DHANALAKSHMI SRINIVASAN INSTITUTE OF RESEARCH AND TECHNOLOGY - ENGINEERING PHYSICS I - PH PART A & PART B 2014 UNIT I CRYSTAL PHYSICS PART B

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1 UNIT I CRYSTAL PHYSICS PART B 1. Calculate No. of. Atoms, Coordination number, atomic radius and Atomic packing Factor for SC, BCC and FCC structures. SC (i) Number of atoms per unit cell : Total number of atoms = 8 = 1 (ii) Atomic Radius : a = r : r = (iii) Co-ordination Number : 1

2 The Co-ordination number of SC is 6. (iv) Atomic Packing Factor : APF = π /6 = 0.5 Therefore 5% Volume of the unit cell is occupies by the atoms. 48% Volume is vacant. BCC (i) Number of atoms per unit cell : = = (ii) Atomic Radius :

3 Atomic radius r = (iii) Co-ordination Number : The Co-ordination number is 8. (iv) Atomic Packing Factor : APF = = 0.68 Therefore 68% volume of the unit cell is occupied by the atoms. 3% volume is vacant. FCC 3

4 (i) Number of atoms per unit cell : = = 4 (ii) Atomic Radius : r = (iii) Co-ordination number : The co-ordination number is = 1. (iv) Atomic Packing Factor : APF = = 0.74 Therefore 74% volume of the unit cell is occupied by the atoms. 6% volume is vacant. 4

5 . Describe the structure of HCP crystal. Give details about its atomic radius, atomic packing factor and axial ratio. (i) Number of atoms per unit cell : = = 6 (ii) Atomic Radius : r = (iii) Co-ordination number : The co-ordination number is = 1. 5

6 (iv) c/a Ratio : OY = OX = = = (v) Atomic Packing Factor : Volume of unit cell = x C APF = = 0.74 Therefore 74% volume of the unit cell is occupied by the atoms. 6% volume is vacant. 3. Write the procedure to find Miller indices and Show that for a cubic structure the inter planar distance d in terms of miller indices and the cell edge a is given by d = a / (h+k+l)1/ (i) Find the intercepts (ii) Take the reciprocals (iii) Reduce into whole numbers. (iv) Write within parantheses. 6

7 ON = d 1 = OM = d = d = 4. Describe the various crystal growth techniques. Crystal growth technique (i) Melt growth CZOCHRALASKI METHOD Basic principle It is a crystal pulling technique from the melt 7

8 Description and working Advantages Disadvantages BRIDGMAN TECHNIQUE It involves selective cooling of the molten material Growth Ex: NaCl, KCl,AgBr Advantages Disadvantages (ii) SOLUTION GROWTH 8

9 Slow cooling method Slow evaporation method Advantages Limitations (iii) EPITAXIAL GROWTH process of growing an oriented single crystal layer on a substrate wafer Vapour Phase Epitaxy construction Process Merits Demerits 9

10 UNIT- PROPERTIES OF MATTER AND THERMAL PHYSICS PART B 1. Describe with necessary theory, the method to determine the Young s modulus of the material of a rectangular bar by uniform bending. Uniform bending: The beam is loaded uniformly on its both ends; the bend forms an arc of a circle. The elevation in the beam is produced.this bending is called uniform bending. Expression for the young s modulus by uniform bending (Elevation): M w a w a P 1 M w( a a ) M P P 1 wa...(1) We know the internal bending moment = y I R g () w a y I R g.(3) 10

11 O A A F F E F E F E.(4) Radius of curvature R l.. (5) 8 x w a 8 y I l x g.(6) The Elevation of point E above A is x w a l 8 y I g. Derive an expression for the internal bending moment of a beam in terms of radius of curvature? Bending of beams: A beam is defined as a rod (or) bar of uniform cross section whose length is very much greater than its other dimensions such as breath and thickness. Expression for the bending moment The moment of couple due to the restoring couple which balances the external couple due to the applied load is called the bending moment. original length PQ= R.(1) extended length = P 1 Q 1 =(R+x)..() Increase in its length = x..(3). 11

