2. Taiz University, YEMEN
• Electromagnetic Radiation carries energy through space
(includes visible light, dental x-rays, radio waves, heat
radiation from a fireplace)
• The wave is composed of a combination of mutually
perpendicular electric and magnetic fields the direction of
propagation of the wave is at right angles to both field
directions, this is known as an ELECTROMAGNETIC WAVE
EM wave move through a vacuum at 3.0 x 108 m/s ("speed of light")
Speed of light in a vacuum c=λ*f
• The refractive index of a medium is defined as the ratio of
the velocity of light in vacuum to the velocity of light in the
medium. The relationship is given by :
4 October 2018 2
)
(
)
(
)
(
medium
a
in
light
of
speed
v
vacuum
a
in
light
of
speed
c
index
refractive
n
3. Taiz University, YEMEN
• A ray of light travels in a straight path unless
deflected by some change in the medium.
• When reflected at any boundary, the angle of
reflection equals the angle of incidence.
• A ray of light travels more slowly in an optically
dense medium than in one that is less dense, and
the refractive index gives a measure of this effect.
• When a ray is incident on the interface between
two dielectrics of differing refractive indices (e.g.
glass-air), refraction occurs as illustrated in the
next figure.
4 October 2018 3
4. Taiz University, YEMEN
4 October 2018 4
Light rays incident on a high to low refractive index interface (e.g. glass–air): (a) refraction; (b) the limiting case of
refraction showing the critical ray at an angle φc; (c) total internal reflection (TIR) where φ > φc
5. Taiz University, YEMEN
• If the dielectric on the other side of the interface has
a refractive index n2 which is less than n1 then the
refraction is such that the ray path in this lower index
medium is at an angle f2 to the normal, where f2 is
greater than f1
• The angles of incidence f1 and refraction f2 are
related to each other and to the refractive indices of
the dielectrics by Snell's law of refraction which
states that:
4 October 2018 5
)
sin(
n
)
sin(
n 2
2
1
1 f
f
6. Taiz University, YEMEN
• When the angle of refraction is 90º and the
refracted ray emerges parallel to the interface
between the dielectrics the angle of incidence is
called critical angle ɸc.
• At angles of incidence greater than the critical
angle, the light is reflected back into the
originating dielectric medium with high
efficiency (around 99.9%). This is known as total
internal reflection (TIR).
• This is the mechanism by which light may be considered
to propagate down an optical fiber with low loss.
4 October 2018 6
7. Taiz University, YEMEN
• As ɸ1 increases, then there is no
refraction
when the incidence angle
ɸ1 = ɸc = Critical angle
ɸ2=90°
By use Snell’s law
n1 sin(ɸ1)= n2 sin(90°)
Thus the critical angle
• Beyond the critical angle, light ray becomes totally internally
reflected
4 October 2018 7
1
2
1
sin
n
n
c
f
The transmission of a light ray in a perfect optical fiber
8. Taiz University, YEMEN
• An optical fiber consists of two main parts: the core and the cladding
• The core is a narrow cylindrical of glass and the cladding is a tubular jacket
surrounding it
• The core has a (slightly) higher refractive index than the cladding
4 October 2018 8
9. Taiz University, YEMEN
• Figure illustrates the transmission of a light ray in an
optical fiber via a series of total internal reflections at the
interface of the silica core and the slightly lower refractive
index silica cladding.
• The light ray shown in figure is known as a meridional ray as
it passes through the axis of the fiber core.
4 October 2018 9
a
c
f
f
, a
c
f
f
,
a
c
f
f
,
Core n1
Cladding n2
a
10. Taiz University, YEMEN
• Only rays with a sufficiently shallow grazing angle (i.e. with
an angle to the normal greater than fc ) at the core-
cladding interface are transmitted by total internal
reflection, it is clear that not all rays entering the fiber core
will continue to be propagated down its length.
• When the ray enters the fiber core at an angle θa to the
fiber axis and is refracted at the air-core interface before
transmission to the core-cladding interface at the critical
angle.
