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Physics – Photonics

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author(s)

These materials represent the collective effort of many teachers across the state. The principal author of this booklet is:

Dr. Greg Wilmoth, B. Sc (Hons)., Ph.D., Dip. Ed., Grad. Dip. Computing (Senior VCE Teacher – Haileybury College).

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To ensure that students are afforded every possible advantage in their examinations, our lectures are prepared and delivered by qualified, currently practising VCE teachers and official VCAA exam markers who possess the knowledge and experience to demonstrate the means by which students can achieve the higher ATAR scores. Further details regarding our teachers (including qualifications and experience) may be obtained at http://www.tsfx.com.au/what-is-tsfx/ourteachers/.

“the essentials” lectures, 2014

The School For Excellence 2014 The Essentials – Physics – Photonics Page 1

STUDY DESIGN

DETAILED STUDY 3.2 — PHOTONICS Students should be able to apply the photon and wave models of light to explain the operation of different light sources and fibre optic wave-guides and their domestic, scientific and industrial uses. Key Knowledge and Skills To achieve this outcome the student should demonstrate the knowledge and skills to: • Describe the production of incoherent light from wide spectrum light sources, including

the Sun, light bulbs, and candles, in terms of the random thermal motion of valence electrons when atoms collide;

• Explain light emission from light emitting diodes (LEDs) as the energy emitted when electrons move from the conduction band of a semiconductor to the valence band (knowledge about n and p type materials is not required);

• Analyse the effect of band gap energy on LED colour, Eg = hf = hc/λ;

• Describe the production of light by coherent light sources (lasers), in terms of light amplification via stimulation from external photons;

• Describe laser light in terms of coherence, wavelength and phase;

• Analyse the operation of fibre optic wave-guides in terms of:

• Light gathering ability using Snell’s Law, critical angle, total internal reflection and acceptance angle;

• Attenuation by Rayleigh scattering;

• Attenuation due to absorption by impurities in the fibre as well as the molecules that make up the fibre;

• Physical characteristics of single mode and multimode optical fibres (step and graded index);

• Causes of and methods to minimise material dispersion and modal dispersion;

• Compare optical fibres that are used for short and long distance telecommunications;

• Explain fibre optic imaging in terms of coherent and incoherent bundles and in terms of composing images using many fibres, and that these fibres represent the pixels that form the image;

• Describe the operation of optical fibres as simple, intensity-based sensors;

• Identify and apply safe and responsible practices when working with photonics equipment.

(From VCAA Physics Study Design 2009 – 2012)


COHERENT AND INCOHERENT LIGHT

COHERENT LIGHT Whenever a charged particle accelerates, electromagnetic radiation is released. An example is a transmitting radio antenna, where electrons are made to vibrate back and forth, hence emitting radio waves of the same frequency. As the oscillating electrons are undergoing continual acceleration, their movement generates the radio waves that radiate out from the aerial. Because of this synchronisation of the radio wave with the electron vibration, the radio wave is said to be coherent. Because the radio wave is coherent, interference effects can often be observed. For example, the radio signal from a transmitter can sometimes reflect off a building and interfere with the signal arriving at the aerial directly. In this case, destructive or constructive interference can occur and the radio signal may fade or grow as the radio receiver is moved from one location to another. Coherent light rays or photons are in phase with each other. They all have peaks and troughs of the electric and magnetic fields occurring at the same time. The photons are of the same wavelength.

Coherent light occurs when photons are emitted from a light source in a completely synchronised fashion. This means that the photons are in phase with each other as they are emitted from the source. The phase-synchronisation of laser light makes it useful for observing wave properties of light such as diffraction and interference.


INCOHERENT LIGHT Incandescent light sources emit light because of their temperature. They are thermal sources. Stars, candles and light bulbs with filaments are all incandescent light sources. As a piece of iron is heated its temperature obviously increases. At first it emits radiation only in the infra-red part of the electromagnetic spectrum. As it gets hotter, it starts to emit red light. It continues to heat up and emits more light in the visible region of the spectrum until it becomes white hot.

Thermal emission of a black body

From Jacaranda Physics 2

As you can see from the diagram, the intensity distribution of thermal radiation usually extends from the ultraviolet, through the visible, to the infrared part of the EM spectrum. The colour of gases in a flame indicates the temperature of the flame. The blue part of a Bunsen burner flame is the hottest. Blue light has a shorter wavelength and more energy than red light. The light emitted by a candle is caused by the thermal motion of the atoms and molecules in the gases of the flame. The atoms in the filament of a light globe vibrate violently when an electric current passes through it. Collisions between outer-shell electrons produce the light. Incandescent globes are a very inefficient way of generating light, as only about 5% of the electrical energy is converted to visible light. Most of the radiation released by the filament is in the infrared part of the electromagnetic spectrum. The radiation emitted by a continuous-wide-spectrum light source has a wide range of frequencies across the electromagnetic spectrum. These light sources emit incoherent light. The photons emitted from the source are completely unsynchronised or out of step with each other. The radiation emitted by a narrow-spectrum discrete-line source, on the other hand, has a few specific frequency values. Each element has its own characteristic pattern of spectral lines. Examples are vapour lamps and light emitted when salts are put into flames. In these cases the light is incoherent. Photons of the same wavelength that are not coherent have peaks and troughs occurring at different times.


