5.3 Losses Between Fiber Joints
233
Drill Problem 5.7 Consider the case when light from a multimode step-index
fiber that has a core radius of 62.5 μm is coupled into a similar fiber that has a
core radius of 50 μm. Show that the coupling loss in going from the larger to
the smaller fiber is 1.94 dB.
If the radii and the index profiles of the two coupled fibers are identical but their
axial numerical apertures are different, then
L F (N A) =
−10 log
N A R (0)
N A E (0)
2
for N A R (0) < N A E (0)
0 for N A R (0) ≥ N A E (0)
(5.36)
Example 5.10 Consider two joined step-index fibers that are perfectly aligned. What
is the coupling loss if the numerical apertures are NA R = 0.20 for the receiving fiber
and NA E = 0.22 for the emitting fiber?
Solution From Eq. (5.36)
L F (N A) = −10 log
0.20
0.22
2
= −10 log 0.826 = −0.828 dB
Finally, if the radii and the axial numerical apertures are the same but the core
refractive-index profiles differ in two joined fibers, then the coupling loss is
L F (α) =
−10 log
α R (α E +2)
α E (α R +2)
for α R < α E
0 for α R ≥ α E
(5.37)
This results because for α R < α E the number of modes that can be supported by the
receiving fiber is less than the number of modes in the emitting fiber. If α R > α E then
all modes in the emitting fiber can be captured by the receiving fiber.
Example 5.11 Consider two joined graded-index fibers that are perfectly aligned.
What is the coupling loss if the refractive index profiles are α R = 1.98 for the receiving
fiber and α E = 2.20 for the emitting fiber?
Solution From Eq. (5.37)
L F (α) = −10 log
α R (α E + 2)
α E (α R + 2)
= −10 log 0.950 = −0.22 dB
233
Drill Problem 5.7 Consider the case when light from a multimode step-index
fiber that has a core radius of 62.5 μm is coupled into a similar fiber that has a
core radius of 50 μm. Show that the coupling loss in going from the larger to
the smaller fiber is 1.94 dB.
If the radii and the index profiles of the two coupled fibers are identical but their
axial numerical apertures are different, then
L F (N A) =
−10 log
N A R (0)
N A E (0)
2
for N A R (0) < N A E (0)
0 for N A R (0) ≥ N A E (0)
(5.36)
Example 5.10 Consider two joined step-index fibers that are perfectly aligned. What
is the coupling loss if the numerical apertures are NA R = 0.20 for the receiving fiber
and NA E = 0.22 for the emitting fiber?
Solution From Eq. (5.36)
L F (N A) = −10 log
0.20
0.22
2
= −10 log 0.826 = −0.828 dB
Finally, if the radii and the axial numerical apertures are the same but the core
refractive-index profiles differ in two joined fibers, then the coupling loss is
L F (α) =
−10 log
α R (α E +2)
α E (α R +2)
for α R < α E
0 for α R ≥ α E
(5.37)
This results because for α R < α E the number of modes that can be supported by the
receiving fiber is less than the number of modes in the emitting fiber. If α R > α E then
all modes in the emitting fiber can be captured by the receiving fiber.
Example 5.11 Consider two joined graded-index fibers that are perfectly aligned.
What is the coupling loss if the refractive index profiles are α R = 1.98 for the receiving
fiber and α E = 2.20 for the emitting fiber?
Solution From Eq. (5.37)
L F (α) = −10 log
α R (α E + 2)
α E (α R + 2)
= −10 log 0.950 = −0.22 dB
