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S.K.P. Engineering College, Tiruvannamalai ISEM

Physics Department 1 Engineering Physics I

SKP Engineering College

Tiruvannamalai – 606611

A Course Material

on

Engineering Physics-I

By

Dr.D.Narayanasamy

Professor

Physics Department



Subject Name:Engineering Physics-I

Year/Sem:I/I

Being prepared by me and it meets the knowledge requirement of the University

curriculum.

Signature of the Author

Name: Dr.D.Narayanasamy

Designation: Professor

This is to certify that the course material being prepared by Dr.D.Narayanasamy is of

the adequate quality. He has referred more than five books and one among them is

from abroad author.

Signature of HD Signature of the Principal

Name: Mrs.Bincy Name: Dr.V.Subramania Bharathi

Seal: Seal:



PH 6151 ENGINEERING PHYSICS I

Lecture : 3 hrs/Week InternalAssessment: 20

Tutorial : 0 hr/week FinalExamination: 80

Practical : - Credits: 3

PREREQUISITE: :Basic knowledge in Physics COURSE OBJECTIVES: To provide a broad foundation in the basics of science and engineering. To provide sensible preparation for other areas of engineering, including

mechanical, electrical, civil engineering and computer science. COURSEOUTCOMES:Upon completion of this course the student will be able to:

CO1 Ability to conduct electricity

CO2 Understand the property of a semiconductor relies on quantum physics to explain the movement of electrons and holes in a crystal lattice.

CO3 Produce a magnetic field in response to an applied magnetic field. Understands the magnetic behavior of dia, para and ferromagnetic materials.

CO4 Study of dielectric properties concerns storage and dissipation of electric and magnetic energy in materials.

CO5 Development of novel metallic-glass-matrix composite materials and studies of the atomic-scale structure to very practical studies of mechanical behavior, including both deformation and fracture and also behavior of some nanomaterials is well understood.



SYLLABUS

UNIT I ULTRASONICS 9

Introduction – Production – magnetostriction effect - magnetostriction generatorpiezoelectric effect - piezoelectric generator- Detection of ultrasonic waves properties – Cavitations - Velocity measurement – acoustic grating - Industrial applications drilling, welding, soldering and cleaning – SONAR - Non Destructive Testing – pulse echo system through transmission and reflection modes - A,B and C –scan displays, Medical applications – Sonograms

UNIT LASERS 9

Introduction – Principle of Spontaneous emission and stimulated emission. Population inversion, pumping. Einsteins A and B coeffcients - derivation. Types of lasers – He-Ne, CO2 , Nd-YAG, Semiconductor lasers (homojunction &heterojunction) Qualitative Industrial Applications - Lasers in welding, heat treatment, cutting – Medical applications - Holography (construction & reconstruction). UNIT III FIBER OPTICS & APPLICATIONS 9

Principle and propagation of light in optical fibres – Numerical aperture and Acceptance angle - Types of optical fibres (material, refractive index, mode) – Double crucible technique of fibre drawing - Splicing, Loss in optical fibre – attenuation, dispersion, bending - Fibre optical communication system (Blockdiagram) - Light sources - Detectors - Fibre optic sensors – temperature &displacement - Endoscope.

UNIT IV QUANTUM PHYSICS 9

Black body radiation – Planck’s theory (derivation) – Deduction of Wien’s displacement law and Rayleigh – Jeans’ Law from Planck’s theory – Compton effect. Theory and experimental verification – Matter waves – Schrödinger’s wave equation – Time independent and time dependent equations – Physical significance of wave function – Particle in a one dimensional box - Electron microscope – Scanning electron microscope - Transmission electron microscope.

UNIT V CRYSTAL PHYSICS 9

Lattice – Unit cell – Bravais lattice – Lattice planes – Miller indices – d spacing in cubic lattice – Calculation of number of atoms per unit cell – Atomic radius – Coordination number – Packing factor for SC, BCC, FCC and HCP structures – NaCl, ZnS, diamond and graphite structures – Polymorphism and allotropy – Crystal defects – point, line and surface defects- Burger vector.

TOTAL PERIOD-45



BEYOND SYLLABUS LEARNING RESOURCES: TEXT BOOKS: 1. R. K. Gaur and S.C. Gupta, ‘Engineering Physics’ Dhanpat Rai Publications, New Delhi(2003) 2. M.N. Avadhanulu and PG Kshirsagar, ‘A Text book of Engineering Physics’, S.Chand and company, Ltd., New Delhi, 2005. REFERENCES: 1. Serway and Jewett, ‘Physics for Scientists and Engineers with Modern Physics’, 6 th Edition, Thomson Brooks/Cole, Indian reprint (2007) 2. Rajendran, V and Marikani A, ‘Engineering Physics’ Tata McGraw Hill Publications Ltd, III Edition, New Delhi, (2004). 3. Palanisamy, P.K., ‘Engineering Physics’ Scitech publications, Chennai, (2007). 4. Jayakumar. S, ‘Engineering Physics’, R.K. Publishers, Coimbatore, (2003). 5. Chitra Shadrach and Sivakumar Vadivelu, ‘Engineering Physics’, Pearson rEducation, New Delhi, (2007). ONLINE RESOURCES :

1. NPTEL TUTORIALS (Internal Server) 2. Online Objective Questions 3. Videos Materials if any (You tube)



CONTENTS

S.No Particulars Page

1 Unit – I 7

2 Unit – II 24

3 Unit – III 52

4 Unit – IV 71

5 Unit – V 97



Unit I Crystal Physics

Part –A 1.What are the co-ordination numbers for SC, BCC & FCC Structures? [CO1- L1-

Dec2012]

Co-ordination number is the number of nearest neighbouring atoms to a particular atom. The co-ordination number for SC is 6 , for BCC it is 8 and for FCC it is12.

2.Definepackingfactor(or)packingdensity(or)densityofpacking.Giveitsunit. [CO1-

L1- Dec2012]

It is defined as the ratio of the volume of atoms per unit cell to the total volume occupied by the unit cell.

APF=No. of atoms present in a unitcell X Volume of one atom

Volume of the unit cell

Since Atomic Packing factor is the ratio, it does not have any unit It is the smallest geometrical structure of a solid from the entire crystal AR

3.What are Miller Indices? [CO1- L1- Jan2011]

Miller Indices are the smallest possible integers which have the same ratios as the reciprocals of the intercepts of the plane concerned on the three axis.

4.What is meant by primitive and Non-primitive Cell ?Give an example. [CO1- L1- May2011]

A primitive cell is the simplest type of unit cell which contains only one lattice point per unit cell.

Example: Simple Cubic(SC).If there are more than one lattice point in an unit cell ,it is called Non-Primitive cell.

Example: BCC & FCC.



5.What are Bravais lattices? [CO1- L1- May2010,Jan 2012]

The 14 possible ways of arranging points in space lattice such that all the lattice points have exactly the same surroundings. These 14 lattices are called the Bravais lattices.

6.A unit cell has the dimensions a=b=c=4.74Å and α≠β≠γ≠90°, what is the crystal structure? [CO1- L1- May 2008]

For a=b=c=4.74Åandα≠β≠γ≠90°,the crystal structure is Trigonal (or) Rhombohedra.

7.Define atomic radius. [CO1- L1- Jan 2011]

It is defined as half of the distance between the nearest neighbours in a crystal of a pure element

8.Define Coordination number. [CO1- L1- Dec 2010] It is the number of nearest neighbouring atoms to a particular atom.

9.Draw the planes for Miller Indices (100), (110), and (111)[ CO1- L1- Dec 2014]



Part – B

1.Show that the atomic packing factor of FCC and HCP are the same.[ CO1- L1-

Dec 2013,Dec 2015]

Face Centered Cubic Structure (Fcc):

In this type of crystal structure, the unit cell has one atom at each corner of the cube and one atom at the center of six face.

Number of atoms per unit cell:

Atomic radius (r):

Atomic radius r =

Packing factor:

Atomic packing factor

Volume of the atoms per unit cell v = 4×

Volume of the unit cell V = a3

Atomic radius r =

Packing factor =

Substituting for and V, we have



=

PF =

PF =

PF ≈ 74 %

Thus, the packing factor is 74 % i.e., 74 % of the volume of the unit cell is occupied by atoms and the remaining 26 % volume of the cell is vacant. Hexagonal Close – Packed Structure (HCP):

A unit cell of a close – packed hexagonal structure is shown in fig.

In this type of crystal structure, the unit cell contains three types of atoms.

i) 12 corner atoms one at every corner of the hexagon. ii) 2 base centered atom, one at center of the top face and another at the center of the bottom face of the hexagon. iii) 3 atoms present in between the top and bottom face of hexagon and these atoms are situated inside the unit cell.

Hexagonal Close – Packed Structure

Here A, B and O are the lattice points and exactly above these atoms at a

perpendicular distance

the next layer atom lies at C.

In ABY, cos 30o =

AY = AB cos 30o =

AY =

---- (1)



Since X is the ortho centered of the triangle ABO,

Therefore, AX =

AY ---- (2)

Substituting for AY, we get

AX =

x

=

x

AX =

--- (3)

In the AXC, AC2 = AX2 + CX2 ---- (4) Substituting the values for

AC = a, AX =

And CX =

in equ (4), we get

a2 =

+

---- (5)

a2 =

+

a2 =

+

= a2

=

=

=

=

---- (6)

Taking square foot on both sides, we have

=

Or

=

= 1.633

1. Packing factor:

The atomic packing factor

Volume of all the atoms in a unit cell (v):

Atomic radius r =

Number of atoms per unit cell n = 6

= 1.633



Volume of all the 6 atoms in the unit cell, v = 6 x

πr3

Substituting for r =

v =

π

v = πa3

Volume of the unit cell (V): Area of the base = 6 x Area of triangle AOB

Area of triangle AOB =

(BO) (AY)

Substituting for BO = a and AY =

we have, =

× a ×

=

×

=

Area of the base = 6 ×

=

×

V =

Packing factor =

Substituting v and V, we have

PF =

Packing factor =

= 0.74 = 74%

Therefore, packing factor is 74% ie., 74% of the volume is occupied by the atoms and the remaining 26% volume is vacant. i.e., the packing factor of FCC and HCP are same.

Packing factor = 74%



2.Define Miller Indices. Show that for a cubic lattice the distance between two

successive plane ( h k l ) is given by

[ CO1- L1- Jan 2010, May

2012,Jan2013]

It is the set of three integers derived from the reciprocal of the intercepts of

the plane made on the three crystallographic axes. ( h k l )

Inter planar distance ( d spacing)

Consider a cubic crystal with ‘a’ as cubic edge and a plane ABC as shown in fig. Let (h k l) be the Miller indices of the plane ABC. OM = D1 be a perpendicular distance of the plane from the origin O. let this OM make angles α′, β′, γ′ with X, Y, and Z axes respectively.

The intercepts can be expressed as reciprocals of Miller indices

i.e., OA: OB: OC =

multiply side ‘a’ in all reciprocals.

Therefore OA =

OB =

OC =

From the right angled triangle OAM



From OBM

From OCM

The law of direction of cosines is

Substituting the values for , and we have

i.e., the distance between the origin and the first plane OM = D1 =

---- (1)

Let us consider the second plane A′B′ C′ parallel to the first plane ABC. Let D2 be the perpendicular distance of the plane from the origin O i.e., ON = D2

OA′: OB′: OC′ =

From the right angled triangle OAN

From OBN

From OCN

The law of direction of cosines is

Substituting the values for , and we have



i.e., the distance between the origin and the second plane ON = D2 =

---- (2)

Inter planar distance (d) Inter planar distance is the perpendicular distance between two parallel successive planes d. Let d is the distance between the two planes ABC and A′B′ C′ . Therefore we can write d = D2 – D1 ---- ( 3)

Substituting the equations (1) and (2) in equation (3)

We get d

d =

---- (4)

The equation (4) represents the inter planar distance. 3.Describe Czochralski method of graving crystal mention its advantages and

limitations .[ CO1- L1- Dec 2014,2015]

Basic principle The Czochralski method is a crystal pulling technique from the melt. The process is based on a liquid-solid phase transition driven by a seed crystal in contact with the melt. Basically, the seed is to be considered as a heat sink, by which the latent heat of solidification escapes and as a nucleation center: the solidified fraction at the surface of the seed will reproduce its single-crystal structure. By raising the seed slowly, a crystal is “pulled” from the melt. Decreasing the melt temperature makes the crystal diameter increasing and vice versa. Description and Working Large, single crystals of Si (for IC fabrication) are grown by the Czochralki method. It involves growing a single-crystal ingot from the melt, using solidification on a seed crystal as illustrated in fig. 1.36. Molten Si is held in a quartz (crystalline SiO2) crucible in a graphite susceptor. It is heated by radio frequency induction coil (RF heating). A small dislocation – free crystal, called a seed, is moved down to touch the melt and then slowly pulled out of the melt.



A crystal grows by solidifying on the seed crystal. The seed in rotated during the pulling stage, to obtain a cylindrical ingot.

Czochralski method for growing an ingot of silicon crystal. The cylindrical ingot forms as it is rotated and pulled slowly from the melt. Advantages

Growth from free surface

Growth of large oriented single crystals

Convenient chemical composition

Control of atmosphere Limitations

High vapor pressure materials

Liquid phase encapsulation

Possible contanmination of the melt by the crucible

No reproductivity of the crystal shape 4..Describe Bridgman Technique of growing crystal. Mention its advantages and

limitations. .[ CO1- L1- Dec 2013,Dec 2015]

A common technique for growing single crystals. It involves selective cooling of the molten material, so that solidification occurs along a particular crystal direction.



