3. Measurement of optical fiber birefringence
using stress method
■ T.A
Song, Du-ree (Fiber Optics Lab., #4318, T.2567, cherry22@kaist.ac.kr
■ Report Due : in 3 weeks after experiment (last team in 1 week)
- You can omit the detailed, cumbersome theory except when necessary.
- Report should be included the HWs.
- ex.) objective, introduction, setup and procedure, result, discussion, reference
■ Grade : attitude(20%), report(80%)
■ Reference
- J. A. Buck, Fundamentals of optical fibers, ch.3, John Wiley & Sons, Inc.(2004)
- J. Dakin and B. Culshaw, Optical fiber sensors : principles...., ch.7, Artech house(1988)
- E. Hecht, Optics 4th ed. Addison-Wesley.
[Lab 1] Fiber Cleaving and Splicing
Most of the optical fiber system is jointed through the fusion splicing, which splices two pieces of optical fiber into
one with arc flame. Before the fusion splicing, each fiber should be cleaved to have a flat facet.
1. Fiber cleaving
① Remove fiber coating with a knife or a stripper.
② Thoroughly clean the bare fiber with a tissue dipped in ethanol.
Dusts can cause rejection-of-operation or arc burning during splicing.
③ Cleave the fiber.
2. Fiber splicing (FSM-30S, Fujikura)
① Select a proper splicing menu (SMF, MMF, EDF...)
② Place the cleaved fibers in the grooves of splicer, then lock down the levers
③ [Set] : Move close to each fiber, showing the lateral surface.
[Set] : Aligning cladding, core, X, Y
[Set] : Fusion splicing with arc flame
[Lab 2] Measurement of optical fiber birefringence using
stress method
1. Theory
1) Jones vector, Jones matrix
The polarization state of light consists of the amplitude and phase of x, y components
expressed as a matrix given by
 Eo x   Eox eix  i
Eo     
e x
i y 
E
e
 Eo y   oy 
 Eo x 
 E ei 
 oy 
, which is called Jones vector.
Light passing through optical system is converted to the another polarization state, and
can be expressed as a 2x2 matrix given by
 A2 x   T11 T12   A1x 


 
 A2 y   T21 T22   A1 y 
J 2  TJ1
, where T denotes Jones matrix. Polarizer and wave plate can be expressed using T.
[Jones vector and Jones matrix]
2) Stress method
Figure 1 Experimental setup
Experimental setup is illustrated in Figure 1. We can express the electric field
transformation using Jones vector and matrix given by
Eout  Pout M 2CM1Pin Ein
, where 𝑃𝑜𝑢𝑡 denotes the output polarization matrix, 𝑃𝑖𝑛 is the input polarization matrix,
𝑀1 represents a transformation matrix when the light propagates the length of x and 𝑀2
represents a transformation matrix when the light propagates the length of (L-x). C
represents a transformation matrix on the stress point.
The polarization maintaining fiber (PM fiber) consists of two orthogonal axes with
different refractive indices. If a linearly polarized light is incident on one of the two
orthogonal axes, it can maintain the initial polarization state. Each Jones matrix is
expressed as follows.
1 0 
M1  
ikx 
0 e 
0 
1
M2  
ik ( L  x ) 
0 e

 cos 
C  R( ) MR( )  
  sin 
sin   1 0  cos 
cos   0 ei   sin 
 sin  
cos  
Where angle θ is the difference between PM axes and stress direction and δ is degree of
phase difference between two PM axes. The phase difference of x and y polarization
component is due to the birefringence. Here, we define Beat length, which is the
propagation length resulting in 2π phase differences.
With the propagation constant 𝑘, we calculate the Beat length LB as
k
2
LB
Then, we can calculate the birefringence ∆n using Beat length as
n 

LB
In this experiment, we will do four kinds of experiments with different polarizer angles.
More specifically, if the input polarization is linear, electric field can be expressed as
follows:
a 
Ein   
b 
1)Polarizer angle of 0° and 45°
1 0 
Pin  

0 0
Eout
,
Pout 
1 1 1
2 1 1
cos2   ei sin 2   eik ( L x ) (1  ei ) cos  sin  
 2

i
2
ik ( L  x )
(1  ei ) cos  sin  
cos   e sin   e
Note that intensity of light can be expressed as
2
2
I  Eout , x  Eout , y 
a
2
2
(1  2sin 2  2   2 sin[k ( L  x)   ])
  c o s 2 (1
 c os ),   sin 



Where θ, δ, a, k, L are constants. The output intensity depends on the length of x ,
then we can measure the output period as we change the length x. Using this period,
we can calculate k. Since δ, θ, a have no relation with the period, we can take
arbitrary values of these. Thus we will get the same result although we use the
simplified matrix C as,
 1 i 



C
 R( ) M (  ) R( )

4
2
4
 i 1 
Then the electric field and the intensity of light are given as
Eout
a 1  ieik ( L x ) 
 

2 1  ieik ( L x ) 
I  a (1  s i n k[ L
(  x )])
2
2. Experimental procedure
1. Set up the experimental equipment.
[Light source : 1319nm Nd:YAG laser
,
Detector : InGaAs photodiode]
2. Set up the two polarizer angles by 0° and 90°, respectively.
It can be set by finding the lowest output voltage on oscilloscope.
3. Connect a photodiode to a chart recorder, and then calibrate the amplitude to a
proper level.
4. Set up the two polarizer angles by 0° and 45°, respectively.
5. Count the number of peaks by rolling the hoop along the rail.
6. Calculate the birefringence of polarization maintaining fiber.
7. Repeat (5), (6) with polarizer angles of −45° and 45°.
 Homework.
1) Polarizer angles of 45° and 0°.
2) Polarizer angles of 45° and -45°
3) Polarizer angles of 0° and 90°