• So, any rays which are incident into the fiber core at an
angle greater than θa will be transmitted to the core-
cladding interface at an angle less than fc and will not be
totally internally reflected.
4 October 2018 10
11. Taiz University, YEMEN
• Thus for rays to be transmitted by total internal reflection within
the fiber core they must be incident on the fiber core within an
acceptance cone defined by the conical half angle θa.
• Hence θa is the maximum angle to the axis that light may enter the
fiber in order to be propagated and is often referred to as the
acceptance angle for the fiber.
• At air-core interface,
where no=1 (the refractive index of air)
At f= fc when θ= θa
Thus
4 October 2018 11
c
a n f
90
sin
sin 1
f
90
sin
sin 1
n
no
c
a n
θ f
cos
sin 1
12. Taiz University, YEMEN
Or
Since
Then
θa is the acceptance angle of the fiber
4 October 2018 12
2
2
2
1
2
1
2
1 1
sin n
n
n
n
n
θa
1
2
sin
n
n
c
f
c
a n
θ f
2
1 sin
1
sin
2
1
2
2
2
1
1
sin n
n
a
13. Taiz University, YEMEN
• Ray theory analysis can be used to obtain a relationship
between the acceptance angle and the refractive indices of
the three media involved, namely the core. cladding and
air.
This leads to the definition of a more generally used term,
the numerical aperture (NA) of the fiber
• The numerical aperture may also be given in terms of the
relative refractive index difference Δ between the core and
the cladding which is defined as:
4 October 2018 13
2
2
2
1
sin n
n
θ
n
NA a
o
2
1
2
1
2
1
2
1
2
2
2
1
2
2 n
n
n
n
n
n
n
n
14. Taiz University, YEMEN
Since for Δ «1
NA determines the light Gathering capabilities of the
fiber
So the acceptance angle
4 October 2018 14
2
1
2
1
n
NA
1
2
2
1
1
2
1
n
n
n
n
n
n
NA
a
1
sin
15. Taiz University, YEMEN
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.
Answer: (a) ɸc=78.5ᵒ
(b) NA=0.3
(c) θa=17.4ᵒ
4 October 2018 15
Example (1)
16. Taiz University, YEMEN
The velocity of light in the core of a step index fiber
is 2.01 × 108 m s−1, and the critical angle at the
core–cladding interface is 80°. Determine the
numerical aperture and the acceptance angle for
the fiber in air, assuming it has a core diameter
suitable for consideration by ray analysis. The
velocity of light in a vacuum is 2.998 × 108 m s−1.
Answer: 0.263, 15.2°
4 October 2018 16
Example (2)
17. Taiz University, YEMEN
• Three common type of fiber in terms of the material used:
– Glass core with glass cladding-all glass or silica fiber
– Glass core with plastic cladding -plastic cladded /coated silica
(PCS)
– Plastic core with plastic cladding -all plastic or polymer fiber
All glass fiber
The refractive index range of glass is limited which causes the
refractive index difference n1-n2 to be small.
This small value then reduces the light coupling efficiency of the
fiber, i.e. large loss of light during coupling.
The attenuation is the lowest compared to the other two fibers
making it suitable for long and high capacity.
Typical size: 10/125 µm, 62.5/125 µm, 50/125 µm and 100/140µm.
4 October 2018 17
18. Taiz University, YEMEN
Plastic Clad Silica (PCS)
• This fiber have higher loss than the all glass fiber and is suitable
for shorter links.
• Normally, the range of refractive index achievable with plastic
fibers are large.
• A larger range for the value of refractive index difference.
• Light coupling efficiency is better than all-glass.
Typical size: 62.5/125 µm, 50/125 µm, 100/140µm and 200µm.
All-plastic fiber
• This type has the highest transmission loss.
• Normally used for very short links.
• Large core size, therefore light coupling efficiency is high
• The core size can be as large as 1mm.
4 October 2018 18
