Download Optical fibres solutions

Chapter 11: Optical fibres
solutions
1.
[2 marks]
(VCE 2004 Sample Q5)
To get the refraction that we see as the ray
enters the fibre n1 must be bigger than air.
To get the total internal reflection that is
shown in the diagram n1 must be bigger
than n2. So nair < n1 > n2.
D (ANS)
2.
[2 marks]
(VCE 2004 Sample Q6)
To do this we use Snell’s law
n1sin  θ1  = n2sin  θ2  or
in this case nair sin  θair  = n1sin  θ1  .
Since nair = 1.00, θair = 9 and θ1 = 6.
We get
1.00×sin  9 = n1sin  6
 n1 =
sin  9 
,
sin  6 
 n 1 = 1.5 (ANS)
3.
[2 marks]
(VCE 2004 Sample Q7)
Glass has a higher refractive index for green
light than it does for red light. So the green
light will bend more as it enters the fibre. If
the green light bends more it will reflect off
the cladding to the right of point P.
 D (ANS)
4.
[2 marks]
(VCE 2004 Sample Q8)
A refractive index can never be below 1.00,
as this is derived from
n=
Speed of light in vacuum
, since the
Speed of light in medium
speed of light in a vacuum is the fastest
anything can travel, n can not be less than 1.
Thus A is incorrect. To have total internal
reflection the core must be greater than the
cladding. Thus C and D are incorrect.
 B (ANS)
5.
[3 marks]
(VCE 2004 Sample Q9)
The majority of the attenuation (loss) that
occurs at 1000nm is caused from Rayleigh
scattering. Thus is caused by slight
fluctuations in the refractive index of the
medium. This causes shorter wavelengths
to be scattered in all directions and lost
from the fibre.
6.
[2 marks]
(VCE 2004 Sample Q10)
Since total internal reflection is how the
light propagates along the fibre it cannot be
a cause of loss. The numerical aperture of a
fibre is a number used to determine how
easy it is to get light in to a fibre and is not a
cause of loss. Dispersion is the spreading of
the pulse and though the light pulse would
have a lower intensity it will have a greater
duration and no overall light would be lost.
In the case of 1450nm this loss is caused by
OH ions in the fibre absorbing light.
D (ANS)
7.
[1 + 2 marks]
a)
decrease
(VCE 2004 Pilot Q5)
b)
The refractive index of the fluid is
greater than the plastic cable, so the will be
no total internal reflection of the light along
the cable. Thus all of the light will exit the
cable and enter the fluid, making the
amount of light emitted at the other end
decrease.
8.
[2 marks]
(VCE 2004 Pilot Q6)
The β ray was still able to propagate along
the fibre so it must be smaller than the
acceptance angle. The γ ray was not able to
propagate along the fibre so it must be
bigger than the acceptance angle.
So β < θa < γ.
 C (ANS)
9.
[2 marks]
(VCE 2004 Pilot Q7)
Modal dispersion is caused by the different
paths that the light can travel down the
fibre. The different paths mean that there is
a difference in arrival time and the pulse of
light will spread. Single mode fibres do not
have modal dispersion and therefore can
have greater data rates over longer
distances.
10.
[2 marks]
14.
[2 marks]
(VCE 2005 Q6)
Since the laser beam is hitting the
hemispherical end at right angles it will not
bend. The acceptance angle was given by
 = 90 - c . The critical angle was given by
sinc =
n2
. So the critical angle was
n1
 1.00 

 1.60 
c = sin -1 
 c = 38.7 0
(VCE 2004 Pilot Q8)
The fibre has a loss of 1dB/km when a light
source of 1.1μm is used. The maximum
distance that the light can travel before
needing a boost is 30km.
11.
[3 marks]
(VCE 2004 Pilot Q9)
An image of the ants is focused on the
surface of the image bundle using a lens.
The light then travels down the optical
fibres until it reaches the other end. The
fibre bundle is coherent, so the optical fibres
remain in their relative positions through
the length of the bundle. An image of the
ants is then viewed at the other end with a
camera.
12.
[2 marks]
(VCE 2004 Pilot Q10)
Therefore α = 90 – 38.7
= 51.30 (ANS)
15.
[2 marks]
(VCE 2005 Q7)
Modal dispersion is caused by the different
paths that the light can travel down the
fibre. If fibre has a smaller diameter there
would be fewer paths and therefore you
would reduce the modal dispersion.
 B (ANS)
16.