It this technique, the melt in a sealed crucible is progressively frozen from one end. this can be achieved by

Moving the cruicible down the temperature gradient (or)

Moving the furnance over the crucible (or)

By keeping both the furnance and the crucible stationary and cooling the furnance so that the freezing isotherm moves steadily through the originally molten charge.

Growth process Platinum crucible filled with high quality nature grown material, to which the desired impurities may be added. The crucible is fixed in the upper furnance until the contents are completely melted. It is then lowered from upper furnance into the lower furnance with help of electrical motor and reduction gearing. Since the pointed tip enters the lower furnance first, the grown material start to crystallize over there. As the crucible continues to be lowered, crystallization proceeds until all the melts become solid crystal. A bulk single pure crystal can be grown in the crucible by lowering the crucible at steady rate and keeping the temperature constant.

Fig. 1.37

Examples of the crystals grown by this method are Sodium chloride NaCl Potassium chloride KCl Calcium fluoride CaF2 Silver bromide AgBr



Advantages

Simple technique

Control over vapor pressure

Containers can be evacuated and sealed

Control of shape and size of growing crystals

Stabilization of thermal gradients

Limitations

Confinement of crystals

Crystal perfection is not better than that of the seed

No visibility

5.Describe the solution growth and vapour growth technique of growing crystals.

Mention its advantages and disadvantages?[CO1- L1- Dec 2012]

(i) Solution growth (ii) Vapour phase Epitaxy(VPE)

(i) Solution Growth

Low temperature solution growth The low temperature solution growth is suitable for the materials which decompose at high temperatures and undergo phase transformation below the melting point. There are two methods of low temperature solution growth. They are

i. Slow cooling method ii. Slow evaporatio0n method

i. Slow cooling method Slow cooling is the easiest method to grow bulk single crystals from solution. this technique needs only a vessel for the solution, in which the crystals grow. The temperature at which crystallization begin is in the range of 45˚C - 75˚C and the lower limit of cooling is the room temperature. ii. Slow evaporation method In this method, the saturated solution is kept at a particular temperature and provision is made for evaporation.



Manson Jar Crystallizer

The basic apparatus (Manson Jar Crystallizer) used for the solution growth technique is shown in Fig Typical growth conditions involve a temperature stabilization of about 0.05˚C and rate of evaporation of few mm3/h. Advantages

(i) This is a simple and convenient method of growing single crystals of large size. (ii) Growth of strain and dislocation free crystals. (iii) Permits the growth of prismatic crystals by varying the growth conditions. (iv) Only method which can be used for substances that undergo decomposition before melting.

Disadvantages (i) The growth substance should not react with the solvent. (ii) This method is applicable for substances fairly soluble in a solvent. (iii) Small crystals are also formed on the walls of the vessel near the surface of the liquid. These tiny crystals fall in the solution and hinder the growth of the crystal. (iv) A variable rate of evaporation may affect the quality of the crystal.

(ii) Vapour Phase Epitaxy (VPE) Epitaxial growth of silicon is exclusively carried out by VPE Construction The process is carried out in reaction chamber consisting of a long cylindrical quartz tube encircled by radio frequency induction coil as shown in fig.The silicon wafers are placed on a rectangular graphite rod called boat. This boat is placed in the reaction chamber and heated inductively to a temperature 1200˚C. The various gases required for the growth of desired epitaxial layers are introduced into the system through a control devices. The reaction is surface-catalysed and silicon is deposited on the wafer surface. However, the deposition temperature is very high. Also, as the reaction is reversible and can proceed in both directions.



Apparatus for growth of silicon epitaxial films Growth process Mostly expitaxial films with specific impurity concentration (P – type (or) N – type) are required. This is accomplished by introducing phosphine (PH3) for n – type doping and Biborane (B2H6) for p – type doping. There us a control console which permits only the required gases at suitable pressure so that it is possible to form an almost step p – n junction layer by this process. Vapour Phase Epitaxy (VPE) of compound semiconductor, for example, GaAs is performed using gallium (Ga) and Arsenic Chloride (AsCl3) as source materials Advantages

Halogen based VPE produces high pure single crystal layers with limited defects.

Hydride based VPE allows proper control over the ratio of group III to group V vapour phase species.

Disadvantages

Higher growth temperature results in higher impurity diffusion. So uniform distribution of impurities cannot be obtained.

Application

This method is mainly used in the manufacture of LED’s and detectors.



6.Write a short note on Diamond Cubic and Graphite structure. .[ CO1- L1- Dec

2010,Jan 2011]

Diamond :

The diamond cubic structure is a very important crystal structure.

In a diamond the carbon atoms are arranged tetrahedrally. Each carbon atom is attached to four other carbon atoms 1.544 Å meter away with a c-c-c band angle of 109.5°.

It is strong, rigid three-dimensional structure that results in an infinite network of atoms. This accounts for diamond’s hardness, extraordinary strength and durability and gives diamond ahigher density than graphite.

Because of its tetrahedral structure, diamond also shows a great resistance to compression. Diamond will scratch all other materials and is the hardest material known.

It is the best conductor of heat conducting upto five times the amount that copper does.

Diamond also conducts sound.

It is an electrical insulator.

Moreover, diamond disperse light. The greater the dispersion, the better the spectrum of colors that is obtained. This property gives rise to the “fire” of diamonds.



The “brilliance” of diamonds stems from a combination of refraction, total internal reflection and dispersion of light.

Since the packing density is low, it is termed as very loosely packed structure.

Graphite:

The carbon atoms in graphite are also arranged in an infinite array, but they are layered. These atoms have two types of interactions with one another. In first, each carbon atom is bonded to three other carbon atoms by covalent bonding with a spacing of about 1.42 Å.

These planar arrays are held together by weak vander waals force known as stacking interactions. The distance between two layers is longer (3.4 Å. these three dimensional structure accounts for physical properties of graphite. Unlike diamond, graphite can be used as a lubricant or in pencils because the layers cleave readily.

It is soft and slippery, and its hardness is less than one.

Graphite also has a lower density than diamond.

The planar structure of graphite allows electrons to move easily within the planes. This permits graphite to conduct electricity and heat as well as absorb light and, unlike diamond, appear black in color.

When the graphite is mixed with oil , it is called oil dag which is used in internal combustion engines.



When graphite is mixed with oil it is called aqua dag, which is used as an oil free lubricant, in food stuff industry.



Unit II Properties Of Matter And Thermal Physics

PART - A 1.What is elasticity? [ CO2- L1- Jan2012]

The property of the body to regain its original shape and size, after the removal of

deforming force is called elasticity.

2.What are elastic bodies?[ CO2- L1- Jan2013]

Bodies which regain its original shape and size after the removal of deforming force

are called elastic bodies.

3.Define stress and its unit.[CO2- L1- May2012]

The restoring force acting per unit area of the body is called the stress.

This restoring force is equal and opposite to the applied force F. Therefore, stress is also defined as the applied deforming force per unit area of the body.

i.e., Stress =

SI unit of stress =

4.What are types of stresses? [CO2- L1- Nov 2008]

It is found that a deforming force may change length or shape or volume of the

body. Accordingly, there are three types of stresses namely (i) Linear stress (ii) Shearing stress (iii) Volume stress.

5.Define strain and its unit. [ CO2- L1- Jan2012

The change in dimension or shape of a body due to the deforming force results in

strain.

It is define as the ratio of change in dimension to the original dimension. i.e.,

Strain =

Strain has no unit.



6.What are types of strains?[ CO2- L1- Jan2012]

According to the changes take place in length, area (Shape) and volume, there are

three types of strains namely,

a) Linear strain (Change in length per unit length) b) Shearing strain (Change in area per unit area) c) Volume strain (Change in volume per unit volume)

7.State Hooke’s law. [ CO2- L1- May2012]

It states that “within elastic limit, the stress developed in the body is directly

proportional to the strain produced in it”.

Stress Strain

Stress = Constant x Strain

In other words, the ratio between stress and strain is a constant. 8.What are types of moduli of elasticity? [ CO2- L1- Jun2011]

i) Young’s modulus corresponding to linear strain. ii) Rigidity modulus corresponding to shearing strain. iii) Bulk modulus corresponding to volume strain.

9.Define Young’s modulus of elasticity and mention its unit. [ CO2- L1- Dec2013]

Within the elastic limit, the ratio of linear stress to linear strain is called Young’s

modulus of elasticity.

Young’s modulus of elasticity (Y) =

Unit: SI unit of stress is and strain has no unit. Therefore, SI unit of

Young’s modulus is



10.Define rigidity modulus and mention its unit. [ CO2- L1- Jun2013]

With in the elastic limit, the ratio of the tangential stress to shearing strain is called rigidity modulus.

It is denoted by the letter n,

Rigidity modulus (n) =

Unit: SI unit of rigidity modulus is

11.Define bulk modulus and mention its unit. [ CO2- L1- Jun2010]

“With in the elastic limit of a body, the ratio of volume stress to volume strain is called bulk modulus of elasticity”.

It is denoted by the letter K

Bulk modulus (K) =

Unit: SI unit of bulk modulus is

12.What is stress-strain diagram? [ CO2- L1- Jan2011]

A graph plotted between strain along the X-axis and stress along the Y-axis is known as stress-strain diagram.



13.What is the use of stress-strain diagram? [ CO2- L1- Jan2012]

The elastic behavior of solid materials are studied by using this stress-strain diagram.

14.Define bending moment of a beam. [ CO2- L1- May201

The moment of the couple due to the static reactions (Restoring couple) which

balances the external couple due to the applied load is called bending moment.

15.What is uniform bending? [ CO2- L1- Jn2011]

The beam is loaded uniformly on its both ends, the bent beam forms an arc of a circle.

The radius of curvature of the bent beam is constant for give load this type of bending

is called uniform bending.

16.What is I shape Girder? [ CO2- L1- Dec2009]

A girder is a metallic beam supported at its two ends by pillars or on opposite walls. It

should be so designed that it should not bend too much or break under its own weight.

The cross section of beam is in the form of letter I.

.

Thermal physics

1.Define coefficient of thermal conductivity and mention its unit. [ CO2- L1-

Dec2010]

It is defined as the quantity of heat conducted per second normally across unit area of

cross-section per unit temperature difference per unit length of the material.

Its unit is watts/metre/Kelvin



2.Derive the unit in which thermal conductivity is measured. [ CO2- L1- APR2009]

Thermal conductivity of material

=

=

3.What is basic principle behind Lee’s disc method in determining the real

conductivity of bad conductor? [ CO2- L1- Jan2012]

The given bad conductor is taken in the form of disc is placed in between the metal disc (Lee’s disc) and steam chamber. The steam I passed through the steam chamber. Heat conducted through bad conductor per second is calculated. Amount of heat lost per second by disc is also calculated. At steady state

From this, thermal conductivity of the bad conductor is calculated.



4.How are heat conduction and electrical conduction analogous to each other? [

CO2- L1- Dec2008]

Heat conduction Electrical conduction

1 Heat is conducted from a point of higher temperature to a point of lower temperature.

Electricity is conducted from a point at higher potential to a point at lower potential.

2 In metals, heat conduction is mainly due to free electrons.

In metals, electrical conduction is due to free electrons.

3 The ability to conduct heat is measured by thermal conductivity.

The ability to conduct electricity is measured by electrical conductivity.

5.What are the characteristics of good and bad conductors? [ CO2- L1- Nov2011]

Good conductors Bad conductors

1 They have high electrical & thermal conductivity

They have very low electrical and thermal conductivity

2 They can be easily heated or cooled They cannot be easily heated or cooled

3 Examples: Metals like iron, copper Examples: Non metals like glass, wood

6.What is thermal resistance? [ CO2- L1- Jan2012]

The thermal resistance of a body is a measure of its opposition to the flow of heat through it.



7.What is radial flow of heat? [ CO2- L1- Jan2011]

In this method, heat flows from the inner side towards the other side along the radius of the spherical shell or cylindrical shell. This method is interesting because there is no loss of heat as in the other methods.

8.Explain why the specimen used to determine thermal conductivity of a bad

conductor should have a larger area and smaller thickness. [ CO2- L1- Apr2013]

For a bad conductor with a small thickness and large area of cross-section, the amount

of heat conducted increases (Large). This will increase accuracy of the measurement.

9.Write down an expression for the amount of heat conducted through a

compound media of two layers. [ CO2- L1- Jan2009]

Amount of heat conducted through a compound media of two layer is

Q =

A – Area of the cross section Temperature at outer face of the material at hot end

Temperature at the outer face of the material at cold end

Thickness of the material at hot end

Thickness of the material at cold end

Thermal conductivity of the material at hot end

Thermal conductivity of the material at cold end

11.Mention the methods to determine thermal conductivity of good and bad

conductors. [ CO2- L1- Jan2012]

Searle’s method – for good conductors like metallic rod

Forbe’s method – for determining the absolute conductivity of metals

Lee’s disc method – for poor conductors

Radial flow method – for bad conductors



12.Define Newton’s law of cooling. [ CO2- L1- Jan2014]

It states that the rate at which a body loses heat is directly proportional to the

temperature difference between the body and that of the surroundings.

PART B

1.Factors affecting elasticity. [ CO2- L1- Jan2012]

The following factors affect the elasticity of the materials.

