸𝑦 = −𝛽2𝐸𝑦 (2)

Substituting into the wave equation (eqn. 3) [Ref. Lecture 4]:

(3)

022

2

22

2

2

y

iy

yEj

c

nE

x

E

02

2

2

2

2

2

2

2

2

2

t

E

c

n

z

E

y

E

x

E yiyyy

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(4)

We have assumed that,

𝜅2 = 𝑘2𝑛12 − 𝛽2

and

𝛼2 = 𝛽2 − 𝑘2𝑛22

Therefore (4) becomes,

|𝑥| < 𝑑2⁄ (Core region) (5)

|𝑥| > 𝑑2⁄ (Cladding region) (6)

02

222

2

2

y

iy

yE

c

nE

x

E

0222

2

2

yiy

yEnkE

x

E

0)( 222

2

2

yi

yEnk

dx

Ed

02

2

2

y

yE

dx

Ed

02

2

2

y

yE

dx

Ed

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The solutions of (5) and (6) can be written as,

𝐸𝑦(𝑥) = {

𝐴𝑐𝑜𝑠𝜅𝑥 + 𝐵𝑠𝑖𝑛𝜅𝑥 ; |𝑥| < 𝑑/2

𝐶𝑒𝛼𝑥 ; |𝑥| < −𝑑/2

𝐷𝑒−𝛼𝑥 ; |𝑥| > 𝑑/2 (7)

Where A, B, C, D are constants.

When the refractive index distribution is symmetric, the solutions are either

symmetric modes or anti-symmetric modes with respect to the symmetry plane of

the waveguide, the y-z plane. Therefore the solutions can be written as:

𝐸𝑦(𝑥) = 𝐴𝑐𝑜𝑠𝜅𝑥 ; |𝑥| < 𝑑/2 Symmetric mode (8)

𝐸𝑦(𝑥) = 𝐵𝑠𝑖𝑛𝜅𝑥 ; |𝑥| < 𝑑/2 Anti-symmetric mode (9)

𝐸𝑦(𝑥) = 𝐷𝑒−𝛼|𝑥|; |𝑥| > 𝑑/2 (10)

𝐸𝑦(𝑥) is continuous at 𝑥 = ±𝑑

2, (tangential components are equal)

𝐸𝑦 (𝑑

2) = 𝐴𝑐𝑜𝑠𝜅

𝑑

2= 𝐷𝑒−𝛼

𝑑

2 ; (11)

𝑑𝐸𝑦(𝑥)

𝑑𝑥 is continuous at 𝑥 = ±

𝑑

2,

𝑑𝐸𝑦(𝑥)

𝑑𝑥= −𝐴𝜅𝑠𝑖𝑛𝜅

𝑑

2= −𝛼𝐷𝑒−𝛼

𝑑

2 ; (12)

Dividing (12) by (11),

𝜅𝑡𝑎𝑛𝜅𝑑

2= 𝛼 ;

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𝜅𝑑

2𝑡𝑎𝑛𝜅

𝑑

2= 𝛼

𝑑

2 ;

𝜌𝑡𝑎𝑛𝜌 = 𝜐; (13)

Similarly for anti-symmetric modes,

−𝜌𝑐𝑜𝑡𝜌 = 𝜐; (14)

Where,

𝜌 = 𝜅𝑑

2 ; and

𝜐 = 𝛼𝑑

2 ;

𝜌2 + 𝜐2 = 𝑑2

4(𝜅2 + 𝛼2)

Putting the values of 𝜅2 and 𝛼2 ,

𝜌2 + 𝜐2 = 𝑑2

4{𝑘2(𝑛1

2 − 𝑛22)}

𝜌2 + 𝜐2 = (𝑉

2)2 (15)

Where 𝑉 = 𝑘𝑑√𝑛12 − 𝑛2

2 is normalized frequency (also called the V number or

V parameter).

Here,𝑑 = 2𝑎; (𝑘 =2𝜋

𝜆);

The transcendental equations (13) and (14) allow the determination of 𝜅 and 𝛽.

It is often convenient to define the dimensionless propagation constant (normalized

guide index),

𝑏 ≡𝛽2 𝑘2−𝑛2

2⁄

𝑛12−𝑛2

2 =𝑛𝑒𝑓𝑓

2 −𝑛22

𝑛12−𝑛2

2 =𝛽2−𝑘2

2

𝑘12−𝑘2

2

Obviously, for a guided mode, we have: 0 < b < 1