The excitation of isolated atoms, however, results in incoherent optical radiation emitted at just a few discrete frequencies or wavelengths. These are the spectral lines we see through a spectroscope when various elements are burnt in a Bunsen flame. With thermal excitation in a lattice, on the other hand, there are many different modes of oscillation. These result in a continuous wide spectrum of incoherent thermal radiation. In a wide-spectrum EM radiation source, the complex nature of possible interactions produces a broad spectrum. Examples include the Sun, candle flames and incandescent light bulbs.


LIGHT EMISSION FROM LEDS LEDs are semiconductor devices that produce incoherent light in a process called electroluminescence, as electrons drop from the conduction band to the valence band. The wavelength of the radiation produced depends on the energy gap between bands. Semi-conductors are made of lattices of silicon (or sometimes germanium) which have four outer shell electrons that are available for bonding. Each atom therefore can form four covalent bonds with four of its nearest neighbours. These electrons are held firmly, but not as strongly as in an insulator. Because there may be a few mobile charges they are referred to as semiconductors. The conductivity of semiconductors can be increased by replacing some of the lattice atoms with atoms of other elements. This is a process called doping. If we replace a few of the atoms with atoms that have one additional proton in the nucleus (hence an additional electron) such as phosphorus, then those dopant atoms will have one more electron than is required for covalent bonding. This resultant electron is loosely bound and only a small amount of energy is needed to free them. This produces an n-type or negative type semiconductor. Alternatively, if we replace a few of the lattice atoms with atoms that have one less proton eg. aluminium, there will be one less electron than is needed for complete covalent bonding of the lattice electrons. This produces a ‘hole’ or an absence of an electron into which an electron from an adjacent atom can move. This results in a p-type or positive type semiconductor. The process of doping increases the conductivity of the semiconductor. When a potential difference is applied, electrons drift through an n-type layer, and holes drift through the p-type layer as shown in the diagrams below.


Diodes contain a junction between an n-type and a p-type semiconductor. When a battery is connected across the diode with the positive terminal connected to the p-type layer, only a small potential difference (known as the threshold voltage) is needed to create a drift of electrons through the material. This is referred to as forward biasing. When a forward potential is applied, some of the electrons will diffuse across the junction from the n-type to the p-type material and fill some of the vacant holes. This process is called recombination. Similarly, some of the holes will diffuse from the p-type to the n-type material at the junction. As a result, around the junction there will be few mobile charges. This region is called the depletion region and is poorly conductive. This charge separation results in an electric field and an electrical potential barrier across the junction that opposes the diffusion of any more mobile charge. For silicon pn junctions the potential barrier (Vb ) is approximately 0.65 V (and 0.25 V for germanium junctions, 1.6 V for gallium arsenide). The barrier potential is closely related to the switch-on voltage of the diode.


The forward-biased pn junction of an LED can also generate incoherent narrow-spectrum light. The forward bias decreases the effective potential barrier across the pn junction and allows a large current to flow across the junction. This means that a large number of electrons are swept from the n-type region through the depletion region into the p-type region, where they can fill one of the many holes. This is called recombination. Similarly, holes migrate from the p-type region into the n-type region, where they can recombine with electrons. Hence the current that is injected across the forward-biased junction gives rise to recombination and a subsequent release of energy. In certain semiconductors (such as gallium arsenide), this energy is released in the form of photons. This process is called electroluminescence and it is the mechanism that allows the light-emitting diode (LED) to generate optical radiation. The energy of the emitted photon(s) is easily calculated if the energy gap or difference in the energy levels is known. The energy of the photon is equal to the difference in energy between the conduction (Ec) and valence bands (Ev), termed the band gap energy (Eg):

λhc

hfEEEE VCgPhoton ==−==

For a light-demitting diode (LED) there is an energy gap of about 1.7–2.5 eV between levels, and this relates to the voltage drop across a forward biased LED. Different semiconductor materials have different bonds holding the atoms of the lattice together, and hence the energy associated with recombination (i.e. the re-forming of incomplete bonds) varies between different semiconductor materials. In energy level terms, we say that different semiconductor materials have different band gaps. This means that the photons emitted from different semiconductor materials have different peak emission wavelengths (i.e. Different colours). LEDs produce light by spontaneous emission (whereas a laser produces light by stimulated emission). When an LED is forward biased, valence electrons move into the depletion layer from the n-type material, and holes move into the depletion layer from the p-type material. When an electron at the bottom of the conduction band of a semiconductor falls into a hole at the top of a valence band, an amount of energy, Eg, is released, where Eg is the energy gap width between the conduction band and the valence band.


EXAMPLE 1 An LED is made with the material gallium phosphide (GaP). It has an energy gap of 2.24 eV. Calculate the wavelength of light it emits.

smceVsheVEg /1000.3,1014.4,24.2 815 ×=×== −

Solution

g

c

E

h=λ

24.2

1000.31014.4 815 ×××=−

m71054.5 −×=

mμ554.0= or nm554 QUESTION 1 Which material is likely to have the larger band gap energy? A Glass

B Copper

C Graphite

D Silver

QUESTION 2 Laser light in terms of photon wavelength and coherence can be described as A Coherent & same wavelength

B Incoherent & different wavelengths

C Coherent & different wavelength

D Incoherent & same wavelength

QUESTION 3 Arrange the following colours in order of increasing wavelength: Orange, green, yellow, blue, red, violet. A Orange, green, yellow, blue, red, violet.