They are 1. Effect of stress 2. Effect of change in temperature 3. Effect of impurities 4. Effect of hammering, rolling and annealing 5. Effect of crystalline nature 1. Effect of stress:

The application of large constant stress or repeated number of cycles of stress acting on a body decreases the elasticity of the body gradually. (Fig 2.8)

Fig 2.8 Elasticity decreases due to application of large constant stress



2. Effect of change in temperature: A change in temperature affects the elastic properties of a material. A rise in temperature usually decreases the elasticity of the material. A carbon filament which is highly elastic at normal temperature becomes plastic when it is at high temperature. Similarly, a decrease in temperature will increase the elastic property. Lead is not a very good elastic material. But at low temperature, it becomes a very good elastic material. However, in some cases like the invar steel, the elasticity is not affected by any change in temperature.

3. Effect of impurities: The elastic property of a material is either increased or decreased due to the addition of impurities (Fig 2.9). It depends upon the elastic or plastic properties of the impurities added. If minute quantities of carbon is added with molten iron, the elastic properties of iron are increased enormously. If more carbon is added, its elastic properties are decreased.Similarly the addition of potassium or copper in gold increases the elastic properties of gold. Fig 2.9 Addition of impurities



4. Effect of hammering, rolling and annealing:

Operations like hammering (Fig 2.10) and rolling (Fig 2.11) help in breaking up the crystal grains into smaller units and results in an increase of their elastic properties. Fig 2.10 Hammering Fig 2.11 Rolling

Operations like annealing (i.e., heating and then cooling gradually) help in forming larger crystal grains. (Fig 2.12) Thus, there is decrease in their elastic properties. Fig 2.12 Annealing

5. Effect of crystalline nature: For a given metal, the modulus of elasticity is more when it is in single crystal form and in polycrystalline state, its modulus of elasticity is comparatively small.

Beam: A beam is a rod or bar of uniform cross-section (Circular or rectangular) whose length is very much greater than its thickness. (Fig 2.13)



Fig 2.13 Beam Bending of Beam: Consider a beam which is bent into an arc of a circle by the application of a load. This beam is made up of a large number of thin plane layers one above the other.

Taking a longitudinal section ABCD of the bent beam, the layers in the upper half are elongated while those in the lower half are compressed. (Fig 2.14)

Fig 2.14 Bending of a beam

In the middle, there is a layer (MN) which is not elongated or compressed due to bending of the beam. This layer is called ‘neutral surface’ and the line (MN) at which the neutral layer intersects the plane of bending is called ‘neutral axis’. It is found that the length of the layers increases or decreases in proportion to its distance away from the neutral axis MN. The layers below MN are compressed and those above MN are elongated. There are a pairs of layers one above MN and one below MN experiencing same forces of elongation and compression due to bending. Each pair of layers forms a couple. This couple is known as internal couple. The resultant of the moment of all these internal couples are called ‘internal bending moment’.



In equilibrium condition, internal bending moment is equal to external bending moment.

1. Bending moment of a beam. Consider a portion ABCD of a bent beam as shown in fig 2.15

Fig 2.15 Bending moment of a beam

P and Q are two points on the neutral axis MN. R is the radius of curvature of the

neutral axis and is the angle subtended by the bent beam at its centre of curvature O. i.e.,

Consider two corresponding points and on a parallel layer at a distance x

from the neutral axis. From the figure 2.15,

PQ = R x … (1) Corresponding length on the parallel layer

Increase in length of

=

= … (2)

Linear strain produced =

= … (3)

If Y is Young’s modulus of the material, then

Y =

Linear stress = Y x Linear strain



= … (4)

If A is the area of cross-section of the layer, then

Force acting on the area A = stress x area

= A

Moment of this force about the neutral axis MN

=

The sum of the moments of forces acting on all the layers

=

=

=

= I is called geometrical moment of inertia of the cross-section of the

beam. The sum of moments of forces acting on all the layers is the internal bending moment which comes into play due to elasticity.

Thus, internal bending moment of the bema =

Note: For a rectangular beam of breadth b and depth (Thickness) d, the geometrical moment of inertia is given by Similarly, for a beam of circular cross section, Where r is the radius of the rod.



2.Depression of a cantilever. [ CO2- L1- Dec 2011]

Cantilever: Definition: It is a beam fixed horizontally at one end and loaded at the other end. Expression for depression produced in the cantilever: Consider a cantilever of length I fixed at the end A and loaded at the free end B by a weight W. the end B is depressed to B’ (Fig 2.16) AB is the neutral axis. BB’ represents the vertical depression at the free end.

Fig 2.16 Depression at the free end of the cantilever. Consider the section of the cantilever P at a distance x from the fixed end A. it is at a distance (l – x) from the loaded end B’. Considering the equilibrium of the portion PB’, there is a force of reaction W at P. External bending moment = W x PB’ = W (l – x)

Internal bending moment =

Where Y – Yong’s modulus of the cantilever. I – Geometrical moment of inertia of its cross-section. R – Radius of the curvature of the neutral axis at P. In the equilibrium position, External bending moment – Internal bending moment

W … (1)



Q is another point at a distance from P

ie., PQ =

‘O’ is the centre of curvature of the are PQ PO = R and

Then, … (2)

` The tangents are drawn at P and Q meeting the line BB’ at C and D. Vertical depression CD = … (3)

From the eqns (2) and (3), we have

… (4)

Substituting for R in eqn. (1), we have

W

W … (5)

Rearranging eqn (5), we have

… (6)

Total depression (= BB’) at the free end is

… (7)

… (8)

Determination of Youngs modulus of the cantilever:

Young’s modulus of the cantilever is determined using the depression produced in the cantilever. Depression at the free end of a cantilever



Y … (9)

For a beam of rectangular cross-section,

where b is breadth and d thickness of the beam

The weight W = Mg where M is the mass suspended at the free end and g is acceleration due to gravity. Substituting for W and I in eqn (9), we have

Y

Young’s modulus From above expression Y is determined. Experimental determination of Young’s modulus of a cantilever: The given bar is fixed rigidly at one end and a weight hanger is suspended at the other end (Fig 2.17) A pin is fixed vertically (using wax) at the free end of the beam. A travelling microscope (T) is focused on the pin. The microscope is adjusted such that the horizontal cross-wire coincides with the tip of the pin. The initial reading in the microscope on the vertical scale is noted. Fig 2.17 Experiment to find Young’s modulus of the cantilever uding depression A suitable mass M is placed on the hanger. The reading in the microscope is again noted. The difference between two readings of the microscope gives the depression y corresponding to load M



The experiment is repeated by increasing the values of M in steps of 50gms. Then, the experiment is also repeated by decreasing the weights. The observations are tabulated as follows:

From these observations, mean depression y corresponding to each value of M is obtained. The length l of the beam, its breadth b (by vernier calipers) and thickness d (by screw gauge) are measured. Young’s modulus of the beam is determined by using the relation.

Load

Microscope readings for depression (y) Mean depression

‘y’ for a load of M kg.

Load increasing

Load decreasing

Mean

Gm Cm Cm Cm Cm

W

W + 50

W + 100

W + 150

W + 200

W + 250

Mean y



3.What is Uniform bending ? Derive an expression for elevation at the centre of a

beam which is loaded at both ends. Describe an experiment to determine the

young’s modulus of a beam by uniform bending (Theory and experiment). [ CO2-

L1- May 2015]

Definition: If the beam is loaded uniformly on its both ends, it forms an arc of a circle. The elevation is produced in the beam. This type of bending is known as uniform bending. Theory of uniform bending: Consider a beam AB arranged horizontally on two knife-edges C and D symmetrically so that AC = BD = a (Fig 2.18) The beam is loaded with equal weights W at each ends A and B. The reactions on the knife edges at C and D are equal to W and they are acting vertically upward.

Fig 2.18 uniform bending External bending moment on the part AF of the beam is W x AF – W x CF = W (AF – CF) = W x AC = W x a = W a … (1)

Internal bending moment = … (2)

Y – Young’s modulus of the beam I – Geometrical moment of inertia of the beam R – Radius of curvature of the beam at F In the equilibrium position, External bending moment = Internal bending moment



W a = … (3)

Since for a given value of W, the values of a, Y and I are constant. R is constant so that the beam bends uniformly into an arc of a circle of radius R. CD = l and y is the elevation of the midpoint E of the beam so that y = EF Then, from the property of a circle (Fig 2.19) EF x EG = CE x ED … (4) EF (2R – EF) =

( CE = ED and EG = 2R – EF)

Fig 2.19 Intersecting chord theorem of a circle

Or … (5)

Substituting the eqn (5) in (3), we get,

If the beam is of rectangular cross-section, then where b is breath and d

is thickness of beam. If M is the mass, the corresponding weight W = Mg,



Then, Y

From which Young’s modulus of the beam is determined. Experiment: A rectangular beam AB of uniform-section is supported horizontally on two knife-edges A and B (Fig 2.20) Two weight hangers of equal masses are suspended from the ends of the beam. A pin is fixed vertically at the mid-point of the beam. A microscope is focused on the tip of the pin.

Fig 2.20 Young’s modulus – uniform bending Initial reading in the microscope in the vertical scale is noted. Equal weights are added to both hangers simultaneously and the reading in the microscope on the vertical scale is noted. The experiment is repeated for decreasing order of the equal masses. The observations are tabulated and mean elevation (y) at the mid point of the bar is determined. The length of the bar between the knife edges ‘l’ is measured. The distance of the one of the weight hangers from the nearest knife edge ‘a’ is measured. The breadth (b) and thickness (d) of the bar are measured by using vernier calipers and screw gauge.

Y



Young’s modulus of the beam is determined by using the relation.

2. THERMAL PHYSICS

1.Describe with the theory Lee’s disc method for determination of thermal

conductivity of bad conductors. [ CO2- L1- May 2016]

1. The thermal conductivity of bad conductors like ebonite or card board is determined by this method.

Description: The apparatus consists of a circular metal disc or slab C (Lee’s disc) suspended by the strings from a stand (fig 3.6) The given bad conductor (Such as glass, ebonite) is taken in the form of a disc (D). This bad conductor has the same diameter as that of the slab and it is placed on the slab

Load

Microscope readings for elevation

Mean elevation ‘y’ for a load of M kg.

Load increasing

Load decreasing

Mean

Gm Cm Cm Cm Cm

W

W + 50

W + 100

W + 150

W + 200

W + 250

Mean y



Fig 3.6 Lee’s disc method A cylindrical hollow steam chamber (A) having the same diameter as that of the slab is placed on the bad conductor. There are holes in steam chamber and slab into which thermometers and are inserted to record the respective

temperatures. Working:

Steam is passed into the steam chamber unit the temperatures in the chamber and the chamber and the slab are steady. When thermometers how steady temperatures, their readings and are noted. The radius ® and thickness (d)

of the disc D are also measured.

i) Observation and calculation: Thickness of the bad conductor = d Radius of the bad conductor = r Mass of the slab (Lee’s disc) (C) = M Steady temperature in the slab =

Steady temperature in the steam chamber =

Thermal conductivity f the bad conductor = K Rate of cooling of the slab at = R

Specific heat capacity of the slab = S Area of the cross section A =

Amount of heat conducted through the disc D per second

Q = … (1)



At this stage all the heat conducted through the bad conductor is completely radiated by the bottom flat surface and the curved surface of the slab C. Amount of heat lost per second by the slab C

Q = Mass x Specific heat capacity x Rate of cooling Q = MSR … (2) At steady state,

Hence, the equations (1) and (2) are equal

… (3)

ii) Determination of rate of cooling R: The bad conductor is removed and the steam chamber is placed directly on the slab. The slab is heated to a temperature of about higher than The steam

chamber is removed and the slab alone is allowed to cool. As the slab cools, the temperatures of the slab are noted at regular intervals of time (0.5 minute) until the temperature of the slab falls to about below

The temperature-time graph is drawn and the rate of cooling at the steady

temperature is determined. Fig 3.7

Fig 3.7 Graph between temperature-time



During the first part of the experiment, the top surface of the slab is covered by the bad conductor. Radiation is taking place only from the bottom surface area and curved surface area of the slab. i.e., Total area =

Where h is the height of the slab C. In the second part of the experiment, heat is radiated from top surface area, bottom surface area and curved sides. i.e., over entire surface area

As the rate of cooling is directly proportional to the surfaces that are exposed (Other condition being equal)

R = … (4)

Substituting for R in equation (3), we have … (5) Thus, thermal conductivity of the bad conductor is determined.

2.Describe the radial flow of heat. [ CO2- L1- May 2016]

In this method, heat flow from inner side towards other side along the radius of the cylindrical shell. This method is interesting because there is no loss of heat as in the other methods.

Cylindrical shell method:

Consider a cylindrical tube of length I, inner radius and outer radius as

shown in fig 3.8. This tube carries steam or some hot liquid. Heat is conducted radially across the walls of the tube. After the steady state is reached, the temperature on the inner surface is and on the outer surface is



This cylindrical tube is imagined to consist of a large number of thin co-axial cylinders of increasing radius. Any such thin imaginary cylinder of thickness ‘dr’ at a distance r form the axis of the cylinder is taken. Fig 3.8 Cylindrical shell method Amount of heat flowing per second through this imaginary cylinder

Q = -KA … (1)

Surface area of the imaginary cylinder A =

Substituting for A in eqn (1), we have

Q = … (2)

After steady state is reached, the amount of heat flowing (Q) through all the imaginary cylinders is same. Rearranging the equation (2), we get

Integrating both sides between their proper limits, we have

=

Q =



3.Describe a method to determine thermal conductivity of rubber. [ CO2- L1-

Jun2011]

Principle:

It is based on the principle of radial flow of heat through a cylindrical shell. Procedure: A bi empty calorimeter with stirrer is weighed It is then filled with two-thirds

of water and again weighed A known length of a rubber tube is

immersed in water in the calorimeter (Fig3.9) The calorimeter is stirred well and the initial temperature is noted. Now, one

end of the rubber tube is connected to a steam generator and steam is passed through it.

