Transcript 14_04_2014 - IB Phys..

IB Physics Topic F
Mr. Jean
April 14th, 2014
The plan:
• Video clip of the day
• Communications
– Fibre Optics
COMMUNICATION
OPTIC FIBRE TRANSMISSION
TOTAL INTERNAL REFLECTION
• https://www.y
outube.com/w
atch?v=EvHIe
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Total Internal Reflection
• When a ray of light travels from a denser
to a rarer medium such that the angle of
incidence is greater than the critical angle,
the ray reflects back into the medium. This
is called total internal reflection.
Optical Fibres
• Consists of a very thin glass
core surrounded by a material
of slightly lower refractive index
called cladding
• The thin fibre can be bent
without breaking and a ray of
light can be sent down the
fibre’s core
• Total internal reflection takes
place at the boundary of the
core and the cladding
Acceptance angle
• The maximum angle of incidence that a
ray can make that will result in total
internal reflection is called acceptance
angle
Material Dispersion
• Light of different wavelengths have
different refractive index and hence come
out of the fibre at different times. This is
called material dispersion
Modal Dispersion
Modal dispersion
• Rays that undergo many internal
reflections are said to follow a high order
mode paths
• Rays undergoing fewer reflections follow
low order mode paths
• Set of rays having same wavelength reach
the end at different times due to different
paths taken. This is called modal
dispersion
Monomode & Multimode fibre
• In multimode fibres, the core has a diameter of
about 100μm and the cladding is about 20μm
thick
• The rays passing through multimode fibres
undergo material as well as modal dispersion
• In monomode fibres, the core has a diameter of
about 8 -10μm and the cladding is about 125μm
thick.
• Rays follow just one path eliminating modal
dispersion
Step Index fibre
• The refractive index of core is constant
• The refractive index of cladding is constant
• The refractive index of cladding is slightly
lower than that of core
Graded index fibre
• Refractive index of core decreases
smoothly from the centre to the outer edge
• Refractive index of cladding is constant
ATTENUATION
• Attenuation in an optic fibre is caused by
the impurities of the glass core. The
amount of attenuation depends on the
wavelength of the light being transmitted.
• Power loss in decibels is defined as
– Power loss = 10log (Pfinal / Pinitial) in dB
• Thus a power loss of 16 decibels means
that the initial power of, say,8.0mW has
been reduced to 0.2mW
VARIATION OF SPECIFIC
ATTENUATION WITH WAVELENGTH
Attenuation & Wavelength
• The specific attenuation ( power loss in dB
per unit length ) actually depends on the
wavelength of the radiation travelling along
the optic fibre
• The graph shows minima at 1310nm and
1550nm, which implies that these are
desirable wavelengths for optimal
transmission
• These are infra red wavelengths
DETECTION
• The light that enters an optic fibre travels
down the length of the fibre and the arrival
of light is registered by a photodiode
• In the absence of any light, falling on the
photodiode, the current is zero
• When light of a specific wavelength falls
on the photodiode, a current flows. The
magnitude of the current is proportional to
the intensity of light
A light detector circuit with a
photodiode
NOISE:
• Source of noise in a cable:
• Random motion of electrons which creates
additional electric fields contaminating the
signal. This increases with temperature.
• Lightning
• Charged particles emitted by the sun
during intense solar activity
NOISE IN OPTICAL FIBRES
• Main source is the dark current of the
photodiode. This is the small current that
flows even when the photodiode is dark
• Signal to noise ratio (SNR) is defined as
• SNR = 10log Psignal / Pnoise
PROBLEM
1. The refractive index of the core of an
optical fibre is 1.50 and that of the
cladding is 1.40.Calculate the
acceptance angle of the fibre.
Ans:330
2. The refractive index of the core of an
optical fibre is 1.50 and the critical angle
of the core- cladding boundary is
750.Calculate the refractive index of the
cladding.
Ans: 1.45
PROBLEM
3. An amplifier amplifies an incoming signal
of power 0.34mW to a signal of power
2.2mW.Calculate the power gain of the
amplifier in decibels.
Ans: 8.1dB
4. A signal of power 12mW is input to a
cable of specific attenuation 4.0 dB/km.
Calculate the power of the signal after it
has travelled 6.0km in the cable.
Ans: 0.048mW
PROBLEM
1. The minimum SNR considered
acceptable for a certain signal is 30dB.If
the power of the noise is 2.0mW,
calculate the least acceptable signal
power.
2. The SNR in a certain signal is 10dB.the
signal passes through an amplifier of gain
6.0dB.What will be the signal to noise
ratio after amplification?