Photonics - Topic 6

Question 1

Symmetric slab waveguide condition for waveguide operation

Answer 1

𝜃_𝑟 < 𝜋/2 − 𝜃_𝑐

where sin𝜃_𝑐 = 𝑛₂/𝑛₁

Question 2

Eq: Numerical aperture

what is it a measure of?

Answer 2

NA = (𝑛₁² − 𝑛₂² ) ^1/2

The NA is a measure of the light-gathering power of the waveguide

Question 3

Waveguide interference condition

Answer 3

( 2𝜋𝑛₁ ·2𝑎 / 𝜆 ) cos(𝜃_𝑚) − 𝜙_𝑚 = 𝑚π

This shows that only certain angles 𝜃_𝑚 are allowed (waveguide modes)

Question 4

Why are not all propagation angles 𝜃𝑟 that meet the TIR condition 𝜃𝑟 < 𝜋/2 − 𝜃_c actually guided along the waveguide?

Answer 4

• Because all waves propagating at angle 𝜃_𝑟 to the axis of the waveguide must be in phase with one another.
  
  • Waves at angle 𝜃_𝑟 must constructively interfere with one another, leading to the waveguide interference condition.

Question 5

Eq: Standing ave mode patterns

Answer 5

2a = q (𝜆 / 2)

where q is an integer and 𝜆 is the wavelength in the guide

Question 6

What are the two cases the wave equation can be solved for?

Answer 6

• The wave equation can be solved for two cases:
  
  • with the electric field direction across the waveguide and parallel to the core-cladding interface (transverse electric or TE),
  • with the magnetic field direction across the waveguide and parallel to the core-cladding interface (transverse magnetic or TM).

Question 7

TE waveguide modes using the wave equation

Answer 7

• Assume that in each layer (𝑖) of the waveguide, the electric field in the 𝑦 direction can be written as: 𝐸_𝑦𝑖(𝑥, 𝑧) = 𝐸_𝑖(𝑥) e^{𝑗(𝜔𝑡 − 𝛽𝑧)}
  
  • As such th wave equqtion becomes: d²𝐸𝑖/d𝑥² + d²𝐸𝑖/dz²+ [𝑛_𝑖²𝑘_𝑜² − 𝛽²]𝐸_𝑖 = 0
  • Since the direction of propagation is z, second differential can be ignored.
  • Resulting in: d²𝐸𝑖/d𝑥² + [𝑛_𝑖²𝑘_𝑜² − 𝛽²]𝐸_𝑖 = 0
  • Depending on the sign of 𝑛_𝑖²𝑘_𝑜² − 𝛽² the solution will be of the form:
    
    • Solution in the core ⇒ 𝐸₁ = e^(𝑝𝑥) , where p isgiven by 𝑝² = 𝛽² - 𝑛_𝑖²𝑘_𝑜²
    • Solution in the cladding layers ⇒ 𝐸₂ = 𝐵cos(ℎ𝑥) or 𝐵sin (ℎ𝑥) where h² = 𝑛_𝑖²𝑘_𝑜² − 𝛽²

Question 8

Symmetric slab waveguide – even modes

Answer 8

Question 9

Symmetric slab waveguide – odd modes

Answer 9

Question 10

Eq: V-parameter

Answer 10

𝑉 = (2𝜋𝑎 / 𝜆) (𝑛₁² − 𝑛₂²)^1/2

Question 11

In terms of the v-parameter what is the condition for a symmetrical slab waveguide to be single-moded?

Answer 11

𝑉 < 𝜋/2

V-parameter is related to the numerical aperture of the waveguide

Question 12

Outline Opitical Fibres

Answer 12

• Made from pure glass (silica, SiO₂)
• refractive index adjusted by doping
• index difference is small
• single mode for v-number <2.405
  
  • 𝑉 = (2𝜋𝑎 / 𝜆) (𝑛₁² − 𝑛₂²)^1/2
  • where a is the radius of the fibre core

Question 13

When does optical fibre become multimodal?

Answer 13

when V>2.4

(2𝜋𝑎 / 𝜆) NA > 2.4

𝜆 < (2𝜋𝑎/2.4) NA

Question 14

Fibe Types

Answer 14

• mm stand for multimode, sm for single mode
• Si-mm
  
  • large core radius
  • able to support he propagation of many tranverse waves
• Gi-mm
  
  • refractive index profile is parabolic
  • reduces modal dispersion as propagation of the hgher order modes is reduced
• Si-sm
  
  • very small core diameter
  • only a sinle transverse mode is supported

Question 15

Advantages of graded index over stepindex multimode fibre

Answer 15

• Gi
  
  • reduces variation in propagation time
  • All rays have same Vg.
  • Gradiating Ri makes rays curve
    *

Question 16

Single-mode and Multi-mode use cases

Answer 16

• SM
  
  • used for long distance, high data rate communication
• MM
  
  • used in cost-sensitive applications (low-rate data signals over short distances)

Question 17

Multi-path dispersion (modal dispersion)

Answer 17

• Gi has lower step index than Si
• But singlemode fibre does not suffer from modal dispersion at all.