The steam is passed continuously till there is a rise of in temperature. The

time taken (t second) for rise in temperature is noted. The final temperature of the water in the calorimeter is also noted.



Fig 3.9 Thermal conductivity of rubber

Observation: Mass of the empty calorimeter with stirrer =

Mass of the calorimeter + water =

Mass of the water =

Initial temperature of the water =

Final temperature of the water =

Rise in the temperature of water =

Time for which steam is passed = t Length of the rubber tube immersed in water = l

Inner radius of the rubber tube =

Outer radius of the rubber tube =

Specific heat capacity of calorimeter =

Specific heat capacity of water =

Heat gained by the calorimeter = Mass x Specific heat capacity x Change in temperature = … (1)

Heat gained by the water = ( ) x … (2)

Total heat gained by the calorimeter and water in t second

Q =

Q = … (3)

The expression for the thermal conductivity (K) in the case of cylindrical shell method is given by

… (4)

Where - Temperature of inner surface of cylindrical shell (Rubber tube) in this case

steam temperature



- Temperature of outer surface of cylindrical shell (Rubber tube) in this case

average temperature of water in the calorimeter ie.,

Substituting eqn (3) in eqn (4), we have

Q =



Unit – III

Quantum Physics Part -A

1.List the physical significance of the wave function? [ CO3- L1- Dec2015]

The wave function Ψ has no direct physical meaning. It is a complex quantity representing the matter wave of electron. It connects particle nature and its associated wave nature statistically. Ψ Ψ*or |Ψ|2 is the probability density function.

Ψ* dτ = 1 if particle is present within the space

Ψ* dτ = 0 if particle is not present within the space

Ψ* dτ = in between 0 and 1 probability of presence in percentage

within the space.

2.What is Compton Effect? Write an expression for Compton Effect? [ CO3- L1-

Dec2014]

When a beam of monochromatic X – rays passes through matter, the scattered beam consists of two types of radiations. One is the primary radiation of same wavelength and other is a modified radiation of slightly longer wavelength. This effect is called as Compton Effect. The shift in wavelength is called as Compton shift. 3.What is quantum theory of light? (or) State Plank’s quantum theory? [ CO3- L1- Dec2012] Quantum theory of light was proposed by Max Plank, in which he shows particle

nature of electromagnetic radiation. Electromagnetic radiation travels in the form of

particles called as quanta. The energy of each quanta E = h ν where h is Plank’s

constant and ν is the frequency of radiation. This theory successfully explains black

body radiation, Compton Effect and photo electric effect etc.



4.What is de – Broglie’s hypothesis? [ CO3- L1- Dec2011]

According to de – Broglie’s wavelength a moving particle is always associated

with waves.

Waves and particles are the two modes through which energy can propagate in

nature.

Since nature loves symmetry, so matter and waves must be symmetric.

The radiation like light can act as a wave as well as particle. Therefore mater should

act as a particle and same time it act as a wave.

It is given by λ=h/mv.

5.What is a perfect black body give an example? [ CO3- L1- Dec2010]

A perfect black body can absorb and emit all electromagnetic radiations. A metallic box with a small opening, coated with carbon black inside the box act as a black body. Radiations sent inside the box through the opening will get absorbed completely. When the box is heated the radiations come out of the opening. 6.What is Wien’s displacement law? [ CO3- L1- Dec2014]

This law states that, the product of the wavelength corresponding to maximum energy

(λm) and absolute temperature is constant.

ie, λm T= constant Wien also showed that the maximum energy Em is directly proportional to the fifth power of the absolute temperature. Em α T 5 Em = const T 5 This law holds good only for short wavelengths and not for longer wavelengths.



7.What is Rayleigh – Jean’s law? ? [ CO3- L1- Dec2013]

This law states that the energy distribution is directly proportional to the absolute temperature (T) and inversely proportional to the fourth power of the wavelength (λ4)

ie Eλ = (8πkT / λ4 ) where k is the Boltzmann’s constant This law holds good only for longer wavelength regions and not for shorter wavelengths.

Part - B Explain Planck’s quantum hypothesis. Derive Planck’s law for black body

radiation and hence deduce Wien’s law and Rayleigh Jean’s law. ? [ CO3- L1-

Dec2009,2012]

1. Planck’s Quantum hypothesis 1. A black body radiation chamber contains large number of oscillating particles. 2. An oscillator cannot absorb or emit energy in a continuous manner. It can absorb or emit energy in the multiples of small unit called quantum This quantum of radiation is called photon. Energy of the photon = , Where h = Planck’s constant.

3. An oscillator vibrating with frequency can emit energy in multiples of

quantas

ie, ε =

= 1, 2, 3, …….. this indicates that oscillators have discrete

energy values. Assume a black body consists of large number of oscillators

Average energy of an oscillator = (1)

Where E = Total energy N = Number of oscillators Number of atomic oscillators in the ground state = N0

From Maxwell’s energy distribution law, the number of oscillators having Energy is given by

Nn = N0 (2)

Let N be the total number of oscillators and N0, N1, N2, N3 be the number of oscillators with energies , , , ……

N = N0 + N1 + N2 + ……..



N = N0 + N0 + N0 + …… (3)

From quantum theory

=

ie, = 0, = , = , …..

Substituting these values in (3)

N = N0 + N0 + N0 +….

Putting x =

N = N0 + N1x + N2x2 + ……..

N = N0 [1 + x + + ……]

N = (4)

Total energy E = ……..

ie E = 0 × N0 + N0 + 2 N0 +…..

Putting x =

E = N0x + 2 N0x2 +……..

E = N0x[ 1 + 2x + 3x2 + …….]

E = (5)

Substituting (4) and (5) in (1) we get

=

ie =

=

=

Substituting x =

= (6)

Number of oscillators per unit volume in the wavelength range and is

given by,

(7)



The energy density of radiation between and is given by,

=

ie, =

=

=

ie, = (8)

This equation represents Plank’s radiation law. Deduction of Wien’s displacement law When is very small, is very large

Hence >> 1

ie,

= (9)

Equation (9) represents Wien’s displacement law. ie Planck’s law reduces to Wien’s law at smaller wavelengths Deduction of Rayleigh – Jean’s law When is very large is small

ie, << T

ie, = 1 +

= .

=

= (10)



This equation (10) represents Rayleigh – Jeans law. Therefore, Planck’s law reduces to Rayleigh – Jeans law at longer wavelength. Hence Planck’s radiation law holds good for the entire black body radiation spectra. 2.What is Compton Effect? Derive an expression for the wavelength of scattered

X – ray photon from a material. [ CO3- L1- Dec2013,2014]

Compton Effect:

When a beam of monochromatic X – rays passes through matter, the scattered beam consists of two types of radiations. One is the primary radiation of same wavelength and other is a modified radiation of slightly longer wavelength. This effect is called as Compton Effect. The shift in wavelength is called as Compton shift.



A monochromatic electromagnetic radiation hit on the rest electron and deflected at angle of θ with less energy. The rest electron recoils with a velocity v and making an angle of φ. θ and φ are calculated with respect to direction of collision. m0 = rest mass of electron m = mass of the recoil electron

= frequency of the incident radiation

= frequency of the scattered radiation

h = Plank’s constant Energy of the system (photon & electron) before collision c2

Energy of the system after collision

Applying the principle of conservation of energy, Energy before collision = Energy after collision

c2

c2 --------------------------------- (1)

Applying principle of conservation of momentum along the direction of incidence,

-------------------------------- (2)

Applying principle of conservation of momentum perpendicular to the direction of incidence,

----------------------------------------- (3)

Squaring and adding the equations (2) & (3)

= h

h

-------------------------------- (4)

Squaring the equation (1)

---------- (5)

Subtracting the equations (5) – (4)

--------- (6)

According to theory of relativity



Multiplying by on both sides

---------------------------- (7)

Comparing the equations (6) & (7)

(1-cosθ)

(1-cosθ)

Multiplying by ‘c’ on both sides

(1-cosθ)

λ’- λ = (1-cosθ)

λ’ = λ+ (1-cosθ)

Is the wavelength of the scattered radiation.

dλ = (1-cosθ)

The change in wavelength d� is also known as Compton wavelength. d� is independent of incident radiation and the nature of the scatter. It only depends on the angle of scattering. CASES:

1. When θ = 0, dλ= 0

2. When θ = , dλ = = 0.024 Å

3. When θ = π, dλ= = 0.048 Å



3.Derive schroedinger’s time independent and time dependent wave equations? [

CO3- L1- Dec2014,2015]

Schroedinger describes the mathematical form and is known as schroedinger wave

equation.

There are two types of wave equations

(i) Time independent wave equation (ii) Time dependent wave equation

Shroedinger Time independent wave equation Consider a system of stationary waves associated with a particle, let x, y, z be the coordinates of the particle and is the wave displacement for the de-Broglie’s wave at

any time‘t’. The classical differential equation of a wave motion is given by

+ + = (1)

ie = (2)

where = + +

The solution of equation (2) is given by,

(x, y, z, t) = (x, y, z)

ie = (3)

here is a function of x, y, z only and it gives the amplitude at a point.

differentiating equation (3) w.r.to. t, we get

=

again differentiating w.r.to. t

=



= (4)

Substituting (4) equation (2) we have

=

+ = 0 (5)

the angular frequency = 2

= 2 (6)

Substituting (6) in (5), we get,

+ = 0 (7)

here the value of . Substituting in equation (7)

+ = 0

+ = 0 (8)

If E is the total energy of the particle, V is the potential energy and ½ mv2 is kinetic energy, then Total energy (E) = Potential energy (V) + Kinetic energy (1/2 mv2) ie E = V + ½ mv2 mv2 = 2[E – V] m2v2 = 2m [E – V] (9)

Substituting (9) in (8)

+ (E – V) = 0

Put =

2 =

ie equation (9) changes to,

+ (E – V) = 0

This equation is known as schroedinger time independent wave equation SCHROEDINGER TIME DEPENDENT WAVE EQUATION From schroedinger time independent equation we have,

= (1)



differentiating equation (1) w. r. to t we get

=

=

=

=

=

Multiplying i on both sides

i =

i =

or E = i (2)

Schrodinger time independent wave equation is given by

+ (E – V) = 0 (3)


+ = 0

=

+ =

=

or H = E (4)

where H is Hamiltonian operator

E is energy operator

This equation (4) is known as schroedinger time dependent equation.



4.Apply schroedinger wave equation to a particle in one dimensional box and

calculate the eigen value and eigen function. [ CO3- L1- Dec2012,May2015]

Consider a particle of mass m moving between two rigid walls of a box at x =0 and x = a along x – axis fig i The potential energy (V) of the particle inside the box is constant and it is zero The walls of the box are infinitely high, and hence the potential energy ‘V’ is infinite outside the walls

Thus we can write the boundary condition as, V(x) = 0 for 0 <x<

V(x) = for 0

The particle cannot come out of the box Also, it cannot exist on the walls of the box. So the wave function

= 0 for 0



To find the wave function of the particle inside the box, let us consider schroedinger time independent one demanding equation.

ie + (E – V) = 0 (1)

Since V = 0 between the walls equation (1) reduces to

+ = 0 (2)

Putting = k2 in equation (2) we get,

+ k2 = 0 (3)

The general solution of equation (3) is given by,

(x) = A sin kx + B coskx (4)

Here A and B are unknown constants. The values of A and B can be obtained by applying the boundary conditions.

Boundary condition (i) = 0 at x = 0

ie, (4) reduces to 0 = A sin 0 + B cos 0 0 = 0 + B Hence B = 0 Boundary condition (ii) = 0 at x = a

Applying this condition in (4) we have, 0 = A sink a + 0 A sin k a = 0 It is found that, either A = 0 or sink a = 0 A cannot be 0 since already B = 0 sinka = 0 Sin ka is 0 when ka = n

Where n = 1, 2, 3, …….

or k = (5)

on squaring (5) we have

k2 = (6)

We know that k2 = (7)

Comparing (6) and (7)

=



= ×

Energy of the particle

En = (8)

Substituting equation (5) in equation (4) we have

=A sin (9)

From (8) it is clear that, for each value of n, there is an energy value. ie the particle in a box cannot possess any arbitrary amount of energy values. In other words its energy is quantized. Each value of En is known as eigen value and the corresponding is called as eigen function.

Normalization of wave function The constant A is determined by normalization. Probability density is given by

We have = A sin

= A sin × A sin

ie, = A2 sin2 (10)

It is certain that the particle is inside the box Thus the probability of finding the particle is given by

=1 (11)


=1

A2 = 1

A2 = 1

A2 = 1

A2 = 1

= 1

A2 =



A = (12)

Substitute (12) in (9) we have

= sin (13)

The expression (13) is known as normalized eigen function. The energy (En) and normalized wave function are shown in fig ii.

5.Describe the principle, construction and working of transmission electron

microscope? [ CO3- L1- Dec2013]

Transmission Electron Microscope (TEM) In this microscope, the image is obtained by transmitting the electrons through the specimen.

