1. SOUNDStationary Waves: If a progressive wave travelling in a medium meets the surface of an...

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1. SOUND

Introduction:

Sound is produced due to the vibrations of the body. These vibrations are transferred to the air medium and

propagated in all directions in the form of waves.

The number of vibrations made in one second is known as frequency of the sound. It is expressed in hertz (Hz).

The range of the frequency between 20 Hz and 20,000 Hz is the audible range, Human ears cannot respond to the

sound below and above this range.

Frequency : below 20 Hz infrasonic

above 20,000 ultrasonic.

Sound is a mechanical wave and hence it requires a medium to propagate. It travels with the velocity of 330 ms-1

in air. Sound waves are reflected and refracted like light waves.

Wave motion:

This continuous movement of the disturbance is called a wave.

If a wave passes in a medium, the particles of the medium vibrate about their mean position. The particles do not

move along with the wave, only the vibrations are transferred from one particle to adjacent particle of the

medium in the form of energy.

There are two types of wave motion. They are

1) Longitudinal wave motion and

2) Transverse wave motion

1) Longitudinal wave motion: If the particles of the medium vibrate parallel to the direction of propagation of the

wave, the wave is known as longitudinal wave.

Examples:

1. The propagation of sound in air

2. The propagation of sound in gas

3. The propagation of sound inside the liquid


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The longitudinal waves travel in a medium in the form of compressions and rarefactions. The place where the

particles of the medium crowded together are called compressions and the places where the particles spread out

are called rarefactions

2) Transverse wave motion: If the particles of the medium vibrate perpendicular to the direction of propagation

of the wave, the wave is known as transverse wave.

Examples :

1. Ripples travelling on the water surfaces.

2. Waves travelling along a rope.

3. Other waves like light waves, heat radiations, radio waves etc.

The transverse waves travel in a medium in the form of crests and troughs. The points where the particles of the

medium displaced maximum in the upward direction are called crests. The points where the particles displaced

maximum in the downward direction are called troughs

Amplitude:

When sound wave propagates in a medium, the maximum displacement of the vibrating particles of the medium

from their mean position is called amplitude.

Wavelength (λ) :

The wavelength is the distance between two consecutive particles of the medium which are in the same

state of vibration.


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It is also defined as the distance travelled by the wave during the time the vibrating particle completes one

vibration.

In longitudinal waves, the wavelength is the distance between two successive compressions or

rarefactions.

In transverse waves, the wavelength is the distance between two successive crests or troughs

Period (T): The time taken by the vibrating particle to make one vibration is called period.

Frequency (n): The frequency is the number of vibrations made by the vibrating particle in one second.

Velocity (v): The distance travelled by the sound wave in one second is known as velocity of sound.

Relation between Wavelength, Frequency and Velocity of a Wave :

Let n be the number of vibrations made by the vibrating particle in one second. It is also known as its frequency.

Time taken for one vibration = period (T) = 1/n.

Let λ be the wavelength of the wave produced

Velocity of the wave is the distance through which the wave advances in the medium in one second

∴ 𝑽𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒐𝒇 𝒘𝒂𝒗𝒆 𝒎𝒐𝒕𝒊𝒐𝒏, 𝑽 =𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒕𝒓𝒂𝒗𝒆𝒍𝒍𝒆𝒅

𝒕𝒊𝒎𝒆 𝒕𝒂𝒌𝒆𝒏

V = λ/T =λ/(1/n) = λ n

∴V = nλ

Stationary Waves:

If a progressive wave travelling in a medium meets the surface of an obstacle, it is reflected. The reflected wave is

superimposed on the incident wave to form a new type of wave called stationary wave

Also, when two identical waves having equal wavelength and amplitude travel in opposite directions they

superimpose on each other forming stationary wave.

At certain points of the medium, the displacement due to the two waves cancel each other and those points

remain at rest. Such points are called nodes (N). At certain other points there is maximum displacement. Such

points are called antinodes (A).


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The distance between two successive nodes or antinodes is λ / 2

The distance between a node and the next antinode is λ / 4

The longitudinal waves also produce the stationary waves.

VIBRATIONS

Free Vibrations : The vibrations of any body with its natural frequency are called free vibrations.

Forced Vibrations : The vibrations of a body with a frequency induces vibrations on another vibrating agent are

called forced vibrations.

Resonance : When the forced vibrations given on the body is equal to its natural frequency of vibrations, the body

vibrates with maximum amplitude. This phenomenon is called resonance.

According to law of vibrations,

𝑛 ∝1

𝑙

𝑇

𝑚 𝑜𝑟 𝑛 = 𝑘

1

𝑙

𝑇

𝑚

Where k – is constant and its value is equal to ½

∴ 𝑛 ∝1

2𝑙

𝑇

𝑚

If ∴ 𝑇 = 𝑀𝑔 =1

2𝑙

𝑀𝑔

𝑚

Note : The linear density,

𝑛 =1

2

𝑀

𝑙2

𝑔

𝑚


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𝑚 =𝑚𝑎𝑠𝑠

𝑙𝑒𝑛𝑔𝑡 𝑕=

𝑣𝑜𝑙𝑢𝑚𝑒 × 𝑑𝑒𝑛𝑠𝑖𝑡𝑦

𝑙𝑒𝑛𝑔𝑡 𝑕=

𝜋𝑟 2𝑙𝜌

𝑙= 𝜋𝑟2𝜌

ACOUSTICS OF BUILDINGS:

Echo : The first reflected sound is known as echo. The sound produced by a source is propagated continuously in

a medium if there is no disturbance. But if it meets the hard surface of an obstacle, it is reflected.

