AP
UNIT-5LASERS AND FIBRE OPTICS Q1) Differentiate between spontaneous and stimulated emissions.A 1)
Q2) What is LASER? Discuss basic phenomena involved in producing laser light?A 2)LASER stands for “Light Amplification by Stimulated Emission of Radiation”.L = LightA = Amplification (by)S = StimulatedE = Emission (of)R = RadiationTheodore H. Maiman of the Hughes Research Laboratory, California, was the first scientist who experimentally demonstrated laser by flashing light through a ruby crystal in 1960. But the basic idea behind the development of laser was given by the great scientist “Albert Einstein” in 1917. Let us discuss Einstein’s theory of the interaction of electromagnetic radiation with matter. He proposed that electromagnetic radiation interacts with matter in the following three steps. Stimulated Absorption Spontaneous Emission Stimulated Emission Stimulated Absorption:Let E1 and E2 be the energies of the ground and excited states of an atom. Suppose, if a photon of energy hν= E1− E2 interacts with an atom present in the ground state, the atom gets excitation from ground state E1 to excited state E2. This process is called stimulated absorption. Stimulated absorption rate depends upon the number of atoms available in the lowest energy state as well as the energy density photons. Stimulated absorption rate ∝ Number of atoms in the ground state∝ The density of photons Spontaneous emission
Spontaneous Emission: Let E1 and E2 be the energies of the ground and excited states of an atom. Suppose, if a photon of energy hν= E1− E2 interacts with an atom present in the ground state, the atom gets excitation from ground stateE1 to excited state E2. The excited atom does not stay a long time in the excited state. The excited atom gets de-excitation after its lifetime by emitting a photon of energy hν= E1− E2. This process is called spontaneous emission. Also Spontaneous means by its own. Here excited atom comes to the ground state on its own so it is named as spontaneous emission.The spontaneous emission rate depends upon the number of atoms present in the excited state.Spontaneous emission ∝ rate number of atoms in the excited state Stimulated Emission: This phenomenon is responsible for producing laser light. Let E1and E2 be the energies of the ground and excited states of an atom. Suppose, if a photon of energy hν= E1− E2 interacts with an atom present in the ground state, the atom gets excitation from ground stateE1 to excited state E2. Let, a photon of energy hν= E1− E2 interacts with the excited atom within their lifetime; the atom gets de-excitation to the ground state by emitting another photon. These photons have the same phase and it follows coherence. This phenomenon is called stimulated emission. Stimulated emission rate depends upon the number of atoms available in the excited state as well as the energy density of photons. Stimulated emission rate ∝ number of atoms in the excited state ∝ Density of photonsQ3) Discuss Properties of Laser? A 3)PROPERTIES OF LASERThe laser light exhibits some peculiar properties compare with the conventional light. Those are 1. Highly monochromatic2. Highly coherence 3. Highly directionality 4. Highly intense5. Laser Speckles 1. Highly monochromaticMonochromatic light means a light containing a single colour or wavelength. The photons emitted from ordinary light sources have different energies, frequencies, wavelengths, or colours. Ordinary light is a mixture of waves having different frequencies or wavelengths. The light waves of laser have a single wavelength or colour.
Figure 5: MonochromaticTherefore, laser light covers a very narrow range of frequencies or wavelengths.Hence The laser light is more monochromatic than that of a conventional light source. This may be due to the stimulated characteristic of laser light. The bandwidth of the conventional monochromatic light source is 1000 Å. But the bandwidth of an ordinary light source is 10 Å. For high sensitive laser source is 10-8 Å.
Figure 6: Intensity- wavelength graph for light and laser 2. Highly coherence Definition: - A predictable correlation of the amplitude and phase at any one point with another point is called coherence.
Figure 7: Incoherence and coherenceTwo waves are said to be coherent, the waves must have In the case of conventional light, the property of coherence exhibits between a source and its virtual source whereas in the case of laser the property coherence exists between any two or more light waves. There are two types of coherenceTemporal coherence Spatial coherence Temporal coherence (or longitudinal coherence): -The predictable correlation of amplitude and phase at one point on the wave train w. r. at another point on the same wave train, then the wave is said to be temporal coherence To understand this, let us consider two points P1 and P2 on the same wave train, which is continuous as shown in the figure.
Figure 8: Wave trainSuppose the phase and amplitude at any one point is known, then we can easily calculate the amplitude and phase for any other point on the same wave train by using the wave equationy= a sin (
(ct-x))Where ‘a’ is the amplitude of the wave and ‘x’ is the displacement of the wave at any instant of time ‘t’.Spatial coherence (or transverse coherence) The predictable correlation of amplitude and phase at one point on the wave train w. r. t another point on a second wave, then the waves are said to be spatial coherence (or transverse coherence)
Figure 9: Spatial coherence3. Highly directionalityThe light ray coming from an ordinary light source travels in all directions, but laser light travels in a single direction. For example, the light emitted from torchlight spreads 1km distance it spreads 1 km distance. But the laser light spreads a few centimetres distance even it travels lacks kilometre distance.The directionality of the laser beam is expressed in terms of divergence ∆θ =
Where r1 and r2 are the radii of laser beam spots at distances of D1 and D2 respectively from the laser source.4. Highly Intense or Brightness We know that the intensity of a wave is the energy per unit time flowing through a unit's normal area. Laser light is highly intense than conventional light. A one mill watt He-Ne laser is highly intense than the sun intensity. This is because of the coherence and directionality of the laser. Suppose when two photons each of amplitude a are in phase with other, then young’s principle of superposition, the resultant amplitude of two photons is 2a and the intensity is 4a2. Since in laser many numbers of photons are in phase with each other, the amplitude of the resulting wave becomes na and hence the intensity of the laser is proportional to n2a2. So 1mW He-Ne laser is highly intense than the sun.In an ordinary light source, the light spreads out uniformly in all directions. If you look at a 100 Watt lamp filament from a distance of 30 cm, the power entering your eye is less than 1/1000 of a watt. If you look at laser beam X (caution: don’t do it at home, direct laser light can damage your eyes) X, then all the power in the laser would enter your eye. Thus, even a 1 Watt laser would appear many thousand times more intense than a 100 Watt ordinary lamp.5. Laser Speckles The term speckle refers to a random granular pattern that can be observed when a highly coherent light beam is diffusely reflected at a surface with a complicated structure. This phenomenon results from the interference of different reflected portions of the incident beam with random relative optical phases.