Principle:-

The electron are allowed to pass through the specimen and the image is formed on the fluorescent screen either by using transmitted electron beam (bright field image) or by using diffracted electron beam (Dark field image) from the specimen



Construction

It consists of an electron gun to produce electrons.

Magnetic condensing lens is used to condense the electron beam ie it adjust the size of electron beam that falls m the specimen.

The specimen is kept in between the magnetic condensing lens and magnetic objective lens.

Magnetic objective lens is used to block the high angle diffracted beam



The aperture is used to eliminate the diffracted beam

The magnetic projector lens is used to increase magnification

The fluorescent screen is used to record the image Working:-

The electron beam produced by the electron gun is made to fall on the specimen using magnetic condensing lens.

Based on the angle of incidence, the beam is partly transmitted and partly diffracted

To increase intensity and contrast of the image, only transmitted beam is used and diffracted beam is eliminated using magnetic objective lens.

The beam is passed through the magnetic objective lens to eliminate diffracted beams.

Finally the beam is passed through projector lens for further magnification

The magnified image is recorded on the fluorescent screen.

The image obtained (Bright field image) is purely due to transmitted beam.

Advantages:- (i) It can produce magnification as high as 1,00,000 times as that of the size of the object. (ii) The focal length of the microscope system can be varied.

Disadvantages: (i) Sample preparation is difficult and time consuming. (ii) Very thin sample is required in TEM (iii) Biological samples can not be studied because they damaged by electron beam (iv) Region analysed is too small (very selective)

Applications:- (i) It is used to determine the complicated structure of the crystal. (ii) It is used in the study of colloids. (iii) In industries it is used to study the structure of textile fibers, surface of the metals etc. (iv) In medical field it is used to study about the structure of virus, bacteria etc.



6.Explain the working of Scanning Electron Microscope (SEM) with neat diagram?

[ CO3- L1- Dec2011,2012]

It is an Improved and latest developed model of an electron microscope. It is used to produce an enlarged three dimensional image of a specimen of very small size even in the order of A. Here, the image is built up by using an electron probe of very small diameter which scans the specimen surface in parallel straight lines as in the case of a television. Principle When the accelerated primary electrons strike the sample, it produces secondary electrons. These secondary electrons are collected by a detector which in turn gives a three dimensional image of the sample. Components The essential parts of the scanning electron microscope are shown in fig. They are:

1. Electron gun 2. Magnetic condensing lenses 3. Scanning Coil 4. Scintillator (electron detector) 5. Photomultiplier 6. CRO



1. Electron gun A high energy electron beam is produced by thermionic emission from directly heated tungsten filament. This electron beam is accelerated by the anode towards the specimen.

2.Magnetic condensing lenses

The electron beam can be converged or diverged by the lenses. These lenses are usually of magnetic type ie., current carrying coils. The focal length can be controlled by the current through the coil of the magnetic lens. The electron beam is focused as a very fine point on the sample.

2. Scanning coil Scanning coil is arranged between the magnetic lens and sample. It is energized by time varying voltage produced by the scanning generator by creating a time varying magnetic field. This magnetic field deflects the electron beam and now the sample is scanned point by point.

3. Scintillator (electron detector) It collects the secondary or scattered electrons and converts into light signal.

4. Photomultiplier The light signal is further amplified by photomultiplier. 5. CRO The amplified voltage is applied to the grid (brightness control) of the Cathode Ray Tube (CRT). The final image is obtained on the CRT screen. Working The stream of electrons produced by the electron gun is accelerated by the anode. These accelerated primary electrons incident on the sample after passing through the condensing lenses and scanning coil. They produce low energy secondary electrons. The scintillator collects these secondary electrons and it produces a large number of photons. These photons are allowed to fall on a photomultiplier tube. The photo multiplier tube produces amplified electric signal corresponding the photon incident on it. The amplified electrical signal is fed into the brightness control of a cathode ray tube (CRT). The final image is obtained on the CRT which contains the topographical features of the surface of the specimen



Unit – IV Acoustics And Ultrasonics

Part A 1.What is loudness? Give the relation between loudness and intensity of

sound. (or) State Weber Fechner law. [ CO4- L1- Dec2013]

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

directly proportional to logarithmic value of intensity

L=K log I

This is also known as Weber-Fechner law. 2.Define sound intensity level and write its unit. [ CO4- L1- Dec2014]

Intensity level (IL) is equal to the difference in loudness, which is given by

IL = L1 - Lo=K log10I1- K log10 Io Whereas L1 is the loudness of any sound of intensity IL, and Lo is the loudness corresponding to the standard reference intensity Io.

IL= K log 10

Unit of intensity is Bel

3.Define absorption co-efficient of a material. [ CO4- L1- Dec2015]

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.

Absorption coefficient (a) =



5.What are units of loudness? Define them. [ CO4- L1- Dec2010]

There are two units of loudness viz., Decibel, Phon and Sone

Decibel: It is the smallest unit compared to Bel. It is the standard unit used to measure

the loudness. One decibel is equal to one tenth of Bel.

Phon: The measure of loudness in phon of any sound is equal to the loudness in

decibels of an equally loud pure tone of frequency 1000Hz.

Sone: The measure of loudness is sone of any sound is equal to the loudness of that

particular sound having a loudness of 40 phon.

6.State the conditions of good acoustics for an auditorium. [ CO4- L1- Dec2012]

Optimum reverberation time should be maintained.

The auditorium must be free from excessive reverberation

There should not be any undesirable focusing of sound in any part of the hall.

Resonance should be avoided and noises should be reduced.

Echoes should be avoided by covering the walls and ceilings with suitable absorbent

materials.

7.What is standard reverberation time? (Or) what is Sabine’s law? [ CO4- L1-

Dec2014]

Sabine’s law states that the standard reverberation time is the time taken by the

intensity of sound to fall to one millionth (10-6) of its initial intensity after the sound

source is cut off.

8.What is Cavitation? Mention its use? (Or)Write a note on Cavitation? [ CO4- L1-

Dec2011]

When ultrasonic waves propagate through a liquid, alternate compressions and rarefactions are generated at any point. Rarefaction results in sudden drop in pressure causing growth of gas bubbles. When these bubbles collapses the pressure and temperature increases. The entire process occurs within one milli second (10-3 s). This phenomenon is known as cavitation. It is used in cold welding, cleaning etc.



9..What is the principle used in the measurement of velocity of blood flow using

ultrasonic waves? [ CO4- L1- Dec2015]

When ultrasound beam strikes the blood vessel, the beam is reflected by particles of blood. Depending on the velocity of blood the reflected beam is Doppler shifted. By calculating the change in frequency, velocity can be measured. 10.What is sonogram? Mention its application? [ CO4- L1- May 2011] When ultrasonic pulses are transmitted into the body, the pulses are reflected at a boundary between two different tissues or a boundary between a tissue and the adjacent fluid. By scanning ultrasonics waves across the body and collecting the echoes from various internal organs, it is possible to image in two dimensions. This type of ultrasonic imaging is known as sonogram.

11.What are the applications of ultrasonics in industries? [ CO4- L1- May2013]

i. Ultrasonics is used in cutting, drilling, welding, soldering etc. ii. They are used to increase the sensitivity of colour in photographs by dispersion of dye in emulsion.

iii. They are also used to remove air bubbles in the liquid metals. iv. Ultrasonics is used in removing paper fibers from paper pulp. v. Ultrasonic are used for flaw detection (NDT).

13. Mention any two medical application of ultrasonics? [ CO4- L1- May2014]

i. Diagnostic application:

Ultrasonics is used for detecting tumors and other defects in human body. It also used in ECG.

ii. Sugerical application : Ultrasonic waves are used to remove kidney stones and brain tumours without any loss of blood. iii. Treatment : Physiotherapist uses ultrasonics to treat severe pain.



ULTRSONICS 1. What is Magnetostriction Effect? [ CO4- L1- Dec2014] When a ferromagnetic rod is placed in a magnetic field parallel to its length, the rod experiences a small change in its length. This is known as Magnetostriction effect. 2. What is the principle of SONAR in ultrasonics? ? [ CO4- L1- Dec2014] SONAR is a device which stands for SOund Navigation and Ranging. The principle of SONAR is based on the echo sounding technique of ultrasonics. In SONAR a sharp ultrasonic pulse is sent through sea water in different directions. The echo signal received from the object is used to find the position of the object. Also by measuring Doppler shift in frequency of the signal received, the velocity of the moving object can be calculated. SONAR is also used to calculate the depth of sea water.

Part-B

1.What is reverberation time? Derive Sabine’s formula for reverberation time.

(Or) Derive expressions for growth and decay of energy density inside a hall and

deduce Sabine’s formula for the reverberation time of the hall. ? [ CO4- L1-

Dec2014,Dec 2016]

Definition: It is defined as time taken for the intensity of sound to decrease to one millionth (1/106) of its initial value after the source had been cut off.

Sabine’s formula for reverberation time

It is givenby second

T=

Where V – Volume of the room or hall in m3

T =



– Absorption coefficients of surface areas of different materials present

in the Hall in O.W.U. = Total absorption of sound i.e., sum of the product of absorption

coefficients and surface areas of the corresponding surfaces in O.W.U. m2 or Sabine Sabine’s formula for reverberation time (Rate of growth and rate of decay): Derivation: Let us consider a small element on a plane wall AB as shown in fig. (a)

Fig (a) Arrangement of an element on a plane wall

Let us consider a wall containing a small element on which the sound energy

is incident. Taking ‘O’ as a mid – point on , two semicircles are drawn with

radii and

Now, consider a small shaded portion between the circles lying between two radii

and drawn at angles and with normal as shown in fig. (a).

Radial length of the shaded portion =

Arc length of the shaded portion =

Area of this shaded portion = ( 1)

Imagine, the whole figure is rotated about the normal through an angle

(Radius of the rotating figure being ). The shaded portion travels through a

small distance (Circumferential length) and thus, traces out an elemental

volume (Fig. b).



Distance travelled by this shaded portion,

Volume traced by the shaded portion, Area x Distance travelled

( 2)

If ‘E’ is the sound energy density i.e., sound energy per unit volume, then, Sound energy present within the volume element

=

= ( 3)

This sound energy from elemental volume is travelling equally in all directions in

total solid angle of 4. Sound energy travels from the volume per unit solid angle

= ( 4)

In this case, solid angle subtended by the area at this element of volume

=

Hence, sound energy in the element of volume that is travelling towards

is given by

=

= ( 5)

Since sound energy will be falling on from all directions, will change from 0

to and will change from 0 to 2. Further, to get the total sound energy

received per second, r will change from 0 to , where is the velocity of sound



(Since sound existing within the distance of 0 to metre from will reach in

one second).

Total sound energy falling on per second

=

=

= (6)

If is the absorption coefficient of the wall AB of which is a part, then the

sound energy absorbed by in one second

=

Total sound energy absorbed per second by the whole enclosure (Entire hall)

=

= ( 7)

Where A = is total absorption of sound in all the surfaces on which sound

energy is incident. Growth and decay of sound energy:

If P is the sound power output, i.e rate of emission of sound energy from the source and V is the total volume of the hall, then, Total sound energy in the hall at a given instant ‘t’ is EV where ‘E’ is the sound energy density at that instant.

Rate of growth or increase in energy per second

= ( 8)

[ Volume of the hall (V) is constant]

Rate of emission of sound energy = by the source

Rate of growth of + sound energy in the room

Rate of absorption of sound energy by the walls

ie. P = V 9



When steady state is reached, and if steady state energy density is

P =

=

Dividing equation (9) by V, we get

(10)

Putting equation (10) may be rewritten as

Multiplying with on both sides of the equation, we get

Integrating on both sides, we obtain (11)

Where K is a constant of integration. The value of K is determined by considering the boundary conditions.

Growth of sound energy:

Sound energy grows from the instant the source begins to emit sound at t = 0 and E = 0 Applying this condition to equation (11), we get

K = (12)

Using equation (12) in equation (11), we get

Or

( 13)



Where is the maximum sound energy density.

Equation (13) expresses the growth of sound energy density ‘E’ with time ‘t’. The

graph shows this variation it indicates that E increases with t, and when t = , E =

Decay of sound energy:

Assume that, when sound energy has reached its steady (Maximum value) state sound energy is cut off.

Then, the rate of emission of sound energy, P = 0 Equation (11) can be written as

Substituting the boundary conditions E = at t = 0 and P = 0 in equation (11), we get

(14)

From equations (11) and (14), we get

(15)

Equation (15) expresses the decay of sound energy density with time after the source is cut off. It is an exponentially decreasing curve as shown in fig.



Expression for reverberation time:

We know that standard reverberation time is the time taken by the sound to fall of its intensity to one – millionth of its initial value after the source is cut off. Now the value of sound energy density before cut off is at standard

reverberation time, it reduces to

[Since E is proportional to I,

Hence, to calculate T, we put

and t = T in equation (15)

Or

Taking log on both sides, we have

Substituting , we get

=

Taking the velocity of the sound, = 340 m/s,

We have,



Or

( )

This equation is in agreement with the experimental values obtained by Sabine.

2.Define absorption coefficient of sound. Describe with necessary theory a

method of measuring the absorption coefficient of a material. ? [ CO4- L1-

Dec2013]

(Or) Using sabine’s formula explain how the sound absorption coefficient of a material

is determined.

Absorption coefficient: Coefficient of absorption (a) of a material is defined as the ratio of sound energy absorbed by its surface to that of total sound energy incident on the surface.

