2.3 Optical Fiber Configurations and Modes
55
(b) From Eq. (2.23) the numerical aperture is
N A =
n
2
1 − n
2
2
1/2 = 0.242
(c) From Eq. (2.22) the acceptance angle in air (n = 1.00) is
θ A = sin
−1 NA = sin
−1 0.242 = 14
◦
Example 2.6 Consider a multimode fiber that has a core refractive index of 1.480
and a core-cladding index difference of 2.0% ( = 0.020). Find (a) the numerical
aperture, (b) the acceptance angle, and (c) the critical angle.
Solution From Eq. (2.20), the cladding index is n 2 = n 1 (1 − ) = 1.480(0.980) =
1.450.
(a) Using Eq. (2.23) then the numerical aperture is
N A = n 1
√
2 = 1.480
√
0.04 = 0.296
(b) Using Eq. (2.22) the acceptance angle in air (n = 1.00) is
θ A = sin
−1 NA = sin
−1 0.296 = 17.2
◦
(c) From Eq. (2.21) the critical angle at the core–cladding interface is
ϕ c = sin
−1 n 2
n 1
= sin
−1 0.980 = 78.5
◦
Drill Problem 2.3 Consider the interface between fiber core and cladding
materials that have refractive indices of n 1 and n 2 , respectively. If n 2 is smaller
than n 1 by 1% and n 1 = 1.450, show that n 2 = 1.435. Show that the critical
angle is ϕ c = 81.9°.
2.3.5 Lightwaves in a Dielectric Slab Waveguide
Referring to Fig. 2.17, the ray theory appears to allow rays at any angle ϕ greater
than the critical angle ϕ c to propagate along the fiber. However, when the interference
effect due to the phase of the plane wave associated with the ray is taken into account,
it is seen that only waves at certain discrete angles greater than or equal to ϕ c are
capable of propagating along the fiber.
55
(b) From Eq. (2.23) the numerical aperture is
N A =
n
2
1 − n
2
2
1/2 = 0.242
(c) From Eq. (2.22) the acceptance angle in air (n = 1.00) is
θ A = sin
−1 NA = sin
−1 0.242 = 14
◦
Example 2.6 Consider a multimode fiber that has a core refractive index of 1.480
and a core-cladding index difference of 2.0% ( = 0.020). Find (a) the numerical
aperture, (b) the acceptance angle, and (c) the critical angle.
Solution From Eq. (2.20), the cladding index is n 2 = n 1 (1 − ) = 1.480(0.980) =
1.450.
(a) Using Eq. (2.23) then the numerical aperture is
N A = n 1
√
2 = 1.480
√
0.04 = 0.296
(b) Using Eq. (2.22) the acceptance angle in air (n = 1.00) is
θ A = sin
−1 NA = sin
−1 0.296 = 17.2
◦
(c) From Eq. (2.21) the critical angle at the core–cladding interface is
ϕ c = sin
−1 n 2
n 1
= sin
−1 0.980 = 78.5
◦
Drill Problem 2.3 Consider the interface between fiber core and cladding
materials that have refractive indices of n 1 and n 2 , respectively. If n 2 is smaller
than n 1 by 1% and n 1 = 1.450, show that n 2 = 1.435. Show that the critical
angle is ϕ c = 81.9°.
2.3.5 Lightwaves in a Dielectric Slab Waveguide
Referring to Fig. 2.17, the ray theory appears to allow rays at any angle ϕ greater
than the critical angle ϕ c to propagate along the fiber. However, when the interference
effect due to the phase of the plane wave associated with the ray is taken into account,
it is seen that only waves at certain discrete angles greater than or equal to ϕ c are
capable of propagating along the fiber.