DEPARTMENT OF PHYSICS

Transcription

1 PANIMALAR INSTITUTE OF TECHNOLOGY (An ISO 9001: 2008 Certified Institution) JAISAKTHI EDUCATIONAL TRUST CHENNAI DEPARTMENT OF PHYSICS PART A (Q&A) - I SEMESTER ENGINEERING PHYSICS I PH6151 (Common to all branches) (ACADEMIC YEAR )

2 UNIT-I CRYSTAL PHYSICS 1. State the values of co-ordination number for HCP structure and diamond structure.(jan 2009,may 2015) Co-ordination number for HCP -12 Co-ordination number for Diamond 4 2. The lattice constant for a FCC structure is A o calculate the interplanar spacing (220) planes (Jan 2009). a=4.938 A o d hkl h 2 a k 2 l 2 = m For a cubic crystal draw the plane with miller (110) and (001) (Jan 2010) 4. An element has a HCP structure. If the radius of the atom is A o. Find the height if the Unit cell.(dec 2010) C=height of a Unit cell, a=lattice constant, r=atomic radius a = 2r a=2 X1.605X10-10 =3.21X10-10 m C= 3.21 X X C = X m

3 5. Calculate the first and second nearest neighbor distance in the body centered cubic unit cell of Sodium, which has lattice constant of 4.3 A o.(june 2010) a 3 r 4 r = r = x m 6. The interplanar distances of (110) planes in a BCC crystal is 2.03 A o. What is the lattice parameter? 10 4r a m d m 7. What are Miller indices? (Apr 02,May 03 & 04, May 11, Jun 11, Dec 2014) It is defined as the reciprocals of the intercepts made by the planes on the crystallographic axes which are reduced to smallest numbers. 8. Name the seven crystal systems? (June 2012) 1. Cubic a=b=c α=β=γ=90 o 2. Tetragonal a=b c α=β=γ=90 o 3. orthorhombic a b c α=β=γ=90 o 4. Monoclinic a b c α=β=90 o γ 90 o 5. Triclinic a b c α β γ 90 o 6. Trigonal a=b=c α=β=γ 90 o

4 7. Hexagonal a=b c α=β=90 o γ=120 o 9. Determine the distance between adjacent atoms of platinum which has an FCC structure where lattice constant is A o. (June 2013) a r = = x m 10. Define unit cell. (May, 03& 04,Nov 03,Jan 2012) It is defined as the smallest geometric figure, the repetition of which gives the actual crystal structure. 11. What are lattice parameters? (Jan 2013) The intercepts of unit cell (a,b,c) and the interfacial angles(α,β,γ) along the three axes are together called the lattice parameters. 12. What is a primitive cell? (Jan 2014) It is the simplest type of unit cell which contains one lattice point per unit cell. 13. Name few techniques of crystal growth from melt. (Jan 2014) (1)Normal freezing (2) crystal pulling (3) Zone melting (4)Flame fusion. 14. A crystal plane cuts at 3a, 4b, and 2c distances along the crystallographic axes. Find the miller indices of a plane. (June 2014) Solution: We know the miller indices of a plane are the smallest integers of the reciprocals of its intercepts. Since the intercepts are 3,4 and,the reciprocals of the intercepts can be written as : : For finding the Miller indices first we have to find the LCM ie.,lcm = 12 Multiplying by LCM,we get : 4 12 : 2 Thus Miller indices can be written as (4 3 0) 15. Define Atomic packing factor or packing density and give its unit. (May 04)

5 It is defined as the ratio of volume of atoms per unit cell to the total volume occupied by the unit cell. It does not have unit. 16. What are the differences between crystalline and non crystalline material. (Dec 2014) Crystalline material Non crystalline material (1) They have a definite and regular (1) They don t have a definite and regular geometrical shapes wich extend throughout the crystal. geometrical shape. (2) They are anisotropic. (2) They are isotropic. (3) They are most stable. (3) They are less stable. (4) Eg. NaCl,KCl (4) glass,rubber. 17. Calculate the value of d-spacing of (321) planes of a cubic cell of lattice constant a = 0.41nm.(May 2015). a d hkl h k l d 321 = 1.095A o What are Bravais lattices? (Jan 2013) Seven types of crystal systems are further subdivided into 14 possible types of space lattice. These 14 types of lattices are called Bravais lattices. a. Cubic --- sc, bcc, fcc b. Tetragonal --- simple,body centered c. Orthorhombic --- simple, body centered Base centered, face centered d. Monoclinic --- simple, base centered e. Triclinic --- simple --- 1

6 f. Trigonal --- simple g. Hexagonal --- simple Write down the relation between atomic radius and lattice parameter of HCP. (Jun 2016) a r Why is diamond called as loosely packed system? The atomic packing factor for diamond cubic structure is 34%, so it has a loosely packed structure Calculate the volume of a FCC unit cell in terms of the atomic radius R. V = a 3 R a 2 4 V =16 2 R 3 UNIT-II PROPERTIES OF MATTER AND THERMAL PHYSICS 1. State Newton s law of cooling. (Jan 2014, June 2014, May 2015) The rate of loss of heat of a body is directly proportional to the temperature difference between the body and its surrounding, of same nature. Let θ 1 be the temperature of the body and θ 2 be the temperature of the surrounding then the rate of cooling is. 2. What is Poisson s ratio? (June 2014)

7 It is defined as the ratio between the lateral strain per unit stress (β) to the longitudinal strain per unit stress (α) with in the elastic limits. Poisson s ratio (σ) = 3. Name the factors which affect the elasticity of a body. (Dec 2014, May 95, Nov 95, Dec 1997) The factors affecting elasticity are (i) Effect of Stress. (iv) Change in temperature. (ii) Presence of Impurities. (v) Effect of Annealing. (iii) Due to nature of crystals. 4. How does Change in temperature affect the elastic property of a material? (May 2015) The elastic property of the materials changes with temperature. Normally the elasticity increases with decrease in temperature and Viceversa. (eg.) : (1) The elastic property of lead increases when the temperature is decreased. (2) The Carbon filament becomes plastic at higher temperature. 5. Explain bending moment of beam. ((Nov 97, Jun 2016) The moment of the couple due to elastic reactions which balances the external couple due to applied load is called the bending moment. 6. What do you infer from the Stress strain diagram? From the Stress-Strain diagram, We can infer (i)the Stress is directly proportional to the Strain within the elastic limit. (ii)it distinguishes the elastic and plastic limit of a material. 7. State Hooke s Law. According to this law, Stress is directly proportional to the Strain produced within the elastic limit. Stress α Strain Stress = E X Strain E = Nm -2 E is the modulus of elasticity.

8 8. Explain Neutral axis.(june 89, Dec 93, Dec 95) The middle layer (or) the filament of a beam which remains unaltered even with the presence of load on the beam is called neutral axis. Filaments which are lying below it are compressed. 9. Define I- Shape girder. The girder is the one in which the upper and lower sections are broadened and the middle section is tapered so that it can withstand heavy loads over it. Since the girder looks like letter I, it is called as I-Shaped girder. 10. Give the applications of I - Shaped girder. I-Shaped girders are used in the construction of bridges over the rivers and useful in the production of iron rails which are employed in railway tracks. 11. A copper wire of 3m length and 1mm diameter is subjected to a tension of 5N. Calculate the elongation produced in the wire if the young s modulus of elasticity of copper is 120 GPa.(Jan 14, Nov 98) Given Data: F=5N; L=3m; Y=120 X 10 9 Pa.; A= π(0.5 X 10-3 ) 2 m 2. Young s modulus = Y= Nm -2 Elongation produced = 1.59 X 10-4 m. 12. Calculate the Poisson s ratio for the material, given Y = x Nm - 2 and n =4.55 x Nm -2. σ = - 1 σ = σ = Define Co-efficient of thermal conductivity. (Dec 1997, Jun 2016)

9 The Co-efficient of thermal conductivity is defined as the amount of heat conducted per second normally across the unit area of cross section, maintained at unit temperature gradient. 14. Why the specimen used to determine thermal conductivity of a bad conductor should have a larger area and smaller thickness? For a bad conductor with a smaller thickness and larger area of cross section, the amount of heat conducted will be more. 15. What is meant by temperature gradient? The rate of fall of temperature with respect to the distance is called as temperature gradient. In general it is called denoted as. The negative sign indicates the fall of temperature with increase in distance. 16. What are the basic entities responsible for thermal conduction of a solid? (Dec 1997) (i) Area of cross section (A) (ii) Temperature difference between the hot and cold layers of the solid (θ 1 -θ 2 ). (iii) Time of conduction(t) (iv) Thickness of the solid (x). 17. Define diffusivity. It is defined as the ratio of thermal conductivity to the thermal capacity per unit volume of the material. Thermal diffusivity (h) = 18. What is the basic principle employed in Lee s disc method for bad conductors? The given bad conductor is taken in the form of disc and is placed in between the disc and steam chamber. The steam is passed through bad conductor per second is calculated. The amount of heat conducted through the bad conductor per second is equal to the amount of heat lost per second by the disc. 19. What is meant by thermal resistance?

10 The thermal resistance of a body is a measure of its opposition to the flow of heat through it. Every body posses some resistive power when it is subjected to heat is termed as thermal resistance. 20. Distinguish between Conduction and Convection. Conduction : It is the process of in which the heat is transferred from hot end to the cold end without the actual movement of the particles. Convection : It is the process in which the heat is transmitted from hot to cold end by actual movement of the particles. UNIT III QUANTUM PHYSICS 1. What is Compton wavelength? Give its value. (Jan 2009,2010, 2011, June 2014) The shift in wavelength corresponding to the scattering angle of 90 o is called Compton wavelength. h Compton shift, 1 cos ) m c When θ = 90 o ; cos θ = = = A o 2. Mention the physical significance of wave function. (June 2009,Jan 2009, 2010, 2011, June2013) 1) The probability of finding a particle in space at any given instant of time characterized by a function ψ(x,y,z) called wave function. 2) It relates the particle and the wave statistically. 3) It gives the information about the particle behaviour. 4) It is a complex quantity. 5) /ψ/ 2 represents the probability density of the particle, which is real and positive.