12 Linear strain = x R....(4) Stress = y x linear strain...(5) Tensile force = y x A R PQ = y A x R I g = The moment of all the forces about the neutral axis = y x A R x A = AK Total moment of all the forces (or) Internal bending moment = Special cases: i) Rectangular cross section: = ii) Circular Cross section: = y b d 1 R y r 4 R (7)... (8) yig R (6) 3. Derive an expression for depression at the free end of a cantilever, due to load. Describe an experiment to determine the Young s modulus of the cantilever material using this expression. Definition It is a beam fixed horizontally at one end and loaded at the other end. Young s modulus Expression for the depression of a cantilever:. 1

13 The external bending moment = w (l x).(1) W.K, the internal bending moment = R = y I g w ( l x ) yig R... ().. (3) d d = dx [ yig / w ( l x )] dx R...(4) w d ( l x ) dx...(5) yig If d is very small then d y ( l x ) d.(6) Sub (5) in (6), w dy l x dx yig ( )...(7) Depression of the cantilever y w l y I 3 3 g...(8) Special cases: (i) Rectangular cross-section: (i) Circular cross-section: wl y r 3 y 4 y 4 w l 3 ybd ( or ) y 3 3 4wl 3 4 r y Experimental determination of young s modulus by cantilever Depression: 13

14 Description: S.No. Load M Loading X10 - m Microscope Reading Unloading Mean X10 - m X10 - m Depression Y (m) M/y (Kgm -1 ) 1 W W+m 3 W+m 4 W+3m Procedure: using calculated. 3 4 gl M y N / m 3 bd y, the young s modulus for the given beam is 4. Write a short note on stress strain diagram. Discuss the factors affecting the elasticity. Stress Strain Diagram: The points found from the diagram. 14

15 Factors affecting elasticity: (i) Effect of stress (ii) Effect of annealing (iii) Change in Temperature (iv) Presence of impurities (v) Due to the nature of crystals. Explain all 5. Give an account of I-Shape Girders. It is one in which the upper and lower sections are broadened and the middle of the section is tapered so that it can withstand heavy loads over it. Explanation: Applications: Advantages: 6. How will you classify three types of elastic modulii? Explain. Explanation: Young s Modulus (Y): Young s modulus(y) = L o n g i t u d i n a l s t r e s s L o n g i t u d i n a l s t r a i n N/m Y = F L A l N/m 15

16 Rigidity modulus (n) : Explanation: Rigidity modulus (n) = T a n g e n tia ls tr e s s S h e a r in g s tr a in N/m n = F A N/m Poisson s Ratio ( ): = = a constant. = L a tera lstra in L o n g itu d in a lstra in Explanation: = L( D d ) LD -ve sign indicates that longitudinal strain and lateral strain are opposite to each other. 7. Describe an experiment to determine the Young s modulus of a beam using bending of beams? Experimental determination of young s modulus by uniform bending : 16

17 S.No. Load (M) Loading X10 - m Microscope Reading Unloading X10 - m Mean X10 - m Depression x (m) M/x (Kgm -1 ) 1 W W+m 3 W+m 4 W+3m the elevation produced is x w a l 8 y I g..(1) moment of inertia 3 bd I g () 1 calculated. M g a l x (4) y b d Young s modulus the value of 3 8 ( / 1 ) M x y 3gal bd M n / m x 3.(5) from tabular column, the young s modulus y of the beam can be 8. Discuss rectilinear flow of heat through a rod. 17

18 Heat flowing (entering) through P in one second Q 1 K A d d x..(1) Heating flowing (leaving) through Q in one second K A d K d x d x d x..() Net gain of gain of heat by element x Q K A d x d x in one second (3) Before the steady state is reached A x S d.(4) d t Heat lost per second due to radiation. x E p.. (5) d s d E p (6) d x K d t K A 9. Derive an expression for the heat conduction through a compound made of two layer when bodies in series and parallel. Heat Conduction through a Compound Media of Two Layers Bodies in series 18

19 Amount of heat flowing through the material (A) per second Q K 1 A( 1 ) x 1.(1) Amount of heat flowing through the material (B) per second Q K A ( ) () x K1A( 1 ) K A( ) x x 1 (3) K x K x K x K x (4) A Q x x K K (5) In general for any number of walls or slabs, the amount of heat conducted is A Q x K 1 Heat conduction through a compound media of two layers Bodies in parallel 19