[2 marks]
(VCE 2005 Q8)
The fibre bundle is 100 fibres down and 100
fibres across, so there are 10,000 fibres in
total.
 D (ANS)
Rayleigh scattering is the major cause of
attenuation in optical fibres for wavelengths
less than 1µm.
 B (ANS)
13.
17.
[2 marks]
(VCE 2005 Q5)
Since air has a refractive index of 1.0 and 1.0
is the smallest refractive index you can
have, the air will be the smallest number. To
have total internal reflection the core must
have a higher refractive index then the
n
cladding, since sinc = 2
n1
 B (ANS)
[4 marks]
(VCE 2005 Q9)
(i)
The physical process is Rayleigh
scattering. When the wavelength is smaller
than the particles Rayleigh scattering will
occur.
(ii)
The physical process is Absorption.
Absorption occurs when light interacts with
electrons in the atoms and the bonds
between the atoms.
18.
[2 marks]
 1.00  sin air   1.48 sin 10
(VCE 2005 Q10)
 sin  air   0.257
A single mode fibre will reduce modal
dispersion and a laser will reduce material
dispersion.
 D (ANS)
 θair  14.90 (ANS)
22.
19.
[3 marks]
(VCE 2006 Q1)
A fibre-optic cable is composed of two
concentric layers called the core and the
cladding.
The refractive index of the core is always
[greater than] the index of the cladding. In a
single-mode fibre, light enters the cable and
propagates along a path which is the
[lowest] order mode. In this mode,
chromatic dispersion and attenuation are
[reduced / increased].
20.
[2 marks]
(VCE 2006 Q5)
Ray 1
[3 marks]
Use Sin  c  
(VCE 2006 Q7)
n2
n1
1.47
1.48
 θc  83.30 (ANS)
Sin  c  
23.
[2 marks]
(VCE 2006 Q8)
The acceptance angle will change since
n 0 Sin  a   n 1 2  n 2 2 . n0 is the refractive
index of the material that the fibre is in
(usually air, 1.00). Since the refractive index
is now 1.33 the acceptance angle will
decrease. Note you could have calculated
these two values and seen the decrease.
For Air
n 0 Sin  a   n 1 2  n 2 2
Ray 2
1.00Sina   1.482  1.472
a  9.890
For water
The light travelling down a step index
multimode fibre can travel many different
paths and thus become separated. This
spreads the light pulse and is known as
modal dispersion. The light travelling down
a graded index fibre travels at different
speeds depending on the distance from the
centre of the fibre, so Ray 1 which has to
travel the greater distance, travels at a
greater average speed. This means that the
two rays don’t separate from one another.
So you reduce modal dispersion.
21.
[2 marks]
(VCE 2006 Q6)
To do this we use Snell’s law
n 1 sin 1   n 2 sin  2  or in this case
n air sin air   n 1 sin 1  .
Since nair = 1.00, n1 = 1.48 and θ1 = 100.
We get
n air sin air   n 1 sin 1 
n 0 Sin  a   n 1 2  n 2 2
1.33Sina   1.482  1.472
a  7.420
The critical angle of an optical fibre doesn’t
change once the fibre has been made.
n 
c  Sin 1  2 
 n1 
So the answers were
The acceptance angle will change
The critical angle will remain the same
24. [2 marks]
(VCE 2006 Q9)
Since the refractive index of Red light is
greater than the refractive index of Infra-red
light, the infra-red light will travel faster
than the red light. So the infra-red light
arrive before the red light.
B (ANS)
25.
[2 marks]
(VCE 2006 Q10)
Red light would have greater UV
absorption and Rayleigh scattering than the
Infra-red light, since red light has a smaller
wavelength than the infra-red light.
26.
[2 marks]
Use
28.
[2 marks]
(VCE 2007 Q5)
n2
n1
1.43
Sin( ) 
1.46
  780
 D (ANS)
Sin( ) 
[2 marks]
(VCE 2007 Q6)
Sin( )  n 1 2  n 2 2
Sin( )  1.462  1.372
   300
B (ANS)
29.
[2 marks]
(VCE 2007 Q7)
The LED emits light that has a range of
wavelengths; different wavelengths will
travel at different speeds in the optical fibre,
so they will arrive at different times. This
will cause the pulse to spread.
30.
[2 marks]
[2 marks]
(VCE 2007 Q9)
Since the red laser has a wavelength of 640
nm, the major cause of attenuation at that
wavelength is Rayleigh scattering.