11 3. An X-ray photon of wavelength 1.24 X 10-3 A o is scattered by a free electron through an angle of 90.Calculate the energy of the scattered photon.(jan 2009) Given : λ = 1.24 X 10-3 A o θ = 90 h m c ' Solution : 1 cos 0 ' h m c 0 1 cos = 1.24 x 10-3 x o + 1 cos90 ' = x m ' ev = MeV 4. In Compton scattering, the incident photon have a wavelength 0.5nm. Calculate the wavelength of scattered radiation if they are viewed at an angle of 45 to the direction of incidence. (June 2009) ' h 1 cos m c ' 0 h m c 0 1 cos = 0.5 x o + 1 cos = A o 5. Explain the principle of TEM? (June 2009,Jan 2014) Electrons are made to pass through the specimen and the image is formed in the fluorescent screen, either by using transmitted beam or by using diffracted beam. 8

12 6. Calculate the equivalent wavelength of electron moving with a velocity of 3 m/s (Jan 2010) h de Broglie wavelength, λ = mv = 31 7 = A o 7. What is Compton effect? Write an expression for the Compton wavelength. (Jan 2010,May 2015) When a photon of energy hυ collides with a scattering element, the scattered beam has two components, namely, one of the same frequency or wavelength as that of the incident radiation and the other has lower frequency or higher wavelength compared to incident frequency or wavelength. This effect is called Compton effect. h 1 cos ) m0c 8. Calculate the de-broglie wavelength of an electron moving with velocity of m/s (June2010) h de Broglie wavelength, λ = mv = 31 7 = A o 9. What is meant by degeneracy and non-degeneracy state? (Jan 2014) Degeneracy state : For various combinations of quantum numbers if we get same eigen value but different eigen functions, then it is called degeneracy state. Example : For three combination of quantum numbers if we get (112), (121) and (211), weget same energy values i.e., E 112 = E 121 = E 211 but different eigen functions i.e., ψ 112 ψ 121 ψ 211

13 Non-Degeneracy state : For various combinations of quantum numbers if we get same eigen value and same eigen functions, then it is called degeneracy state. 10. An electron is accelerated by a potential difference of 2kV. Calculate the debroglie s wavelength of matter wave associated with it.(june 2010) de Broglie wavelength h 2mev λ = = x m λ = A o 11. Find the eigen values and eigen functions of an electron moving in a one dimensional box of infinite height and 1 A o of width. Given that m=9.11 and h= (June 2010) 2 2 n h Energy of an electron in a deep potential well, E = 2 8ml E = E 1 = ev 34 ( ) 2 E 2 = 2 2 E 3 = ev = ev ev = ev ψ 1(x) = ψ 2(x) = ψ 3(x) =

14 12. Define Black Body and Black body radiation? (Dec 2010,June 2013) A perfect black body is the one which absorbs and also emits the radiation completely. In practice no body is perfectly black. We have to coat the black colour over the surface to make a black body. Black body is said to be a perfect absorber, since it absorbs all the wavelength of the incident radiation. The black body is a perfect radiator, because it radiates all the wavelength absorbed by it. This phenomenon is called black body radiation. 13. What is meant by wave function.(dec 2010) Wave function (ψ) is a variable quantity that is associated with a moving particle at any position (x,y,z) and at any time t. It relates the probability of finding the particle at that point and at that time. 14. Calculate the energy in ev of a photon of wavelength 1.2 A o.(jan 2011) E = 56 ev 15. What is Rayleigh Jeans law of radiation (Jan 2010, 2012,May 2011) It is defined as The energy is directly proportional to the absolute temperature and inversely proportional to the fourth power of wavelength. ie., E T/ 4 (or) E = 8K B T/ 4 where K B = Boltzmann Constant Limitation : It holds good only for longer wavelength. 16. Compare the resolution and magnification of optical microscope with that of electron microscope.(may 2011,Dec 2014) Magnification of optical microscope is 2000 times as that of the size of the object, whereas that of electron microscope is 1,00,000 times. Resolution of optical microscope is poor whereas that of electron microscope is good.

15 17. Calculate the debroglie s wavelength of an alpha particle accelerated by an electric field of potential 1000V.Given that mass of proton is Kg, mass of neutron is Kg and the charge on a proton is 1.6 C. Assume h= (May 2011) de Broglie wavelength h 2mev λ = = x m 18. State Wein s displacement law of black body radiation. (Jan 2010, June2011, Dec 2014) It is defined as, The product of the wavelength ( m ) of maximum energy emitted and the absolute temperature (T) is a constant. ie., m T = constant Also E max T 5 E m = constant.t 5 Limitation : It holds good only for shorter wavelength. 19. Mention any two limitations of TEM. (June 2011) (i) Specimen should be very thin (ii) 3- dimensional image cannot be obtained. 21. Calculate the de-broglie wavelength of a proton whose kinetic energy is 1MeV(June 2011) Given : E = 1MeV = 1 x 10 6 x 1.6 x J Solution : h 2mE = x 10 6 x

16 λ = x m. 22. Find the lowest energy of an electron confined to move in a one dimensional box of length 1 A o. Express the result in ev. (June 2012,Jan 2014) 2 2 n h Energy of an electron in a deep potential well, E = 2 8ml E = = x J 34 ( ) = ev = 37.64eV 22. What is meant by normalization of wave function? (June 2012) Normalisation is the process by which the probability of finding a particle inside any potential well can be done. 23. Write Planck s radiation formula. (Jan 2013) = * In a Compton scattering experiment the incident photon have a wavelength of 3 A o. What is the wavelength of the scattered photons if they are viewed at an angle of 60 to the direction of incidence? (Jan 2010) ' h m c 0 ' 1 cos h m c 0 1 cos

17 = 3 x o + 1 cos60 = A o Calculate the de-broglie wavelength of a proton and an electron accelerated by a potential of 150V. (May 2015) 8 h 2mev x m For proton: λ = = x m For electron: λ = = Given that the wave function of a particle in a one dimensional box is given by ψ=.. Evaluate the probability of finding the particle in the region. (May 2015) P = P = P = 2k.* + P = [ ] P = - e -6 +e -4

18 P = A particle of mass one microgram takes 100s to travel from one end to other end of a one dimensional box of width 1mm.Assume that the potential inside and at the walls of the box to be zero and infinity respectively. Determine the quantum number described by this motion.(may 2015) 2 2 n h Energy of an electron in a deep potential well, E = 2 8ml E = n 2 8ml = = 2 h 2. n 2 = n = / m Why should the wave function of a particle be normalized? (Jan 2016) The wave function of a particle is to be normalized in order to find the particle within the given limit. UNIT - IV ACOUSTICS 1. Define reverberation time of an auditorium. (Dec 97, 98) The persistence of audible sound, even after the source has stopped to emit the sound is called reverberation. The time during which the sound persists in the hall is called as reverberation time. 2. Define absorption coefficient of a material. (April 98, Dec 14) The absorption coefficient of a material is defined as the ratio of the sound energy absorbed by the surface to that of the total sound energy incident on the surface.

19 Absorption coefficient (a) = 3. What is loudness and state Weber Fechner law? (Jan 2005, Jan 2014) Loudness of sound is defined as the degree of sensation produced on the ear. This cannot be measured directly. So that it is measured in terms of intensity. Loudness is proportional to the logarithmic value of intensity. L α log I L = K log I. This is also known as Weber Fechner law. 4. What is meant by echelon effect? If there is a regular repetition of echoes of the original sound received by the observer due to the presence of flight of stairs or set of railings, then the effect is called echelon effect. 5. Give the importance of Sabine s law for a good auditorium. i) The Sabine s law can be used to calculate the reverberation time of an auditorium. ii) It is also used to find the absorption coefficient of any unknown material. iii) The reverberation time should be maintained with an optimum value for a good auditorium. 6. What is meant by optimum reverberation time? Give its value for concert halls and theatres. Optimum reverberation time is the persistent time of sound in hall, without causing echoes. For concert halls it should be 0.5 seconds. For small theatres it should be between 1.1 to 1.5 seconds and for large theatres it should be between 1.5 to 3 seconds. 7. The intensity of sound produced by roaring of a lion at a distance of 5m is 0.01 Wm -2. Calculate the intensity level in decibel. (Jan 2014, May 2015, Jun 2014) The intensity level (or) relative intensity

20 1 I L = 10 log 10 I 0 = 10 log 10 = 100 db I State the conditions of good acoustics for an auditorium. i) Optimum reverberation time should be maintained. ii) The auditorium should be free from excessive reverberation. iii) There should not be any regions of poor audibility anywhere inside the hall. iv) Resonance should be avoided and noises should be reduced. 9. State Sabine s law. It states that the reverberation time is the time taken by the sound to fall from one millionth of its original intensity, after the source of sound is stopped. 0 (i.e) I = 6 If V is the volume of the hall, is the average absorption coefficient and S is the total surface area, the reverberation time can be related as T = 10. Give the difference between loudness and intensity. I 10 S.NO LOUDNESS INTENSITY 1. It is the degree of sensation produced in the ear. It refers to external measurement. 2. It is physiological quantity. It is a physical quantity. 3. It is difficult to measure. It can be easily measured. 11. The volume of a hall is 475 m 3. The area of wall is 200 m 2, area of the floor and ceiling each is 100 m 2. If absorption coefficient of the wall, ceiling and

21 floor are 0.025, 0.02 and 0.55 respectively, calculate the reverberation time for the hall. (Dec 95) Given data: V = 475 m 3 a 1 = a 2 = 0.02 a 3 = 0.55 S 1 = 200 m 2 S 2 = 100 m 2 S 3 = 100 m 2 Solution: Reverberation time T = 0.167V as = = 0.167X 475 (200X X X 0.55) = seconds ULTRASONICS 1. What is the principle behind the magnetostriction oscillator? (or) What is meant by magnetostriction effect? (Jan 2011) Magnetostriction effect is the principle behind the magnetostriction oscillator. When an alternating magnetic field is applied to the ferromagnetic rod, the rod is thrown into longitudinal vibrations, producing ultrasonic waves at resonance. 2. What is the principle behind the piezo electric generator? (or) What is inverse piezo electric effect? (Jan 2012, 2013) If an alternating voltage is applied to one pair of opposite faces of the crystal, alternatively mechanical contractions and expansions are produced across the other opposite faces. This phenomenon is known as inverse piezo electric effect.

22 3. Are the ultrasonic waves electromagnetic waves? Give proper reasons. (Dec 97, Jan 2014) Ultrasonic waves are not electromagnetic waves because they are sound wave which does not consist of electric and magnetic vectors as in electromagnetic waves. 4. Mention the properties of ultrasonic waves. (Jan 2010, 2011, Jun 2016) 1. They are highly energetic. 2. They can travel over long distances. 3. It produces heating effect. 4. They are reflected, refracted, absorbed similar to ordinary sound waves. 5. Mention the (industrial) applications of ultrasonics. (Dec 2003, Dec 2014) a. They are used in welding and soldering. b. They are used in drilling holes in hard steel plates. c. They are used for cleaning clothes and tiny parts of watches. d. Measurement of depth of the sea. 6. What is meant by acoustic grating? (Jan 2009, Jan 2013) When ultrasonic waves are passed through a liquid, the density of the liquid varies layer by layer due to variation in pressure and hence the liquid will act as a acoustic grating. Now, when a monochromatic source of light is passed through acoustic grating, the light gets diffracted. By using the condition for diffraction, the velocity of ultrasonic waves can be determined. 7. Why not ultrasonic s be produced passing high frequency alternating current through a loud speaker? (Nov 2002, Jan 2013) (1) Loud speaker coil cannot vibrate at such high frequency. (2) Inductance of the speaker coil becomes so high and practically no current flows through it.