Measurement of absorption coefficient:

Consider a hall of volume ‘V’ and the surface area ‘s’ without any sound absorbing material in the hall, the reverberation time T1 is measured (Fig. a)

Reverberation time T1 = ( 1)

i.e., (2)



With an absorbing material inside the hall (Fig. b), the corresponding reverberation time is noted and is given by

( 3)

Where a1 – absorption coefficient of the absorbing material and s1 is its surface area

( 4)

Subtracting equation (4) from equation (2), we get

( 5)

From equation (5), the absorption coefficient of the sound absorbing material is given by

( 6)

( 7)

By knowing the value of the surface area of the absorbing material s1 volume of the hall V, T1 and T2, absorption coefficient is determined using expression (7).



3.Explain the principle, construction and working of a Magnetostriction generator. Magnetostriction Effect: ? [ CO4- L1- Dec2011,2015,2016]

When a ferromagnetic rod is placed in a magnetic field parallel to its length, the rod experiences a small change in its length. This is known as Magnetostriction effect.

Principle: It works on the principle of magnetostrictive effect. When an alternating magnetic field is applied to a ferromagnetic rod, the rod experiences contraction and expansion alternatively. Ultrasonic waves are produced at the ends of the rod.

Construction: A circuit diagram for a magnetostriction generator is shown in fig

A ferromagnetic rod is clamped between two knife edges. Construct a heartly oscillator with L1, C1, L2 and NPN transistors. Coil L1 is wound on the right end of the rod. C1 is a variable capacitor, L1 and C1 from the tank circuit for the electronic oscillator. L1 C1 connected to collector circuit of an NPN transistors. The frequency of this oscillator can be adjusted by variable capacitor C1 Coil L2 is wound on the left end of the rod.



L2 is connected to the base of the transistor which act as a feedback circuit.

Working: When the battery is switched on, the oscillator circuit L1C1 produces alternating current of frequency.

------------- (1)

L1 = inductance of coil C1 = capacitance of the variable capacitor. This alternating current in the L1 produces an alternating magnetic field on the ferromagnetic rod. Now the rod begins to vibrate due to magnetostriction effect. The oscillator in the L1C1 circuit is damped. To maintain sustained oscillations, e.m.f. generated in the coil L2 is applied to the base of the transistor. It acts as the feedback circuit. By adjusting the variable capacitor C the frequency of alternating current is made equal to the natural frequency of the rod. Natural frequency of the rod is given as

-------------- (2)

l = length of the rod E = Young’s modulus of the material of the rod ρ = Density of the rod

At resonance condition, the rod vibrates vigorously and produces intense ultrasonic waves.

i.e. frequency of the rod = frequency of the alternating current.

=

Advantages: It is a simple oscillator and production cost is low. Large output power can be generated.

Disadvantage: This oscillator can produce sound frequencies upto 3MHz Same frequency oscillations cannot be generated. Loss of energy due to hysteresis and eddy current.



4. What is acoustic grating? With a neat diagram explain, how it is used to

determine the velocity of ultrasonic sound in a liquid? ? [ CO4- L1-

Dec2011,2013]

(or)

Explain the method to find the velocity of ultrasonic waves passing through a liquid.

If ultrasonic waves passes through a liquid get reflected, longitudinal stationary waves are formed. These waves have fixed nodal and anti nodal planes. If a monochromatic light is passed perpendicular to these waves, the liquid medium act as a diffraction grating and hence diffraction is obtained. This type of grating is known as acoustic grating.

Experimental Arrangement:

The experimental arrangement is shown in fig (i)

In an ultrasonic cell (glass cell) the given liquid is taken. Ultrasonic transducer is fixed inside the cell and it generates ultrasonic waves. These waves get reflected from the reflector placed on the opposite side. A standing wave pattern is formed. These waves have fixed nodal and anti nodal planes. At nodal planes density is maximum and anti nodal planes density is minimum. This periodic variation of density, changes the refractive index of

the medium. i.e., refractive index increases with increase in density. Hence a diffraction grating is formed due to the standing wave pattern. When a monochromatic light is passed through the acoustic grating diffraction occurs. Condition for the diffraction is given by



--------------------- (1)

d = distance between any two successive nodes or anti nodes or grating element n = order of the diffraction λ= wavelength of the monochromatic light θn = angle of diffraction for nth order. Let λm be the wave length of ultrasonic waves in the medium fig (ii)

Then ------------------ (2)

Sub equation (2) in (1)

sin = nλ

-------------------- (3)

If the frequency of ultrasonic waves is f, then the velocity of ultrasonic waves in the given medium is

------------------ (4)

Sub (3) in (4)

This is the expression for velocity of ultrasonic waves in liquid.

Uses:

This method is useful in measuring velocity of ultrasonic waves in liquids and gases at different temperatures. Free volume and compressibility of the liquid can be calculated.



5.What is Piezo electric effect? Describe the principle, construction and working of a Piezo electric generator to produce ultrasonic waves? ? [ CO4- L1- Dec2013]

Piezo electric effect: When mechanical stress is applied to one pair of opposite faces of a quartz crystal, then equal and opposite electrical charges are developed on the other pair of opposite faces of the crystal. This is known as Piezo electric effect.

Principle: In this generator ultrasonic waves are produced by the principle of inverse Piezo electric effect. Inverse Piezo electric effect:

When an electric field is applied to one pair of opposite faces of a quartz crystal, expansion or contraction is developed across the other pair of opposite faces of the crystal. This is called inverse Piezo electric effect. If we apply an alternating voltage to the quartz crystal, we get alternate contraction and expansion ie the crystal is set in to mechanical vibrations and produce ultrasonic waves at resonance.

Quartz Crystal: In this generator a quartz crystal is used for the generation of ultrasonic waves. The natural quartz crystal is shown in fig (i).

X – Cut Crystal: When the crystal is cut perpendicular to both the X- axis and the optic axis is called X - cut crystal fig (ii). X -cut crystal are used to produce longitudinal ultrasonic waves.

Y - Cut Crystal: When the crystal is cut perpendicular to both the Y- axis and the optic axis is called Y - cut crystal fig (iii). Y -cut crystal are used to produce transverse ultrasonic waves.



Construction: The circuit diagram of a Piezo electric generator is shown in fig.

Q is a thin slice of quartz crystal and it is placed in between two metal plates A and B. The plates are connected to ends of the coil L2 Coils L, L1 and L2 are inductively coupled. The tank circuit LC is connected to the collector circuit. C is a variable capacitor. L1 is connected between the base and the emitter of the NPN transistor which act as a feedback circuit.

Working : When the battery is switched on the tank circuit LC produces alternating voltage with frequency

L = inductance of coil C = capacitance of the variable capacitor. An alternating e.m.f. is induced in the coil L2 and the same is applied to the crystal.

The oscillations in the tank circuit are damped. To maintain sustained oscillations, necessary positive feedback is provided with L2 to the base of the transistor.



The crystal expands and contracts due to inverse Piezo electric effect. It generates ultrasonic waves. The capacitance of the capacitor C is adjusted such that frequency of the applied voltage is equal to the natural frequency of the crystal.

Natural frequency of the crystal is given by,

------------- (2)

P = 1, 2, 3, …….. (Stands for fundamental, first over tone second over tone etc) l = length or thickness of the crystal E = Young’s modulus of the material of the crystal ρ= Density of the crystal

At resonance condition, the crystal vibrates with large amplitude and produce intense ultrasonic waves.

i.e. natural frequency of the crystal = frequency of the tank circuit.

=

Advantages: Ultrasonic frequencies as high as 500 MHz can be obtained with this generator. Stable single frequency waves can be generated. It is more efficient than magnetostriction generator.

Disadvantages: Piezo electric crystals are very expensive. Cutting and shaping are not easy.

6. Explain in detail various scanning modes using ultrasonic waves. ? [ CO4- L1-

Dec2009]

The most common scan displays are A – Scan, B – Scan and C – Scan

A – Scan display:- A – Scan gives only one dimensional information. The transducer is kept only in fixed position. It is also known as amplitude mode display. In this mode the intensity of reflected signals from the various parts of specimen is given to the Y



– plate and time base is connected to X – plate of CRO, so that displayed as vertical spikes along horizontal base line as shown in fig i The height of the vertical spikes corresponds to the strength of the echo the position of the spikes along x-axis (time axis) corresponds to the depth of the flaw from the transducer. ie it gives the Total time taken by the ultrasonic sound to travel from transmitter to receiver.

B – Scan display:- B – Scan or Brightness mode display gives a two dimensional image. In B – Scan the transducer is moved from one to another end of the surface of the specimen The reflected echo from each point is displayed as dots on the screen (fig ii) The brightness and size of the dot depends on the intensity of the reflected signals. X axis gives the position of the transducer. Thus B – Scan provides exact image of the internal structure of the specimen.

\



C– Scan display In C-scan display, the transducer is moved over the surface of the specimen in zigzag fashion or closely placed parallel lines (fig iii) the output ultrasonic probe is connected to an X – Y plotter. It plots the cross sectional view of the specimen with size of the defect. If we scan the specimen in three axis, we can calculate the volume of the defect and depth.

7. Describe the various non-destructive testing methods? ? [ CO4- L1- Dec2010]

Non – destructing testing (NDT) is a method of testing a material without damaging it. Ultrasonic inspection is one of the most widely used methods of NDT. It uses high frequency ultrasonic waves for flaw detections. It is based on the principle of echo reflection of ultrasonic sound at the interfaces.

Construction The basic block diagram for the ultrasonic flaw detector is shown in fig It consists of a high frequency generator and a cathode ray oscilloscope (CRO). It uses two separates probe, one for transmitting the ultrasonic wave and other two receive them after passing through the specimen.

Working The ultrasonic wave is sent from transmitter probe. It strikes the upper surface of the specimen and makes a sharp spike at the left hand side of the CRO screen. If the specimen is good without any defect this sound waves strikes the bottom



surface of the specimen and it get reflected. This is indicated by a spike in the right hand side of the CRO. It is the property of ultrasonic waves that it gets deflected whenever there is a change in medium. In case a defect exists inside the specimen the ultrasonic waves striking the defect will be reflected. This reflected beam reaches the receiver probe and is indicated by a spike on the CRO screen in between the left and right spikes.



The time interval (t) between transmission signal and reflected signal is measured from the CRO.

If the velocity of sound waves in a material is known, then the distance of defect from the surface of the specimen (d) is calculated by using the relation

The exact size and shape of the flaw can be found by examining the specimen from all the directions. 8.Explain how ultrasonic waves are used in ultrasound imaging system in

medical field. (Or) Explain how the internal organs of the human body are

visualized using sonogram. ? [ CO4- L1- Dec2014]

Sonogram

Sonogram is an instrument used to monitor and visualize the image of the interior parts of the body by using high frequency sound waves. It provides high speed, high resolution, high accuracy, and high contrast ultrasonic images using digital image processing unit.

Working:

1. The high frequency ultrasonic waves are transmitted into the human body by using ultrasonic transducer. 2. The reflected ultrasonic waves from the interior parts of the body are received by the receiver circuit as shown in fig 1.33.



3. The received signals are converted into digital signal by digital processing unit and it is stored in memory device. 4. The digital signals are properly amplified and sent to screen of the monitor.

5. The monitor produces the proper image of the interior parts of the body.

Applications:

1. To monitor the health and development of foetus. 2. It is used to confirm pregnancies, to ensure the fetuses are developing normally, to determine age, and to indicate delivery dates more accurately.

3. To monitor the interior parts of the body more accurately.

4.

9. Detail the applications of ultrasonics in medical, engineering and industries. ? [

CO4- L1- Dec211]

Medical applications:

1. Ultrasonics are used to find velocity of blood flow and the movement



of heart the human body. 2. It is used to detect tumours in human body. 3. While treating the patient, it is used to gather the blood in columns separated by plasma. 4. It is used for bloodless brain surgery and painless extraction of teeth. 5. It is used to decrease the strength of the scar by affecting the directions of elastic fibers, which are responsible for the scar. 6. It is used to study the development of fetuses and to determine age without any risk to the mother or the child.

Biological applications: 1. Small animals like frog, fish, rat etc., are killed when exposed by ultrasonic waves. 2. Ultrasonic waves can kill bacteria. So, they are used for sterilizing milk.

Chemical applications: 1. Ultrasonic waves are used to form stable emulsions even in liquids like the mixture of water and oil or water and mercury. They finds the applications in the preparation of photographic films, face creams etc. 2. They are used to make the liquid gels such as Aluminium Hydroxide.

3. Ultrasonic act as a catalytic agent and accelerate chemical reactions. Ultrasonic waves accelerate crystallization process also.

Engineering applications: 1. Sound signaling:

The ultrasonic waves are used to identify and land the ships. Therefore, the sound signaling is used to identify the warship which is useful to military purpose.

2. Depth sounding: Echo sounding is used to determine the distance of the object in the sea. The depth of the sea can be directly calibrated using the instrument fathometer or echometer.



Industry:

1. Ultrasonic welding and soldering: The properties of some metals can be changed by heating and therefore, they cannot weld by electric or gas welding. In such cases, the sheets are welded together at room temperature using ultrasonic waves. It is also used to solder metals like Aluminium without using flux.

2. Ultrasonic drilling and cutting: It is a vibratory process. Ultrasonic waves are used for making holes in hard and brittle materials like ceramics, diamond, glasses etc.

3. Ultrasonic cleaning and drying: It is one of the cheap techniques. It is used to clean different parts of the machine, electrical components like resistors, capacitors, watches etc. which cannot be easily cleaned by other technique.