Reverberation :

The sound produced in a hall suffers multiple reflections before it becomes inaudible. As a result of these

reflections, the listener continues to receive sound, even if the source of sound is cut off. This prolonged

reflection of sound in a room even after the sound source has been stopped is called reverberation. It is the

persistence of sound due to multiple reflections from the walls, floor and ceiling of a hall. The reverberation is also

called multiple echoes.

Reverberation time:

If a building is to be acoustically correct, its reverberation time must be in optimum level. It should not be too long

or too short. if it is too short, then the room becomes dead in sound aspect. If it is too long, then the

reverberation will be there inside the building for long duration. This time is known as reverberation time.

Sabine formula :

Sabine derived an equation for the reverberation time. Where V is the volume of the hall, α is the coefficient of

absorption of each reflecting surface present in the hall and A is the area of the each second sound absorbing

surface present in the hall.

𝑇 =0.16𝑉

α A𝑠𝑒𝑐𝑜𝑛𝑑

Coefficient of absorption of sound energy:

The co-efficient of absorption of sound energy of any surface is defined as the ratio of the sound energy absorbed

by the surface to the total sound energy incident on the surface. Let α be the coefficient of absorption of sound

energy of a surface, then

α = 𝑇𝑕𝑒 𝑠𝑜𝑢𝑛𝑑 𝑒𝑛𝑒𝑟𝑔𝑦 𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑 𝑏𝑦 𝑡𝑕𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒

𝑇𝑕𝑒 𝑡𝑜𝑡𝑎𝑙 𝑠𝑜𝑢𝑛𝑑 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 𝑜𝑛 𝑡𝑕𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒

The reverberation time value depends on the use for which the building is designed. The acceptable limit for

reverberation time is


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For speech, 0.5 s

For music, between 1.0 and 2.0 s

For small theatres, between 1.0 and 1.5 s

For larger theatres, up to 2.3 s

Practice Problems:

1. A wire 50cm long and of mass 6.5 × 10−3kg is stretched so that it makes 80 vibrations per second. Find

the stretching tension.

Sol:

𝑚 =𝑚𝑎𝑠𝑠

𝐿𝑒𝑛𝑔𝑡 𝑕=

6.5×10−3

0.5= 13 × 10−3𝑘𝑔𝑚−1

𝑛 =1

2𝑙

𝑇

𝑚 𝑜𝑟 𝑛2 =

1

4𝑙2

𝑇

𝑚

∴ Tension 𝑇 = 4𝑙2𝑚2 𝑛 = 2 × 0.5 × 0.5 × 13 × 10−3 × 80 × 80

= 83.2 N

2. The density of a sonometer wire of radius 0.3mm is 7800 𝑘𝑔𝑚−3. Find its linear density

Linear density 𝑚 = 𝜋𝑟2𝜌

Sol:

𝑚 = 3.14 × 0.3 × 10−3 2 × 7800

= 2204.28 × 10−6

= 2.204 × 10−3𝑘𝑔 𝑚−1


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2. Lattices

A regular three dimensional arrangement of points in space is called a crystal lattice. There are only 14 possible

three dimensional lattices. These are called Bravais Lattices. The following are the characteristics of a crystal

lattice.

a) Each point in a lattice is called lattice point or lattice site.

b) Each point in a crystal lattice represents one constituent particle which may be an atom. a molecule (group

of atoms) or an ion.

c) Lattice points are joined by straight lines to bring out the geometry of the lattice.

Unit cell is the smallest portion of a crystal lattice which, when repeated in different directions, generates the

entire lattice.

A unit cell is characterized by:

I. Its dimensions along the three edges, a, b and c. These edges may or may not be mutually perpendicular.

II. angles between the edges, 𝛼(between b and c) 𝛽 (between a and c) and 𝛾 (between a and b). Thus, a unit

cell is characterized by six parameters. a, b, c, 𝛼 , 𝛽 and 𝛾.

Unit cells can be broadly divided into two categories, primitive and centred unit cells.

a) Primitive Unit Cells

When constituent particles are present only on the corner positions of a unit cell. It is called as primitive

unit cell.

b) Centred Unit Cells

When a unit cell contains one or more constituent particles present at positions other than corners in

addition to those at corners, it is called a centred unit cell. Centred unit cells are of three types:

I. Body – Centred Unit Cells: Such a unit cell contains one constituent particle (atom, molecule or ion)

at its body – centre besides the ones that are at its corners.

II. Face – Centred Unit Cells: Such a unit cell contains one constituent particle present at the centre of

each face, besides the ones that are at its corners.

III. End – Centred Unit Cells: In such a unit cell, one constituent particle is present at the centre of any

two opposite faces besides the ones present at its corners.