23 8. Mention any two medical applications of ultrasonics. (May 2004, June 2005) (1) Ultrasonics are used for detecting, tumours and other defects in human body. (2) Ultrasonics are used to remove kidney stones without any loss of blood. 9. Can we use a copper rod in a magnetostriction generator? Why? (Jan 2011) No, copper rod cannot be used to produce to ultrasonics in magnetostriction generator, because it is not a ferromagnetic material. 10. Why are ultrasonics not audible to human? (Jan 2011) The audible range of frequencies for human beings is between 20 Hz to 20,000 Hz. Since the frequency of ultrasonic wave is above 20,000 Hz is not audible to human. 11. What is sonogram? Mention its application. (Jan 2010, 2011) Sonogram is a technique which is used to record the sounds produced due to the pumping action of the heart, using ultrasonics. It also provides the information on heart ratio, rhythmicity, blood pumping and valve action. 12. List the four methods of detecting ultrasonic waves. (June 2013) (i) Kundt s tube method (ii) Sensitive flame method (iii) Thermal detector (iv) Piezo electric detector 13. What is SONAR? Mention two applications of it. (Jan 2009) SONAR is a device which stands for Sound Navigation and Ranging. Its principle is based on echo sounding technique of ultrasonics. 14. Calculate the frequency to which a piezo electric oscillator circuit should be tuned so that a piezo electric crystal of thickness 0.1 cm vibrates in its fundamental mode to generate ultrasonic waves. Young s modulus = 80 GPa and Density of crystal = 2654 kg/m 3. (April 2003, Jan 2006, Jan 2009,June 2012) Given: t = 0.1X10-2 m E=80GPa = 80 X 10 9 Pa ρ=2654 kgm -3

24 Solution: f = 1 2t E = x 10 6 Hz 15. A quartz crystal in an ultrasonic interferometer produces stationary waves of frequency 1.5 MHz. If the distance between 6 consecutive nodes is 2.75 mm, find the velocity of ultrasonic waves. (Jan 2005) Given : Frequency of ultrasonics υ u = 1.5MHz. Distance between 6 consecutive nodes = 2.75mm From figure, 5d = 2.75mm d = 5.5 x 10-4 m From figure, λ u = 2d = 2 x 5.5 x 10-4 = 1.1 x 10-3 m Velocity of ultrasonic waves, v = frequency x wavelength v = υ u x λ u = 1.5 x10 6 x 1.1 x 10-3 m/sec

25 v = 1650 m/sec. 16. Find the depth of a submerged submarine if an ultrasonic wave is received after 0.33 sec. Find the time of transmission. Given: the velocity of ultrasonic waves in sea water = 1440 m/s. (Jan 2013) Given Solution: t = 0.33sec v = 1440m/s Depth of sea, d = m 17. What is the principle of A scan display in ultrasonics?( May 2015) A scan is an amplitude mode display. It gives one dimensional information about the specimen. It is used to detect the position and size of the flaws with the help of change in its amplitude. UNIT V PHOTONICS 1. Define population inversion and meta stable state.( Jan 2009, Dec 2014, Jun 2016) The state of achieving more number of atoms in the higher energy level than that of lower energy level is called population inversion. Metastable state is the state which lies between the excited state and ground state. 2. What are the advantages of heterojunction semiconductor laser over homojunction semiconductor laser? (Jan 2009) Power output is very high Produces continuous wave output It has low threshold frequency

26 It has longer life time 3. Give the principle of semiconductor laser. (May 2009) The electron in conduction band combines with a hole in valence band and hence the recombination of electrons and hole produces energy in the form of photon. This photon further stimulates the emission of another photon. 4. Mention any four applications of laser in medicine. (May 2009, Jan 2013) Used to treat cancer in human. Used to Shatter kidney stones Used to cut bones Used to drill holes 5. Why is population inversion necessary to achieve lasing? (Dec 2010) Because only when N 2 > N 1, there is more chance for stimulated emission to take place which gives rise to intense laser. 6. What are the different methods of achieving population inversion? (Jan 2010) Optical pumping Direct electron excitation Inelastic atom-atom collision Direct conversion Chemical process 7. What are the conditions for laser action? (Jun 2010, 2012) a) Population inversion must be achieved. b) Stimulated emission should predominant over spontaneous emission. 8. Can a two level system be used for the population of laser? Why? (Jan 2011, Jan 2014) No, because for population inversion to be achieved, at least three levels are required. 9. Name the properties of laser which are making it suitable for industrial applications. (May 2011, June 2012) i) High penetrating power iii) Not absorbed by water ii) Highly intense iv) Highly directional. 10. What are the characteristics of laser? (June 2011) Directionality

27 Intensity Monochromatic Coherence 11. Give the wavelengths of light emitted by CO 2 laser. (Jun 2011) 10.6 μm and 9.6 μm are the wavelengths emitted by CO 2 laser. 12. Explain stimulated emission. (Jan 2013) An atom in the excited is induced to return to ground state thereby resulting in two photons of frequency and energy is called as stimulated emission. 13. Calculate the wavelength of radiation emitted by an LED made up of a semi conducting material whose E g = 2.8 ev. (June 2013) Wavelength (λ) = λ = X X 3 X X 1.6 X = A o (blue color) 14. What are the roles played by N 2 (Nitrogen) and He (Helium) in CO 2 laser? (June 2014) In CO 2 laser, nitrogen helps to increase the population of atoms in the upper level of CO 2 and helium helps to depopulate the atoms in the lower level of CO 2 and also to cool the discharge tube. 15. What are Einstein s coefficients? (Jan 2009) The Einstein s coefficients A and B accounts for spontaneous and stimulated emission/absorption probabilities of light by a system of particles. It also explains the importance of metastable states. 16. What are the differences between homojunction and heterojunction laser. (Jan 2009,2010)

28 Homojunction laser Heterojunction laser 1. It is made by single crystalline material. 1. It is made by different crystalline materials. 2. Power output is low. 2. Power output is high. 3. Pulsed output. 3. Continuous output. 4. Cost is less. 4. Cost is more. 5. Life time is less. 5. Life time is more. 6. Eg: GaAs, InP 6. Eg: GaAs/ GaAlAs 17. Differentiate between stimulated emission and spontaneous emission of radiation (Jan 2012, 2013) Stimulated emission Spontaneous emission (1) Emission of light photon takes place (1) Emission of light photon takes by an external photon. place spontaneously. (2) It is not a random process. (2) It is a random process. (3) It is a controllable process. (3) It is a uncontrollable process. (4) Photons get multiplied through chain (4) Photons do not get multiplied reaction. through chain reaction. (5) More intense. (5) Less intense. FIBRE OPTICS 1. Distinguish between step index and graded index fibre. (Jan 2009, Jun 2016) Step index fiber Graded index fiber

29 The difference in refractive index between core and cladding is obtained in single step. Distortion is more Their difference in refractive index between core and cladding increases gradually from centre towards centre interface. Distortion is less 2. An optical fiber has a core of refractive index of 1.50 and a cladding refractive index of Calculate the critical angle at the core cladding interface. (Jan 2009) Φ c = = = 78 o Define acceptance angle in optical fiber.( May 2009, June 2010) Acceptance angle is the maximum angle to the axis at which light may enter into the fibre so that it can have total internal reflection. 4. What are the types of sensors used in fibre optic communication? (May 2009) (i) (ii) Intrinsic sensors fibre itself acts as sensing element Extrinsic sensors separate sensing system. 5. What are the different types of optical fibers? (Dec 2010) (i) based on the material glass and plastic fiber (ii) based on number of modes single mode and multi mode fiber (iii) based on the refractive index step index and graded index fiber. 6. Explain dispersion in optical fiber. (Dec 2010) When an optical signal is sent into the fiber, the pulse spreads / broadens as it propagates through the fiber. 7. A signal of 100 mw is injected into a fibre. The out coming signal from the other end is 40 mw. Find the loss in db. ( Jan 2010)

30 Power loss (P L ) = 10 log = 10 log = 10 log (0.4) = db. 8. Give the applications of fibre optical system. (June 2010, Dec 2014) Used for long distance communication Used in computer networks Used as optical sensors A large no. of telephone signals can be passed through optical fibre without any interference. 9. What is called mode of propagation in optical fibers? (Jan 2011) It represents the number of possible directions or path of propagation of light through the optical fibers. 10. Mention the properties of detectors used in fibre optic communication. (Jan 2011) (i) Ability to convert optical signal into electrical signal (ii) Fast response time (iii) Zero dark current (iv) Cost effective 11. Name any two light sources and detectors used in fibre optic communication system. (May 2011) Light sources (i) light emitting diode (ii) laser diode Detectors (i) PIN photo diode (ii) Avalanche photo diode.

31 12. An optical fiber has a Numerical aperture of 0.2 and a cladding refractive index of Find the acceptance angle for the same fiber in water. The refractive index of water is (May 2011) i m = sin -1 NA = sin = 11 o Define attenuation in optical fibre and mention its unit. (June 2011) Attenuation is the ratio of optical power output from a fiber of length L to power input. 14. Calculate NA and i m of an optical fiber having n 1 = 1.49 and n 2 = ( June 2011) NA = n n 2 i m = sin -1 NA = ( ) ( ) = sin = = 22 o State the principle of total internal reflection. (June 2012, Jan 2013) When light travels from denser medium to rarer medium, when incident angle exceeds critical angle, the ray is totally reflected back. This is called as total internal reflection. 16. Mention any two medical applications of optical fibers. (June 2012) Fibre optic endoscopes used in diagnosis To visualize inner organs Used in cardioscopy, laproscopy, etc. 17. What are the conditions for total internal reflection? (June 2013) (i) Light should travel from denser to rarer medium (ii) Angle of incidence should be greater than critical angle, Φ i > Φ c (iii) Refractive index of core should be greater than refractive index of cladding, n 1 > n 2

32 18. Write four major advantages of fiber optic communication over other communication systems. (Jan 2014) (ii) It is possible without electricity. (iii) Not affected by lightening (iv) Suitable to any environment (v) Easy maintenance, long life & economical. 19. Define Numerical Aperture. (June 2014) Numerical Aperture is defined as sine of the acceptance angle of the fiber. NA = sin i m = n 1 2 n What are the two types of coherence? (May 2011, May 2015) i) Temporal coherence ii) Spatial coherence 21. Draw the block diagram of fiber optic communication system. ( May 2015)

33 22. What are the uses of optical fibers? (Dec 2014) used in diagnosis To visualize inner organs To find pores, cracks in machineries Used in coagulation in industries. 23. Why is intermodal dispersion reduced in graded index fibres?(jan 2016). There is variation in refractive index from core to cladding, hence the light rays travel as skew rays in graded index fibre, so intermodal dispersion is reduced. 24. Calculate the total number of guided modes propagating in the multimode step index fiber having a diameter of 60 m and NA of 0.25 operating at a wavelength of 2.7 m The no. of modes for a step index fiber 2 d NA 4.9 N = = = N step = 151 modes. PANIMALAR INSTITUTE OF TECHNOLOGY (An ISO 9001: 2008 Certified Institution) JAISAKTHI EDUCATIONAL TRUST CHENNAI

34 DEPARTMENT OF PHYSICS PART B (Q&A) I SEMESTER ENGINEERING PHYSICS I PH6151 (ACADEMIC YEAR ) UNIT-I

35 CRYSTAL PHYSICS SEVEN TYPES OF CRYSTAL SYSTEM THE CRYSTAL SYSTEM Crystals are classified into 7 Crystal systems on the basis of lattice parameters. The basic crystal systems are Cubic Tetragonal Orthorhombic Monoclinic Triclinic Rhombohedral or Trigonal Hexagonal SEVEN CRYSTAL SYSTEMS S.No. Crystal system Axial length Interfacial angles Example 1. Cubic a = b = c α = β = γ = 90 NaCl, CaF 2, Au, Cu 2. Tetragonal a = b c α = β = γ = Orthorhombic a b c α = β = γ = 90 Ordinary white, tin, Indium, SnO 2 Sulphur, Topaz, BaSO 4, KNO 3 4. Monoclinic a b c α = β = 90 ; γ 90 Na 2 SO 4, FeSO 4, Gypsum 5. Triclinic a b c α β γ 90 CuSo 4, K 2 Cr 2 O 7 6. Rhombohedral a = b = c α = β = γ 90 Calcite, Sb, Bi.

