2.4 Modes in Circular Waveguides
61
Solution From Eq. (2.27) the NA is
N A = V
λ
2πa
= 26.2
1.30 μm
2π × 25 μm
= 0.22
Drill Problem 2.5 Consider a fiber that has a core refractive index of 1.480, a
cladding index of 1.476, and a core radius of 4.4 μm. Using Eq. (2.27), show
that the wavelength at which this fiber becomes single-mode with V = 2.405
is λ c = 1250 nm.
The V number also can be used to express the number of modes M in a multimode
step-index fiber when V is large (see Sect. 2.6 for modes in a graded-index multimode
fiber). For the multimode step-index case, an estimate of the total number of modes
supported in such a fiber is
M =
1
2
2πa
λ
2
n
2
1 − n
2
2
=
V
2
2
(2.30)
Example 2.8 Consider a multimode step-index fiber with a 62.5 μm core diameter and a core-cladding index difference of 1.5%. If the core refractive index is
1.480, estimate the normalized frequency of the fiber and the total number of modes
supported in the fiber at a wavelength of 850 nm.
Solution From Eq. (2.27) the normalized frequency is
V =
2πa
λ
n 1
√
2 =
2π × 31.25 μm × 1.48
0.85 μm
√
2 × 0.015 = 59.2
Using Eq. (2.30), the total number of modes is
M =
V
2
2
= 1752
Example 2.9 Consider a multimode step-index optical fiber that has a core radius
of 25 μm, a core index of 1.48, and an index difference = 0.01. How many modes
are in the fiber at wavelengths 860, 1310, and 1550 nm?
Solution
(a) First, from Eq. (2.27) at an operating wavelength of 860 nm the value of V is
V =
2πa
λ
n 1
√
2 =
2π × 25 μm × 1.48
0.86 μm
√
2 × 0.01 = 38.2
61
Solution From Eq. (2.27) the NA is
N A = V
λ
2πa
= 26.2
1.30 μm
2π × 25 μm
= 0.22
Drill Problem 2.5 Consider a fiber that has a core refractive index of 1.480, a
cladding index of 1.476, and a core radius of 4.4 μm. Using Eq. (2.27), show
that the wavelength at which this fiber becomes single-mode with V = 2.405
is λ c = 1250 nm.
The V number also can be used to express the number of modes M in a multimode
step-index fiber when V is large (see Sect. 2.6 for modes in a graded-index multimode
fiber). For the multimode step-index case, an estimate of the total number of modes
supported in such a fiber is
M =
1
2
2πa
λ
2
n
2
1 − n
2
2
=
V
2
2
(2.30)
Example 2.8 Consider a multimode step-index fiber with a 62.5 μm core diameter and a core-cladding index difference of 1.5%. If the core refractive index is
1.480, estimate the normalized frequency of the fiber and the total number of modes
supported in the fiber at a wavelength of 850 nm.
Solution From Eq. (2.27) the normalized frequency is
V =
2πa
λ
n 1
√
2 =
2π × 31.25 μm × 1.48
0.85 μm
√
2 × 0.015 = 59.2
Using Eq. (2.30), the total number of modes is
M =
V
2
2
= 1752
Example 2.9 Consider a multimode step-index optical fiber that has a core radius
of 25 μm, a core index of 1.48, and an index difference = 0.01. How many modes
are in the fiber at wavelengths 860, 1310, and 1550 nm?
Solution
(a) First, from Eq. (2.27) at an operating wavelength of 860 nm the value of V is
V =
2πa
λ
n 1
√
2 =
2π × 25 μm × 1.48
0.86 μm
√
2 × 0.01 = 38.2