In all, there are seven types of primitive unit cells.


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Crystal system Possible Variations

Axial distances or edge lengths

Axial angles Examples

Cubic Primitive, Body – centred, Face – Centred

a = b = c 𝛼 = 𝛽 = 𝛾 = 90° NaCl, Zinc blende, Cu

Tetragonal Primitive, Body – centred

a = b ≠ c 𝛼 = 𝛽 = 𝛾 = 90° WHite tin, SnO2, TiO2, CaSO4

Orthorhombic Primitive, Body – centred, Face – centred, End – centred

a ≠ b ≠ c 𝛼 = 𝛽 = 𝛾 = 90° Rhombic sulphur, KNO3, BaSO4

Hexagonal Primitive a = b ≠ c 𝛼 = 𝛽 = 90° 𝛾 = 120°

Graphite, ZnO, CdS,

Rhombohedral or Trigonal

Primitive a = b = c 𝛼 = 𝛽 = 𝛾 ≠ 90° Calcite (CaCO3), HgS (Cinnabar)

Monoclinic Primitive, End – centred

a ≠ b ≠ c 𝛼 = 𝛾 = 90 𝛽 ≠ 90

Monoclinic sulphur, Na2SO4, 10H2O

Triclinic Primitive a ≠ b ≠ c 𝛼 ≠ 𝛽 ≠ 𝛾 ≠ 90 K2Cr2O7,CuSO4, 5H2O,H2BO3

Primitive Cubic Unit Cell

Primitive cubic unit cell has atoms only at its corner. Each atom at a corner is shared

between eight adjacent unit cells as shown in Fig. four unit cells in the same layer and

four unit cells of the upper (or lower) layer. Therefore, Only 1

8th of an atom (or molecule

or ion) actually belongs to a particular unit cell.

In all, since each cubic unit cell has 8 atoms on its corners, the total number of atoms in

one unit cell is 8 ×1

8= 1 atom.

A body-centred cubic (bcc) unit cell has an atom at each of its corners and also one atom at its body centre. Body centre wholly belongs to the unit cell in which it is present. Thus in a body-centered cubic (bcc) unit cell:


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(i) 8 corners ×1

8 per corner atom = 8 ×

1

8 = 1 atom

(ii) 1 body centre atom = 1 × 1 = 1 atom

Total number of atoms per unit cell = 2 atoms

Face Centred Cubic Unit Cell:

A face-centred cubic (fcc) unit cell contains atoms at all the corners and at the centre of all the faces of the cube. Thus, in a face-centred cubic (fcc) unit cell:

(i) 8 corners atoms × 1

8 atom per unit cell 8 ×

1

8 = 1 atom

(ii) 6 face-centred atoms × 1

2 atom per unit cell = 6 ×

1

2 = 3 atoms

Total number of atoms per unit cell = 4 atoms


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3. ULTRASONICS

Sound waves of frequencies above the audible range (above 20,000 Hz or 20kHz ) are known as ultrasonic waves

or simply ultrasonics. These sound waves are not audible to human ear.

Production of Ultrasonics

Ingeneral, there are two important methods to produce ultrasonic waves. They are.

a) Magnetostriction generator or oscillator

b) Piezo – electric generator or oscillator

Magnetostriction Effect

Statement

When a ferromagnetic rod is placed in a magnetic field parallel to its length, then there is small change in length

of the rod . This is known as magnetostriction effect.

The change in length (increase or decrease) of the rod depends on the strength of the magnetic field

applied and the nature of the rod.

The change in length does not depend on the direction of the magnetic field applied (i.e., parallel and anti

parallel to the length of the rod).

The change in length is very small.

Magnetostriction Materials

Example: Iron, Nickel, Cobalt or alloys of these materials.

Advantages:

Simple construction, low cost

Magnetostrictive materials are not expensive.

Disadvantages:

This oscillator can produce ultrasonic sound waves of frequencies upto 3MHz (3 × 106Hz) only.

The natural frequency of the rod depends on temperature. So, it cannot withstand high temperature.

Sound waves frequency

0-20Hz Infrasonic

frequency

20Hz – 20,000Hz Audible

frequency

Above 20,000Hz

Ultrasonic frequency

𝒏 =𝟏

𝟐𝒍

𝒀

𝝆

The natural frequency of vibration of the

rod is given by

Where l – Length of the rod

Y – Young’s modulus of the rod,

𝝆 − Density of the rod


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Practice Problems:

Calculate the natural frequency of 40mm length of a pure iron rod. Given that the density of pure iron is

7.25 × 103𝑘𝑔𝑚−3 and its Young’s modulus is 115 × 109𝑁𝑚−2 Can you use it in magnetostriction oscillator to

produce ultrasonic waves?

Given data

Length of the iron rod (l ) = 40mm = 40 × 103𝑚

Density of pure iron (𝜌) 7.25 × 103𝑘𝑔𝑚−3

Young’s modulus of iron (Y) = 115 × 109𝑁𝑚−2

Solution

The natural frequency of the rod is given by

𝒏 =𝟏

𝟐𝒍

𝒀

𝝆

Substituting the given values, we have

n = 1

2×40×10−3

115×109

7.25×103

n = 49784 Hertz

n = 49.784kHz.

