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1K views11 pagesProblem F m1
17ec82 problems
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Shreemahathma HdCopyright
© © All Rights Reserved
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Example 2.1
A silica optical fiber with a core diameter large enough to be considered by ray theory
analysis has a core refractive index of 1.50 and a cladding refractive index of 1.47.
Determine: (a) the critical angle at the core-cladding interface; (b) the NA for the
fiber; (c) the acceptance angle in air for the fiber.
Solution: (a) The critical angle @ at the core-cladding interface is given by
Eq. (2.2) where:
9.5 sir ie
12 = sin Frat
= 785°
(b) From Eq. (2.8) the NA is:
NA =(n} =n}! =(1.50?- 147)
2.25 — 2.16)!
30
(©) Considering Eq. (2.8) the acceptance angle in air @, is given by:
6,= sin" NA = sin 0,30
= 174°
20 Optical fiber waveguides Chapter 2
Example 2.2
A typical relative refractive index difference for an optical fiber designed for long-
distance transmission is 1%. Estimate the NA and the solid acceptance angle in air
for the fiber when the core index is 1.46. Further, calculate the critical angle at the
core-cladding interface within the fiber. It may be assumed that the concepts of geo-
metric optics hold for the fiber.
Solution: Using Eq. (2.10) with A= 0.01 gives the NA as:
For small angles the solid acceptance angle in air Cis given by:
= n0?=zsin’ 0,
Hence from Eq. (2.8):
= m(NAY' = x 0.04
).13 rad
Using Eq. (2.9) for the relative refractive index difference A gives:
Aare
ny
Hence
n
A=1-001
ny
=0.99
From Eq. (2.2) the critical angle at the core-cladding interface is:
ae
in’ —? = sin’ 0.99
*
=81.9° Example 2.3
An optical fiber in air has an NA of 0.4. Compare the acceptance angle for meridional
rays with that for skew rays which change direction by 100° at each reflection.
Solution: The acceptance angle for meridional rays is given by Eq. (2.8) with
No = 1 as:
@,= sin" NA = sin! 0.4
= 236°
The skew rays change direction by 100° at each reflection, therefore y = 50°. Hence
using Eq. (2.17) the acceptance angle for skew rays is:
@,, = sin! (| = sin! ( os
. cos ¥ cos 50°
= 38.5°
In this example, the acceptance angle for the skew rays is about 15° greater than
the corresponding angle for meridional rays. However, it must be noted that we have
only compared the acceptance angle of one particular skew ray path. When the light
input to the fiber is at an angle to the fiber axis, it is possible that y will vary from
zero for meridional rays to 90° for rays which enter the fiber at the core—cladding
interface giving acceptance of skew rays over a conical half angle of 7/2 radians. Example 2.4
A multimode step index fiber with a core diameter of 80 [tm and a relative index dif-
ference of 1.5% is operating at a wavelength of 0.85 um. If the core refractive index
is 1.48, estimate: (a) the normalized frequency for the fiber; (b) the number of guided
modes.
Solution: (a) The normalized frequency may be obtained from Eq. (2.70) where:
6
2n x 40 x 10° x 1.48 (2x 0.015)'=75.8
_ 2m 5
Ve aay = 085 x 10°
(b) The total number of guided modes is given by Eq. (2.74) as:
V? 5745.6
M. = —=—
ce? 2
= 2873
Hence this fiber has a V number of approximately 76, giving nearly 3000 guided
modes. Example 2.5
A graded index fiber has a core with a parabolic refractive index profile which has
a diameter of 50 um. The fiber has a numerical aperture of 0.2. Estimate the total
number of guided modes propagating in the fiber when it is operating at a wave-
length of | jm.
Solution: Using Eq. (2.69), the normalized frequency for the fiber is:
2n 2nx 25x 10° x 0.2
V=— a(NA) = *
(oo 1x10°
=31.4
The mode volume may be obtained from Eq. (2.95) where for a parabolic profile:
86
—=247
4
M,=—=
z
|
Hence the fiber supports approximately 247 guided modes. Example 2.6
Estimate the maximum core diameter for an optical fiber with the same relative
refractive index difference (1.5%) and core refractive index (1.48) as the fiber given
in Example 2.4 in order that it may be suitable for single-mode operation. It may be
assumed that the fiber is operating at the same wavelength (0.85 tm). Further, esti-
mate the new maximum core diameter for single-mode operation when the relative
refractive index difference is reduced by a factor of 10.
Solution: Considering the relationship given in Eq. (2.96), the maximum V value
for a fiber which gives single-mode operation is 2.4. Hence, from Eq. (2.70) the core
radius a is:
ae VA __2.4 x 0.85 x 10
2m\(2A)) 27x 1.48 x (0.03)!
=1.3um
Therefore the maximum core diameter for single-mode operation is approximately
2.6 Um.
Reducing the relative refractive index difference by a factor of 10 and again using
Eq. (2.70) gives:
2.4 x 0.85 x 10%
a>—@_ = 4.0 pm
2x 1.48 x (0.003)?
Hence the maximum core diameter for single-mode operation is now approximately
8 um. Example 2.7
A graded index fiber with a parabolic refractive index profile core has a refractive
index at the core axis of 1.5 and a relative index difference of 1%. Estimate the maxi-
mum possible core diameter which allows single-mode operation at a wavelength of
1.3 um.
Solution: Using Eq. (2.97) the maximum value of normalized frequency for
single-mode operation is:
V=2.4(1 + 2/a)? = 2.4(1 + 2/2)?
=2.4V2
The maximum core radius may be obtained from Eq. (2.70) where:
ae _24V2 x 1.3 x 10°
2mn,(2A)! 2m x 1.5 x (0.02):
=3.3 um
Hence the maximum core diameter which allows single-mode operation is approx-
imately 6.6 im. Example 2.8
Determine the cutoff wavelength for a step index fiber to exhibit single-mode opera-
tion when the core refractive index and radius are 1.46 and 4.5 um, respectively,
with the relative index difference being 0.25%.
Solution: Using Eq. (2.98) with V. = 2.405 gives:
Le 2nan,(2A)* _ 274.5 x 1.46(0.005)'
~ 2405 — 2.405
= 1.214 um
= 1214 nm
Hence the fiber is single-moded to a wavelength of 1214 nm. Example 2.9
Given that a useful approximation for the eigenvalue of the single-mode step index
fiber cladding W is [Ref. 43]:
W(V) = 1.1428V — 0.9960
deduce an approximation for the normalized propagation constant b(V).
Solution: Substituting from Eq. (2.68) into Eq. (2.71), the normalized propagation
constant is given by:
_,_ V-W)_W
AY)=1 v6
Then substitution of the approximation above gives:
~ (.1428V—- 0.9960)"
Vv;
02000)
BV)
= (1.1428 -
The relative error on this approximation for b(V) is less than 0.2% for 1.5