4. The ultrasonics are used to form alloys of uniform compositions.

Other applications: 1. They are used to increase the sensitivity of colours in photographs by dispersion of dye in the emulsion. 2. They are used to remove air bubbles in the liquid metals and convert them into fused metals. 3. Low frequency ultrasonics are used in sorting paper fibers from the paper pulp.

4. They are used in sound navigation purposes (SONAR).



Unit –V Photonics And Fibre Optics

Part-A 1.Write the differences between spontaneous emission and stimulatedemission?[

CO5- L1- Dec2009,2012]

Spontaneous emission Stimulated emission

1 Emission of light radiation is not triggered by external influence.

Induced emission of light radiations caused by incident photons.

2 Emitted photon travels in random direction.

Emitted photon can be made to travel in particular direction.

3 Emitted photon cannot be controlled.

Emitted photon can be controlled.

4 This process is a key factor for ordinary light.

This process is a key factor for laser operation.

2. What is the principle of laser action? ? [ CO5- L1- Dec2013]

Stimulated emission process is a key factor for the laser action. This can be multiplied through chain reaction. This multiplication of photons through stimulated emission leads to coherent, powerful, monochromatic, collimated beam of light radiation. 3. What are the three important components of any laser device? [ CO5- L1-

Dec2010]

1. Active medium 2. Pumping source 3. Optical resonator



4.What are differences between homojunction and heterojunction lase . ? [ CO5-

L1- Dec2009]

S.No. Homojunction laser Heterojunction

1 Homojunction laser is made by a single crystalline material.

Heterojunction laser is made by different crystalline materials.

2 Power output is low. Power output is high.

3 Pulsed output (sometimes continuous). Continuous output.

4 It has high threshold current density. It has low threshold current density.

5 Cost is less. Cost is more.

6 Life time is less. Life time is more.

7 Examples: GaAS InP

Examples: GaAs/GaA1As InP / InA1P

5. What is meant by population inversion? [ CO5- L1- Jun2010]

If the number of atoms in the excited state is more than the number of atoms in the lower energy state, it is called population inversion. 6. Whether two level laser is possible? Why? (or) Can a two-level system be used for the production of laser? Why? [ CO5- L1-

Jan2011]

If two energy levels are involved, by direct pumping the higher level cannot be made more populated than the lower level. Hence laser action is not at all possible in a two level system.



7.Define population inversion and metastable state. ? [ CO45L1- Dec2014]

When the number of atoms (i.e.,population of atoms) in any excited state is higher than their number in the lower energy state, that condition is called population inversion. For this to happen, the life time of higher energy state must be longer such a state is called (metastable state). If only population inversion is achieved light amplification by stimulated emission of radiation can occur. 8.What are the different methods of achieving population inversion? (or) What is meant by laser pumping? ? [ CO5- L1- Jan2012]

The process of exciting the atoms of the active medium to a higher energy state from their lower energy state by supplying energy in relevant form is called pumping. Different pumping methods to achieve population inversion are Optical pumping Electrical discharge method Direct Conversion Inelastic collision between atoms 9. State the properties of laser beam. (or) Give the important characteristics of laser light. [ CO5- L1- Jan2012] The important characteristics of laser beam are High monochromaticity (i.e. highly single wavelength) High degree of coherence High directionality (i.e. less divergence) and

High brightness It is highly powerful. It is capable of travelling over long distance without any energy loss. Laser beam is not easily absorbed by the water.

10. List any four applications of lasers in engineering. [ CO5- L1- Jan2009] Material processing such as cutting, drilling, welding, etching, surface hardening etc. (Ruby laser, CO2 laser, Nd: YAG laser)

For pollution monitoring and remote sensing (CO2 laser, Nd: YAG laser) Non destructive testing of components using holographic interferometery (He-Ne laser) Used in compact discs for storing and retrieving the data.

11. State any four applications of lasers in the field of medicine. [ CO5- L1- Jan2012]



In medical field lasers are used as scalpel for bloodless surgery (CO2 laser, ND: YAG laser) Fiber optic endoscopes with lasers as light sources are very useful in treatment of internal organs (e.g. Nd: YAG lasers in treatment of gastrointestinal bleeding)

PART-B 1. For atomic transitions derive Einstein relations and hence deduce the expression for the ratio of spontaneous emission rate to the stimulated emission rate? [ CO5- L1- Jan2012,Dec2016] Let us consider an assembly of atoms (material) at an absolute temperature T which is exposed to monochromatic radiation (a stream of photons with energy hν). There are three different processes occur in the medium. PROCESS: 1 STIMULATED ABSORPTION An atom in the ground state with energy E1 absorbs an incident photon of energy hν and goes to excited state (higher energy state) with energy E2 (fig). This process is known as stimulated absorption. This process is possible only when the incident photon energy hν is equal to the energy difference between ground state and excited state (E2 - E1).

In a system of atoms, the rate of stimulated absorption is directly proportional to number of atoms (N1) in state E1 and the energy density of incident radiation. Thus, the rate

of stimulated absorption is given by

----------------------- (1)

where is proportionality constant and also called as Einstein coefficient of

stimulated absorption. This process is an upward transition. PROCESS: 2 SPONTANEOUS EMISSIONS



The atom in the excited state E2 (higher energy state) return to ground state E1 (lower energy state) by emitting a photon of energy hν (equal to E2- E1) without the action of any external agency (fig) after finishing its lifetime in this state. While the excited atom returns to the ground state, the difference in energy is emitted as photon. This process is spontaneous and uncontrollable.

The rate of spontaneous emission is directly proportional to the number of atoms in the excited state (N2). Hence the rate of spontaneous emission is given by

------------------------- (2)

Here is proportionality constant and it is also called as Einstein co-efficient of

spontaneous emission. This process is a downward transition. PROCESS: 3 STIMULATED EMISSION Einstein discovered that there must be another mechanism by which an atom in excited state can return to ground state. The incident photon having energy (E2-E1) interact with the atom which do not finish its life time in excited state, so that the atom comes down to ground state with the emission of photon along with the incident photon. These two photons are coherent in nature. This process is called the stimulated emission and this is the key factor for the operation of a laser.

The rate of stimulated emission is directly proportional to the number of atoms in the excited energy level (N2) and the energy density of incident radiation .



---------------- (3)

Here is proportionality constant and it is also called as Einstein co-efficient of

stimulated emission. This process is a downward transition. Under equilibrium condition, the numbers of upward and downward transitions per unit volume per second are equal.

ie.,

Substituting from equations (1), (2) and (3) we have

----------------------------- (4)

Rearranging the equation (4) we have,

Therefore ---------------------- (5)

Dividing numerator and denominator by , we have

--------------------- (6)

On substituting the value of = in equation (6), we have

--------------------- (7)

The Planck’s radiation formula for energy distribution in terms of frequency is given by

------------------------------------ (8)

Comparing the equations (7) and (8), we get

;

and ------------------------------ (9)

These equations are known as Einstein’s relations. The ratio of spontaneous emission rate to the stimulated emission rate is given by,



----------------------------- (10)

In practice, the absorption and the emission process occur simultaneously. Even for sources operating at a high temperature and low frequency and hence .

This confirms that spontaneous emission predominates over stimulated emission. From equation (10) it is clear that to make R smaller has to be made larger. If we

increase , both stimulated emission and stimulated absorption occur.

Let us consider the ratio of stimulated emission rate to stimulated absorption rate.

(As B21= B12)

At thermal equilibrium stimulated absorption predominates over stimulated emission. In order to increases stimulated emission N2>N1 and this condition is called population inversion. Hence for Laser action we must Create population inversion and Increase the energy density of interacting radiation. 2 (a). Explain the modes of vibration of CO2 molecule? [ CO5- L1- Jan2011,DEC2015] (or) Describe the construction and working of CO2 laser with necessary diagrams? ENERGY STATES OF CO2 MOLECULE A carbon dioxide molecule has a carbon atom at the centre with two oxygen atoms attached. There are three independent modes of vibrations. They are:

Symmetric stretching mode Bending mode Asymmetric stretching mode

1.SYMMETRIC STRETCHING MODE

In this mode of vibration, carbon atom is at rest and both oxygen atoms vibrate simultaneously along the axis of the molecule departing or approaching towards a fixed carbon atom (fig.i).

2.BENDING MODE



In this mode of vibration, oxygen atoms and carbon atom vibrate perpendicular to molecular axis and vibrating opposite to each other. (fig.ii).

3.ASYMMETRI STRETCHING MODE

In this mode of vibration, oxygen atoms and carbon atom vibrates asymmetrically, i.e., oxygen atoms move in one direction while carbon atom moves in the opposite direction in the same axis. (fig.iii).

CARBON DIOXIDE (CO2) LASER It is a four – level molecular gas laser. Principle The active medium is a gas mixture of CO2, N2 and He. The laser transition takes place between the vibrational energy level of CO2 molecules. Construction It consists of a quartz discharge tube 5m long and 2.5 cm in diameter (fig. iv) this discharge is filled with the gaseous mixture of CO2 (active medium), helium and nitrogen with suitable partial pressures under the pressure of 1:4:5



The terminals of the discharge tube are connected to a D.C. power supply. The ends of this tube are fitted with NaCl Brewster windows so that the laser light generated will be polarized. The optical resonator is formed with two concave mirrors one fully reflecting and other partially reflecting. Working The energy levels of nitrogen and carbon dioxide molecules are shown in fig. v When the electrical discharge occurs in the gas, the electrons collide with nitrogen molecules and they are raised to excited states. This process is represented by the equation

Electron with kinetic energy

Same electron with lesser energy

Now, molecules in excited collide with CO2 atoms in ground state and excite them to

higher vibrational levels.

This process is represented by the equation

Carbon dioxide atoms in ground state

Carbon dioxide atoms in excited state



Since the excited level of nitrogen is very close to E5 level of CO2 atoms, population in meta stable state E5 level increases. As soon as population inversion is reached, any of the spontaneously emitted photon will trigger laser action in the tube. There are two possible types of laser transitions.

Transitions E5 – E4 This transition will produce a laser beam of wavelength 10.6 μm Transitions E5 – E3This transition will produce a laser beam of wavelength 9.6 μm

Normally 10.6 μm transition is more intense than 9.6 μm transitions. The power output from this laser is 10 KW. The He increases the laser efficiency at 10.6 μm by speeding up the transition from E4 to E2. So it maintains a large population inversion between E5 to E4. Applications Material processing such as cutting, drilling, welding, surface hardening etc.,

In the field of medicine CO2 laser is used for blood less surgery. CO2 lasers are used for pollution monitoring and remote sensing.

Demerits 1. The contamination of oxygen by carbon mono oxide reduces the efficiency. 2. Accidental exposure damages our eyes since it is invisible to our eyes. 3. Corrosion may occur at the reflecting plates.

3. Describe the construction and working of Nd-YAG laser with energy level diagram. [ CO5- L1- Jan2011,DEC2015] Or In detail explain the principle, construction and working of a four level solid state laser. [A.U. Chennai May 2009] Nd – YAG laser is a Neodymium based laser. It is a four level solid state laser. Principle The active medium Nd-YAG rod is optically pumped by krypton flash tubes. The neodymium ions (Nd3+) are raised to excited levels. During the transition from metastable state to ground state, a laser beam of wavelength 1.064 is emitted.

Construction The construction of Nd-YAG laser is shown in fig.1. A small amount of yttrium ions (Y3+) is replaced by neodymium ions (Nd3+) in the active element of Nd-YAG crystal.



This active element is cut into a cylindrical rod. The ends of the rod are highly polished and they are optically flat and parallel. This cylindrical rod (laser rod) and a pumping source (flash tube) are placed inside an elliptical reflector cavity.

The optical resonator is formed by using two external reflecting mirrors. One mirror (M1) is 100% reflecting while the other mirror (M2) is partially reflecting. Working The energy level diagram for Nd-YAG laser is shown in fig.2. These energy levels are those of neodymium (Nd3+) ions.

1.When the krypton flash lamp is switched on, the neodymium atoms are raised from ground level E0 to upper levels E3 and E4 (pump bands) due to absorption of light radiation.



2.The neodymium atoms make a transition from these energy levels to level E2 by non-radiative transition. E2 is a metastable state. 3.The neodymium ions are collected in the level E2 and the population inversion is achieved between E2 and E1. 4.An ion makes a spontaneous transition from E2 to E1, emitting a photon of energy This emitted photon will trigger a chain of stimulated photons

between E2 and E1. 5.The photons thus generated travel back and forth between two mirrors and grow in strength. After some time, the photon number multiplies more rapidly. 6.After enough strength is attained (condition for laser satisfied), an intense laser

light of wavelength 1.06 is emitted through the partial reflector.

This corresponds to transition from E2 to E1.

Advantages 1.It has high energy output. 2.It is much easy to achieve population inversion in this laser. 3.It has very high repetition rate of operation.

Applications

1.It is widely used in welding, drilling, cutting etc., 2.It is used in medical applications such as endoscopy, surgery, dermatology etc.,

4. With suitable diagram explain the construction and working of homojunction Ga-As laser. [ CO5- L1- Dec 2013,2011,DEC2014]Definition It is a specially fabricated p-n junction diode. This diode emits laser light when it is forward-biased. Principle When a p-n junction diode is forward-biased electrons from n-region and holes from p-region cross the junction and recombine with each other at junction. During the recombination process, the light radiation (photons) is released from a certain specified direct band gap semiconductors like Ga-As. This light radiation is known as recombination radiation.