Figure 10: Laser SpecklesEven minor changes of the conditions, such as of the illuminated spot or the direction of the incident laser beam, can change the detailed shape of a speckle pattern.When laser light that has been scattered off a rough surface falls on another surface, it forms an "objective speckle pattern". If a photographic plate or another 2-D optical sensor is located within the scattered light field without a lens, a speckle pattern is obtained whose characteristics depend on the geometry of the system and the wavelength of the laser. Q4) Derive Einstein’s coefficient A and B? A 4)The distribution of atoms in the two energy levels will change by absorption or emission of radiation. Einstein introduced three empirical coefficients to quantify the change of population of the two levels. Let N1 be the number of atoms per unit volume with energy E1 and N2 be the number of atoms per unit volume with energy E2. Let ‘n’ be the number of photons per unit volume at frequency ‘υ’ such that hυ= E1− E2. Then, the energy density of photons ρ(υ) = nhυ
When these photons interact with atoms, both upward (absorption) and downward (emission) transition occurs. At the equilibrium, the upward transitions must be equal to downward transitions.Upward Transition Stimulated absorption rate depends upon the number of atoms available in the lowest energy state as well as the energy density photons.We have seen above that Stimulated absorption rate ∝ N1 i.e. Number of atoms in the ground state ∝ ρ(υ) i.e. Density of photons spontaneous emission Stimulated absorption rate = B12N1ρ(υ) ………(1)Where B12 is the Einstein coefficient of stimulated absorption.Downward transition The spontaneous emission rate depends upon the number of atoms present in the excited state. Spontaneous emission rate ∝ N2 i.e. number of atoms in the excited state Spontaneous emission rate = A21N2 ………(2)Where A21 is the Einstein coefficient of spontaneous emission.Stimulated emission rate depends upon the number of atoms available in the excited state as well as the energy density of photons. Stimulated emission rate ∝ N2 i.e. number of atoms in the excited state ∝ ρ(υ) i.e. Density of photonsStimulated emission rate = B21N2ρ(υ) ………(3)If the system is in equilibrium the upward transitions must be equal to downward transitions.upward transitions = downward transitionsB12N1ρ(υ) = A21N2 + B21N2ρ(υ) ………(4)B12N1ρ(υ) - B21N2ρ(υ) = A21N2 (B12N1- B21N2) ρ(υ) = A21N2 ρ(υ) =
………(5)Divide with B21N2 in numerator and denominator in the right side of the above equation,ρ(υ) = 
= 
………(6)ρ(υ) = 
= 
=
………(7)We know from Maxwell Boltzmann distribution law
= 
………(8) And also from Planck’s law, the radiation densityρ(υ) = 
………(9)Comparing the two equations (7) and (9)
=
and 
=1 ………(10)The above relations are referred to as Einstein relations.From the above equation for non-degenerate energy levels, the stimulated emission rate is equal to the stimulated absorption rate at the equilibrium condition.
= 
………(11)Q5) What is the relationship between B21 and B12?a) B12 > B21
b) B12 < B21
c) B12 = B21
d) No specific relationA 5)C is the correct answer.
B21 is the coefficient for the stimulated emission while B12 is the coefficient for stimulated absorption. Both the processes are mutually reverse processes and their probabilities are equal. Therefore, B12 = B21.Q6) Write Characteristics of Optical Fibre?A 6)It has a large bandwidth. The optical frequency of 2 x 1014 Hz can be used and hence the system has higher bandwidth. Thus optical fibres have greater information-carrying capacity due to greater bandwidth. In an optical fibre system transmission losses are as low as 0.1 db/km. Optical fibre is of small size and lightweight as compared to electrical fibre. Optical fibre communication is free from electromagnetic interference. Optical fibre does not carry high voltage and current hence they are safer than electrical cable. Optical Fibre is flexible and has high tensile strength. Thus can be bent or twisted easily. Q7) Explain the principle on which optical fibre is based? Also explain critical angle?A 7)Principle: Optical Fibre works on the principle of Total Internal Reflection.Total internal reflection:- When the light ray travels from denser medium to rarer medium the refracted ray bends away from the normal. When the angle of incidence is greater than the critical angle, the refracted ray again reflects into the same medium. This phenomenon is called total internal reflection. The refracted ray bends towards the normal as the ray travels from rarer medium to denser medium. The refracted ray bends away from the normal as it travels from denser medium to rarer medium.
Figure 6 – Total Internal ReflectionWhen light passes from a medium with one index of refraction (m1), to another medium with a lower index of refraction (m2), it bends or refracts away from an imaginary line perpendicular to the surface (normal line). As the angle of the beam through m1 becomes greater with respect to the normal line, the refracted light through m2 bends further away from the line.At one particular angle (critical angle), the refracted light will not go into m2, but instead will travel along the surface between the two media (sine [critical angle] = n2/n1 where n1 and n2 are the indices of refraction [n1 is greater than n2]). If the beam through m1 is greater than the critical angle, then the refracted beam will be reflected entirely back into m1 (total internal reflection), even though m2 may be transparent.In physics, the critical angle is described with respect to the normal line. In fiber optics, the critical angle is described with respect to the parallel axis running down the middle of the fiber. Therefore, the fiber-optic critical angle = (90 degrees - physics critical angle).In an optical fiber, the light travels through the core (m1, high index of refraction) by constantly reflecting from the cladding (m2, lower index of refraction) because the angle of the light is always greater than the critical angle. Light reflects from the cladding no matter what angle the fiber itself gets bent at, even if it's a full circle.Because the cladding does not absorb any light from the core, the light wave can travel great distances. However, some of the light signal degrades within the fiber, mostly due to impurities in the glass. The extent that the signal degrades depends upon the purity of the glass and the wavelength of the transmitted light When the angle of incidence (θ1) is progressively increased, there will be progressive increase of refractive angle (θ2). At some condition (θ1) the refractive angle (θ2) becomes 90o to the normal. When this happens the refracted light ray travels along the interface. The angle of incidence (θ1) at the point at which the refractive angle (θ1) becomes 90 degree is called the critical angle. It is denoted by θc. The critical angle is defined as the minimum angle of incidence (θ1) at which the ray strikes the interface of two media and causes an angle of refraction (θ2) equal to 90o. Figure 6 shows critical angle refractionHence at critical angle θ1= θc and θ2= 90o .Using Snell‘s law: n1 sin θ1 = n2 sin θ2 n1 sin θc = n2 sin90osin θc = n2 / n1 θc = sin-1 (n2 / n1)The actual value of critical angle is dependent upon combination of materials present on each side of boundary.Q8) Discuss the construction, working, and application of gas Laser? OrDiscuss the construction, working, and application of He-Ne Laser?OrDiscuss four-level laser?A 8)He-Ne Laser The first He-Ne gas laser was fabricated in 1961 by Ali Javan, Bennett, and Herriott at Bell Telephone Laboratories. others. Helium-Neon laser is a type of gas laser in which a mixture of helium and neon gas is used as a gain medium. Helium-Neon laser is also known as He-Ne laser. The helium-neon laser was the first continuous-wave laser ever constructed. The helium-neon laser operates at a wavelength of 632.8 nanometres (nm), in the red portion of the visible spectrum.Ruby laser is a pulse laser, even it has high intense output. For continuous laser beams, gas lasers are used. Using gas lasers, we can achieve high coherence, high directionality, and high monochromaticity beam. The output power of the gas laser is generally in few milliwatts. CONSTRUCTIONThe helium-neon laser consists of three essential components: Pump source (high voltage power supply) Gain medium (laser glass tube or discharge glass tube) Resonating cavity Pump source The gain medium of a helium-neon laser is made up of a mixture of helium and neon gas contained in a glass tube at low pressure. In the He-Ne gas laser, the He and Ne gases are taken in the ratio 10:1 in the discharge tube. Gain medium In He-Ne laser 80cm length and 1cm diameter discharge are generally used. The out power of these lasers depends on the length of the discharge tube and the pressure of the gas mixture. Therefore, to achieve population inversion, we need to pump electrons from the lower energy state of the helium. In He-Ne laser, neon atoms are the active centres and have energy levels suitable for laser transitions while helium atoms help in exciting neon atoms.