36 7. Hexagonal a = b c α = β =90 ;γ =120 Quartz, Zn, Mg. BRAVAIS LATTICES The 14 different types of lattices in three dimensional spaces are called Bravais lattices. Simple Lattice - P Body Centered Lattice - I Face Centered Lattice - F Base Centered Lattice- C S.No. SYSTEM SYMBOL 1. Cubic-3 P, I, F 2. Tetragonal-2 P, I 3. Orthorhombic 4 P, C, I, F 4. Monoclinic -1 P 5. Triclinic 2 P, C 6. Rhombohedral 1 P 7. Hexagonal 1 P Total - 14 Bravais Lattice 1. SIMPLE CUBIC (SC) STRUCTURE

37 NO.OF ATOMS PER UNIT CELL The simple cubic structure shares 8 atoms, one in each of its 8 corners. Therefore the total number of atoms in a cubic structure is one. ATOMIC RADIUS It is defined as half the distance between nearest neighbors in a crystal. It is denoted by r 2r = a r = a / 2 CO-ORDINATION NUMBER In simple cubic structure each corner atom is surrounded by 6 other atoms of adjacent unit cells as shown in Fig. The corner atom is surrounded by two atoms (1 and 2) along X - X axis. Similarly (3

38 and 4) two atoms along Y - Y axis. Two more atoms (5 and 6) along z z axis. Thus the coordination number of SC is 6. ATOMIC PACKING FACTOR Atomic packing factor is the ratio of the volume of the atoms occupying the unit cell to the volume of the unit cell No. of atoms per unit cell = 1 Volume of one atom = Radius of atom in simple cubic Volume of an atoms per unit cell (v) = 1 X * + Volume of the unit cell = V = length X breadth X height

39 V = APF = * + APF = APF = 0.52 This means that 52% of the space in the unit cell is occupied and the remaining 48% is unoccupied. Hence Simple Cubic is a loosely packed structure. 2. BODY-CENTERED CUBIC (BCC) STRUCTURE NO.OF ATOMS PER UNIT CELL The total number of atoms per unit cell is two. *( ) + ATOMIC RADIUS From Fig. It is seen that one complete atom is located at the centre of the unit cell. It is also found that one eight of an atom is located at each corner of the unit cell, making another atom. Thus, there is a total of 1 (at the center) + 1 from the eight corners of the unit cell.

40 From Fig., the diagonal, (DF) 2 = (4r) 2 = (DG) 2 +(FG) 2. (DG) 2 = (DC) 2 +(CG) (1) But, DG) 2 = a 2 + a 2. (DG) 2 =2a (2) Substituting (2) in (1) (4r) 2 = 2a 2 +a 2. [FG=a] (4r) 2 =3a (3) CO-ORDINATION NUMBER In BCC, there are eight surrounding unit cells. For any corner atom, there are 8 body centered atoms from the neighbors for any corner atom. Therefore the coordination number for BCC is 8.

41 ATOMIC PACKING FACTOR Atomic packing factor is the ratio of the volume of the atoms occupying the unit cell to the volume of the unit cell We know that, r = a 3 4 Therefore volume of the unit cell, V = a 3. Number of atoms in the unit cell = 2 Volume of two atoms = Volume occupied by the atoms per unit cell = v = * + v= Volume of the unit cell, V = a 3

42 APF = APF = APF = 0.68 In this case 68% of the space is occupied and 32% of the space is unoccupied. Hence BCC is a closely packed structure when compared to SC. 3. FACE CENTERED CUBIC (FCC) STRUCTURE NO.OF ATOMS PER UNIT CELL Each corner atom is shared by 8 other unit cells No. of corner atoms per unit cell = = 1 Each Face centered atom is shared by 2 other unit cells No. of. Face centered atoms per unit cell = = 3 Therefore Total no. of. atoms per unit cell = 1+ 3 = 4 ATOMIC RADIUS It is defined as half the distance between nearest neighbours in a crystal. It is denoted by r

43 CO-ORDINATION NUMBER The Coordination number for FCC is 12. In this case, the nearest neighbours of any corner atom are the face centered atoms of the surrounding unit cell. Any corner atom has 4 such atoms in its own plane, 4 in a plane above it and 4 in a plane below it. Therefore the coordination number for FCC is 12. ATOMIC PACKING FACTOR Atomic packing factor is the ratio of the volume of the atoms occupying the unit cell to the volume of the unit cell

44 We know that, Volume of the unit cell, V = a 3 Number of atoms in the unit cell = 4 ( ) APF = APF = = 0.74 In the case of FCC, 74% of the unit cell is occupied by the atoms. Hence FCC is a closely packed structure.

45 4. HEXAGONAL CLOSELY PACKED STRUCTURE (HCP) NUMBER OF ATOMS PER UNIT CELL The HCP unit cell has 6 atoms per unit cell. Three atoms form a triangle in the middle layer. There are atom sections on both the top and bottom layers making two more atoms. There is a one-half an atom in the center of both the top and bottom layers making one more atom. Thus the total number of atoms in the HCP unit cell is 3+2+1=6. ATOMIC RADIUS The atomic radius of HCP is r = a / 2. Provided that r is the distance between the successive atoms. COORDINATION NUMBER In the bottom layer, the central atom has 6 nearest neighboring atoms in the same plane. Then at a distance of c/2 from the bottom layer, there are two layers one is above and other is below the bottom layer containing 3 atoms in each layer. Thus the total number of nearest neighboring atoms is =12. ATOMIC PACKING FACTOR Ratio Let I, J, K, L, M, N be the corner atoms, O be the base centred atom, P, Q, T be the middle atoms.

46 Draw a perpendicular line OR from O which meets at the point R in IN. IN Triangle IRO IF S IS the orthocenter for triangle INO then it is found that the length of Sub the value of OR from eqn (1) We get In Triangle SOP OS =.. (1) OS = OS =.. (2) OP 2 =OS 2 + SP 2.. (3) We know OP = a, SP = Sub I eqn.. (3) a 2 =( ) ( ) On re arranging we get

47 a 2 - = = = APF: Number of atoms per unit cell 6 Volume of the unit cell, v A PF = ( ) c Volume of atoms in the unit of cell Volume of the unit cell, V base area height Therefo re the atomic packing factor, Volume of the unit cell, V base area height PF A Area of triangle ION = Area of triangle ION = = Area of base = 6 x Area of triangle ION = 6 x = APF =

48 Since APF = * + APF = = 0.74 In the case of HCP, 74% of the unit cell is occupied by the atoms. Hence HCP is a closely packed structure. 5. DIAMOND CUBIC STRUCTURE Silicon (Si), Germanium (Ge), Carbon (C) and Tellurium (Te) posses this structure, which is a combination of two interpenetrating FCC sub lattices. This structure is loosely packed structure since each atom has only 4 nearest neighbours. And also the packing factor for this structure is found to be 34% NO. OF ATOMS PER UNIT CELL No. of corner atoms per unit cell = = 1 No.of. Face centered atoms per unit cell = = 3 No. of. atoms inside the unit cell = 4 Therefore Total no. of. atoms per unit cell = = 8 Co-ordination No = 4 and atomic radius = = (XT) 2 + (TZ) 2 + (ZY) 2 ( ) ( ) ( )

49 xy = 2r ATOMIC PACKING FACTOR APF = Volume of the unit cell = V = APF = APF = APF = ( ) APF = 0.34

50 34% of the unit cell is occupied by the atoms and remaining 66% is vacant, hence it is a loosely packed structure. 6. GRAPHITE STRUCTURE Diamond structure is a combination of two interpenetrating FCC structure. The atoms of diamond are united by covalent bonds. Thus it acts as a good electrical insulator and it is hard one. But in the case of graphite, carbon atoms are arranged in regular hexagons in flat parallel layers such that each atom is linked by neighbor atoms. Thus there is strong bonding between different layers, hence they can be easily separable from each other.

51 Fig. This is the cause of softness and lubricating action of graphite. The Fig shows the graphite structure. The carbon atoms are arranged in layer sheet molecular structure. In graphite each carbon atoms are covalently bonded to others in the same plane. The bond angle is 120, so the carbon atoms will form six-membered rings that link up to form planes or flat sheets of carbon atoms. The bond length is 1.42Ǻ. APPLICATION It is used to make the objects such as turbine blades for jet engines and tennis racquets. 7. MILLER INDICES The miller indices are defined as the reciprocals of the intercepts made by the plane on the crystallographic axes when reduced to smallest numbers. They are commonly written as (h k l). The following procedure is used to find the Miller Indices of a plane or face: Determine the intercepts of the plane in terms of multiples or sub-multiples of lattice parameters a, b, c Take reciprocals. Reduce it to the smallest integers (termed h k l) having the same ratio. Enclose these integers in brackets (h k l) to get Miller indices. Use a bar in Miller indices if the intercept is on the negative side.

52

53

54 Important features of Miller indices are as follows, All equally, spaced parallel planes have the same index numbers (hkl) A plane parallel to one of the co-ordinate axes has an intercept of infinity. If the miller indices of two planes have the same ratio (i.e., 422 or 211), then the planes are parallel to each other. 8. RELATION BETWEEN THE INTERPLANAR SPACING, MILLER INDICES AND THE CUBE EDGE Consider a plane ABC, as shown in Fig, cutting all the 3 axes. Let N be a point on the plane. Let ON be normal to the plane from the origin. Let us assume that h, k, l are the indices of the plane. Then the intercepts on the x, y, z axes are Let ON = d (i.e.) the perpendicular distance to the plane from the origin. Let α, β, γ are the angles made by the length ON with X,Y,Z axes respectively. ; ; ; We know that Cos 2 α + Cos 2 β+ Cos 2 γ =1

55 9. METHODS OF CRYSTAL GROWTH: BRIDGMAN TECHNIQUE: PRINCIPLE: The selective cooling of the molten material is to form single crystal by solidification along a particular direction. EXPERIMENTAL SET UP: The material to be grown in the form of a single crystal is taken in a cylindrical crucible. The crucible is made of platinum. The crucible is suspended in the upper furnace Upper furnace is hot zone. Lower furnace is cold zone. A pulley is used to move the container up and down to heat or cool the melt.

56 WORKING: The material is taken in the crucible. Upper furnace is switched on. The material in the crucible is completely melted into a molten liquid. The crucible is slowly lowered from upper furnace into lower furnace. Since the pointed tip enters the lower furnace first, the melt at this point starts to solidify to form crystal As the crucible is continuously lowered, the solidification of melt continues to form crystal. The process is repeated till the crystal grows to a larger size. ADVANTAGES: Relatively cheaper. Simpler technology. DISADVANTAGES: Confinement of crystals. No visibility of material during growth. CZOCHRALSKI METHOD OF CRYSTAL GROWTH FROM THE MELT

57 PRINCIPLE: It is a crystal pulling technique of growth of crystal by a gradual layer by layer condensation of melt. It is based on liquid solid phase transition initiated by a seed crystal. WORKING: The material to be grown in the form of a single crystal is taken in a crucible. The material is heated by means heating coils. Thus, the melt is obtained in the crucible. A seed crystal is introduced into the melt by means of a crystal holder. A small portion of seed crystal is initially melted. The temperature is then suitably adjusted for the growth of seed crystal by the solidification of melt into a thin layer of crystal. The seed crystal is rotated and gradually pulled out. Thus, a single crystal grows on the seed crystal. ADVANTAGES: This technique provides crystal free from crystal defects.