20 Amount of heat flowing through the first material (A) in one second. Q 1 K A x 1.. (1) Q Amount of heat flowing through the second material (B) in one second. K A 1 x () Q. The total heat flowing through these materials per second is the sum of these two heats Q 1 and Q Explain Lee s disc method of determining thermal conductivity of bad conductor. Lee s Disc Method for Bad Conductors this method. Description KA x The thermal conductivity of a bad conductor like, ebonite or card board is determined by 0

21 Working Observation and calculation Amount of heat conducted through the specimen per second Q K 1 1 r K A.. (1) d d Amount of heat lost per second by the slab C Q = MSR () M SR d K W / m / K r 1.. (3) (ii) Determination of rate of cooling R. R d r h dt r h.(4) K d d r h M S dt r r h 1 (5) 11. Describe a method of determining thermal conductivity of rubber. Principle: It is based on the principle of radial flow of heat through a cylindrical shell. 1

22 Procedure: Theory: Heat gained by the calorimeter = W 1 S 1 Heat gained by the water = (W - W 1 ) S 1 1 Total heat gained by the calorimeter and water in t seconds Q W W S W S t Substituting the expression for (Q) in the formula K r Q log e r1 l 1 We get from this experiment K r W 1S1 W W1 S 1 log e lt s 1 r 1

23 UNIT III QUANTUM PHYSICS PART B 1. State Planck s Hypothesis. Derive Planck s law for black body radiation and hence deduce Wien s Displacement law and Rayleigh Jean s law. Assumptions: Planck s Radiation Law: = 1 The total no. of oscillators N = N 0 + N1 + N +.+ N r The total energy of the oscillators = 0 N 0 + E N 1 +EN +.+ re N r = N = The total no. of oscillators = = = After Substituting the value of E = hγ = The no. of oscillators per unit volume within the range of frequency + is = Total energy per unit volume = No. of oscillators per unit volume x Average energy of the oscillator = h

24 In terms of wavelength Wein s displacement law Shorter wavelength If is less 1/ will be greater so Rayleigh Jean s displacement law - Longer wavelength If is greater 1/ will be lesser. What is Compton Effect? Derive the equation for Compton shift. Compton Effect When a photon of energy h ν collides with a scattering element, the scattered beam has two components, one of the same frequency as that of the incident radiation and the other has lower frequency compared to incident frequency. This effect is called Compton Effect. The shift in wavelength is called Compton shift. 4

25 x-rays scattered from target containing very loosely bound electrons Wavelength of scattered x-rays found to be different from that of incident X-rays AND to depend on detection angle : Total energy before and after collision Total momentum before and after collision along X component Total momentum before and after collision along Y component From eqn From eqn 3 Squaring and adding eqn 5 From eqn 3 Squaring on both sides 5

26 Subtracting eqn 6 from 8 We know Squaring on both sides and rearranging eqn 10 Multiplying c on both sides Equating eqn 1 and 9 we get Since λ = c/v finally we get We know Compton shift Δλ = h / m o c ( 1- cos θ ) Case i) when θ = 0, Δλ = 0 Case ii) when θ = 45 o, Δλ = Case iii) when θ = 90 o, Δλ =

27 3. Explain the experimental verification of Compton Effect with the diagram. Experimental verification of Compton Effect. A beam of monochromatic X-rays of wavelength λ is allowed to fall on a scattering material (Fig a). The scattered X-rays are received by a Bragg spectrometer. The intensity of scattered X-ray is measured for various scattering angles. The graph is plotted (intensity Vs wavelength) as shown in Fig. (b) It is found that the curves have two peaks, one corresponding to unmodifiedd radiation and other corresponding to modified radiation. The difference between two peaks on the wavelength axis gives the Compton shift. The curves show that the greater the scattering angle, the greater is Compton shift in accordance with the expression. The change in wavelength dλ=0.043 A at θ=90 is found to be in good agreement with the theoretical value A. Thus, the Compton Effect is experimentally verified. 4. Derive Schrödinger s time independent and time dependent Schrödinger wave equation. The classical differential eqn of a wave motion is given by