Resonating cavityTwo reflecting mirrors are fixed on either end of the discharge tube, in that, one is partially reflecting and the other is fully reflecting. The fully silvered mirror will completely reflect the light whereas the partially silvered mirror will reflect most of the light but allows some part of the light to produce the laser beam.WORKING When the electric discharge is passing through the gas mixture, the electrons accelerated towards the positive electrode. During their passage, they collide with He atoms and excite them into higher levels. 23s1 and 21s0 form the ground state of the He atom. In higher levels, 23s1 and 21s0, the lifetime of He atoms are more. So there is a maximum possibility of energy transfer between He and Ne atoms through atomic collisions. When He atoms present in the levels 23s1 and 21s0 collide with the Ne atom's present ground state, the Ne atoms get excitation into higher levels 4s and 5s. Due to the continuous excitation of Ne atoms, we can achieve the population inversion between the higher levels 4s and 5s and lower levels 3p and 4p. The various transitions 5s to 4p, 4s to 3p, and 5s to 3p leads to the emission of wavelengths 3.93μm, 1.51μm, and 6328 Å or 632.8μm.
The first two correspondings to the infrared region while the last wavelength is corresponding to the visible region. The Ne atoms present in the 4s level are de-excited into 3s level, by spontaneously emitting a photon of around wavelength 6000 Å. When a narrow discharge tube is used, the Ne atoms present in the level 3s collide with the walls of the tube and get de-excited to the ground state energy level.ADVANTAGES OF HELIUM-NEON LASERThe helium-neon laser emits laser light in the visible portion of the spectrum. High stability Low cost Operates without damage at higher temperatures DISADVANTAGES OF HELIUM-NEON LASERLow efficiency Low gain Helium-neon lasers are limited to low power tasks APPLICATIONS OF HELIUM-NEON LASERSHelium-neon lasers are used in industries. Helium-neon lasers are used in scientific instruments. Helium-neon lasers are used in the college laboratories Q9) Discuss Applications of Lasers in various fields?A 9)Applications of lasers because of the unique property of laser beams such as coherence, monochromaticity, directionality, and high intensity, they are widely used in various fields like 1. Communication 2. Computers 3. Chemistry 4. Photography 5. Industry 6. Medicine 7. Military 8. Scientific Research1. Communication In the case of optical communication semiconductors, laser diodes are used as optical sources and their bandwidth is (1014Hz) is very high compared to radio and microwave communications. More channels can be sent simultaneously Signal cannot be tapped as the bandwidth is large, more data can be sent. A laser is highly directional and less divergence, hence it has greater potential use in space crafts and submarines. It is used in optical fiber communications to send information over large distances with low loss. Laser light is used in underwater communication networks. Lasers are used in space communication, radars, and satellites.2. Computers In LAN (local area network), data can be transferred from memory storage of one computer to another computer using laser for short time. Lasers are used in CD-ROMS during recording and reading the data. Lasers are used in computer printers.3. Chemistry Lasers are used in molecular structure identification Lasers are also used to accelerate some chemical reactions. Using lasers, new chemical compounds can be created by breaking bonds between atoms are molecules. 4. Photography Lasers can be used to get a 3-D lens with less photography. Lasers are also used in the construction of holograms. 5. Industry Lasers can be used to blast holes in diamonds and hard steel. Lasers are also used as a source of intense heat Carbon dioxide laser is used for cutting drilling of metals and non-metals, such as ceramics plastics glass, etc. High power lasers are used to weld or melt any material. Lasers are also used to cut teeth in saws and test the quality of the fabric. It is used to cut glass and quartz, used in electronic industries for trimming the components of Integrated Circuits (ICs). Lasers are used for heat treatment in the automotive industry. Laser light is used to collect information about the prefixed prices of various products in shops and business establishments from the bar code printed on the product. Ultraviolet lasers are used in the semiconductor industries for photolithography. Photolithography is the method used for manufacturing printed circuit board (PCB) and microprocessor by using ultraviolet light. It is also used to drill aerosol nozzles and control orifices within the required precision. 6. Medicine Pulsed neodymium laser is employed in the treatment of liver cancer. Argon and carbon dioxide lasers are used in the treat men of the liver and lungs. Lasers are used in the treatment of Glaucoma. Lasers are used in endoscopy to scan the inner parts of the stomach. Lasers are used in the elimination of moles and tumors which are developing in the skin tissue and hair removal. It is also used for bloodless surgery.Lasers are used to destroy kidney stones, in cancer diagnosis and therapy also used for eye lens curvature corrections. Lasers are used to study the internal structure of microorganisms and cells. It is used to create plasma. Lasers are used to remove caries or decayed portion of the teeth.7. Military Lasers can be used as a war weapon. High energy lasers are used to destroy the enemy air-crofts and missiles. Lasers can be used in the detection and ranging likes RADAR. Laser range finders are used to determine the distance to an object. The ring laser gyroscope is used for sensing and measuring a very small angle of rotation of the moving objects.Lasers can be used as a secretive illuminator for reconnaissance during the night with high precision. 8. Scientific research Lasers are used in the field of 3D-photography Lasers are used in Recording and reconstruction of the hologram. Lasers are employed to create plasma. Lasers are used in Raman spectroscopy to identify the structure of the molecule and to count the number of atoms in a substance. Lasers are used in the Michelson- Morley experiment. A laser beam is used to confirm Doppler shifts in frequency for moving objects. A laser helps in studying the Brownian motion of particles. With the help of a helium-neon laser, it was proved that the velocity of light is the same in all directions. Lasers are used to measure the pollutant gases and other contaminants of the atmosphere. Lasers help in determining the rate of rotation of the earth accurately. Lasers are used for detecting earthquakes and underwater nuclear blasts. A gallium arsenide diode laser can be used to set up an invisible fence to protect an area. Q10) When mean optical power launched into an 8 km length of fibre is 12 μW, the mean optical power at the fibre output is 3 μW. Determine Overall signal attenuation in dB.The overall signal attenuation for a 10 km optical link using the same fibre with splices at 1 km intervals, each giving an attenuation of 1 dB.A 10)Given :z=8kmP(0) = 120 μWP(z) = 3 μWOverall attenuation is given by,αp (dB/km ) = 10. log 
= 16.02Overall attenuation for 10 km,Attenuation per km = αp (dB/km )= 
= 