B Violet, green, blue, yellow, orange, red.

C Violet, blue, green, yellow, orange, red.

D Red, orange, yellow, green, blue, violet.


QUESTION 4 An LED is made from a material that has an energy band gap of 1.96 eV. What will be the approximate wavelength of the light emitted from this LED? Express your answer in nanometres. A 624 nm

B 634 nm

C 644 nm

D 654 nm

QUESTION 5 An indium phosphate LED emits light of wavelength 930 nm. What is the energy band gap of this material in electron volts. A 1.04 eV

B 1.14 eV

C 1.24 eV

D 1.34 eV

QUESTION 6 An LED made from gallium phosphide emits light of wavelength 554 nm. What is the size of its band gap energy in electron volts and Joules? A 2.14 eV / 3.6 x 10-19J

B 2.14 eV / 3.4 x 10-19J

C 2.24 eV / 3.6 x 10-19J

D 2.24 eV / 3.4 x 10-19J


LASER LIGHT The term LASER is an acronym for ‘Light Amplification by Stimulated Emission of Radiation’. A laser is a device that generates a beam of light of coherent electromagnetic radiation. The radiation can be infra-red, visible light or ultraviolet. The following conditions must be met for LASER operation: • There must be more atoms in the excited state than in the ground state. This is achieved by ‘pumping’ the lasing material with energy from an external source. This

energy is provided by either bombarding with electrons (electrical pumping) or with photons (optical pumping).

• The excited state that is used must be metastable. This means that it must be a

relatively stable energy level where the electron takes longer than normal to de-excite and drop to a lower level. In this situation, stimulated emission will occur before spontaneous emission.

• The emitted photons must be used to continue the chain reaction of photon emissions.

This is achieved by placing mirrors at the end of the laser tube. One mirror reflects all the photons back and the other mirror is partially silvered so that some light can escape which forms the LASER beam. The mirrors cause the photons to be reflected back and forth through the lasing medium, stimulating further emissions. The process is also called amplification because one photon produces many identical photons.

A photon with the same energy as the band gap can stimulate an electron to fall from the conduction band to the valence band, releasing another identical photon. This is called stimulated emission and is fundamental to the operation of lasers.

From Heinemann Physics 12

As the emitted photons reflect back and forth between the mirrors, further stimulated emissions occur and a laser beam is produced.


There are many types of LASERS with great variation in emitted wavelength and power output.

Type of LASER Description Uses

Gas lasers

Use gases as the lasing medium,

such as He-Ne and CO2.

Low power lasers for barcode

scanning, high power CO2 lasers for cutting metal and surgery.

Semiconductor Lasers

Low power diode lasers.

Used in CD players and printers.

Infrared diodes used for hair removal and optical fibre

communication.

Excimer Lasers

Use a mixture of reactive gases as

the lasing medium. Produce an ultraviolet beam.

Eye and heart surgery, some industrial applications.

Dye Lasers

Uses a coloured liquid and is

controlled by prisms and gratings to produce a tuneable

frequency of light.

When the wavelength needs to be selected in order to pass through a coloured medium.

Solid State Lasers

Use crystals such as ruby

to emit infrared.

Eye surgery and high precision welding.

In summary: A laser contains a lasing medium in which most of the atoms are kept in an excited metastable state by ‘pumping’ the atoms with energy from bombarding electrons or photons. When an incident photon of the right energy collides with an excited electron, the electron returns to the ground state and releases a photon that is in phase with the incident photon. This is called stimulated emission. The photons reflect back and forth between two mirrors and collide with other electrons, thus building up a chain reaction of released photons that are all in phase. The photons are synchronised with each other. Some of this coherent light passes through the partially reflective mirror, forming the laser beam. The light emerges from the partially reflective mirror as a coherent, monochromatic parallel beam. The beam is parallel because photons that do not bounce between the mirrors are quickly lost through the sides of the medium without being amplified. It is dangerous to look into a LASER light beam as the light energy is concentrated into a small area. All types of LASERs, including ultraviolet, visible and infrared, can quickly cause permanent damage to the retina of the eye. Never look directly into a laser beam or into an optical fibre.


A helium-neon laser emits light with a wavelength of 633 nm. QUESTION 7 What is the frequency of this light? A 4.74 x 1014 Hz

B 4.74 x 1011 Hz

C 3.74 x 1014 Hz

D 3.74 x 1011 Hz

QUESTION 8 Calculate the energy of a single photon. A 2.84 x 10-19 J

B 2.94 x 10-19 J

C 3.04 x 10-19 J

D 3.14 x 10-19 J

QUESTION 9 If the laser had a power rating of 2.5 mW, how many photons would it be emitting each second? A 6.96 x 1012 photons

B 6.96 x 1015 photons

C 7.96 x 1012 photons

D 7.96 x 1015 photons

QUESTION 10 If the laser was aimed at the sky and turned on for just one tenth of a second, how long (in km) would the resultant laser pulse be? A 30 km

B 3.0 x 104 km

C 3.0 x 107 km

D 3.0 x 1010 km


In a CO2 laser, the energy levels between which the excited electrons travel are 0.11 eV apart. QUESTION 11 What is the energy of the emitted photons in eV and in joules? A 0.11 eV / 1.76 x 10-20J

B 0.11 eV / 1.1 x 10-20J

C 0.011 eV / 1.76 x 10-19J

D 0.011 eV / 1.76 x 10-20J

QUESTION 12 Determine the wavelength (in nm) of the light that is produced by the laser. A 1.03 x 104 nm

B 1.03 x 107 nm

C 1.13 x 104 nm

D 1.13 x 107 nm

QUESTION 13 Visible light has wavelengths ranging from around 400 to 700 nm. Does the CO2 laser produce visible light? A Ultraviolet light

B Infrared light

C Visible light

D X rays

The following information refers to Questions 14 to 16. An excimer argon-fluoride laser is electrically pumped and produces UV light of wavelength 1.93 x 10–7 m.