28 is a Laplacian s operator. The solution of the eqn is Differentiating the eqn 3 with respect to t, we get Again Differentiating with respect to t, We know Substitute equation 6 in 5 Substitute in equation 7 If E is the total energy, V is the potential energy and is the kinetic energy then, 8

29 Rearranging the above equation and multiplying by m on both sides we get Subs eqn 9 in 8 we get This is Schroedinger s time independent equation. Time dependent wave equation Schrödinger time dependent wave equation is derived from time independent wave equation.the solution of classical differential equation is given by eqn 1 Differentiating equation 1we get Multiplying i on both sides We know 9

30 On substituting the value of in eqn 5 Where H Hamiltonian operator and E Energy operator 5. Derive the expression for energy levels of a particle enclosed in one-dimensional potential box of width. Schrödinger s wave eqn. one dimension box is Since V = 0 between the walls, eqn. 1 reduces to The general eqn of eqn is given by here A and B are constants. This can be obtained by applying the boundary conditions. 30

31 Boundary condition (i) B = 0 Boundary condition (i) Asin ka = 0 Sin ka = 0 Sin ka = 0 when ka takes the valus of nπ (4) (5) ( 6) Normalization of wave function The constant A is determined by normalization of wave function as follows: Probability density given by It is certain that the particle is somewhere inside the box. Thus the probability of finding the particle inside the box of length a is given by 31

32 This expression is known as normalized eigen function. 6. Describe the principle, construction and working of Scanning Electron Microscope with neat sketch. Principle: When the accelerated primary electrons strike the sample, it produces secondary electrons. These secondary electrons are collected by a detector which in turn gives a 3-dimensional image of the sample. Components: 1. Electron gun.. Magnetic condensing lenses 3. Scanning coil 4. Scintillator (electron detector) 5. Photomultiplier 6. CRO 3

33 Working: Advantages: Applications: 7. Explain the principle, construction and working of Transmission Electron Microscope with neat sketch. Principle The electrons are allowed to pass through the specimen and the image is formed on the fluorescent screen either by using transmitted electron beam. Components: Construction: a) Electron gun b) Magnetic condensing lenses c) Fluorescent screen or CCD Working: Advantages: 33

34 Applications: 8. Explain how the matter waves are experimentally evidenced using G.P Thomson s experiment. Experimental arrangement: Working: 34

35 UNIT 4 ACOUSTICS & ULTRASONICS PART B 1. What is reverberation time? Using Sabine s formula explains how the sound absorption coefficient of a material is determined. Reverberation Time Definition The time duration for which a sound persists even after the source of sound is cut off is called reverberation time. Standard reverberation time: Standard reverberation time is defined as time taken for the sound intensity to fall to onemillionth (10-6 ) of its initial intensity after the sound source is cut off. I I 10 6 Reverberation time 0.165V T1 (1) as i.e., 1 as T V () T 0.165V as a s 1 1 (3) Where a 1 - absorption coefficient of the absorbing material and s 1 is its surface area 35

36 1 as a1s T V subtracting equation () from equation(4) We get 1..(4) 1 1 a 1s1 T T V 1 (5) From equation (5), the absorption coefficient of the sound absorbing material is given by a V s 1 1 T 1 T1.(6) a V T1 T s1 T1T.(7). Derive expressions for growth and decay of energy density inside a hall and hence deduce Sabine s formula for the reverberation time of a hall. SABINE S FORMULA FOR REVERBERATION TIME [Rate of growth and rate of decay] Derivtation Fig (a) Arrangement of an element ds on a plane wall Area of this shaded portion = rd dr.(1) 36

37 Distance travelled by this shadedd portion, dv r sin d dr d..() Er sin dr d d (3) EdV Er sin drdd 4 4 (4) = Eds 4 sin cos d d dr..(5) Total sound energy absorbed per second by the whole enclosure (entire hall) = Ev ads 4 EvA = 4 (6) GROWTH AND DECAY OF SOUND ENERGY Rate of growth or increase in energy per second = d EV dt de V. dt.(7) Rate of emission Of sound energy = by the source Ie., Rate of growth of sound energy in he room de EvA P V. dt th 4 Rate of absorption + of sound energy by the walls..(9) 37