 C (ANS)
(VCE 2006 Q11)
The light from the laser is able to remain
inside the stream of water because of total
internal reflection. The light hits the waterair junction at an angle greater than the
critical angle and is reflected back into the
water. It continues to bounce back and forth
down the water stream.
27.
31.
(VCE 2007 Q8)
A Laser will have a very small range of
wavelengths so the wavelengths that make
up the pulse will only have a slight
difference in their speed and the amount of
dispersion will be greatly reduced.
32.
[2 marks]
(VCE 2007 Q10)
The best laser to use would be the infrared
laser with 1700 nm. This material has less
attenuation at 1700 nm than at 640 nm.
33.
[2 marks]
(VCE 2007 Q11)
When the concrete beam bends the angle
the light hits the surface of the corecladding boundary will vary and the
amount of light that is internally reflected
will change. The light was sent in at an
angle just smaller than the critical angle, so
any bending of the fibre will result in a
change in the light intensity reaching the
detector.
34.
[2 marks]
(VCE 2008 Q5)
n2 is the refractive index of the cladding,
1.41.
n1 is the refractive index of the core, 1.48.
n
Sinc  2
n1
1.41
Sinc 
1.48
c  720
C (ANS)
35.
[2 marks]
(VCE 2008 Q6)
From the information supplied in the
question the ray hits the core-cladding
boundary at an angle of 600, this means that
the ray must have refracted to an angle of
300 after entering the fibre (as shown
below).
n2 = 1.30
n0 = 1.00
α
600
n1 = 1.50
[2 marks]
(VCE 2008 Q8)
The light is being sent at 50 mW and the
smallest signal that can be received is 20
mW. This means that you can lose 30 mW
and still receive the signal. When the
wavelength is 1300 nm the loss of the signal
is 2.5 mW/km. To find how far you could
go before you lost 30 mW you would just
divide 30 by 2.5. You get 12 km.
 C (ANS)
0
30
38.
n 0 Sin0  n 1 Sin1
1.00  Sin0  1.50  Sin(300 )
Sin0  0.75
0  48.50
 B (ANS)
36.
37.
[2 marks]
(VCE 2008 Q7)
Since the critical angle is a property of the
optical fibre
n
Sinc  2
n1
1.30
Sinc 
1.50
c  60.10
The critical angle will not change unless the
refractive index of the core or cladding is
changed.
The best way of working out what happens
to the acceptance angle is to calculate it
when the refractive index of n0 = 1.33.
When the fibre is in air n0 = 1.00.
n 0 Sina  n 12  n 2 2
[2 marks]
(VCE 2008 Q9)
Graded index fibres have a graduated
refractive index, it is higher in the middle of
the fibre and smaller towards the edges.
This means that the mode that have the
shortest distance to travel (Down the
middle of the fibre) travel the slowest. The
modes then stay together and the modal
dispersion is reduced in the graded index
fibre. The other properties of the fibre
(absorption, Rayleigh scattering, material
dispersion) are the same for graded index
and step index.
 C (ANS)
39.
[2 marks]
(VCE 2008 Q10)
Material dispersion is caused because
different wavelength of light travel at
different speeds through the fibre. The
shorter the wavelength the slower it travels.
An LED has a range of wavelengths so the
light tends to travel at different speeds. A
laser has a very small range of wavelengths,
so the light from a laser travels at the same
speed along the fibre and the material
dispersion is greatly reduced.
 A (ANS)
1.00  Sina  1.502  1.302
a  48.40
40.
n 0 Sina  n 12  n 2 2
You would not get enough resolution from
1 pixel or from 5 pixels.
When looking inside a pipe, the amount of
detail required is not that great, so you only
need around a 1000 pixels.
 C (ANS)
1.33  Sina  1.502  1.302
0  34.20
The acceptance angle decreases as the
refractive index of n0 increases.
 B (ANS)
[2 marks]
(VCE 2008 Q11)
41.
[2 marks]
(VCE 2008 Q12)
The important thing to consider in long
distance communication is the amount of
attenuation (Loss of signal), not how the
signal was lost (absorption, scattering). Red
light has more attenuation then the infrared,
shown on the graph, so infrared is the best
choice.
 C (ANS)
42.
[2 marks]
(VCE 2008 Q13)
Material dispersion is found in both multimode and single mode fibres and it can only
be reduced by using a light source that has a
very small range of wavelengths.
Having a multi-mode fibre means that there
are many modes in the fibre and therefore
more modal dispersion. A single mode fibre
reduces the modal dispersion.
 A (ANS)