Construction



The construction of semiconductor laser is shown in fig.1. The active medium is a p-n junction diode made from a single crystal of gallium arsenide. This crystal is cut in the form of a platelet having a thickness of 0.5mm.

The platelet consists of two parts having an electron conductivity (n-type) and hole conductivity (p-type). The photon emission is stimulated in a very thin layer of pn junction (in the order of few microns). The electrical voltage is applied to the crystal through the electrode fixed on the upper surface. The end faces of the junction diode are well polished and parallel to each other. They act as an optical resonator through which the emitted light comes out. Working The energy level diagram of semiconductor laser is shown in fig.2. When the pn junction is forward-biased with large applied voltage, the electrons and holes are injected into junction region in considerable concentration. The region around the junction contains a large amount of electrons in the conduction band and a large amount of holes in the valance band. If the population density is high, a condition of population inversion is achieved. The electrons and holes recombine with each other and this recombination’s produce radiation in the form of light.



When the forward-biased voltage is increased, more and more light photons are emitted and the light production instantly becomes stronger. These photons will trigger a chain of stimulated recombination’s resulting in the release of photons in phase. The photons moving at the plane of the junction travels back and forth by reflection between two sides placed parallel and opposite to each other and grow in strength.

After gaining enough strength, it gives out the laser beam of wavelength 8400 .

The wavelength of laser light is given by

Where Eg band gap energy in joule.

Advantages 1.It is very small in dimension, simple and compact. 2.It has high efficiency. 3.Laser output can be easily increased by controlling the junction current.

Disadvantages

1.Laser output beam has large divergence. 2.The purity and the monochromaticity are poor than other types of laser. 3.It has poor coherence and poor stability. 4.Threshold current density is very large.

Applications

1.It is widely used fiber optic communication. 2.It is used in laser printers and CD writing and reading. 3.It is also used as a pain killer.



5. Describe the construction and working of a heterojunction Ga-As laser. [ CO5- L1- Dec 2009] A p-n junction made up of different materials in two regions i.e., n-type and p-type is known as heterojunction. Principle When the p-n junction diode is forward biased, electrons from n-region and holes from p-region recombine with each other at the junction. During recombination process, light is released from certain specified direct band gap semiconductors. Construction This laser consists of five layers as shown in fig.1. A layer of Ga-As p-type (3rd layer) acts as active region. This layer is kept between two layers having wider band gap viz. GaA1As-p-type (2nd layer) and GaA1As n-type (4th layer).

The electrical current is applied to the crystal through the electrodes fixed on top and bottom layer. The end faces of the junctions of 3rd and 4th layer are well polished and parallel to each other. They act as an optical resonator.



Working The energy band diagram is shown in fig.2.

When the p-n junction is forward biased, the electrons and holes are injected into the junction region. The region around the junction contains large amount of electrons in the conduction band and holes in the valence band. Thus the population inversion is achieved. At this stage, some of the injected charge carries recombines and produce radiation in the form of light. When the forward biased voltage is increased, more and more light photons are emitted and the light intensity is more. These photons can trigger a chain of stimulated recombination’s resulting in the release of photons in phase. The photons moving at the plane of the junction travels back and forth by reflection between two sides and

grow in its strength. A coherent beam of laser having wavelength nearly 8000

emerge out from the junction region. Advantages

1.It produces continuous wave output. 2.The power output is very high.



Disadvantages

1.It is very difficult to grow different layers of p-n junction. 2.The cost of this laser is very high.

Applications

1.This type of laser mostly used in optical communication. 2.It is widely used in computers, especially on CD-ROMs.

6. With a neat diagram explain the material processing using laser? Principle[ CO5- L1- Dec2012,DEC2013] When the material is exposed to laser light, then light energy is converted to heat energy. Due to heating effect, the material is heated then melted and vapourised. This technique of laser heat treatment is used in engineering applications like surface hardening, alloying, cutting, welding, drilling and perforating holes in the materials and hence this process is called material processing. Ruby laser, Nd-YAG laser and Co2 laser are used for this purpose. Construction The instrumentation for material processing is shown in fig.1.

It consists of a laser source to produce laser beam, shutter to control the intensity of the laser beam and an assembly of lens to effectively focus the laser to the specimen. Separate control arrangement is made for removing the molten material, smokes, fumes etc., with help of a shielding gas jet. The powder feeder is used to feed the metal powder.



Working Laser source is switched on. The light reflected by the plane mirror is made to pass through the shutter. The intensity of the laser beam is controlled by the shutter and allowed to fall on the lens assembly. This lens assembly focuses the light effectively on the specimen. Now the specimen get heated, giving rise to smokes, fumes and molten material. These smokes, fumes and molten material are removed by shielding gas jet and this in turn makes the laser beam to continuously fall on the specimen. Thus the material can be drilled, cut, put holes etc., using this technique effectively and easily. In case of alloying, welding etc., the powder feeder will be used to spray the metal powder over the specimen. Advantages

1.There is no damage to the material 2.Time consumption is less 3.In normal welding process heat spread all over the surrounding and affect the other areas of the material but in laser welding very small area get affected. Therefore after welding we can use the material directly. 4.Any dissimilar metals can be welded.

7. Define acceptance angle and numerical aperture. Derive the expression sin

[ CO5- L1- Jan2011,DEC2015]

ACCEPTANCE ANGLE The maximum angle with respect to axis of fiber through which a ray of light can enter into the fiber and propagate is called acceptance angle of the fiber. NUMERICAL APERTURE The site of the acceptance angle of the fibre is known as numerical aperture (NA). It denotes the light gathering capability of the optical fiber. DERIVATION:



Let us consider the propagation of light in an optical fibre shown in Fig.

The incident ray AO enters into the core at an angle to the fibre

axis.

This incident ray is refracted along OB at an angle to the fibre

axis in the core.

The refracted ray falls on the interface of core and cladding at the critical angle of incidence (90 - ) and it moves along BC.

Any light ray which enters into core at an angle of incidence less than have angle of refraction less than . Therefore, the angle of

incidence (90 ) at the interface is more than the critical angle of

incidence.

Thus, the light ray makes total internal reflection and back into the core.

The light ray which enters at an angle of incidence greater than

at O incident at B at an angle less than the critical angle. Now the light is refracted into the cladding region and it is escaped.

Let and be the refractive indices of the core and cladding

and the refractive index of the surroundings.

Now, applying Snell’s law of refraction at the point of incidence of the ray AO into the core, we have,

sin =

sin = . . . . (2)

At the point B on the interface of core and cladding. Angle of incidence = 90 -



Applying Snell’s law of refraction again, we have,

or

or cos . . . . . (3)

Substituting for from equation (3) in equation (2), we get

sin

sin

sin . . . . (4)

sin . . . . (5)

This angle is called as acceptance angle.

If the medium surrounding the fibre is air, then

. . . (6)

Numerical aperture is given by NA = sin . . . (7)

Substituting for sin from equations (4), we have

NA = . . . . (8)

If the medium surrounding the fiber is air, then

NA = . . . . (9)

CONDITION FOR PROPAGATION OF LIGHT Let be the angle of incidence of an incident ray. Then, the light ray will propagate

only if

or sin

or sin



i.e. sin

This is the condition for propagation of light within the fibre. . 8. How are fibres classified? Explain the classification in detail. [ CO5- L1- Jan2010,DEC2014] Types Of Optical Fibres The optical fibres are classified into three major types based on

a) Material b) Number of modes and c) Refractive index profile

A general classification of optical fibers is shown in fig.

MATERIAL CLASSIFICATION: The optical fibres are generally classified into two types on the basis of materials. They are

Glass fibre and Plastic fibre Glass fibre If the optical fiber is made from fusing mixture of silica glasses and metal oxides, it is known as glass fiber. Example: The glass fibers can be made by any one of the following combination of core and cladding. Core - GeO2 - Si O2, cladding - SiO2 Plastic fibre If the fibre is made up of plastics, then it is called plastic fiber. Example: The plastic fibre is made by any one of the following combinations of core and cladding.



core – polystyrene, cladding – methyl methacrylate (a)CLASSIFICATION BASED ON THE NUMBER OF MODES

Depending on the number of modes of propagation, the optical fibres are classified into two types

i. Single-Mode (SM) fibre ii. Multi Mode (MM) fibre

(b) CLASSIFICATION BASED ON REFRACTIVE INDEX PROFILE: Based on the variation in the refractive index of the core and cladding, the optical fibers are classified into two types. They are

1. Step-index fibre 2. Graded-index fibre.

In general optical fibres are classified in to three types: (i) Step – index single – mode fibre

The basic structure of the step-index single-mode fibre is shown in fig. It consists of thin core of uniform refractive index of a higher value. This core is surrounded by a cladding of uniform refractive index lesser than that of the core.

A typical step-index single-mode fibre has a core diameter of 5 to 10 and an

external diameter of cladding of 50 to 125

The refractive index changes abruptly (or in step) at the core-cladding boundary. Its refractive index profile takes the shape of a step (fig.) Due to its small core diameter, only a single-mode is allowed. Here the rays are called meridianal rays, since the ray crosses the axis of the fiber for every total internal reflection and it takes zigzag path.



ADVANTAGES:

It has very high capacity

Because of single mode propagation lost due to inter modal dispersion does not exist.

The band width of this fiber is higher than multi mode fiber.

About 80 % of the optical fibers manufactured are of this type. DISADVANTAGES:

The manufacturing and handling of this fiber is very expensive.

(ii) Step-index multimode fibre: The geometry of normal cross-section of a typical step-index multimode fibre is shown in fig. Its core has a much larger diameter which makes it easier to support propagation of a large number of modes. A typical step-index multimode fibre has a core diameter of 50 to 200 and an

external diameter of cladding 125 to 300

It has a core material with uniform refractive index and a cladding material of lesser refractive index than that of the core. There is a sudden increase in the value of refractive index from cladding to core. Thus, its refractive index profile takes the shape of a step (fig).



Because of larger diameter of the core, the propagation of many modes are possible. Here the rays are called meridianal rays, since the ray crosses the axis of the fiber for every total internal reflection and it takes zigzag path. Advantages:

Since LED’s are used as the source of light, they are easier to operate.

LED’s have longer life than laser diode making this more suitable in many applications.

They are less expensive and require less circuitry. Disadvatages:

They suffer intermodal dispersion loss. Inter Modal Dispersion: The ray which enters into the fiber near the axis of the fiber will reach the end of the fiber first while the ray which enters into the fiber away from the axis of the fiber will reach the end of the fiber little bit late. This will cause broadening of pulse width leads to over lapping of pulses. This is called



Inter model dispersion. Remedy:

We have to increases the pulse interval so that we can avoid the over lapping. (iii)Graded-index Multimode Fibre (GRIN)

Here, the refractive index of the core varies radially from the axis of the fiber. The refractive index of the core is maximum along the fiber axis and it gradually decreases towards core-cladding interface. Thus it is called graded index fiber.The geometry of normal cross-section of a typical graded-index fibre is shown in fig.

A typical graded-index multimode fibre has a core diameter of 50 to 200 an

external diameter of cladding 100 to 250 .

The refractive index of the core is maximum at the axis of the fibre and it gradually decreases towards the cladding. The refractive index profile is shown in fig. If the diameter of the core is high, the intermodal dispersion loss must be high. But, because of the gradual decrease in the refractive index of the core, the intermodal dispersion loss is minimized. The light ray propagates in helical path. They are called as skew rays since they never touch the axis of the fiber throughout propagation. Advantages:

Intermodal dispersion can be reduced in this fiber.

It is a high quality fiber.

It has low attenuation. Disadvantages:

It is most expensive of all types of fibers.



Its fabrication is difficult 9.Explain with a neat block diagram, the working of fibre optical communication

system. [ CO5- L1- Jan2010,Dec2015]

Fibre Optic Communication System:

A fibre optic communication system is used to transfer an information from a source to a distant user. Principle: The basic principle of optic fibre communication is transmission of information over a required distance by the propagation of optical signal through optical fibres. It involves deriving an optical signal from electrical signal at transmitting end and converting optical signal back to electrical signal at receiving end.

Description A basic block diagram of a fibre optic communication system is shown in fig. The main parts of the fibre optic communication system are

1.Information signal source 2.Transmitter 3.Light source 4.Propagation medium (optical fiber) 5.Receiver

1. Information signal source The information signal source may be voice, music, digital data or analog voltage and video signals. Here, it is an analog information. 2. Transmitter:



It consists of a drive circuit and a light source. The drive circuit transfers the electric input signals (information signal) into digital pulses. 3. Light source The light source generates optical energy which acts as the information carrier. Laser or LED is used for light sources. This light source converts digital pulses into optical pulses. 4. Propagation medium The optical fibre is used as a propagation medium. It acts as a waveguide and transmits optical pulses towards receiver by the principle of total internal reflection. 5. Receiver: It consists of a photo detector, an amplifier and a signal-restoring circuit. The photo detector detects optical signal and converts back into an electrical signal. Further, the signals are amplified by an amplifier. These electrical signals are decoded i.e., converted from digital to analog signals. Working An analog information such as the voice of a telephone user produces electrical signals in analog form. This analog electrical signal is converted into digital signal which is in the form of electrical pulses. The electrical signal modulates the light emitted by an optical source (such as an LED or laser diode). Now, this optical signal is fed into the fibre. At the receiving end, the optical signal passing through the fibre is fed into a photo detector. The photo detector detects optical signal and converts it into pulses of electric current. This digital signal is once again converted into analog signal. This analog signal contains the same information voice as transferred from transmitting end. By this way, the information is transmitted from one end to other end.

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