= 2.00 dB/km Attenuation in 10 km link = 2.00 x 10 = 20 dBIn 10 km link there will be 9 splices at 1 km interval. Each splices introducing attenuation of 1 dB.Total attenuation = 20 dB + 9 dB = 29 dBQ11) Explain the terms acceptance angle, acceptance cone?A11)Acceptance angle Definition:- Acceptance angle is defined as the maximum angle of incidence at the interface of air medium and core medium for which the light ray enters into the core and travels along with the interface of core and cladding. Let n0 be the refractive indices of airn1 be the refractive indices of coren2 be the refractive indices of cladding
Let a light ray OA is an incident on the interface of air medium and core medium with an angle of incidence θ0 The light ray refracts into the core medium with an angle of refraction θ1 and the refracted ray AB is again incident on the interface of core and cladding with an angle of the incident (90- θ1) If (90- θ1) is equal to the critical angle of core and cladding media then the ray travels along with the interface of core and cladding along the path BC. If the angle of the incident at the interface of air and core θ1< θ0 then (90- θ1) will be greater than the critical angle. Therefore, The total internal reflection takes place.According to Snell’s law at point An0 Sin θ0 = n1 Sin θ1Sin θ0= (n1 / n0) Sin θ1 ………(1)According to Snell’s law at point Bn1 Sin(90- θ1) = n2 Sin90 ………(2)n1 Cosθ1 = n2 as (Sin90=1)Cosθ1 = n2 /n1Sinθ1 = (1-Cos2 θ1)1/2Sinθ1= (1- (n2 /n1)2)1/2Sinθ1= ( n12- n22 )1/2/ n1 ………(3)We know Sin θ0= (n1 / n0) Sin θ1 from equation (1)Substitute the value of Sinθ1 from equation (3)Sinθ0= (n1 / n0) *( n12- n22 )1/2/ n1On simplificationSinθ0= ( n12- n22 )1/2/ n0θ0=Sin-1 ( n12- n22 )1/2/ n0 Acceptance Angle is θ0=Sin-1 ( n12- n22 )1/2/ n0 ………(4)Acceptance ConeAcceptance angle is the maximum angle that a light ray can have relative to the axis of the fibre and propagate down the fibre. Thus, only those rays that are incident on the face of the fibre making angles less than θ0 will undergo repeated total internal reflections and reach the other end of the fibre. Hence, larger acceptance angles make it easier to launch light into fibre.
Figure: 27In three dimensions, the light rays contained within the cone having a full angle 2θ0 are accepted and transmitted along with the fibre as shown in figure 27. Therefore, the cone is called the acceptance cone. Light incident at an angle beyond θ0 refracts through the cladding and corresponding optical energy is lost.Q12) Derive expressions for the numerical aperture and fraction change in refractive index change of an optical fibre. A 12)Numerical apertureDefinition: -Numerical aperture is defined as the light gathering capacity of an optical fibre and it is directly proportional to the acceptance angle. Numerically it is equal to the sin of the acceptance angle.NA = Sin(acceptance angle)NA = Sin {Sin-1 (( n12- n22 )1/2/ n0)} from equation (4)NA = (( n12- n22 )1/2/ n0) ………(5)If the refractive index of the air medium is unity i.e. n0=1 put in (5)NA = ( n12- n22 )1/2 ………(6)Fractional change in refractive index∆= (n1- n2)/ n1n1∆ = (n1- n2) ………(7)from equation (6), we haveNA = {( n1- n2 )( n1+n2 )}1/2NA = { n1∆ (n1+n2 )}1/2 as n1∆ = (n1- n2) by Eq(7)NA = { n1∆ 2n1}1/2 n1 ≈ n2, so n1+n2 =2n1NA = n1{2∆}1/2 This gives the relation between Numerical aperture and Fractional change in refractive index.Q13) What are the advantages and disadvantages of an optical fibre over normal wire?A 13)Advantages of fibre optic communicationOptical fibre communication has more advantages than conventional communication. 1. Enormous Bandwidth 2. Low Transmission Loss 3. Electric Isolation 4. Signal Security5. Small Size and Less Weight6. Immunity Cross Talk 1. Enormous bandwidth: - The information-carrying capacity of a transmission system is directly proportional to the frequency of the transmitted signals. In the coaxial cable transmission, the bandwidth range is up to around 500MHz only. Whereas in optical fibre communication, the bandwidth range is large as 105 GHz. 2. Low transmission loss:- The transmission loss is very low in optical fibres (i.e.KmdB/2.0) than compare with the conventional communication system. Hence for long-distance communication fibres are preferred. 3. Electric isolation:- Since fibre optic materials are insulators, they do not exhibit earth and interface problems. Hence communicate through fibre even in an electrical dangerous environment. 4. Signal security:- The transmitted signal through the fibre does not radiate, unlike the copper cables, a transmitted signal cannot be drawn from fibre without tampering with it. Thus the optical fibre communication provides 100% signal security. 5. Small size and less weight:- The size of the fibre ranges from 10μm to 50μm, which is very small. The space occupied by the fibre cable is negligibly small compared to conventional electrical cables. Optical fibres are light in weight.6. Immunity cross-talk:- Since the optical fibres are dielectric waveguides, they are free from any electromagnetic interference and radio frequency interference. Since optical interference among different fibres is not possible, cross talk is negligible even many fibres are cabled together. Disadvantages of Optical FibreThe disadvantages of optical fibre include the followingThe main disadvantages of these cables are installation is expensive and difficult to fix together.The optical fibre cables are very difficult to merge & there will be a loss of the beam within the cable while scattering. Fibre optic cables are compact and highly vulnerable while fittingThese cables are more delicate than copper wires.Special devices are needed to check the transmission of fibre cables.Q14) Write a note on the applications of an optical fibre?A 14)Application of optical fibre Optical fibres are extensively used in a communication system.Optical fibres are in the exchange of information between different computers Optical fibres are used for the exchange of information in cable televisions, space vehicles, submarines, etc. Optical fibres are used in the industry in security alarm systems, process control, and industrial auto machine. Optical fibres are used in pressure sensors in biomedical and engine control. Optical fibres are used in medicine, in the fabrication of endoscopy for the visualization of internal parts of the human body. Sensing applications of optical fibres are Displacement sensor, Fluid level detector, Liquid level sensor, Temperature/pressure sensor, and Chemical sensors Medical applications of optical fibres are Gastroscopy, Orthoscopic, Couldoscope, Peritonescope, and Fibrescope.Q15) Explain how the optical fibres are classified?A15)TYPES OF OPTICAL FIBRESThe types of optical fibres depend on the refractive index, materials used, and mode of propagation of light.The classification based on the materials used is as follows:Plastic Optical Fibres: The polymethylmethacrylate is used as a core material for the transmission of light. Example: Core: polymethyl methacrylate : Cladding: Co- PolymerCore: Polystyrene : Cladding: Methyl methacrylateGlass Fibres: It consists of extremely fine glass fibres. Example: Core: SiO2 Cladding: SiO2Core: GeO2- SiO2 Cladding: SiO2The classification based on the mode of propagation of light is as follows:Mode of propagation:Light propagates as electromagnetic waves through an optical