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nr( + δ r)cos( +α δα ) = n r( )cos( )α (8.107)

Now, taking the leading terms of a Taylor expansion of the LHS of this equa-
tion leads us to:

 dn r ()   ( + )   α 2 
αδα
2
  nr () + dr r δ   1 − 2   ≈ n r()1 − 2   (8.108)

 
Further simplification of this equation gives to first order in the variations:

δα ≈ 1 dn r() δ ≈r 1 ( −nn r) δ = −z n ( 2 δz r ) (8.109)
2
αnr() dr n 02 2
0
which can be expressed in matrix form as:

 rz( + δ z)  1 0   rz()
  =  2    (8.110)
α z ( + δ z)  − nzδ 1  α z ()
2
The total variation in the values of the position and slope of the ray can be
obtained by taking the product of the two matrices in Eqs. (8.106) and (8.110),
giving:

nz)

 rz( + δ z)  1 −( δ 2 δ z  rz()
  =  2    (8.111)
α z ( + δ z)  − nz δ 1  α z ()
2
2
Equation (8.111) provides us with the required recursion relation to numeri-
cally iterate the progress of the ray inside the fiber. Thus, the ray distance
from the fiber axis and the angle that it makes with this axis can be computed
at any z in the fiber if we know the values of the ray transverse coordinate and
its slope at the entrance plane.
The problem can also be solved analytically if we note that the determinant
of this matrix is 1 (the matrix is unimodular). Sylvester’s theorem provides
the means to obtain the following result:

 sin( nz)
r()
 rz() cos( nz) 2  0
  =  2 n    (8.112)
()
 α z ()  2   α 0
 − n sin( n z) cos( n z)
2
2
2
Homework Problems
Pb. 8.32 Consider an optical fiber of radius a = 30µ, n = 4/3, and n =
0
2
3
–1
10 m . Three ray enters this fiber parallel to the fiber axis at distances of 5µ,
10µ, and 15µ from the fiber’s axis.
© 2001 by CRC Press LLC