58 It can produce large size single crystal. This method enables easy control of atmosphere during growth. DISADVANTAGES: It may produce contamination of melt by the crucible. UNIT II PROPERTIES OF MATTER AND THERMAL PHYSICS 1. BENDING MOMENT OF A BEAM Beam: A beam is defined as a rod or bar (circular/rectangular) of uniform cross section whose length is very much greater than its other dimensions, such as breadth and thickness. Bending Of Beam: The moment of the couple due to the elastic reaction (restoring couple) which balances the external couple due to the applied load is called the bending moment. Expression for the Bending Moment Assumptions: 1) The length of the beam should be large compared to other dimensions. 2) The load (forces) applied should be large compared to the weight of the beam. 3) The cross-section of the beam remains constant and hence the geometrical moment of inertia I g also remains constant. 4) The shearing stresses are negligible. 5) The curvature of the beam is very small.

59 Let PQ be the small segment taken on the neutral axis AB and the extended length be P'Q'. be the angle subtended by the arc PQ. Original length PQ R (1) Extended length P'Q' (R x) (2) Increase in length = P'Q' PQ Or Increase in length = (R x)r Therefore increase in its length x (3) Linear strain = {increase in length / original length} Linear Strain = = (4) Young s modulus of the material Y = stress/linear strain. Substituting eqn. (4) in (5) we get, Stress = Y. linear strain (5)

60 Stress = If Ais the area of cross section of the filament P'Q', then Tensile force on the area ( A) = Stress. Area Tensile force = A Moment of force = force. Perpendicular distance Moment of force = A.x = A Moment of all the forces about the neutral axis = A Where I g = A=AK 2 Where, A total area of the beam, K- radius of the gyration Total moment of all forces Or Internal bending moment = Special cases: (i) Rectangular cross section If b is the breadth and d is the thickness of the beam, Then Area = bd and K 2 = Bending moment of rectangular cross-sectional bar = (i) Circular cross section If r is the radius then area, and K 2 = Bending moment of a circular cross-sectional bar = (ii) STRESS STRAIN DIAGRAM

61 Consider a body which is subjected to a uniformly increasing stress. Due to the application of stress, the change in dimension of the body takes place i.e., The strain is developed. If we plot a graph between stress and strain we get a curve as shown below and is called as stress and strain diagram. (i)the body obeys Hooke s law upto the region OA called as elastic range. (ii)the maximum elastic limit i.e., yield point B is crossed, the strain increases rapidly then the stress (iii) The body remains partly elastic and partly plastic which is represented by the curve BC. (iv) Even the small external force is applied, the body will take a new path CD and remains as plastic called as plastic range, where D is called as ultimate strength. (v) After this, the body will not come to its original state and the body acquires a permanent residual strain and it breaks down at a point called as breaking stress, indicated by dotted line. 2. DEPRESSION OF CANTILEVER. CANTILEVER: A beam fixed at one-end and loaded at its other free end as is known as Cantilever.

62 THEORY: Let us consider a beam fixed at one-end and loaded at its other free end as shown in figure. Due to the load applied at the free end, a couple is created between the two forces. (i) (ii) Force (load OW') applied at the free end towards downward direction and Reaction (R) acting in the upward direction at the supporting end. This external bending couple tends to bend the beam in the clockwise direction. But, since one end of the beam is fixed, the beam cannot rotate. Therefore the external bending couple must be balanced by another equal and opposite couple, created due to the elastic nature of the body called as internal bending moment. Under equilibrium condition, External bending moment = Internal bending moment DEPRESSION OF A CANTILEVER-LOADED AT ITS ENDS THEORY AND ITS DERIVATION: Let 'l' be the length of the cantilever OA fixed at 'O'. Let OW' be the weight suspended (loaded) at the free end of the cantilever. Due to the load applied the cantilever moves to a new position OA' as shown in the given diagram O

63 C OW ' Let us consider an element PQ of the beam of length dx, at a distance OP = x from the fixed end. Let 'C' be the centre of curvature of the element PQ and let 'R' be the radius of curvature. Due to.the load applied at the free end of the cantilever, the external couple is created between the load W at 'A" and the force of reaction at 'Q'). Here, the arm of the couple (Distance between the two equal and opposite forces) is (l - x). External bending moment = (1) We know Internal bending moment = (2) Under equilibrium condition External bending moment = Internal bending moment Therefore Eqn.(1) = Eqn.(2) (i.e.) R = (3) Two tangents are drawn at points P and Q which meet the vertical line AA at T and S respectively The smallest depression from T to S = dy Angle between the two tangents = d The angle between CP and CQ is also d (i.e) We can write the arc length PQ = Rd = dx (Or) d (4) PCQ = d Substituting eqn.(3) and (4), we have [ ] (5) From we can write

64 If d is very small then (6) Substituting (5) in (6) (7) Total depression at the free end of the cantilever can be derived by integrating Eqn. (7) within the limits 0 to. Therefore * + y = ( ) y = SPECIAL CASES: Depression of the cantilever at free end y = (8) (i) Rectangular Cross section If b is the breadth and d is the thickness of the beam then Substituting I g in Eqn. (8), we can write The depression at free end for a rectangular cross section (ii) Circular Cross section If r is the radius of the circular cross section, then

65 Substituting I g in Eqn. (8), Depression Produced is 3. ELEVATION AT THE CENTER OF THE BEAM WHICH IS LOADED AT ITS ENDS. THEORY: Let us consider a beam pf negligible mass, supported symmetrically on the two knife edges A and B as shown in the figure given below. Let the length between A and B be 'l'. Let equal weights W, be added to either end of the beam C and D. Let the distance CA = BD = a. Due to the load applied the beam bends from position F to E into an arc of a circle and produces as elevation 'x' from position F to E. Let 'w' be the reaction produced at the points A and B which acts vertically upwards as shown in figure. Consider a point 'P' on the cross section of the beam. Then the forces acting on the parts. PC of the beam is (i) Force W at 'c' and (ii) Reaction W at A

66 Let the distance PC = a 1 and PA = a 2, then The External bending moment about 'P' is Mp =W a 1 - W a 2 Here, the clockwise moment is taken as negative and anti-clockwise moment is.taken as positive. External bending moment about P can be written as M = W (a 1 a 2 ) M = Wa... (1) A B We know internal bending moment =...(2) Under equilibrium condition: External bending moment = Internal bending moment

67 We can write eqn.(1) = eqn.(2) Wa =...(3) Since for a given load (W) Y, Ig, a and R are constant. The bending is called as Uniform Bending Here it is found that the elevation 'x' forms an arc of the circle of radius 'R' as shown in figure. From the Δ AFO we can write OA 2 =AF 2 +FO 2 Since OF= FE, therefore we can write OA 2 =AF 2 +FE 2 (or) AF 2 =OA 2 -FE 2 Rearranging we can write AF 2 =FE [(OA 2 /FE)-FE] AF=l/2; FE= x; OA=R Eqn. (4) can be written as l 2 / 4 = x[2r x] = 2xR X 2 If the elevation is very small X 2 can be neglected Therefore we can write (l 2 /4) = 2xR (Or) R = (l 2 / 8x) (Or) Wa = 8YIgx / l 2 Therefore the elevation of point E above A is 4. I - SHAPED GIRDERS, APPLICATIONS OF I-SHAPED GIRDERS.

68 DEFINITION: The girders with upper and lower section broadened and the middle section tapered, so that it can withstand heavy, loads over it is called as I shape girders. Since the girder look like letter I as shown in the figure. They are named as I shape girder. EXPLANATION: 1. In general any girder supported at its two ends as on the opposite walls of a room, bends under its own weight and a small depression is produced at the middle portion. This may also be caused when loads are applied to the beams. 2. Due to the depression produced, the upper parts of the girder above the neutral axis contracts, while the lower parts below the neutral axis expands. 3. The stresses have a maximum value at the top and bottom. The stresses progressively decrease as it approaches towards the neutral axis. Therefore the upper and lower surfaces of the girder must be stronger than the intervening part. Thus the girders are made of I shape and are called as I-shaped girders. MINIMIZATION OF THE DEPRESSION:

69 1. Here, the depression can be minimized by either decreasing the load (W) (or) the length of the girder (l) (or) by increasing the young's modulus (or) the breadth (b) (or) the thickness of the girder. 2. Since the length of the beam 'I' is the fixed quantity, it cannot be decreased. Therefore the breadth and thickness may be adjusted by making the girder of large depth and small breadth (since thickness increases by b 3 ). 3. Thus the volume of the girder is increased and hence the depression produced is reduced. Therefore for stability, the upper part and the lower part are made broader than the centre part and hence forming I-shape called as I-shape girders. The depression can also be reduced by properly choosing the materials of high young s modulus APPLICATIONS OF I-SHAPED GIRDER 1. They are used in the construction of bridges over the rivers. 2. They are very much useful in the production of iron rails which are employed in Railway tracks. 3. They are used as supporting beams for the ceilings in the construction of buildings. THERMAL PHYSICS 1. CO-EFFICIENT OF THERMAL CONDUCTIVITY OF A BAD CONDUCTOR - LEE S DISC METHOD DESCRIPTION: The given bad conductor (B) is shaped with the diameter as that of the circular slab (or) disc 'D'. The bad conductor is placed in between the steam chamber (S) and the disc (or) slab (D) in which thermometer is inserted to measure the temperatures. The total arrangement is hanged over the stand as shown in fig. WORKING:

70 Steam is passed through the steam chamber till the steady state is reached (i.e.) the thermometer show constant temperature. Let the temperature of the steam chamber (hot end) and the disc or slab (cold end) be 1 and 2 respectively. The radius r of the Disc D and its thickness are also noted. Thickness of the bad conductor = x meter Radius of the bad conductor = r meter Mass of the disc = m Kg Steady state temperature of the slab = Steady state temperature in the steam chamber = Thermal conductivity of the bad conductor = K Amount of heat conducted by the bad conductor per second = Since the Area of cross section =

71 Amount of heat conducted per second = (1) Amount of heat lost by the slab per second = m x s x Rate of cool = msr c (2) Amount of heat conducted per second = Amount of heat lost by the slab per second (3) TO FIND THE RATE OF COOLING The bad conductor is removed and the steam chamber is directly placed over the slab and heated. When the temperature of the slab attains 5 C higher than, the steam chamber is removed and the slab is allowed to cool. A graph is plotted by taking time along X-axis and temperature along Y-axis. The rate of cooling of the slab alone = CASE (I): The steam chamber and the bad conductor are placed over the slab, in which the radiation takes place from the bottom surface of area of the slab and the sides of the slab area.(2πrh) (4) CASE (II):

72 The heat is radiated by the slab alone i.e. from the bottom and the top surface of the slab and through the sides of the slab. ( ) (5) From eqn. (4) and (5) ( ) Substituting eqn. (6) in eqn.(3) ( ) Thus the thermal conductivity of the bad conductor can be calculated using the above formula.