fibre. All waves, having ray directions above the critical angle will be trapped within the fibre due to total internal reflection. However, all such waves do not propagate through the fibre. Only certain ray directions are allowed to propagate. The allowed directions correspond to the modes of the fibre. In simple terms, modes can be visualized as the possible number of paths of light in an optical fibre. Single-Mode Fibres: These fibres are used for long-distance transmission of signals. In general, the single mode fibres are step-index fibres. These types of fibres are made from doped silica. It has a very small core diameter so that it can allow only one mode of propagation and hence called single-mode fibres.The cladding diameter must be very large compared to the core diameter. Thus in the case of single-mode fibre, the optical loss is very much reduced. The structure of a single-mode fibre is given below.Structure:Core diameter : 5-10μmCladding diameter : Generally around 125μmProtective layer : 250 to 1000μmNumerical aperture : 0.08 to 0.10Bandwidth : More than 50MHz km.Application:Because of high bandwidth, they are used in long-haul communication systems.
Multimode Fibres:
Spontaneous emission | Stimulated emission |
1.The spontaneous emission was postulated by Bohr | 1.The stimulated emission was postulated by Einstein |
2. Additional photons are not required in spontaneous emission | 2. Additional photons are required in stimulated emission |
3.One photon is emitted in spontaneous emission | 3.Two photons are emitted in stimulated emission |
4.The emitted radiation is poly-monochromatic | 4.The emitted radiation is monochromatic |
5. The emitted radiation is Incoherent | 5. The emitted radiation is Coherent |
6. The emitted radiation is less intense | 6. The emitted radiation is high intense |
7.The emitted radiation has less directionality | 7.The emitted radiation has high directionality |
8. Example: light from sodium or mercury lamp | 8. Example: light from the laser source.
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(ct-x))Where ‘a’ is the amplitude of the wave and ‘x’ is the displacement of the wave at any instant of time ‘t’.Spatial coherence (or transverse coherence) The predictable correlation of amplitude and phase at one point on the wave train w. r. t another point on a second wave, then the waves are said to be spatial coherence (or transverse coherence)
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Where r1 and r2 are the radii of laser beam spots at distances of D1 and D2 respectively from the laser source.4. Highly Intense or Brightness We know that the intensity of a wave is the energy per unit time flowing through a unit's normal area. Laser light is highly intense than conventional light. A one mill watt He-Ne laser is highly intense than the sun intensity. This is because of the coherence and directionality of the laser. Suppose when two photons each of amplitude a are in phase with other, then young’s principle of superposition, the resultant amplitude of two photons is 2a and the intensity is 4a2. Since in laser many numbers of photons are in phase with each other, the amplitude of the resulting wave becomes na and hence the intensity of the laser is proportional to n2a2. So 1mW He-Ne laser is highly intense than the sun.In an ordinary light source, the light spreads out uniformly in all directions. If you look at a 100 Watt lamp filament from a distance of 30 cm, the power entering your eye is less than 1/1000 of a watt. If you look at laser beam X (caution: don’t do it at home, direct laser light can damage your eyes) X, then all the power in the laser would enter your eye. Thus, even a 1 Watt laser would appear many thousand times more intense than a 100 Watt ordinary lamp.5. Laser Speckles The term speckle refers to a random granular pattern that can be observed when a highly coherent light beam is diffusely reflected at a surface with a complicated structure. This phenomenon results from the interference of different reflected portions of the incident beam with random relative optical phases.
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………(5)Divide with B21N2 in numerator and denominator in the right side of the above equation,ρ(υ) = = ………(6)ρ(υ) = = = ………(7)We know from Maxwell Boltzmann distribution law = ………(8) And also from Planck’s law, the radiation densityρ(υ) = ………(9)Comparing the two equations (7) and (9)= and =1 ………(10)The above relations are referred to as Einstein relations.From the above equation for non-degenerate energy levels, the stimulated emission rate is equal to the stimulated absorption rate at the equilibrium condition.= ………(11)Q5) What is the relationship between B21 and B12?a) B12 > B21b) B12 < B21
c) B12 = B21
d) No specific relationA 5)C is the correct answer.
B21 is the coefficient for the stimulated emission while B12 is the coefficient for stimulated absorption. Both the processes are mutually reverse processes and their probabilities are equal. Therefore, B12 = B21.Q6) Write Characteristics of Optical Fibre?A 6)
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= 16.02Overall attenuation for 10 km,Attenuation per km = αp (dB/km )= = = 2.00 dB/km Attenuation in 10 km link = 2.00 x 10 = 20 dBIn 10 km link there will be 9 splices at 1 km interval. Each splices introducing attenuation of 1 dB.Total attenuation = 20 dB + 9 dB = 29 dBQ11) Explain the terms acceptance angle, acceptance cone?A11)Acceptance angle Definition:- Acceptance angle is defined as the maximum angle of incidence at the interface of air medium and core medium for which the light ray enters into the core and travels along with the interface of core and cladding. Let n0 be the refractive indices of airn1 be the refractive indices of coren2 be the refractive indices of cladding
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These fibres are used for short-distance transmission of signals. The multi-mode fibres are useful in manufacturing both for step-index and graded-index fibres. The multi-mode fibres are made by multi-component glass compounds such as Glass – Clad Glass, Silica – Clad – Silica, doped silica, etc. Here the core diameter is very large compared to single-mode fibres, so that it can allow many modes to propagate through it and hence called Multi-mode fibres. The cladding diameter is also larger than the diameter of the single-mode fibres. The structure of the multimode fibre is as shown in the figure above.Structure:Core diameter : 50-350μmCladding diameter : 125μm - 500μmProtective layer : 250 to 1100μmNumerical aperture : 0.12 to 0.5Bandwidth : Less than 50MHz km. The total number of modes possible for such an electromagnetic waveguide is
Because of its less bandwidth, it is very useful in short-haul communication systems.The classification based on the refractive index is as follows:Step Index Single-mode FibresIt consists of a core surrounded by the cladding, which has a single uniform index of refraction. Step index-single mode fibres: A single-mode step-index fibre consists of a very thin core of uniform refractive index surrounded by a cladding of refractive index lower than that of the core. The refractive index abruptly changes at the core-cladding boundary. Light travels along a side path, i.e., along the axis only. So zero-order modes are supported by Single Mode Fibre.