The natural frequency of the rod is greater than 20 kHz. Hence, this rod can be used to produce ultrasonic waves.

Piezo – Electric Effect:

When a mechanical stress is applied to one pair of opposite faces of a quartz crystal, then and opposite

electrical charges are developed on the other pair of opposite faces of the crystal. This is known as 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 (mechanical stress)

is developed across the other pair of opposite faces of the crystal.

This is known as inverse Piezo – electric effect.

Piezo – electric materials

Quartz, Rochelle salt, tourmaline and zinc blende are the best

examples for piezo electric materials.

𝒏 =𝑷

𝟐𝒍

𝒀

𝝆

The natural frequency of vibration of the

rod is given by

Where P=1,2,3..(stands for fundamental,

1st overtone, 2nd overtone, etc.

l – Length of crystal

Y – Young’s modulus of the crystal

𝝆 − Density of the Crystal


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Advantages:

a) This oscillator is more efficient than magnetostriction

oscillator.

b) Ultrasonic frequencies up to 500MHz (500 × 106) can be

produced by this oscillator.

Disadvantages

a) The cost of piezo – electric quartz is very high.

b) Its cutting and shaping are very complex and difficult.

Acoustic Grating

When the ultrasonic waves passes through a

transparent liquid, longitudinal stationary waves are formed

due to alternating compression and rarefaction.

Now, the liquid medium behaves like a diffraction

grating. When a monochromatic light is passed through this

grating, the diffraction pattern is obtained. Such a grating is

known as acoustic grating.

Properties of Ultrasonic Waves

1) Ultrasonic waves are sound waves of short

wavelength with very high frequency.

2) They have high energy content and high

penetrating power

3) Just like ordinary sound waves, they get reflected,

refracted and absorbed.

4) They can travel over long distances without much

loss of energy.

5) They produce heat when they pass through a

substance.

6) If an arrangement is made in a liquid to from

stationary ultrasonic waves, it acts as a diffraction

grating. It is called acoustic grating.


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Non – Destructive Testing (NDT)

It is a method of testing a material without destructing or damaging the material. In this method, radiation

like X-rays or ultrasonics is passed through the material. The interaction of radiation with the material is studied.

The grating formula is given by

𝑑 𝑠𝑖𝑛𝜃𝑛 = 𝑛𝜆 … (1)

d – Distance between any two successive nodes or antinodes of stationary waves.

n – Order of diffraction

𝜆 − Wavelength of the monochromatic light used

𝜃𝑛 - Angle of diffraction for nth order.

Let 𝜆𝑚be the wavelength of ultrasonic waves in the liquid medium, then

2𝑑 = 𝜆𝑚

𝑑 =𝜆𝑚

2 … (2)

Substituting eqn (2) in eqn (1), we have 𝜆𝑚

2𝑠𝑖𝑛𝜃𝑛 = 𝑛𝜆

𝜆𝑚 =2𝑛𝜆

𝑠𝑖𝑛𝜃 𝑛 … (3)

Knowing the wavelength of monochromatic light and measuring 𝜃𝑛 , the wavelength of ultrasonic waves (𝜆𝑚 ) in a given liquid medium is calculated. Velocity of ultrasonic waves in liquid If the frequency of the ultrasonic waves v is known, then the velocity of ultrasonic waves int eh liquid is found from the relation 𝑣 = 𝑣𝜆𝑚 … (4) Substituting for 𝜆𝑚 from eqn (3) in eqn (4),

we have 𝑣 =2𝑛𝜆𝑣

𝑠𝑖𝑛𝜃 𝑛 … (5)


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Working:

The high frequency generator excites the transmitting probe and it produces ultrasonic sound pulses.

These sound pulses are sent into specimen. They strike the upper surface of the specimen and produce a sharp

pip (echo) at the left hand side of the CRO screen (Shown in fig).

The Ultrasonic sound pulses pass through the different interfaces of

the specimen.

If the specimen is good without any defects, this sound wave

strikes the bottom surface of the specimen and reflects back. This

produces a pip on the right hand side of the CRO screen.

The ultrasonic waves get reflected whenever there is a

change in medium.

Whenever a defect is present between top and bottom

surfaces of the specimen, the most of the ultrasonic sound pulses

strike this defect and they are reflected. The reflected sound pulses

from the defect reach the receiver probe earlier than back echo.

This is indicated by a pip (echo) on the CRO screen in between left

and right pips.

Practice Problems:

1. Calculate the velocity of ultrasonic waves in a liquid in an acoustic grating experiment using the following

data.

Solution:

We know that

𝑣 =2𝑣𝑛𝜆

𝑠𝑖𝑛𝜃𝑛


𝒅 = 𝒗𝒕

𝟐

If the velocity (v) of sound waves in the

specimen is known, then the depth of defect

from the surface of the specimen (d) is

determined by using the relation.


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𝑣 =2 × 100 × 106 × 1 × 600 × 10−9

sin 5°

=2 × 6 × 10

0.0872

𝑣 = 1376𝑚𝑠−1

2. An Ultrasonic source of 0.09 MHz sends down a pulse towards the seabed which returns after 0.55 sec.

The velocity of sound in water is 1800m/s. Calculate the depth of the sea and wavelength of the pulse.