QUESTION 14 Can we see this laser light?

A Yes

B No


QUESTION 15 What is used to stimulate the atoms and molecules in the gas mixture? A Neutrons

B Protons

C Photons

D Electrons

QUESTION 16 Calculate the energy of the photons (in eV). A 6.0 eV

B 6.4 eV

C 6.8 eV

D 7.2 eV

QUESTION 17 An LED is made with the gallium fluoride. What is the energy gap of gallium fluoride in electron volts if it emits light with a wavelength of 630 nm? (Use h = 4.14 × 10−15 eV s and c = 3.00 × 108 m s−1.) A 1.04 eV

B 1.14 eV

C 1.24 eV

D 1.97 eV

QUESTION 18 How is laser light different from light from an incandescent globe? Which one of the following statements is incorrect A Laser light is incoherent (out of phase), whereas incandescent light is in coherent (out

of phase). B Laser light is monochromatic whereas the incandescent globe produces a wide range

of wavelengths C Laser light is coherent (in phase), whereas incandescent light is in coherent (out of

phase). D Laser light produces a parallel beam whereas the incandescent light spreads out in all

directions.


PRINCIPLES OF OPTICAL COMMUNICATION

LIGHT GATHERING ABILITY The light-gathering ability of an optical fibre is an indication of the range of angles at which light can enter the end of the core of an optical fibre and be propagated along the fibre. The bigger the range of angles, the greater the light-gathering ability of the fibre. The light-gathering ability of an optical fibre depends on the following factors: • The diameter of the core – Large-diameter cores have a greater light gathering ability than small-diameter cores. • The critical angle for light meeting the core–cladding interface – The larger the critical angle, the greater the light-gathering ability of the fibre.

TOTAL INTERNAL REFLECTION When a light ray passes from one medium into another, it changes its direction of propagation. This change of direction is called refraction. The change of direction takes place at the actual boundary between the two media which is often called the interface.

The amount of refraction occurring at any boundary depends upon the extent to which the speed of light has been altered, i.e. The ratio of the two speeds of light in the two different media. Therefore:

2

1

2

1

2

1

1

221 sin

sinλλ

θθ ====→ v

v

n

nn

Interface

Medium 1

Incident ray

Refracted

ray

Normal

Medium 2

θ1

θ2


If a light ray passes from a material of high n to a material of low n, the ray will be bent away from the normal. As the angle of incidence is increased, the angle of refraction also increases. This continues until, at a certain angle of incidence called the critical angle, Cθ , the angle of refraction will

be almost 90° and the transmitted ray travels just along the interface. This principle applies to all situations where light travels into a medium of lower refractive index, eg. from glass into air as seen in optical fibres. The critical angle can be found for any boundary between two media by using Snell’s law.

1

2

sinsin

n

n

r

i =

If the incident angle is equal to the critical angle, i.e. i = Cθ , then r = 90°.

The above equation becomes:

1

2

90sinsin

n

no

c =θ

Since sin90° = 1; therefore the critical incident angle can be given by:

1

2sinn

nc =θ

high n

low n


EXAMPLE 2 What is the critical angle for light passing from glass of index of refraction of 1.52 to water of index of refraction 1.33? Solution

oc .

.

n

nsin 61

521331

1

2 ===θ

A light beam entering at an angle greater than the critical angle for the medium used will be reflected along a glass fibre to emerge from the other end. Light entering the fibre with a small angle of incidence will have a large critical angle at the core-cladding boundary and will be totally internally reflected along the fibre. A typical optical fibre has a glass core doped with germanium surrounded by a cladding of slightly lower refractive index. Because the refractive index of the core is only slightly higher than that of the cladding, the critical angle of this interface will be large. On the outside of the cladding, another layer – made of a sponge-like, shock-absorbing plastic – is used to coat the fibre. This coating is called the buffer. It seals the glass surface and protects it from forming small cracks. This structure ensures that the fibre is flexible. There are many advantages with using optical fibres for communication: • High bandwidth. The capacity of optical fibres is many orders of magnitude greater than that of the best copper cable. • Low signal attenuation. The conducting core is pure enough to transmit high-speed signals for 80 km before requiring a repeater, whereas boosters are needed roughly every 50 km for copper media. • Size. Optical fibres are lighter, smaller and more flexible than copper cable. A large number of optical fibres can be carried in a cable as thick as a coaxial cable. • Electrical isolation. Optical fibres do not suffer from ‘crosstalk’ or radio interference. • Security. It is very difficult to ‘tap’ into an optical fibre. In order for this to occur, the

fibre must be broken, in which case the tap will be detectable as no signal is detected at the other end.