Figure: 23Step index-Multimode fibres A multimode step-index fibre consists of a core of uniform refractive index surrounded by a cladding of refractive index lower than that of the core. The refractive index abruptly changes at the core-cladding boundary. The core is of large diameter. Light follows zigzag paths inside the fibre. Many such zigzag paths of propagation are permitted in Multi-Mode Fibre. The Numerical Aperture of a Multi-mode fibre is larger as the core diameter of the fibre is larger
Figure: 24Graded Index Fibres: The refractive index of the optical fibre decreases as the radial distance from the fibre axis increases. GRIN fibre is one in which the refractive index varies radially, decreasing continuously in a parabolic manner from the maximum value of n1, at the center of the core to a constant value of n2 at the core-cladding interface. In graded-index fibre, light rays travel at different speeds in different parts of the fibre because the refractive index varies throughout the fibre. Near the outer edge, the refractive index is lower. As a result, rays near the outer edge travel faster than the rays at the center of the core. Because of this, rays arrive at the end of the fibre at approximately the same time. In effect light rays that arrive at the end of the fibre are continuously refocused as they travel down the fibre. All rays take the same amount of time in traversing the fibre. This leads to small pulse dispersion.
Figure: 25
For a parabolic index fibre, the pulse dispersion is reduced by a factor of about 200 in comparison to step-index fibre. It is because of this reason that first and second-generation optical communication systems used near parabolic index fibres.Q16) Explain optical fibre as a dielectric wave guide? Also define term Meridional rays and Skew Rays? A 16) Light propagates inside an optical fiber by virtue of multiple TIRs at the core-cladding interface. The refractive index of the core glass is greater than that of the cladding. This meets the first condition for a TIR. All the light energy that is launched into the optical fiber through its tip does not get guided along the fiber. Only those light rays propagate through the fiber which is launched into the fiber at such an angle that the refracted ray inside the core of the optical fiber is incident on the core-cladding interface at an angle greater than the critical angle of the core with respect to the cladding. 1. Optical fiber is basically a solid glass rod. The diameter of rod is so small that it looks like a fiber. 2. Optical fiber is a dielectric waveguide. The light travels like an electromagnetic wave inside the waveguide. The dielectric waveguide is different from a metallic waveguide which is used at microwave and millimeter wave frequencies.3. In a metallic waveguide, there is a complete shielding of electromagnetic radiation but in an optical fiber the electromagnetic radiation is not just confined inside the fiber but also extends outside the fiber. 4. The light gets guided inside the structure, through the basic phenomenon of total internal reflection. 5. The optical fiber consists of two concentric cylinders; the inside solid cylinder is called the core and the surrounding shell is called the cladding. (See Fig 12)
Figure 20: Schematic of an optical fiber6. For the light to propagate inside the fiber through total internal reflections at core-cladding interface, the refractive index of the core must be greater than the refractive index of the cladding. That is n1>n2.SIMPLE RAY MODEL
Figure 21: (optical fiber with core, cladding and total internally reflected ray)For propagation of light inside the core there are two possibilities.1. A light ray is launched in a plane containing the axis of the fiber. We can then see the light ray after total internal reflection travels in the same plane i.e., the ray is confined to the plane in which it was launched and never leave the plane. In this situation the rays will always cross the axis of the fiber. These are called the Meridional rays. (Figure 21)2. The other possibility is that the ray is not launched in a plane containing the axis of the fiber.For example if the ray is launched at some angle such that it does not intersect the axis of the fiber, then after total internal reflection it will go to some other plane. We can see that in this situation the ray will never intersect the axis of the fiber. The ray essentially will spiral around the axis of fiber. These rays are called the Skew rays. So it can be concluded that if the light is to propagate inside an optical fiber it could be through two types of raysa) Meridional rays: The rays which always pass through the axis of fiber giving high optical intensity at the center of the core of the fiber. b) Skew Rays: The rays which never intersect the axis of the fiber, giving low optical intensity at the center and high intensity towards the rim of the fiber. Q17) Discuss the principle, construction and working of Optical Fibre? A 17)A cable that is used to transmit the data through fibres (threads) or plastic (glass) is known as an optical fibre cable. This cable includes a pack of glass threads that transmits modulated messages over light waves. Principle: Optical Fibre works on the principle of Total Internal Reflection.Total internal reflection: - When the light ray travels from a denser medium to a rarer medium the refracted ray bends away from the normal. When the angle of incidence is greater than the critical angle, the refracted ray again reflects into the same medium. This phenomenon is called total internal reflection. The refracted ray bends towards the normal as the ray travels from a rarer medium to a denser medium. The refracted ray bends away from the normal as it travels from denser medium to rarer medium.
Figure – Total Internal Reflection
Figure: Optical FibreConstruction of Optical Fibre:It consists of a very thin fibre of silica or glass or plastic of a high refractive index called the core. The core has a diameter of 10 um to 100 um. The core is enclosed by a cover of glass or plastic called cladding. The refractive index of the cladding is less than that of the core (which is a must condition for the working of the optical fibre). The difference between the two indicates is very small of order 10-3. The core and the cladding are enclosed in an outer protective jacket made of plastic to provide strength to the optical fibre. The refractive index can change from core to cladding abruptly (as in step-index fibre) or gradually (as in graded-index fibre).