Solution:

Depth of the sea 𝑑 =𝑣𝑡

2=

1800 ×0.55

2

d = 495m

Wavelength of the ultrasonic pulse is

𝜆 =𝑢

𝑣=

1800

0.09×106 = 20000 × 10−6

𝜆 = 0.02𝑚


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4. LASERS & FIBRE OPTICS

Introduction:

LASER stands for Light Amplification by Stimulated Emission of Radiation.

Laser is a device which emits a powerful, monochromatic, collimated beam of light. The emitted light waves from

laser source are coherent in nature.

Laser light emerges as a narrow beam which can travel long distance without much loss of intensity and

energy. Actually, the laser amplifies the light waves.

Interaction of light radiation with materials

Consider an assembly of atoms in a material which is exposed to light radiation (a stream of photons with

energy hv.)

In general, three different processes occur when the light radiation interacts with material. They are

1. Stimulated absorption

2. Spontaneous emission

3. Stimulated emission

Process – 1: Stimulated absorption

An atom in ground state with energy E1 absorbs and incident photon of energy hv and excited to higher

energy state with energy E2 .

The excited atoms do not stay in the higher energy state for a longer time. It is the tendency of atoms in

excited state to come to the lower energy state.

Thus, the atoms in excited state quickly return to ground state by emitting a photon of energy hv.

The emission of photons takes place in two ways, namely

a) Spontaneous emission

b) Stimulated emission

Process – 2: Spontaneous emission

The atom in the excited state E2 (higher energy state) returns to ground state E1 (lower energy state) by

emitting a photon of energy hv (∆𝐸 = 𝐸2 − 𝐸1 ) without the influence of any external agency


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Such emission of light radiation which is not triggered by any external influence is called spontaneous emission.

It is a random and also uncontrollable process.

Process – 3: Stimulated emission

Such kind of forced emission of photons by the incident photons is called stimulated emission. It is also

known as induced emission. It plays a key factor for the working of a laser.

Spontaneous emission Stimulated emission

1. Emission of light radiation is spontaneous i.e., without any external agency

Emission of light radiation is stimulated (triggered) by incident photons.

2. Emitted photons travel randomly in all directions

Emitted photons travel in particular direction

3. Emitted photons cannot be controlled. They are not coherent

Emitted photons can be controlled and they are coherent.

4. This process is a key factor for ordinary light emission

This process is a key factor for laser operation.

Population Inversion

It is a situation in which the number of atoms in higher energy state is more than that in lower energy

state.

The state of achieving more number of atoms in higher energy state than the that of lower energy state

(i.e., 𝑁2 > 𝑁1) is known as population inversion.

Conditions for Population inversion:

There must be atleast two energy levels (𝐸1 𝑎𝑛𝑑 𝐸2)

There must be a source to supply the energy to the medium.

The atoms must be continuously raised to the excited state.


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Characteristics of Laser:

Laser is basically a light source. Laser light has the following important characteristics

1. High directionality

2. High intensity

Laser source emits light as a narrow beam and its energy is concentrated in a small region (spot).

This concentration of energy gives a high intensity to the laser light.

3. Highly monochromatic

Ordinary light spreads over a wavelength range of the order of 100nm. But a laser beam has very

little spreading of the order of 1nm. Thus, laser beam is highly monochromatic. ie., it emits only

one colour of light.

4. Highly coherent

The light emitted from a laser source consists of wave trains. These wave trains have same

frequency, phase and direction. So, they are coherent.

S.No. Ordinary Light Laser Light

1. Light emitted is not monochromatic Light emitted is highly monochromatic.

2. Light emitted does not have high degree of coherence

Light emitted has high degree of coherence.

3. Emitted light spreads in all directions (not directional)

Emitted light spreads only in one direction (directional)

4. Light is less intense and bright Laser light is more intense and bright

Types of Lasers:

Based on the type of active medium, the laser systems are broadly classified into the following types:

S.no Type of Laser Examples

1. Solid state laser Ruby, Nd:YAG lasers

2. Gas Laser He-Ne, CO2, Argon lasers

3. Liquid Laser SeOCl2, Europium chelate lasers

4. Due laser Rhodamine 6G, Coumarin dye lasers

5. Semiconductor laser GaAs, GaAsP, GaAlAs, InP Lasers.


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Comparison table of Different Types of Lasers

S.No. Characteristics Nd-YAG laser CO2 Laser Semiconductor laser

1. Type Solid state laser Molecular gas laser Semiconductor laser

2. Active medium Yitrium Aluminium

Garnet (Y3Al5O12)

Gas mixture of CO2, N2 and

He

Pn junction

3. Pumping method Optical pumping Electrical discharge

method

Direct conversion

4. Optical resonator Ends of the polished

rods in silver

Metallic concave mirror of

gold or silicon coated with

aluminium

End faces of the

junction diode

5. Power output 2×104 W 10kW 1 mW

6. Nature of output Pulsed Continuous or pulsed Pulsed or continuous

wave from

7. Wavelength 1.06𝜇m 9.6 𝜇m and 10.6 𝜇m 8300-8500 Å

Fibre Optics:

The development of lasers and optical fibres has brought about a revolution in field of communication

systems.