38 P E m 4 va..(8) 4P E m. va.(9) Dividing equation (9) by V, we get de EvA P = dt 4 V V.(10) t = t 4Pe (11) Ee K va Where K is a constant of integration. The value of K is determind by considering the boundary conditions. Rate Growth of Sound Energy K 4P...(1) va E E m 1 e t..(13) E m Growth Energy (E) density E E m 1 e t 38 Time(t)

39 Decay of sound Energy Fig(C ) Growth and sound with time E K= m.. (14) E Ee t.. (15) Fig (d ) Decay of sound with time Expression for reverberation time T V as 3. Discuss the factors affecting the acoustics of buildings and explain the factors to be followed to overcome it. The factors that affect the acoustics of a building are: 1. Optimum reverberation time.loudness 3. Focussing 4.Echo 5.Resonance 6.Noise (short form OLFERN) 1. Opitimum reverberation time 39

40 A satisfactory or preferred value of the reverberation time. reverberation time is called optimum The satisfactory reverberation times are: Speeches Music Theaters second - 1 to second to 1.5 second Remedy. Loudness Loudness is the degree of sensation produced on the ear Remedy 3. Foucussing The intensity of sound will be maximum at such points and zero at other places. This is called focusing effect. Remedy Fig (b) Focussing Effect 40

41 4. Echoes Remedy 5. Resonance Remedy 6. Noise the sound is reflected as a distinct repetition of direct sound. This reflected sound is called an echo. Due to the interference between original sound and the created sound, the original sound is disorted. Unwanted sound is called noise. There are three types of noises: i. Inside noise ii. Air- borne noise iii. Structure borne noise i) Inside noise Noise produces a disturbing and displeasing effect on the ear. Hence, noise should be avoided and controlled. Noise that is produced inside the same room or in an adjacent room is known as inside noise. Example of inside noise Remedy ii) Air-borne noise Remedy Noises that originate outside and come through open windows, doors and ventilators are known as air- borne noises. iii) Structure borne noise Noises that are conveyed through the structure of a building are called structure- borne noises. Remedy 41

42 4. Describe the production of ultrasonic waves by Magnetostriction oscillator method. Give the merits and demerits of this method. Principle When an alternating magnetic field is applied to a rod of ferromagnetic material such as nickel, iron, cobalt, then the rod is thrown into longitudinal vibrations producing ultrasonic waves at resonance. Construction Working Resonance condition. Merits Demerits n = f : 1/l Y/ ρ = 1/π L1/C 5. Describe the production of ultrasonic waves by piezo electric oscillator method. Give the merits and demerits of this method. 4

43 Principle It works on the principle of inverse piezo electric effect: Construction Working Resonance condition. n = f P/l Y/ ρ = 1/π L1/C Advantages Disadvantages 6. Draw a block diagram of ultrasonic flaw detector for NDT. Mention its advantages and disadvantages. Ultrasonic flaw detector- Principle When there is change in the medium, ultrasonic waves gets reflected. This property of ultrasonic is used as principle in flaw detection technique Pulse echo system It is based on the principle of echo reflection of ultrasound at the interfaces. Working 43

44 Reflection mode Let d be the distance of flaw from the transducer; v be the velocity of sound in the test specimen and t be the time of flight. Then the total distance traveled by the sound in the specimen=d. The following diagram shows such echo pattern indicating the presence of a defect. Transmission mode Advantages Limitations 44

45 7. Explain the ultrasonic imaging system with a neat block diagram. A-scan or Amplitude mode display: Amplitude mode display gives only the one dimensional information about the specimen. In this, a single transducer is used to transmit and receive the pulse from the specimen. The received echo signal from the specimen is given to the Y-plate and time base is connected to X-plate of CRO. they are displayed as vertical spikes along horizontal base line. The height of the vertical spikes corresponds to the strength of the echo from the specimen. The position of the vertical spikes from the left to right along the X-axis corresponds to the depth of the specimen. Transmitted Defect pulse Reflected pulse B-scan or Brightness mode display: Penetration depth Distance = velocity X time This type of mode display gives a two dimensional image. The principle of B scan is same as that of the A scan except with a small difference. B scan the transducer can be moved rather than keeping in a fixed position. As a result each echo s are displayed as dots on the screen. The distance between the two dots gives the penetration depth. Thus B scan provides exact information about internal structures of the specimen. V Penetration depth 45