QUESTION 19 What is the critical angle for a glass fibre of index 1.50 which is immersed in water of index 1.33. A 54.5o

B 59o

C 62.5o

D 65o


QUESTION 20 The relationship between the refractive index of an optical fibre and its wavelength is shown below.

What is the speed of light for the wavelength 1300 nm in the optical fibre? (Use c = 2.9978 x 108 ms-1) A 1.618 x 107 ms-1

B 1.618 x 108 ms-1

C 1.718 x 107 ms-1

D 1.718 x 108 ms-1

QUESTION 21 The core of an optical fibre with a refractive index of 1.48 has a cladding layer of 1.42. Calculate the critical angle for the core-cladding interface in this fibre. A 71.9o

B 73.6o

C 75.2o

D 78.1o


ACCEPTANCE ANGLE

The acceptance angle 1θ , is the maximum angle at which a ray can enter an optical fibre and then propagate by total internal reflection.

From Heinemann Physics 12 The acceptance angle corresponds to the cone of rays that may enter the fibre, as shown in the diagram above. This cone of acceptance has an apex angle of 2θ1. Any ray that is incident upon the fibre within this cone will be totally internally reflected through the core of the fibre.


NUMERICAL APERTURE The numerical aperture of an optical fibre is a measure of how readily it will capture light. This value depends upon the critical angle inside the fibre, which in turn depends upon the difference in refractive indices of the core and cladding.

221sin claddingcoreext nnnapertureNumerical −== θ

The larger the difference in refractive indices, the more light can get into the fibre. The numerical aperture has a value between zero and one. A typical multimode fibre has a numerical aperture of 0.2 to 0.3. If the fibre is immersed in water instead of in air, then the external refractive index has a value greater than one. In this case, the numerical aperture is unchanged, but the acceptance angle, θ1, is smaller.

EXAMPLE 3 A step-index optical fibre has a doped-glass core of refractive index 1.500 and a surrounding cladding of refractive index 1.480. (a) Calculate the numerical aperture for this fibre. (b) Calculate the acceptance angle for this fibre. (i) If the fibre is in air ( n = 1.000). (ii) If the fibre is immersed in water ( n = 1.330). Solution

(a) 1sinθextnNA = 22claddingcore nn −=

244.0480.1500.1 22 =−= (b) (i) In air: 1sinθextnNA =

244.0sin 1 =θextn

244.0sin 1 =θ

o1.141 =θ (ii) In water: 1sinθextnNA =

244.0sin 1 =θextn

330.1244.0sin 1 =θ

o6.101 =θ The acceptable angle of the fibre is o1.14 in air, but only o6.10 in water.


QUESTION 22 The diagram shows a simple optic light pipe. The core has a refractive index of 1.50 and the cladding an index of 1.40.

(a) What is the maximum angle θi for which the light will be totally reflected? (b) What is the acceptance angle θa, the maximum angle at which light entering the pipe will be transmitted down its length? (c) At what maximum angle θe does the light emerge at the far end?


(d) Proportionately, how much further will light travel if it zig-zags down the length of the pipe at this angle than if it travels straight down the centre? (e) What is the numerical aperture of this fibre?


QUESTION 23 For an optical fibre with a core of absolute refractive index of 1.48 and cladding of 1.42, calculate the numerical aperture for this fibre. A 0.417

B 0.405

C 0.395

D 0.370

QUESTION 24 Calculate the acceptance angle for this fibre. A 19.0o

B 20.5o

C 21.8o

D 24.7o


ATTENUATION Attenuation (T) is a measure of power loss of the signal. If the length of an optical fibre is known, the attenuation rate can be determined.

Attenuation rate (dB/km) = ( )

( )T dB

distance km

EXAMPLE 4 An input signal at 6.0 mW experiences an attenuation of 11.4 dB along a 35 km optic fibre. Calculate the attenuation rate along the fibre.

Attenuation rate (dB/km) = ( )

( )T dB

distance km 354.11= = 0.33 dB / km

The two main factors responsible for attenuation are:

Rayleigh scattering of the light.

Absorption of the light.


QUESTION 25 A 45 km length of optical fibre experiences a signal loss of 13.2 dB. What is the fibre attenuation per kilometre? A 0.26 dB

B 0.29 dB

C 1.32 dB

D 13.2 dB

QUESTION 26 An optical fibre system can experience a maximum signal loss of 16 dB. The attenuation rate of the selected optical fibre is 0.20 dB km-1. What is the maximum length of the optical fibre that could be used? A 80 km

B 32 km

C 16 km

D 3.2 km

QUESTION 27 The following diagram represents the attenuation of a particular fibre.

(a) What is the attenuation at a wavelength of 1400 nm?

A 3.5 dB/km B 2.5 dB/km C 2 dB/km D 1.5 dB/km


(b) What wavelengths are best for transmitting signals along optical fibres made from this material?

A 700 nm – 850 nm B 1300 nm & 1450 nm -1600 nm C 1250 nm – 1275 nm D 1350 nm – 1450 nm

BANDWIDTH The bandwidth of an optical fibre is the range of frequencies at which light signals or pulses can be efficiently transmitted. Bandwidth is equal to the highest rate at which information can be sent along the fibre. If an optical fibre has a bandwidth of 10 Gbps, it can carry up to ten billion bits (pulses) per second.