To have an efficient communication system, the information carried by the light waves should need a

guiding medium through which it can be transmitted safely.

This guiding medium is called as optical fibre. The communication through optical fibre is known as light –

wave communication or optical communication.

Currently in most part of the world, fibre optics is used to transmit voice, video and digital data signals

using light waves from one place to other place.

Optical Fibre:

The optical fibre is a wave guide.

It consists of an inner cylinder made of glass or plastic called core. The core has high refractive index n1. This core

is surrounded by a cylindrical shell of glass or plastic called cladding.

The cladding has low refractive index n2. This cladding is covered by a jacket. It protects the fibre from the

moisture and abrasion.


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The light is transmitted through this fibre by the principle of total

internal reflection. The fibre guides the light waves to travel over a long

distance without much loss of energy.

Optical fibre cable:

A bundle of optical fibres consists of thousands of individual fibre wires as thin as human hair, measuring

0.004mm in diameter is known as optical fibre cable.

Guiding mechanism:

The light which enters from one end of the fibre strikes on the interface of core and cladding at greater

angle of incidence. The light beam undergoes total internal reflection and it passes along the length of the fibre.

Most of the light propagate along the length of fibre and it comes out at the other end of the fibre. Thus,

optical fibre acts as a wave guide.

Total internal reflection in optical fibre.

Case – 1

The incident ray AO makes and angle 𝜃1 with normal in the medium of refractive index 𝑛1 . This incident

ray is refracted into the medium of refractive index 𝑛2 .

Case – 2

If the angle of incidence (𝜃1) is increased for a certain value equal to critical angle (𝜃𝑐 ), then 𝜃2is 90°.

i.e., If 𝜃1 = 𝜃𝑐 , then 𝜃2= 90°.

In this case the incident ray along BO is refracted at the interface and it just emerges along the boundary of

separation OB’.


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Case - 3

If the angle of incidence (𝜃1) is greater than the critical angle of incidence (𝜃𝑐 ), then the incident light ray

‘CO’ is reflected back into the same medium by total internal reflection.

Expression for critical angle 𝜽𝒄

For the refraction of light, a relation between the angle of incidence (𝜃1) and angle of refraction (𝜃2) is

given by Snell’s law.

𝑛1𝑠𝑖𝑛𝜃1 = 𝑛2𝑠𝑖𝑛𝜃2

for total internal reflection,

𝜃1 = 𝜃𝑐 𝑎𝑛𝑑 𝜃2 = 90°

𝑛1𝑠𝑖𝑛 𝜃𝑐 = 𝑛2𝑠𝑖𝑛90°

𝑠𝑖𝑛 𝜃𝑐 =𝑛2

𝑛1𝑠𝑖𝑛90°

𝑠𝑖𝑛 𝜃𝑐 =𝑛2

𝑛1 [∵ sin 90° = 1]

𝜃𝑐 = 𝑠𝑖𝑛−1 𝑛2

𝑛1

Conditions for total internal reflection

Total internal reflection the walls of optic fibre can occur only in the following two conditions:

1. The glass around the centre of fibre (core) should have higher refractive index (𝑛1) than that of the

material (cladding) surrounding the fibre (𝑛2)

2. The light should incident at an angle (between the path of the ray and normal to the fibre wall) greater

than the critical angle 𝜃𝑐

Acceptance angle:

The maximum angle 𝜃0 at which a ray of light can enter through one end of the fibre and still be totally

internally reflected is called acceptance angle of the fibre.

𝜃0 = 𝑠𝑖𝑛−1 𝑛1

2 − 𝑛22

𝑛0

If the medium surrounding the fibre is air, then 𝑛0 = 1

𝜃0 = 𝑠𝑖𝑛−1 𝑛12 − 𝑛2

2

Numerical Aperture (NA)

The sine of the acceptance angle of the fibre is known as numerical aperture (NA). It denotes the light

gathering capability of the optical fibre.

It is given by

𝑁𝐴 = 𝑠𝑖𝑛𝜃0

Substituting for 𝑠𝑖𝑛𝜃0 from eqn (4) we have,


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If the medium surrounding the fibre is air, then 𝑛0 = 1

𝑁𝐴 = 𝑛12 − 𝑛2

2

Condition for propagation of light

This is the condition for propagation of light within the fibre.

Fractional Index change (∆):

Fractional index change

∆=𝑅𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑣𝑒 𝑖𝑛𝑑𝑒𝑥 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑐𝑜𝑟𝑒 𝑎𝑛𝑑 𝑐𝑙𝑎𝑑𝑑𝑖𝑛𝑔

𝑅𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑣𝑒 𝑖𝑛𝑑𝑒𝑥 𝑜𝑓 𝑐𝑜𝑟𝑒 𝑜𝑓 𝑎𝑛 𝑜𝑝𝑡𝑖𝑐𝑎𝑙 𝑓𝑖𝑏𝑒𝑟

i.e., ∆=𝑛1−𝑛2

𝑛1

Relation between NA and ∆ :

𝑵𝑨 = 𝒏𝟏 𝟐∆

Thus an increase in the value of ∆ increase NA and this enhances the light gathering capacity of the fiber. We

cannot increases ∆ to a very large value since it leads to what is called ‘intermodal dispersion’ which causes signal

distortion.