Figure Representation of Optical FibreWorking of Optical FibreWhen a ray of light is incident on the core of the optical fibre at a small angle, it suffers refraction and strikes the core-cladding interface, As the diameter of the fibre is very small hence the angle of incidence is greater than the critical angle. Therefore, the ray suffers total internal reflection at the core-cladding interface and strikes the opposite interface. At this interface also, the angle of incidence is greater than the critical angle, so it again suffers total internal reflection. Thus, the ray of light reaches the other end of the fibre after suffering repeated total internal reflections along the length of the fibre. At the other end, the ray suffers refraction and emerges out of the optical fibre. We can see that the light travels in the core in a guided manner. Hence the communication through the optical fibre is sometimes referred to as an optical waveguide.Q18) The refractive indices of core and cladding materials of a step index fibre are 1.48 and 1.45, respectively. Calculate numerical aperture?A 18) Let the refractive index of core, n1 = 1.48and the refractive index of cladding , n2 = 1.45Numerical aperture (NA) 
Q19) A step-index fiber has a core index of refraction of n1 = 1.425. The cut-off angle for light entering the fiber from air is found to be 8.50o. (a) What is the numerical aperture of the fiber? (b) What is the index of refraction of the cladding of this fiber? (c) If the fiber were submersed in water, what would be the new numerical aperture and cut-off angle?
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Q19) A step-index fiber has a core index of refraction of n1 = 1.425. The cut-off angle for light entering the fiber from air is found to be 8.50o. (a) What is the numerical aperture of the fiber? (b) What is the index of refraction of the cladding of this fiber? (c) If the fiber were submersed in water, what would be the new numerical aperture and cut-off angle? A 19) (a) we know that n0 = nair = 1.0003. The numerical aperture is therefore NA = n0 sin θ0max = (1.0003) sin (8.50o) = 0.148.(b) The index of refraction of the cladding can be found from the numerical aperture: n12 - n22 = NA2.
n22 = n12 - NA2 = (1.425)2 - (0.1479)2 = 2.0088
n2 = 1.417.(c) We know that that the n0 = nwater = 1.33. Since the numerical aperture is a property of the fiber and only depends upon n1 and n2, it will not change when the medium outside the fiber changes. The cut-off angle, however, will have to change if the numerical aperture is to be unaffected by a change in n0: NA = 0.148.
sin θ0max = NA/n0
θ0max = sin-1(NA/n0) = sin-1(0.1479/1.33) = = sin-1(0.1112) = 6.38o.Q20) Let n = 1, n1 = 1.46 and n2 = 1.45 in the diagram of the optical fiber system
Find
a) the critical angle θc at the core - cladding interface.
b) the numerical aperture N.A. of the optical fiber
c) the angle of acceptance αmax of the the optical fiber system.
A 20)a) θc = sin-1 (n2 / n1) = sin-1 (1.45 / 1.46) = 83.29 °
b) N.A. = √(n21 - n22) = √(1.462 - 1.452) = 0.17
c) αmax = sin-1√(1.462 - 1.452) = 9.82 ° Q 21) Let n = 1, n1 = 1.46 and n2 = 1.45 in the diagram of the optical fiber
the angle of incidence of a light ray on the outside - core interface be α = 5°. Find
a) angle of refraction β at the outside - core interface.
b) angle θ
c) and explain why this light ray will be reflected at the core - cladding interface and hence guided along the fiber.
A 21)a) Angle β is found using Snell's law at the outside - core interface as follows
n sin(α) = n1 sin(β)
Substitute the given parameters from Q 20) to obtain
β = sin-1 ( sin(5°) / n1) = 3.42 °
b) Angle θ is complementary to angle β hence
θ = 90 - 3.42 = 86.58 °
c) The angle of incidence θ = 86.58 ° at the core - cladding interface is larger that the critical angle θc = 83.29 ° calculated in Q 20) above and will therefore be totally reflected at this interface and hence guided along the fiber.
Also angle α = 5° is smaller to αmax = 9.82 ° Q22) Discuss the construction, working, and application of Solid-State Laser? OrDiscuss the construction, working, and application of Ruby Laser?OrDiscuss three-level laser?A 22)Ruby laser Ruby laser is a three-level solid-state laser and was constructed by Maiman in 1960. Ruby laser is one of the few solid-state lasers that produce visible light. It emits deep red light of wavelength 694.3 nm.Construction A ruby laser consists of three important elements: laser medium, the pump source, and the optical resonator.Laser MediumRuby (Al2O3+Cr2O3) is a crystal of Aluminium oxide, in which 0.05% of Al+3 ions are replaced by the Cr+3 ions. The colour of the rod is pink. The active medium or laser medium in the ruby rod is Cr+3 ions. In ruby laser, 4cm length and 5mm diameter rod is generally used. The ruby has good thermal properties.
The pump sourceThe ruby rod is surrounded by a xenon flash tube, which provides the pumping light to excite the chromium ions into upper energy levels. The ruby rod was surrounded by a helical xenon flash lamp.We know that population inversion is required to achieve laser emission. Population inversion is the process of achieving a greater population of a higher energy state than the lower energy state. To achieve population inversion, we need to supply energy to the laser medium i.e. to ruby crystal. Xenon flash tube emits thousands of joules of energy in few milliseconds, but only a part of that energy is utilized by the chromium ions while the rest energy heats the apparatus. A cooling arrangement is provided to keep the experimental set up at normal temperatures.Optical resonatorBoth the ends of the rods are highly polished and made strictly parallel. The ends are silvered in such a way, one becomes partially reflected the laser beam was emitted through that end and the other end fully reflected to reflect all the rays of light striking it. Working of ruby laser:Consider a ruby laser medium consisting of three energy levels E1, E2, E3 with N number of electrons. We assume that the energy levels will be E1 < E2 < E3. The energy level E1 is known as the ground state or lower energy state, the energy level E2 is known as the metastable state, and the energy level E3 is known as the pump state.Let us assume that initially most of the electrons are in the lower energy state (E1) and only a tiny number of electrons are in the excited states (E2 and E3).The energy level diagram of chromium ions is shown in the figure. The chromium ions get excitation into higher energy levels by absorbing 5500Å of wavelength radiation. The excited chromium ions stay in the level E3 for a short interval of time (10-8 to 10-9 Sec). After their life, most of the chromium ions are de-excited from E3 to E1 and a few chromium ions are de-excited from E3 to E2.
The transition between E3 and E2 is non-radioactive i.e. the chromium ions give their energy to the lattice in the form of heat. In the Metastable state, the lifetime of chromium ions is 10-3 sec. The lifetime of chromium ions in the Metastable state is 105 times greater than the lifetime of chromium ions in a higher state. Due to the continuous working of the flash lamp, the chromium ions are excited to a higher state E3 and returned to the E2 level. After a few milliseconds, the level E2 is more populated than the level E1 and hence the desired population inversion is achieved. The state of population inversion is not a stable one. The process of spontaneous transition is very high. When the excited chromium ion passes spontaneously from E3 to E2it emits one photon of wavelength 6943 Å. The photon reflects back and forth by the silver ends and until it stimulates an excited chromium ion in the E2 state and it to emit a fresh photon in phase with the earlier photon. The process is repeated again and again until the laser beam intensity is reached a sufficient value. When the photon beam becomes sufficiently intense, it emerges through the partially silvered end of the rod. The wavelength 6943 Å is Drawbacks of the ruby laser The laser requires high pumping power The efficiency of a ruby laser is very small It is a pulse laser Application of ruby laser Ruby lasers are in optical photography Ruby lasers can be used for the measurement of plasma properties such as electron density and temperature. Ruby lasers are used to remove the melanin from the skin. Ruby laser can be used for the recording of holograms. Q23) Define normalized frequency Or V-number ?