MODES A mode is a permitted pathway that a ray of light can follow as it is transmitted along an optical fibre. Fibres can have between 1 and 100 000 modes. For a particular wavelength of light, the number of modes supported by a fibre depends on the diameter of its core and the refractive index values of the core and cladding. Modes are distinguished by the number of internal reflections they make as they pass along the core. A high-order mode makes many internal reflections. A low-order mode makes very few internal reflections.

RAYLEIGH SCATTERING Rayleigh scattering is the same process that causes a red sky at sunset. Blue light is scattered more than red light by molecules and particles in the atmosphere because blue light has a shorter wavelength than red light. Rayleigh scattering is caused by variations in the composition and structure of the fibre. During the manufacturing process of the optical fibre, non-uniform mixing of the glass produces very tiny variations in density as it is drawn into a fibre and cools. This causes scattering because when light passes from one refractive index to another some of the light is reflected. Rayleigh scattering increases as the frequency increases or the wavelength decreases. The degree to which it occurs depends upon the relative size of the irregularities and the wavelength of light involved. The larger the irregularities are, the greater the scattering. Generally Rayleigh scattering is caused by particles that are smaller than the wavelength of light. The amount of attenuation is proportional to 4−λ . If the wavelength is doubled, attenuation reduces by 24 which is a factor of 16. Therefore red light scatters far less than blue light. As a result, it is advantageous to use light of longer wavelengths in fibre optics, in order to minimise this scattering of the signal.


QUESTION 28 The following diagram shows the optical loss of a fibre due to Rayleigh scattering.

(a) At what wavelength is an optical loss of 1.0 dB km-1 due to Rayleigh scattering?

A 0.98 µm B 1.2 µm C 0.75 µm D 10 µm

(b) Estimate the optical loss for this fibre at a wavelength of 850 nm.

A 0.8 dB km-1

B 1.8 dB km-1

C 1.5 dB km-1

D 1.7 dB km-1


ABSORPTION As light travels along a fibre, impurities in the glass absorb different components from the beam. Captured energy is converted to heat and is lost from the signal. Particular atoms are capable of absorbing certain wavelengths of light, according to their electron-orbital structure. Attenuation varies with wavelength.


Two major peaks in attenuation are observed. Strong resonances will occur for water contamination and for metallic impurities originating from the metallic crucibles used to melt the glass during manufacture. Note the rapid increase in attenuation in the infrared region, due to absorption by the Si-O and Ge-O bonds found in the glass core. The level of attenuation generally decreases with increasing wavelength, due to reduced Rayleigh scattering, up until about 1.5 µm. The useful minima for signal losses are found around wavelengths of 1.3 µm (1300 nm) and 1.55 µm (1550 nm). As a result, these wavelengths are most commonly employed in optical-fibre systems.


TYPES OF OPTICAL FIBRES

STEP GRADED MULTIMODE FIBRES Step-index fibres are the most common type of optical fibre, but they are not the most efficient. This is primarily because of modal dispersion. Multimode fibres transmit many modes in each fibre. They are used over short distances – for example, in computer networks. The step-index fibre is a multimode fibre, which means it will support more than one propagating mode. It usually has a core diameter of 50 µm, which is large enough to allow the cable to easily be coupled to light sources. The refractive index is constant in value throughout the glass core and then abruptly decreases at the core-cladding interface. All light rays inside the core travel at the same velocity, due to the constant refractive index in this region. Because the higher-order modes take longer to travel the length of the fibre, any sharp pulse of light entering this fibre will become more spread out as it travels further. This effect limits how close the pulses can be before they overlap and become indistinguishable from each other. The spreading out – or loss of definition – of the pulses is referred to as modal dispersion.


In this simple type of fibre, modal dispersion can amount to about 100 ns (or 10–7 s) per km: a very sharp pulse of light would be spread over 100 nanoseconds after a kilometre of travel. This means that less than 107 pulses can be sent along a kilometre of this cable each second. That may seem a lot, but it is less than 10 MHz, which in today’s terms corresponds to a fairly slow rate of information transfer: Compare this figure with the speed of a typical home computer, which would easily be 1000 MHz.


GRADED INDEX MULTIMODE FIBRES Graded-index fibres are used to reduce the effects of modal dispersion. A graded-index fibre has a core whose refractive index is gradually reduced from the centre to the edge. Light travelling down a graded-index fibre bends as it travels through the different layers of the core. It does not reflect sharply at an intersection between the core and cladding as happens with step-index fibres. The graded-index multimode fibre, like the step-index fibre, carries a number of modes of propagation and has a diameter of 50 µm. However, the refractive index is high in value at the core of the fibre and gradually decreases in value extending outwards to the core-cladding boundary. Light travelling through the centre of the core has the slowest velocity, because the refractive index is high in this region. This has the effect of slowing the travel time of lower-order modes. Higher order modes travel greater distances, but with increased velocity due to the lower refractive index towards the edge. The profile of the core’s refractive index is parabolic, which ensures that all modes transit the fibre in approximately the same time. As a result, modal dispersion is reduced to about 1 ns per kilometre of signal transmission.


The graded refractive index in this type of multimode fibre results in the lowest-order mode travelling the shortest distance at the slowest velocity. Higher order modes travel the greatest distance at a faster velocity. Hence, all modes transit in approximately the same time.