Practice Problems:

1. Compute the numerical aperture and acceptance angle of an optical fibre from the following data.

Refractive index of core 𝑛1 = 1.55

Refractive index of cladding 𝑛2 = 1.50

Surrounding medium (air) 𝑛0 = 1

Solution:

Numerical aperture is given by

𝑁𝐴 = 𝑛12 − 𝑛2

2

𝑁𝐴 = 1.552 − 1.502 = 0.394

𝑁𝐴 = 𝑛1

2 − 𝑛22

𝑛0

𝑠𝑖𝑛𝜃𝑖 < 𝑁𝐴


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NA = 0.394

Acceptance angle is given by

𝜃0 = 𝑠𝑖𝑛−1(𝑁𝐴)

𝜃0 = 𝑠𝑖𝑛−10.39 = 23°

𝜃0 = 23°

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 fibres is shown in fig.

Differences between single mode fiber and multi mode fiber

S.No. Single mode fiber Multimode fiber

1. In single mode fiber, only one mode can

propagate through the fiber

Multimode fiber allows a large number of paths or modes for

the light rays travelling through it.

2. It has smaller core dia and the difference

between the refractive index of core and

cladding is very small.

It has larger core dia and refractive index difference is larger

than the single mode fiber.

3. No dispersion i.e., degradation of signal

during travel in fiber.

There is signal degradation due to multimode dispersion

4. Fabrication is difficult and costly Fabrication is less difficult and not costly.

Optical fibers

Material No. of modes Refractive index profile

Glass

fiber

Plastic

fiber

Single

mode

fiber

Multi

mode

fiber

Step

index

fiber

Graded Index

fiber (GRIN

fiber)


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Differences between step index and graded index fiber

S.No. Step index fiber Graded index fiber

1. The refractive index of the core is uniform

throughout and undergoes an abrupt (or) step

change at the cladding boundary

The refractive index of the core is made to vary gradually

such that maximum it is at the centre of the core.

2. The diameter of the core is about 50 – 200𝜇𝑚

in the case of multimode fiber and 10 𝜇𝑚 in

the case of single mode fibre.

The diameter of the core is about 50 𝜇𝑚 in the case of

multimode fibre.

3. Attenuation is more for multimode step index

fiber but for single mode step index fiber. it is

very less

Attenuation is less

4. Numerical aperture is more for multimode

step index fiber but for single mode step index

fiber it is very less.

Numerical aperture is less

Splicing of Fibre:

We join two ends of copper wires by a process called soldering.

Similarly, two optical fibres, used for communication are joined by a process called splicing.

To carry the information’s (using light) for a long distance, the fibres should be connected with one another.

The process of joining two fibres is called splicing

Dispersion:

When an optical signal (or) pulse is sent into the fiber, the pulse spreads or broaden as it propagates

through the fiber. This phenomenon is called as dispersion.

Types of dispersion:

There are three types of dispersion in optical fiber. They are

1. Chromatic dispersion

Characteristics:

The chromatic dispersion is directly proportional to the frequency bandwidth of the

transmitted pulse.


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2. Waveguide dispersion and

3. Intermodal dispersion

Characteristics:

This kind of dispersion is completely absent in single mode fibres.

Advantages of Fibre – Optic Communication

1. Extremely high bandwidth

2. Ease of handling

3. Less cross – talk

4. Noise – free transmission

5. Economical

6. Safety

7. Longer life span

8. Ease of maintenance

Disadvantages of Optical Fibres

1. Price

2. Special skills

3. Tapping

4. Caution

Practice Problems:

1. In an optical fibre, the core material has refractive index 1.6 and refractive index of clad material is 1.3.

What is the value of critical angle? Also calculate the value of angle of acceptance cone.

Solution:

Critical angle is given by

sin 𝜃𝑐 =𝑛2

𝑛1

= 1.3

1.6= 0.8125

∴ 𝜃𝑐 = 54.3°

Acceptance angle

𝑠𝑖𝑛𝜃0 = 𝑛12 − 𝑛2

2

𝜃0 = 𝑠𝑖𝑛−1 𝑛12 − 𝑛2

2

= 𝑠𝑖𝑛−1 1.62 − 1.32

𝜃0 = 𝑠𝑖𝑛−1(0.87)

𝜃0 = 60.5°

Angle of acceptance cone = 2𝜃0 = 121°


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2. Calculate the fractional index change for a given optical fibre if the refractive indices of the core and the

cladding are 1.563 and 1.498 respectively

Solution

Fractional index change ∆=𝑛1−𝑛2

𝑛1

= 1.563−1.498

1.563

= 0.065

1.563= 0.0415

∆ = 0.0415


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5. Quantum Physics

Compton Effect

Compton effect refers to the change in the wavelength of scattered X – rays by a material.

Statement

When a beam of X – rays is scattered by a substance of low atomic number, the scattered X – ray radiation

consists of two components, one component has the same wavelength λ as the incident ray and the other

component has a slightly longer wavelength λ’.