A 23)An optical fibre is characterized by one more important parameter, known as V-number which is more generally called normalized frequency of the fibre. It is given by the relation V – number determines how many modes a fibre can support, It is given by, V =
NAWhere d is the diameter of the core, l is the wavelength of light used NA is the numerical aperture of the fibre.V =
Or V =
If V ≤ 2.405, then the fibre is single mode fibre (SMF) If V > 2.405, then the fibre is multimode fibre (MMF) Q24) Consider a multimode step index fibre with n1 = 1.53 and n2 = 1.50 and λ= 1μm. If the core radius = 50 μm then calculate the normalized frequency of the fibre (V) and the number of guided mode.A 24)d = 2r = 2 x 50 x 10-6 m V = 
NA =
= 
=94.72= normalized frequencyTotal number of guided mode = M = V2/2 = 4486.Q 25) When mean optical power launched into an 8 km length of fibre is 12 μW, the mean optical power at the fibre output is 3 μW. Determine Overall signal attenuation in dB.The overall signal attenuation for a 10 km optical link using the same fibre with splices at 1 km intervals, each giving an attenuation of 1 dB.A 25)Given :z=8kmP(0) = 120 μWP(z) = 3 μWOverall attenuation is given by,αp (dB/km ) = 10. log 
= 16.02Overall attenuation for 10 km,Attenuation per km = αp (dB/km )= 
= 
= 2.00 dB/km Attenuation in 10 km link = 2.00 x 10 = 20 dBIn 10 km link there will be 9 splices at 1 km interval. Each splices introducing attenuation of 1 dB.Total attenuation = 20 dB + 9 dB = 29 dB
n22 = n12 - NA2 = (1.425)2 - (0.1479)2 = 2.0088
n2 = 1.417.(c) We know that that the n0 = nwater = 1.33. Since the numerical aperture is a property of the fiber and only depends upon n1 and n2, it will not change when the medium outside the fiber changes. The cut-off angle, however, will have to change if the numerical aperture is to be unaffected by a change in n0: NA = 0.148.
sin θ0max = NA/n0
θ0max = sin-1(NA/n0) = sin-1(0.1479/1.33) = = sin-1(0.1112) = 6.38o.Q20) Let n = 1, n1 = 1.46 and n2 = 1.45 in the diagram of the optical fiber system
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a) the critical angle θc at the core - cladding interface.
b) the numerical aperture N.A. of the optical fiber
c) the angle of acceptance αmax of the the optical fiber system.
A 20)a) θc = sin-1 (n2 / n1) = sin-1 (1.45 / 1.46) = 83.29 °
b) N.A. = √(n21 - n22) = √(1.462 - 1.452) = 0.17
c) αmax = sin-1√(1.462 - 1.452) = 9.82 ° Q 21) Let n = 1, n1 = 1.46 and n2 = 1.45 in the diagram of the optical fiber
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a) angle of refraction β at the outside - core interface.
b) angle θ
c) and explain why this light ray will be reflected at the core - cladding interface and hence guided along the fiber.
A 21)a) Angle β is found using Snell's law at the outside - core interface as follows
n sin(α) = n1 sin(β)
Substitute the given parameters from Q 20) to obtain
β = sin-1 ( sin(5°) / n1) = 3.42 °
b) Angle θ is complementary to angle β hence
θ = 90 - 3.42 = 86.58 °
c) The angle of incidence θ = 86.58 ° at the core - cladding interface is larger that the critical angle θc = 83.29 ° calculated in Q 20) above and will therefore be totally reflected at this interface and hence guided along the fiber.
Also angle α = 5° is smaller to αmax = 9.82 ° Q22) Discuss the construction, working, and application of Solid-State Laser? OrDiscuss the construction, working, and application of Ruby Laser?OrDiscuss three-level laser?A 22)Ruby laser Ruby laser is a three-level solid-state laser and was constructed by Maiman in 1960. Ruby laser is one of the few solid-state lasers that produce visible light. It emits deep red light of wavelength 694.3 nm.Construction A ruby laser consists of three important elements: laser medium, the pump source, and the optical resonator.Laser MediumRuby (Al2O3+Cr2O3) is a crystal of Aluminium oxide, in which 0.05% of Al+3 ions are replaced by the Cr+3 ions. The colour of the rod is pink. The active medium or laser medium in the ruby rod is Cr+3 ions. In ruby laser, 4cm length and 5mm diameter rod is generally used. The ruby has good thermal properties.
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A 23)An optical fibre is characterized by one more important parameter, known as V-number which is more generally called normalized frequency of the fibre. It is given by the relation V – number determines how many modes a fibre can support, It is given by, V =
NAWhere d is the diameter of the core, l is the wavelength of light used NA is the numerical aperture of the fibre.V = Or V = If V ≤ 2.405, then the fibre is single mode fibre (SMF) If V > 2.405, then the fibre is multimode fibre (MMF) Q24) Consider a multimode step index fibre with n1 = 1.53 and n2 = 1.50 and λ= 1μm. If the core radius = 50 μm then calculate the normalized frequency of the fibre (V) and the number of guided mode.A 24)d = 2r = 2 x 50 x 10-6 m V = NA = = =94.72= normalized frequencyTotal number of guided mode = M = V2/2 = 4486.Q 25) When mean optical power launched into an 8 km length of fibre is 12 μW, the mean optical power at the fibre output is 3 μW. Determine Overall signal attenuation in dB.The overall signal attenuation for a 10 km optical link using the same fibre with splices at 1 km intervals, each giving an attenuation of 1 dB.A 25)Given :z=8kmP(0) = 120 μWP(z) = 3 μWOverall attenuation is given by,αp (dB/km ) = 10. log = 16.02Overall attenuation for 10 km,Attenuation per km = αp (dB/km )= = = 2.00 dB/km Attenuation in 10 km link = 2.00 x 10 = 20 dBIn 10 km link there will be 9 splices at 1 km interval. Each splices introducing attenuation of 1 dB.Total attenuation = 20 dB + 9 dB = 29 dB 
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