SINGLE MODE FIBRES Single-mode fibres are used to transmit one mode. This eliminates the effects of modal dispersion, although material dispersion will still be a factor. They are used in telephone and cable television networks. Single-mode fibres are step-index fibres. A single-mode fibre has a core with a very small diameter of 5–10 um. The cladding is usually 125 um thick. Single-mode fibres have a bandwidth of 50–100 GHz km−1. The core diameter is so small that only the lowest-order or fundamental mode can propagate. The core refractive index is constant, with an abrupt decrease at the core-cladding interface. Because only a single mode travels through the fibre, there is zero modal dispersion. Clearly, single-mode fibres are advantageous to use because they suffer no modal dispersion. Unfortunately they are more expensive to make, and coupling light into and out of such a small diameter core is very difficult. Single-mode fibres are used increasingly for long-haul transmission systems, but multimode fibres are satisfactory for many short distance and low-bit-rate applications.


DISPERSION Dispersion is the spreading out of a light signal or pulse as it travels down a fibre. It limits the bandwidth or the information-carrying capacity of the fibre. There are two main types of dispersion that affect optical fibres: Modal dispersion and material dispersion.

MATERIAL DISPERSION Material dispersion occurs because different wavelengths of light travel at different speeds in a material. Wavelengths at the red end of the spectrum travel faster through a material than do those near the blue end of the spectrum. Material dispersion in optical fibres occurs because light sources do not emit one single wavelength of light – they emit a range of wavelengths. LEDs emit light with about a 35 nm range of wavelengths. For example, an 860 nm LED will actually emit light in the range of about 852–78 nm. Lasers emit light with a 2–3 nm range of wavelengths. Material dispersion means that the longer wavelength part of the signal travels faster through the optical fibre than does the shorter wavelength part. The longer wavelengths will arrive at the destination before the shorter wavelengths. This means that the signal will be spread out which causes the dispersion of the input pulse. Material dispersion is of greater concern in single-mode fibres. In multimode fibres, modal dispersion is so significant that material dispersion is not considered to be a significant factor.


MODAL DISPERSION

Modal dispersion occurs only in multimode fibres. It happens because parts of a light pulse follow different paths (called modes) through the fibre. This means they arrive at different times at the other end of the fibre. This has the effect of spreading the pulse out over time. Note that a high order mode makes many reflections from side to side within the fibre and therefore travels much further than a low-order mode travelling straight down the centre of the fibre. This means that a low-order mode will arrive at the end of the fibre before the high-order mode. This is why the output pulse is spread out over time. Modal dispersion can be reduced in three ways: (a) Use a smaller diameter fibre. This means there will be fewer possible modes. (b) Use a graded-index fibre. (c) Use single-mode fibres. Modal dispersion significantly reduces bandwidth.


QUESTION 29 A computer network uses a step-index fibre with a refractive index for the core of 1.5 and 1.4 for the cladding. (a) What is the numerical aperture for the fibre.

A 0.63 B 0.71 C 0.54 D 0.42

(b) What is the acceptance angle for the fibre?

A 90 o B 66 o C 33o D 28 o


BENDING LOSSES The term macrobending describes the situation in which a bend in a fibre causes light to hit the core-cladding interface at an angle less than its critical angle. As a result, it is refracted through the cladding and lost from the fibre. Light is also lost through a process called microbending. Microbends are tiny kinks in the fibre, within the plastic jacket. As the fibre cools during the manufacturing process, the jacket may shrink more than the fibre and thus kink the fibre. Alternatively, these tiny variations may form when the fibre is placed under stress. Their presence can cause light to be incident at angles less than the critical angle, as with macrobending, and similarly be lost as a result.

This principle has some practical applications. Optical fibres can be incorporated into structures that experience bending forces so that light sent down the fibre is close to the critical angle eg. blades of wind turbines. As the wind strength increases, the turbine blades bend more, resulting in some of the light striking the core/cladding interface at an angle less than the critical angle. Hence some light is lost from the core. This loss can be measured by an optical sensor at the end of the fibre. The intensity of light at the end of the fibre can be related to the extent of bending. Threshold light levels can then feedback to the control unit which would tilt the blades into the wind and reduce the contact surface area. Similar principles are involved in measuring the flex of wings in aircraft.


FIBRE OPTIC IMAGING BUNDLES In addition to communication, optical fibres are used in medical imaging equipment, industrial diagnostic tools and smart structures for buildings and the aerospace industry. Optical fibres can be bundled together to obtain images of hard-to-get-at places. It is important that both ends of the individual fibres are in the same position relative to each other, otherwise the image will be jumbled up. Such bundles of fibres are said to be coherent. Other non-coherent bundles carry light to illuminate the region immediately in front of the probe. By binding a large number of individual optical fibres together in such a way that their arrangement is kept constant, a fibre-optic bundle is produced. This bundle can be used to transmit an image. A fused-fibre bundle is made up of a large number of individual optical fibres, arranged in an ordered array. Heating a bundle of individual optical fibres to a very high temperature fuses the cladding, so that the cores make up an ordered fibre bundle. The fused bundle is then carefully drawn out into a thin, flexible glass thread. The resulting miniature fused-fibre bundle has a diameter of around one quarter of a millimetre and contains approximately 10 000 glass cores, each with a diameter of a few microns and surrounded by a thin glass cladding. These fibres form a 10 000-pixel ordered array with good resolution. What could be some applications of such a bundle?


The following diagram represents three different modes (numbered 1, 2 and 3) of laser light travelling along a step-index multimode optical fibre.