This change in the wavelength of scattered X – rays is known as Compton shift. The phenomenon is called

Compton effect.

Theory of Compton Effect (derivation)

the change in wavelength is given by

It is found that the change in wavelength (dλ) does not depend on the wavelength of the incident

radiations and the nature of the scattering substance. But it depends only on the angle of scattering (𝜃)

When 𝜽 = 𝟎 When 𝜽 = 𝟗𝟎° When 𝜽 = 𝟏𝟖𝟎°

dλ = 0 dλ = 0.0243Å dλ = 0.0486Å

Thus, the change in wavelength is maximum at 𝜃 = 180°

Importance of Compton effect:

Historically, Compton effect provided the direct confirmation of the particle nature of electromagnetic

radiation. It had conclusively showed that photons carry energy and momentum like any material particle.

Further, it proved that the momentum as well as the energy of electromagnetic radiation is quantized.

Compton effect showed that the photon description applies not only to visible light but also to X – rays.

Conclusion:

Hence, Compton effect explains the elastic collision of two particles, i.e., electron and photon. It also

proves that the particle nature of light radiation and the validity of the quantum concept.

𝐦𝐜𝟐 = 𝐡 𝐯 − 𝐯′ + 𝐦𝟎𝐜𝟐

𝐝𝛌 =𝐡

𝐦𝟎𝐜(𝟏 − 𝐜𝐨𝐬𝛉


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de-BROGLIE WAVELENGTH:

The waves associated with the matter particles are called matter waves or de-Broglie Waves.

𝜆 =𝑕

𝑝=

𝑕

𝑚𝑣

This equation is known as de-Broglie’s wave equation.

de-Broglie wavelength in terms of energy

de-Broglie wavelength 𝜆 =𝑕

2𝑚𝐸

𝜆 =𝑕

2𝑚𝑒𝑉


𝑕 − 6.625 × 10−34𝐽𝑠, 𝑒 − 1.6 × 10−19𝐶

𝑚 − 9.1 × 10−34𝑘𝑔

𝜆 =6.625×10−34

2×9.1×10−31 ×1.6×10−19 ×𝑉

𝜆 = 12.25×10−10

𝑉𝑚𝑒𝑡𝑟𝑒

𝜆 =12.25

𝑉Å


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6. INTERFERENCE FRINGES

The study about of light is called optics. The study of optics acoustic impedances of the media. This

property is used is divided into geometrical optics and physical optics, in ultrasonic testing. Geometrical optics

deals with the applications of the rectilinear propagation of light to different optical light must travel in straight

lines. Optical instruments like microscope, telescope and camera work on this principle. Reflection and Refraction

can be explained by this principle.

Interference, Diffraction and polarization can be explained only on the basis of wave theory of light. That

is, these phenomena could be explained only it we assume that light is a form of wave motion. The physical optic

deals with these phenomena exhibited by light on account of its wave aspect.

Interference:

Interference is the optical phenomenon in which brightness and darkness are produced by two exactly

similar light Waves meeting. When two light waves, of same frequency and having constant phase difference,

coincide in space and time, there is a modification in the intensity of light. The resultant intensity at any point

depends upon the amplitudes and the phase relationships of the two waves." This modification in the intensity

distribution resulting from the superposition of two waves of light is called' interference and the pattern of bright

and dark fringes produced is called interference pattern.

Conditions for interference:

1. To produce interference, we require two light sources. That two light sources must be coherent i.e. two sources

must send out waves having 1 constant phase difference or same phase,.

2. The two sources must be perfectly monochromatic emitting light, of a single wavelength.

3. The two sources must be as near as possible and the screen must be as far as possible from them.

4. The frequency and amplitude of the waves from these two coherent sources must be the same.

5. The two sources must be narrow.

6. The two waves must be propagated along the same direction to get coincidence.

7. The interfering waves are polarised, then their state of polarisation must be same.

The two coherence sources can be obtained from the same single soured. Any change inphase at the time

of emission will affect these two sources (which are derived from a single source) equally and the light starting

from them will always be in phase.


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Fringewidth 𝛽:

𝛽2 − 𝛽; =0.20

𝑑𝜆

where

d is the distance b/w the two virtual sources 𝑆1 and 𝑆2

Determination of d: To measure d, the distance b/w the two virtual sources by a direct method is difficult.

Hence a convex lens is introduced b/w the biprism and the eyepiece and it is moved until the two images of the

slit are obtained in the eyepiece.

𝛽2 − 𝛽1 =0.20

𝑑1𝑑2𝜆

Since 𝛽2, 𝛽1 , 𝑑1and 𝑑2 are all known, the wavelength of light can be calculated.

Displacement of fringes due to the introduction of mica sheet: With a thin mica sheet of refractive index ‘𝜇’ is

placed in the path of light from one of the two sources, the central bright fringe is shifted to some other points

due to the change in the path difference

Displacement of any maximum or central bright fringe by introducing mica sheet of thickness T is given by

𝑆 =𝐷

𝑑 𝜇 − 1 𝑡

or 𝑡 =𝑠𝑑

𝐷 (𝜇−1)

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