46 C-scan or T.M.scan or Time mode scan: Its used to obtain the information about the moving object. It combines certain features of A-scan and B- scan. In T.M scan the transducer is held stationary as in A- scan and echo s appear as dots as in the B- scan. X-axis indicates the dots at relevant location or position of the defect depending on the depth of the reflection. The Y-axis indicates the movement of the object. Therefore when the object moves, the dots also move at a low speed. Transmitted pulse Trace of a stationary object Movement Trace of a moving object Location or position 8. Write a short note on Sonogram with neat sketch. Sonogram is an instrument used to monitor and visualize the image of the interior part of the body using high frequency sound waves. Working: Transducer in contact with human body Receiver circuit Digital processing unit Applications: Memory device Monitor 46

47 Diagnosis: It is found that when the heart of the fetus is moving towards the transducer, the shift in frequency is higher. if the heart of the fetus is moving away from the transducer the frequency shift is lower. Thus from the Doppler shift in frequency the movement of the fetus heart can be found. 9. Describe the method of determining the velocity of ultrasonic waves using acoustic grating. Experimental Determination of Velocity of Sound in a Liquid:. d sin θ n = nλ.(1) λ m = d ().. λ m = nλ / sinθ..(3) Velocity of ultrasonics (v) = ν u x λ m (4) V = ν u nλ / sinθ (5) 47

48 UNIT V PHOTONICS AND FIBRE OPTICS PART B 1. Derive the equation for Einstein s coefficients. R 1 = B 1 N 1 E (1) R =A 1 N () R 3 =B 1 N E (3) In steady state R 1 =R +R 3 = N A 1 /N B 1 [N 1 B 1 /N B 1-1] E=A 1 /B 1 {1/N 1 /N (B 1 /B 1-1)) (4) B 1 =B 1 E=A 1 /B 1 (1/N 1 /N -1) (5) \ E= A 1 /B 1 (1/e hv/kt -1) (6) E = 8πhv 3 /c 3 (1/e hv/kt ) (7) A 1 /B 1 =8πhv 3 /c 3 Significance of Einstein coefficient relation: At thermal equilibrium, the probability of spontaneous emission increases rapidly with the energy difference between two states.. Describe the principle, construction and working of Nd: YAG Laser. Principle : A more modern example is Nd:YAG which consists of Yttrium Aluminium Garnet (Y 3 Al 5 O 1 ) with neodymium Nd 3+ impurity in Yttrium sites. It is this impurity which does the work. Essentially a 4 level system with a laser transition of = 1.06μm (1.17eV).Pumping is by optical flash, using a light pulse of duration 1ms. Description : 48

49 Working : 3. Explain the principle, construction and the working of CO Laser. Principle : The Nitrogen atoms are initially raised to excited state. The nitrogen atoms delivers the energy to CO atoms. Then transition takes place between the vibrational energy levels of the CO atoms and hence laser beam is emitted. Fundamental modes of vibrations (i) (ii) (iii) Symmetric stretching mode Bending Mode Asymmetric stretching mode 49

50 Construction : Working 4. Explain the principle, construction and working of Semi-Conducting laser 50

51 Homo-junction laser: If a p-n junction formed in a single crystalline material, then it is called as homo-junction laser. Eg) single crystal of (GaAs) Hetero-junction laser: If p-n junction is formed with different semiconducting materials, then it is known as heterojunction laser. It is also called modern laser diode. Eg) Ga-Al-As. Homo-junction semiconductor laser: Principle: When a p-n junction is forward biased the electrons from the n-region and the holes from the p-region cross the junction and recombine with each other. During the recombination process, the light radiation is released from a certain specified direct band gap semiconductors like GaAs. That photon emitted stimulates other electrons and holes to recombine. As a result, stimulated emission takes place which produces laser. Construction: Working: The wavelength of laser beam is given by 51