73 2. THE EXPERIMENTAL METHOD - THE THERMAL CONDUCTIVITY OF RUBBER BASED ON THE RUBBER TUBE METHOD. DESCRIPTION: Let us consider a thick cylindrical tube of length l inner radius r 1 and outer radius r 2.The steam can be passed through the center of the shell. Steam is allowed to pass through the axis of the cylindrical shell. The heat flows from the inner surface to the outer surface radially. Temperature at the inner surface = Temperature at the outer surface = CALCULATION: The cylinder may be considered to consist of a large number of co-axial cylinders of increasing radii. Consider such an elemental cylindrical shell o f the thickness dr at a distance r from the axis. Temperature at the inner and outer surface of the elemental shell be. Amount of heat conducted per second Q = Area of cross-section A = 2

74 Rearranging we have The thermal conductivity of the whole cylinder can be got by integrating Eqn.(1) within the limits r 1 to r 2 and and, [ ] = Rearranging we get, = ( ) (Or) ( ) Wm -1 K -1 Thus the thermal conductivity of the given material can be found using the above formula. 3. CONDUCTION OF HEAT BODIES IN SERIES Consider a composite slab or compound wall of two different materials, A and B of thermal conductivity K 1 and K 2 respectively. Let the thickness of these two layers A and B be d 1 and d 2 respectively as shown in the given figure. Let the temperature of the end faces be which is unknown. Heat will flow from face- A to face- B through surface of contact only if after steady state is reached, heat flowing per second (Q) through every layer is same

75 Amount of heat flowing per second through A is Amount of heat flowing per second through B is Where A is Area of cross-section of both layers, which is same. Since the amount of heat flowing through A and B are equal, we can write Equation (1) = Equation (2) Rearranging we get [ ]

76 Eqn.(3) gives us the expression for temperature at the interface of the compound media which is having two layers. By substituting the value of in Eqn.(1) we get [ ] * + * + * + By Re-arranging we get Eqn.(4) gives the amount of heat conducted by the two layers in serie

77 BODIES IN PARALLEL Consider a composite slab or compound wall of two different materials, A and B of thermal conductivity K 1 and K 2 respectively. Let the thickness of these two layers A and B be d 1 and d 2 respectively as shown in the given figure. The opposite faces of the material are kept at temperatures respectively. Let A 1 and A 2 be the areas of cross section of the materials. Then Amount of heat flowing through the first slab The Amount of heat flowing through the second slab The Total heat flowing through two slabs per second Substituting Eqn.(1) and (2) in Eqn.(3), we get [ ] * Eqn.(4) gives the Amount of heat flowing through compound wall of two layers in Parallel.

78 UNIT III QUANTUM MECHANICS PLANCK S RADIATION LAW To derive the Planck s radiation law, let us consider N Number of oscillators with their total energy as Then, the average of an oscillator is given by (1)

79 If,,, are the oscillators of energy 0,1E,2E,3E, re respectively then we can write (i) The total number of oscillators (2) and (ii) Total energy of oscillators ----(3) According to Maxwell s distribution formula, the number of oscillator having an energy re is given by (4) Where is called Boltzmann constant and r = 0, 1, 2, 3, The number of oscillators,,, can be got as follows: (i) For ; (ii) For ; (iii) For ; (iv) For ; Similarly for ; The total number of oscillators can be got by substituting the values of,,, in eqn. (2). N = N = [ ] (5) We know, 1 + x + x 2 + x 3 + =. therefore we can write eqn.(5) as The total number of oscillators, N * + (6) Similarly by substituting the values of,,, in eqn. (3), the total energy can be written as

80 We know,1 +2 x + 3x 2 + = Equation (7) becomes [ ]..(7) The total energy of oscillators,.(8) [ ( ) ] Substituting the values of equation (6) and (8) in equation (1) we get The average energy of the oscillator, = [ ( ) ] ( ) [ ] = ( ).(9) Substituting the value of E = h in eqn (9), we get.(10) The eqn. (10) represents the average energy of the oscillator. The number of oscillators per unit volume within the range of wavelength λ and λ + dλ is given by

81 N =..(11) Energy density, Energy density, (13) [ ].(14) Equation (14) represents Planck s radiation law in terms of wavelength. The Planck s radiation law in terms of frequency is, E ( ) Wien s Displacement law This law holds good only at shorter wavelength λ is small, will be greater [ ] This is Wien s Displacement law Rayleigh Jean s law This law holds good only at longer wavelength λ is large, will be Lesser +

82 Neglecting higher terms This is Rayleigh Jean's law G.P.THOMSON S EXPERIMENT (1) It is used to study the wave nature of electrons. (2) A high energy electron beam is produced by the cathode and excited with a potential upto 50,000 volts. (3) A narrow electron beam is obtained by passing through a slit and is accelerated to fall on the gold foil. (4) The photograph of the beam from gold foil is recorded on a photographic plate. (5) The photographic plate is developed to obtain a diffraction pattern. (6) The pattern confirms that the diffraction is only due to presence of electrons. (7) If the electron behaves as a particle, then it would have been scattered. (8) As the electron behaves as a wave, here the diffraction pattern is obtained. (9) Thus this experiment proves the wave nature of electrons.

83 SCHRODINGER TIME INDEPENDENT WAVE EQUATION The equation which describes the wave nature of the particle in mathematical form is known as Schrodinger wave equation. Let us consider a particle of mass m moving with a velocity, v Classical differential equation + De Broglie wavelength = Schrodinger wave equation The Classical differential equation is given by - wave function of particle. (1) The solution of eqn. (1) is given by.(2) - Amplitude of the wave. Differentiating twice eqn.(2) w.r.to t, (3) Substitute (3) in (1) (.(4)

84 We know that.(5) Substitute (5) in (4) = 0 (6) Now introducing the wave nature of the particle λ in eqn.(6).(7) We know that, Total energy E = P.E + K.E E = mv 2 + V 2 (E V) = mv 2 Multiply by m on both sides 2m (E V) = m 2 v 2.(8) Substitute eqn. (8) in (7) We know that, ħ = ħ ħ..(9) Eqn. (9) is Schrodinger time independent wave equation

85 PARTICLE IN A BOX Let us consider a particle (electron) of mass, m moving with a velocity, v. The electron is inside the box and it can t come out. The electron is travelling along the x-axis from x = 0 to x = l Potential energy (V) of the electron inside the box is zero. Potential energy (V) of the electron outside the box is infinite. V = 0, 0 V =, when 0 The Schrodinger time independent wave equation is ħ (V = 0) where ħ The solution for the above equation is where A & B are constants which can be found by applying the boundary conditions (i) At x = 0, 0 = B = 0 (ii) At x = l, 0 = 0 = A Also = x

86 We know that ħ Also ħ ħ is Energy of particle called Eigen value. ground state If n = 2. First excited state Normalization of wave equation: The probability density of finding a particle is given by

87 [ ] [ ] [ ] ( ) This is Eigen function i.e., wave function of a particle. SCHRODINGER TIME DEPENDENT WAVE EQUATION The solution of above eqn. is given by Differentiating w.r.to t, = We know that, E = h ħ (ħ )

88 Multiply and divide by i on RHS, we get ħ = ħ E - Energy operator We know that Schrodinger time independent wave equation is ħ Substitute in above equation ħ ħ ħ ħ - ( ħ ) ħ This is Schrodinger time dependent wave equation. H Hamiltonian operator E Energy operator SCANNING ELECTRON MICROSCOPE Principle: When accelerated primary electrons strikes the sample, it produces secondary electrons. These secondary electrons are collected by a collector which in turn gives a 3-D image of a sample. Construction: (1) Electron gun is used to produce electron beam. (2) Magnetic condensing lens is used to condense the electron beam.

89 (3) Scanning coil is arranged between magnetic condensing lens and the sample. (4) Collector is used to collect the secondary electrons and convert into electrical signals. These signals are fed into CRO through photomultiplier tube. Working: Electrons are produced by the electron gun. Primary electrons are made to fall on the sample through the condensing lens and scanning coil. Primary electrons on striking the sample produce secondary electrons. To collect the secondary electrons, a very high voltage is applied to the collector.

90 Collected electrons produce scintillations, convert into electrical signals and are amplified by the photomultiplier and are fed into the CRO. By the same procedure, electron beam scans the sample from left to right and againfrom left to right similar to how we read a book and the whole picture of the sample is obtained in the CRO screen. Advantages: Specimen of large thickness can be examined. It is used to produce 3D image of object. Magnification is 3,00,000 times greater than the size of the object. It has large depth of focus. Disadvantages: Resolution is limited to about nm. Applications: It is used to examine structure of very large specimen in 3D view. SEM finds applications in physics, chemistry, biology, engineering and industries. TRANSMISSION ELECTRON MICROSCOPE Principle: Electrons are made to pass through the specimen and the image is formed in fluorescent screen, either by using transmitted beam or by using diffracted beam. Construction: Electron gun is used to produce electron beam. Magnetic condensing lens is used to condense the electron beam. Specimen is placed between the condensing lens and objective lens. Aperture is used to eliminate the diffracted beam. Magnetic projector lens is placed above the screen in order to achieve higher magnification. Image can be recorded by a fluorescent screen.

91 Working: Stream of electrons are produced by the electron gun and is made to fall on the specimen using magnetic condensing lens. Based on the angle of incidence, beam is partly transmitted and partly diffracted. Both the transmitted beam and the diffracted beam are recombined at the E-wald sphere. The combined image is called phase contrast image. To eliminate the diffracted beam, the resultant beam is passed through the magnetic objective lens and aperture. The final image is obtained only due to the transmitted beam and final image is recorded in fluorescent screen and it is called as bright field image. If the image is obtained only due to the diffracted beam, it is called as dark field image.

92

93 Advantages: It is used to examine the specimen of size upto 0.2 nm. It has high resolution. Magnification is 10,00,000 times greater than the size of the object. Resolving power is 1Å to 2 Å. Disadvantages: Specimen should be very thin. 3D image cannot be obtained. Structural changes occur during sample preparation. For biological samples, electrons interact with the sample and it will damage the sample. Applications: It is used in nano sciences to find the internal structures of nanomaterials. It is used in thin film technology, metallurgy, biochemistry, etc. It is used to find 2D image of very small biological cells, virus, bacteria, etc. It is used to study the composition of paints, papers, alloys, etc. UNIT IV ACOUSTICS 1. FACTORS AFFECTING ACOUSTICS OF A BUILDING The factors affecting the acoustics of a building are as follows:

94 Un optimized reverberation time Very low or very high loudness Improper focusing of sound to a particular area, which may cause interference. Echoes produced inside the buildings Resonance caused due to matching of sound waves Unwanted sound from outside or inside the building, so called noise may also affect the acoustics of buildings. OPTIMUM REVERBERATION TIME AND ITS REMEDY The reverberation time plays a vital role in the auditorium for clear audibility of sound. If the reverberation time is very high then it produces echoes in the hall and if the reverberation time is very low, the sound will not be clearly heard by the audience. The optimum reverberation time can be achieved by the following steps: By having the full capacity of audience in the auditorium By choosing absorbents like felt, fiber, board, glass etc. Reverberation time can be optimized by providing windows and ventilators Reverberation time can also be optimized by decorating the walls with pictures. LOUDNESS Loudness is the degree of sensation produced on the ear; it varies from observer to observer. The loudness will be very low in some area and will be very high in some area. REMEDIES Loudspeakers should be placed at the places where we have low loudness The loudness can also be increased by making reflecting surfaces, wherever necessary Loudness can be increased by constructed low ceilings Absorbents are placed at the places where we have high loudness. FOCUSSING AND INTERFERENCE EFFECTS

95 In some places of a hall, the sound will not be heard properly, which is due to the presence of convex or concave surfaces in the hall. REMEDIES Avoiding curved surfaces. Covering the curved surfaces by suitable absorbents. ECHOES The echoes are formed when the time interval between the direct and reflected sound waves are about 1/15 th of a second. This effect occurs due to the reason that the reflected sound waves reaches the observer later than the direct sound. REMEDY The echo can be avoided by lining the surfaces with suitable sound absorbing materials and by providing enough number of doors and windows. RESONANCE Resonance occurs when a new sound note of frequency matches with standard audio frequency. This will create disturbance to the audience. REMEDIES The resonance effect can be avoided by providing proper ventilation and by adjusting the reverberation time to the optimum level. Now a days the resonance is completely eliminated by air conditioning the halls NOISE

96 Noise is an unwanted sound which leads to displeasing effect to the ear. Types of noises: Airborne noise Structure borne noise Inside noise REMEDIES By making the hall air conditioned By providing carpets By allotting proper places for doors and windows By breaking the continuity of the interposing layers by some acoustical insulators.