QUESTION 30 Which mode is the highest order mode? A Mode 1

B Mode 2

C Mode 3 QUESTION 31 Which mode is the fundamental mode? A Mode 1

B Mode 2

C Mode 3 QUESTION 32 This optical fibre is likely to be significantly influenced by: A Material dispersion

B Modal dispersion

C Attenuation

D Absorption


QUESTION 33 If the light travelling along this optic fibre was from an LED, which additional problem will the fibre experience? QUESTION 34 If a single pulse of laser light travelling through 1 km of this fibre was spread out over 50 nanoseconds, what would be the maximum bandwidth for this fibre? A 20 kHz

B 2 MHz

C 20 MHz

D 50 GHz QUESTION 35 The optic fibre described previously is replaced with a graded-index multimode fibre. Redraw the three original modes to show how they would travel along this new fibre.


QUESTION 36 What is the main structural difference between this new graded-index fibre and the original step-index fibre? QUESTION 37 What is the main advantage of this new graded-index fibre over the original step-index fibre? QUESTION 38 Which optic fibre is the best to use for long distance telecommunications? A Step-index multimode

B Step-index single mode

C Graded-index multimode

D Graded-index single mode

SOLUTIONS


FOR ERRORS AND UPDATES, PLEASE VISIT

WWW.TSFX.COM.AU/VCE-UPDATES QUESTION 1 Answer is A

Glass is an insulator whereas the other materials are conductors. Insulators have larger band gap energies than conductors.

QUESTION 2 Answer is A

Laser light is coherent, meaning the photons produced by a laser are all in phase with each other. The light is monochromatic, meaning the photons produced by a laser all have the same wavelength.

QUESTION 3 Answer is C

Violet, blue, green, yellow, orange, red.

QUESTION 4 Answer is B

gE

ch=λ

( )96.1

1831014.4 815 ×××=−

λ

nm634=λ

QUESTION 5 Answer is D

λch

Eg =

( )9

815

109301031014.4

−

−

××××=gE

eVEg 34.1=

QUESTION 6 Answer is C

eVch

Eg 24.210554

100.31014.49

815

=×

×××== −

−

λ

This value can be directly converted to Joules by multiplying by 1.6 x 10-19, which gives an energy value of 3.6 x 10-19J.


QUESTION 7 Answer is A 4.74 x 1014 Hz

QUESTION 8 Answer is D 3.14 x 10-19 J

QUESTION 9 Answer is D 7.96 x 1015 photons

QUESTION 10 Answer is B 3.0 x 104 km

QUESTION 11 Answer is A 0.11 eV, 1.76 x 10-20 J

QUESTION 12 Answer is C 1.13 x 104 nm

QUESTION 13 Answer is B (it produces infrared photons.)

QUESTION 14 Answer is B (wavelength is in the UV spectrum and can’t be seen)

QUESTION 15 Electrons.

QUESTION 16 Answer is B 6.4 eV

QUESTION 17 Answer is C 1.34 eV


Laser light is monochromatic whereas the incandescent globe produces a wide range of wavelengths. Laser light produces a parallel beam whereas the incandescent light spreads out in all directions. Laser light is coherent (in phase), whereas incandescent light is in coherent (out of phase).

QUESTION 19 Answer is C 62.5o

QUESTION 20 Answer is D 81.718 10 /m s×


1

2sinn

nc =θ

1

2

n

n=

oc 6.73=θ


QUESTION 22

(a) 5.14.1sin =Cθ , hence o

C 69=θ . Therefore oC 21901 =−= θθ .

(b) 1θ is the refracted angle at the end; 538.0sin5.1sin 1 == θθ a , hence oa 5.32=θ .

(c) eθ is the same as aθ which is 32.5o.

(d) The maximum extra distance is given by 071.1cos

1

1

=θ

. Therefore the longest path

that can be taken is 7.1% longer. (e) The numerical aperture is 538.0sin 1 =θn .


( )22

21 nnNA −=

( )22 42.148.1 −=NA

417.0=NA

QUESTION 24 Answer is D

( )4171.0sin

4171.0sin1−=

==

A

A NA

θθ

o

A 7.24=θ


Fibre attenuation = km

dB.

45213

= 0.29 dB km 1−


Max. distance = kmdB

ratenattenuatio

losssignal80

2.016

==

QUESTION 27 (a) Answer is C 2 dB / km (b) Answer is B 1300 nm and 1450 – 1600 nm

QUESTION 28 (a) Answer is A 0.98 μm (b) Answer is B 1.8 dB km-1


QUESTION 29 (a) Answer is C 0.54 (b) Answer is C 33o

QUESTION 30 Answer is A Number 1 is the highest order mode. QUESTION 31 Answer is C Number 3 is the fundamental mode. QUESTION 32 Answer is B

Modal dispersion (because the three modes are spread out).

QUESTION 33

Material dispersion.

QUESTION 34 Answer is C 20 MHz

timebandwidth

1=

910501

−×=bandwidth

bandwidth = 20 MHz

QUESTION 35

QUESTION 36

The refractive index of the graded-index fibre is greater in the centre of the core and decreases towards the cladding. The step-index fibre has constant refractive index in the core.

QUESTION 37

There is much less modal dispersion in the graded-index fibre than in the step-index fibre.


Step-index single mode fibre.

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