52 Eg = hν = hc / λ λ = hc / Eg Where Eg is the band gap energy in joule. Characteristics: Type Active medium Active centre Pumping method Optical resonator Power output Nature of output Wavelength emitted Band gap - Homo-junction semiconductor laser - p-n junction diode - Recombination of electrons and holes - Direct pumping - Junction of diodes - 1mw - Continuous wave form Å Å -1.44eV Advantages: Disadvantages: Hetero-junction semiconductor laser: Principle: When a p-n junction is forward biased the electrons from the n-region and the holes from the p-region cross the junction and recombine with each other. During the recombination process, the light radiation is released from a certain specified direct band gap semiconductors like GaAs. The photon emitted during recombination stimulates other electrons and holes to recombine. As a result, stimulated emission takes place which produces laser. Construction: 5

53 Working: Advantages: Disadvantages: 5. Explain, in detail, the different types optical fibers and compare their performance. Types of optical fibres: It can be classified based on three categories material, mode and refractive index. 53

54 Based on material 1. Glass /glass fibres (glass core with glass cladding). Plastic/plastic fibres (plastic core with plastic cladding) Based on modes Single mode fibre Cladding Core Multimode fibre Cladding Core Based on refractive index 54

55 Step index fibre Graded index fibre Step index single mode fibre n core diameter 5-10m Distance from the axis Refractive index profile Cladding diameter m Mode of propagation Step index multimode fibre 55

56 n Core diameter 50-00m Distance from the axis Cladding diameter m Geometrical dimensions Graded index multimode fibre n core diameter 50-00m Cladding diameter m Distance from the axis Refractive index profile Geometrical dimension 6. Derive expressions for the acceptance angle and numerical aperture. 56

57 Applying snells law of refraction at the point of incidence of the ray AO into the core, we have no sinөo = n 1 sin Ө 1 sin Ө 1 = 1-cos Ө 1 ] At the point B on the interface of core and cladding, Angle of incidence Өc = 90- Ө 1 Applying Snell s law of refraction again, we have n 1 sin (90- Ө 1 ) = n sin 90 cos n n 1 sin 0 1 n 0 n 1 n If the surrounding medium is air or vacuum, n 0 = 1. sin 0 n n 1 0 sin n1 n 0 is called the acceptance angle or half angle of the acceptance cone. Acceptance angle:the maximum angle at or below which the light ray can suffer total internal reflection is called as acceptance angle. The cone is referred as acceptance cone. 57

58 Numerical aperture: It is defined as the sin of the acceptance angle of the fiber ie NA = sin 0 Or NA = n 1 n 7. Discuss the working of fiber optical communication system with a block diagram. Principle: The transmission of information over the required distance by the propagation of optical signal through optical fibers. Construction: The main parts of the fiber optic communication system are, Information signal source Transmitter Light source Propagation medium (optical fiber) Receiver Transmitter Optical Optical fiber Drive circuit 58 Light source

59 Analog signal Electrical Signal signal Receiver Analog signal Electrical Signal restorer Photo detector optical Signal signal 8. Describe in detail of the intrinsic and extrinsic sensors. Fiber optic sensors: A sensor is a transducer which converts one form of energy into another. Types of sensors Intrinsic sensor or active sensor Extrinsic sensor or passive sensor Intrinsic sensor or active sensor: In intrinsic sensor or active sensors, the physical parameters to be sensed directly acts on the fiber itself to produce the changes in the transmission characteristics. Examples: Pressure sensor, Liquid level sensor Extrinsic sensor or passive sensor: In extrinsic sensor or passive sensors, separate sensing element is used and the fiber acts as a guiding media to the sensors. Examples: Displacement sensor,laser Doppler velocimeter sensor Temperature sensor: 59

60 Temperature sensor is a sensor used to sense and measure the temperature of an object. Principle: It is based on the principle of interference between the beams emerging out from the reference fiber and the fiber kept in the measuring environment. Construction : Beam splitter Laser source L1 Reference fiber L3 Test fiber L Interference pattern Working: Displacement sensor: Principle: 60

61 Light is sent through a transmitting fiber and is made to fall on a moving object. The reflected light from the target is sensed by a detector. Construction: Working: 9. Describe the construction and working of medical endoscope Fiber optic endoscope:. Principle and construction: Usually in the endoscope, there are two fiber namely inner fiber and outer fiber. The inner fiber is used to illuminate the inner structure of the object under study. The outer fiber is used to collect the reflected light from that area and using this can be seen the inner structure of the object. Working: 61