97 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I ULTRASONICS 1.MAGNETO-STRICTION GENERATOR PRINCIPLE: Magnetostriction effect: When an alternating magnetic field is applied to the ferromagnetic rod, the rod is thrown into longitudinal vibrations, producing ultrasonic waves at resonance. CONSTRUCTION The experimental arrangement is shown infigure. A ferromagnetic rod is clamped in the middle. Coil L1 is wound at one end of the rod along with a variable capacitor C. Coil L2 wound at the other end of the rod is connected to the base of the transistor. The battery connected between emitter and collector provides necessary biasing. WORKING The rod is magnetized by passing direct current. When the battery is switched on, the transistor produces current and is passed to the coil L 1. The rod vibrates due to magnetostriction effect. Due to converse magnetostriction effect, an emf is induced in coil L 2 which is fed to the base of the transistor. The frequency of the oscillatory circuit is adjusted by the capacitor C and when this frequency is equal to the frequency of the vibrating rod, resonance occurs. At resonance, the rod vibrates vigorously producing ultrasonic waves. 97

98 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I RESONANCE CONDITION Frequency of the oscillator = Frequency of the rod where l is the length of the rod E is the Young s modulus of the rod ρ is the density of the rod. ADVANTAGES Design is very simple. Cost is low DISADVANTAGES It cannot produce frequencies greater than 3MHz. The frequency of oscillations depends on temperature. 2. PIEZO ELECTRIC GENERATOR PRINCIPLE: Inverse piezoelectric effect: When an alternating voltage is applied to the top and bottom faces of a quartz crystal, the other two faces of the crystal undergoes contractions and expansions. CONSTRUCTION: The circuit diagram is shown in figure 98

99 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I Piezo electric oscillator consists of primary and secondary circuits. Primary circuit: It consists of battery, transistor,capacitor, coils L 1 and L 2. Coil L 1 is connected to the base and coil L 2 is connected to the collector of the transistor. Capacitor C connected parallel to L 1 forms the tank circuit. Secondary circuit: It consists of a coil L 3 which is connected to the metallic plates A and B. The quartz crystal is placed between two metal plates A andb. WORKING When battery is switched on, the transistor produces current. Current flows through coils L 1 and L 2. Due to transformer action, current is induced in coil L 3. Due to inverse piezoelectric effect the crystal begins to vibrate. The frequency of the oscillatory circuit is adjusted by the capacitor C 1 and when this frequency is equal to the frequency of the vibrating crystal, resonance occurs. At resonance, the crystal vibrates vigorously producing ultrasonics. Resonance condition 99

100 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I Frequency of the oscillator = Frequency of the vibrating crystal where l is the length of the crystal E is the Youngs modulus of the crystal ρ is the density of the crystal P = 1,2,3 for fundamental, first over tone, second overtone atc ADVANTAGES It can produce frequencies upto 500 MHz. It is not affected by temperature and humidity. DISADVANTAGES The cost is very high. The cutting and shaping of quartz crystal is very complex. 3. ACOUSTIC GRATING MEASUREMENT OF VELOCITY OF ULTRASONIC WAVES. PRINCIPLE: When ultrasonic waves are passed through a liquid, the density of the liquid varies layer by layer due to the variation in pressure and hence the liquid will act as a diffraction grating, so called acoustic grating. Under this condition, when a monochromatic source of light is passed through the acoustical grating, the light gets diffracted. CONSTRUCTION: It consists of a glass tank filled with the liquid. Piezoelectric crystal is fixed at the top of the glass tank. 100

101 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I WORKING: Crystal is at rest: When the light is focused onto the tank filled with the liquid, a single peak is observed which shows that there is no diffraction. Crystal is set into vibration: Ultrasonic waves are produced and are reflected by the walls of the glass tank forming stationary waves.at nodes, the density of the liquid becomes more and at antinodes the density of the liquid becomes less. When light is passed through the liquid, it gets diffracted and a diffraction pattern is obtained. The grating formula is given by, d sin θ n =n λ -----(1) where n = 0, 1, 2, 3, is the order of diffraction, λ is the wavelength of monochromatic light used d is the distance between any two successive nodes or antinodes of stationary wave. 101

102 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I Wavelength of ultrasonics, Substitute (2) in (1), we get λ m =2d -----(2) λ m = Velocity of ultrasonics (v) = frequency of ultrasonics x wavelength of ultasonics v= υ u 4. ULTRASONIC FLAW DETECTOR/NON-DESTRUCTIVE TESTING/ PULSE ECHO SYSTEM PRINCIPLE: 102

103 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I Whenever there is a change in medium, then the ultrasonic waves will be reflected. DESCRIPTION : It consists of a transducer coupled to the specimen. A pulse generator is connected to the transducer. An amplifier is connected between the transducer and CRO. WORKING: The pulse generator generates high frequency waves and is applied to the transducer. (Pulse A) Ultrasonic waves produced by the transducer travel through the specimen and is reflected by the other end. Reflected ultrasonics are received by the transducer and recorded in the CRO. (Pulse B) If there is any defect on the specimen like a small hole, then we get another pulse (Pulse C) between pulses A & B. From the height of the pulse, the depth of the hole can be determined. ADVANTAGES: Cost is low; High speed. Safest method. It shows internal defects. 103

104 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I DISADVANTAGES: Only trained technicians can perform this testing. Difficult to find defects having complex shapes. 5. ULTRASONIC SCANNING METHODS: Based on the position of the transducer and the output displayed in CRO, we can classify the scanning methods into three types. i) A- scan ii) iii) B-scan C-scan or T-M scan A-Scan or Amplitude mode display It gives 1-D information about the specimen. Single transducer is used to transmit and receive pulses from the specimen, Transducer is fixed. Vertical spike is displayed in the CRO. The height of the vertical spike gives the strength of the echo. The position of the spike along the horizontal axis gives the depth of the flaw. B-Scan or Brightness mode display 104

105 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I It gives 2-D information about the specimen. Transducer is movable. Echoes are displayed as dots. The brightness of the dot gives the intensity of echo pulse. The size of the dot gives the strength of the reflected pulse. T.M Scan (or) Time motion Scan (or) C-Scan It gives 3-D information about the specimen. Transducer is fixed. Ultrasonic probe is moved over the surface of the specimen in either zigzag fashion or in closely placed parallel lines. The intensity of echo is recorded as a shading with blank spaces corresponding to defect regions. 105

106 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I 6. SONOGRAM- RECORDING OF MOVEMENT OF HEART Principle: It works under the principle of Doppler Effect, (i.e) there is a change in frequency between the incident sound waves on the fetus and the reflected sound waves from the fetus. Construction: 1. It consists of RFO (Radio Frequency Oscillator) to produce 2MHZ of frequency and RFA (Radio Frequency Amplifier) to amplify the received signals. 2. To mix the transmitted and received signals the mixer is used. 3. Loudspeaker and CRO helps to hear and view the soundwaves. Working: The transducer is fixed over the mother s abdomen. RFO is switched on. Ultrasonic waves are made to incident on the fetus. The reflected waves are received by the transducer. Both the incident and reflected signals are mixed by the mixer and filtered. Change in frequency is measured. The movement of heart can be viewed by CRO (or) can be heard by the loudspeaker. UNIT V 106

107 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I 1. CO 2 LASER Principle of CO 2 laser: LASER laser beam. The transition between various vibrational levels of CO 2 molecule leads to the construction of Fundamental modes of vibration of CO 2 molecule: There are three fundamental modes of vibration 1. Symmetry stretching mode 2. Bending mode 3. Asymmetric stretching mode Symmetry stretching mode: Here the carbon atoms are stationary and the oxygen atoms vibrate along the axis of the molecule as shown in Fig. Bending mode: Here the atoms will vibrate perpendicular to the molecular axis as shown in Fig. Asymmetric stretching mode: Here all the three atoms will vibrate. Here the oxygen atoms vibrate in the opposite direction to the vibration direction of carbon atom as shown in fig. Construction: It consists of a discharge tube filled with CO 2, nitrogen and helium gases in the ratio 1:2:3. Nitrogen helps to increase the population of atoms in the upper level CO 2, while helium helps to depopulate the atoms in the lower level of CO 2. The discharge is produced by D.C. excitation. NaCl/Brewster windows are placed at ends of discharge tube. Mirrors M 1 and M 2 forms optical resonator. 107

108 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I Working: 1. DC source is switched ON. 2. Due to ionization, electrons are produced. 3. The energy of electron is absorbed by nitrogen atoms and it goes to excited state. e - + N 2 N 2 * 4. Since excited level of nitrogen is very close to E 5 level of CO 2 molecule, CO 2 molecules are excited by energy transfer and population inversion is achieved. N 2 * +CO2 CO * 2 + N 2 5. Transition from E 5 to E 4 emits a photon of wavelength 9.6μm. 6. Transition from E 5 to E 3 emits a photon of wavelength10.6μm. 7. Normally 10.6μm transition is more intense than 9.6μm 8. When the gas flow is longitudinal, the power output is W/m and when the gas flow is perpendicular, the power output is 10 KW/m. 9. This type of CO 2 laser is known as TEA laser (Transversely Excited Atmospheric Pressure Laser). 108

109 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I Energy Level Diagram Advantages or Merits of CO 2 Laser: It is the first molecular laser. The output of the laser is continuous. It has high output power. Disadvantages or demerits of CO 2 laser: The contamination of carbon monoxide and oxygen will have some effect on the laser action. Corrosion problem may occur in the reflecting plates. Application and uses of CO 2 Laser: High power CO 2 laser find application in materials processing, welding, drilling, cutting, soldering etc., because of their very high output power. 109

110 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I It is used in the treatment of lever and lungs. 2. FOUR LEVEL SOLID STATE LASER. Nd-YAG laser: Principle: Nd-YAG laser is a four level laser in which Yttrium Aluminum garnet is doped with neodymium Nd 3+. Due to optical pumping these ions are raised to excited levels. During the transition from metastable state to ground state, the laser beam of wavelength μm is emitted. Construction: The active medium is a rod which has Yttrium Aluminum Garnet [Y 3 Al 5 O 12 ] doped with neodymium Nd 3+. This rod is placed inside a highly reflecting elliptical cavity as shown in fig. Optical pumping is made by placing the xenon flash lamp near by the laser rod. The flash tube may be switched ON and controlled with the help of a capacitor. Construction diagram Working: The xenon flash lamp is switched ON and the light is allowed to fall on the laser rod. The intense white light excites the neodymium (Nd 3+ ) ions from the ground state to various energy levels above E 2. Hence the atoms are raised to higher energy E 3 as illustrated in the energy 110

111 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I level diagram. From these energy levels the ions make non-radioactive decay and are gathered in a state called as Meta stable state (E 4 ), until the population inversion is achieved. Once the population inversion is achieved, the stimulated emission builds up rapidly between E 4 and E 1. Thus laser beam is emitted. Applications of Nd-YAG Laser: Nd-YAG laser find many applications in range finders and illuminators. They also find applications in resistor trimming, scribing, micro machining operations as well as welding, drilling etc., They find applications in medicine field like Endoscopy, Urology, neurosurgery, ENT, Gynecology, Dermatology, Dental surgery and General surgery. 4. HOMO JUNCTION LASER Homojunction Semiconductor Laser: A pn junction made up of same semiconductor materials known as homo junction laser. E.g. GaAs laser. 111

112 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I Principle Electroluminescence - The electron in the conduction band combines with a hole in the valence band and hence the recombination of electron and holes produces energy in the form of light. Construction: The active medium is a p-n junction diode made from a single crystalline material (GaAs). The junctions of the p and n are well polished to form an optical resonator. The upper and lower electrodes fixed in the p and n region are used for the flow of current to the diode during biasing. WORKING: 1. Due to forward biasing, the electrons from the n type and the holes from p type is injected across the pn junction. 2. Population inversion is achieved in the pn junction. 112

113 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I 3. The electron from conduction band recombines with holes in valence band across the pn junction to emit photons spontaneously. 4. These photons will stimulate the electrons to undergo more recombination of electrons and holes to produce stimulated photons. 5. The photons travels back and forth between two sides, after gaining enough strength, the laser beam is emitted across pn junction. Wavelength of laser emitted is 8300Å Å (near IR region). The wavelength of laser light is given by E g = hυ = hc/λ λ = hc/ E g where E g is the band gap energy in Joule. ADVANTAGES: It is easy to manufacture the diode, low cost. Gives continuous wave output. DISADVANTAGES: USES It produces low power output. The output wave is pulsed and will be continuous only for some time. Used in long distance communication. 113

114 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I Used in laser printers. 4. HETERO JUNCTION LASER A pn junction made up of different materials in two regions that is n type and p- type is known as heterojunction laser. E.g. GaAlAs laser. PRINCIPLE: Electroluminescence - The electron in the conduction band combines with a hole in the valence band and hence the recombination of electron and holes produces energy in the form of light. CONSTRUCTION: 1. This laser consists of five layers as shown in the fig. 2. A layer of Ga-As p type (third) layer acts as active region. 3. Third and fourth layers acts as p-n junction and they act as optical resonator. WORKING: The working of the hetero junction laser is also similar to that of the homo junction laser. 114

115 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I 1. Due to forward biasing, the electrons from the fourth layer and the holes from second layer are injected across the pn junction (third layer). 2. This process is continued until population inversion is achieved in the pn junction. 3. The electron from conduction band recombines with holes in valence band across the pn junction to emit photons spontaneously. 4. These photons will stimulate the electrons to undergo more recombination of electrons and holes to produce stimulated photons. 5. The photons travels back and forth between two sides, after gaining enough strength, the laser beam is emitted across third and fourth layers. The wavelength of laser light is given by E g = hυ = hc/λ λ = hc/e g where E g is the band gap energy in Joule. The wavelength of the output is nearly 8000Å. Advantages: It produces continuous wave output. 115

116 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I The power output is very high. Disadvantages: It is very difficult to grow different layers of pn junction. The cost of this laser is very high. 5. EINSTEIN S COEFFICIENT OF STIMULATED AND SPONTANEOUS EMISSIONS. Let us consider an atom that has only two energy levels E 1 and E 2. When the atom is exposed to (light) photons of energy E 2 E 1 = hυ. Three distinct processes takes place. Stimulated Absorption: Stimulated Absorption Spontaneous emission Stimulated emission An atom in the ground state with energy E 1 absorbs a photon of energy hυ and goes to the excited state (Higher state) with energy E 2 as shown in fig. This process is called Stimulated Absorption. 116

117 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I The rate of absorption (N 12 ) is proportional to the following. (i.e) N 12 α Energy density of incident radiation (Q) α No of atoms in the ground state (N 1 ) N12 α Q N1 (Or) N12 = B12 Q N (1) unit time. Spontaneous emission: Where B 12 is a constant which gives the probability of absorption transition per The atoms in the excited state returns to the ground state by emitting a photon of energy E = (E 2 - E 1 ) = hυ, spontaneously without any external triggering as shown in fig. This process is known as Spontaneous emission. The rate of spontaneous emission is N 21 (SP) α N 2 N 21 (SP) = A 21 N (2) 117

118 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I Where A 21 is a constant which gives the probability of transitions per unit time. spontaneous emission Stimulated emission: The excited state can also return to the ground state by external triggering (or) inducement of photon thereby emitting a photon of energy equal to the energy of the incident photon, known as stimulated emission. Thus results in two photons of same energy, phase difference and of same directionality as shown in Fig. The rate of stimulated emission is given by N 21 (ST) α Q N 2 N 21 (ST) = B 21 Q N (3) where B 21 is a constant which gives the probability of stimulated emission transitions per unit time. EINSTEIN S THEORY: Einstein s theory of absorption and emission of light by an atom is based on Planck s theory of radiation. Also under thermal equilibrium, the populations of energy levels obey the Boltzmann s distribution law, 118

119 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I (i.e.) under thermal equilibrium, The rate of absorption = the rate of emission N12 = B12 Q N (1) N21 (SP) = A 21 N (2) N21 (ST) = B 21 Q N (3) (i.e.) Eqn. (1) = Eqn. (2) + Eqn. (3) B12 Q N1 = A21 N2 + B21 Q N2 Q [B 12N 1 - B 21 N 2 ] = A 21 N 2 ( ) On substituting (from Boltzmann distribution equation) is ( ) We know that Planck s black body radiation formula for energy distribution Comparing the above two equations, we get 119

120 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I Taking A 21 = A and B 12 = B 21 = B The constants A and B are called as Einstein s coefficients. FIBRE OPTICS 1. FIBER OPTICAL COMMUNICATION SYSTEM. Its purpose is to transfer information from source to a distance user. The basic optic fiber communication is the transmission of information by the propagation of optical signal through optical fibers over the required distance. CONSTRUCTION: A basic block diagram of fiber optic communication system is shown in fig. 120

121 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I Fiber optic communication system consists of Information signal source Transmitter Propagation medium. Light source. Photo detector. Receiver WORKING: Analog information such as voice is fed into analog to digital converter. These digital signals are fed into voltage to current converter. Current is fed into LED to produce light signal. The light signal is passed into fiber optic cable and light is propagated by total internal reflection. Light is fed into PIN, where it is converted into electrical signal. This signal is again converted into voltage and then amplified by an amplifier. The amplified signal is converted into analog signal The output is same as input signal, thus the information is transferred one end to the other. 2. MEDICAL ENDOSCOPE 121

122 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I FIBER OPTIC ENDOSCOPE: The fiber optic endoscope is a tubular optical instrument. It is used to view the internal parts of human body which are not visible to the naked eye. CONSTRUCTION & WORKING: Usually in the endoscope, there are two fibers namely 1. Inner fiber and 2. Outer fiber. The inner fiber illuminates the inner structure of the object under study. The outer fiber is used to collect the reflected light from that area and from that we can view the inner structure of the object. A lens is used to focus the light. WORKING: Light from the source is passed through the outer fiber and light is illuminated on the internal parts of the body. The object reflects the light and enters the inner fiber. The reflected light falls on the receiving end or telescope, finally the whole picture is viewed through the eyepiece. Cross sectional view consists of i) outer fiber ii)inner fiber iiii) instrumentation channel iv) irrigation channel. Instrumentation channel is used to insert surgical instruments needed for operation. 122

123 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I Irrigation channel is used to blow air, send some liquids like glucose, water etc. to clean the blood in the affected parts during surgery. ADVANTAGES OF OPTICAL FIBER COMMUNICATION: Easy handling, less cross talk. Noise free transmission, Economical Safe, Longer life span and Easy maintenance 3. TEMPERATURE SENSOR AND DISPLACEMENT SENSOR. They are two types of Fiber optic sensors. Intrinsic fiber optic sensors (temperature sensor) Extrinsic Fiber optic sensors (displacement sensor). Intrinsic Fiber optic sensors: The quantity to be measured directly takes place on the fiber itself Extrinsic sensors or Passive sensors: fiber. The interaction between the light and quantity to be measured takes place outside the TEMPERATURE SENSORS: 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. 123

124 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I It consists of a laser source to emit light. A beam splitter, made of glass plate is inclined to an angle of 45 o with respect to the direction of the laser beam. WORKING: Reference fiber which is isolated from the environment. Test fiber kept in the environment to be sensed. A monochromatic source of light is emitted from the laser source. The beam splitter kept at 45 o inclination divides the beam emerging from the laser source into two beams (I) Main beam and (II) Splitted beam The main beam passes through the lens L 1 and is focused onto the reference fiber which is isolated from the environment to be sensed. The beam after passing through the reference fiber then falls on the lens L 2. The splitted beam passes through the lens L 3 and is focused onto the test fiber kept in the environment to be sensed. The splitted beam after passing through the test fiber is made to fall on the lens L 2. The two beams after passing through the fibers, produces a path difference due to the change in parameters such as pressure, temperature etc., in the environment. Therefore a path difference is produced between the two beams, causing the interference as shown in fig. Thus the change in temperature can be accurately measured with the help of the interference pattern obtained. DISPLACEMENT SENSORS: 124

125 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I PRINCIPLE: Light is sent through a transmitting fiber and is made to fall on a moving target. The reflected light from the target is sensed by a detector. With respect to intensity of light reflected from it the displacement of the target is measured. WORKING: It consists of a bundle of transmitting fiber coupled to the laser source and a bundle of receiving fiber coupled to the detector. Light from the source is transmitted through the transmitting fiber and is made to fall on the moving target. The light reflected from the target is made to pass through the receiving fiber and the same is detected by the detector. Based on the intensity of the light received, the displacement of the target can be measured. If the received intensity is more, then we can say that the target is moving towards the sensor and if the intensity is less, we can say that the target is moving away from the sensor. 125

126 PANIMALAR INSTITUTE OF TECHNOLOGY I SEM PART A Q & A PHYSICS - I 4. ACCEPTANCE ANGLE AND NUMERICAL APERTURE. along Let us consider an optical fiber into which the light is injected. The light ray travels AO and enter into the core at an angle θ o to the axis of the fiber. If θ is greater than the critical angle at the core cladding interface, the light ray get total internally reflected and propagates through the fiber i.e., the light will stay inside the fiber. The light ray is refracted along OB at an angle θ r in the core. It further proceeds to fall at critical angle of incidence θ = (90- θ r ) on the interface between core and cladding at B. CONDITION FOR PROPAGATION: n 1 and n 2 are the refractive indices of the core and cladding, n o is the refractive index of the surrounding. Now applying Snell s law of refraction at the point of entry of the ray AO into the core, we have, 126