Prasanna

Students can Download Chapter 12 Atoms Notes, Plus Two Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Notes Chapter 12 Atoms

Introduction
What is the arrangement of +ve charge and the electrons inside the atom? In other words, what is the structure of an atom?

Alpha-particle Scattering And Rutherford’s Nuclear Model Of Atom
Rutherford’s scattering experiment:

Experimental arrangement:
α particles are incident on a gold foil (very small thickness) through a lead collimator. They are scattered at different angles. The scattered particles are counted by a particle detector.

Observations:
Most of the alpha particles are scattered by small angles. A few alpha particles are scattered at an angle greater than 90°.

Conclusions

1. Major portion of the atom is empty space.
2. All the positive charges of the atom are concentrated in a small portion of the atom.
3. The whole mass of the atom is concentrated in a small portion of the atom.

Rutherford’s model of atom

1. The massive part of the atom (nucleus) is concentrated at the centre of the atom.
2. The nucleus contains all the positive charges of the atom.
3. The size of the nucleus is the order of 10^-15m.
4. Electrons move around the nucleus in circular orbits.
5. The electrostatic force of attraction (between proton and electron) provides centripetal force.

1. Alpha-particle trajectory and Impact parameter:
The impact parameter is the perpendicular distance of the initial velocity vector of the a particle from the centre of the nucleus.

It is seen that an α particle close to the nucleus (small impact parameter) suffers large scattering. In case of head-on collision, the impact parameter is minimum and the α particle rebounds back. For a large impact parameter, the α particle goes nearly undeviated and has a small deflection.

2. Electron orbits (Rutherford model of atom):
In Rutherford atom model, electrons are revolving around the positively charged nucleus. The electro-static force of attraction between the positive charge and negative charge provide centripetal force required for rotation.
For a dynamically stable orbit,
Centripetal force = Electrostatic force of attraction
F_c = F_e

Thus the relation between the orbit radius and the electrons velocity,

Total energy of electron of Hydrogen atom (Rutherford model atom):
From eq. (1), we get

∴ Kinetic energy of electron
KE = \(\frac{1}{2}\)mv² ……….(3)
Substituting eq.(2) in eq. (3) we get
KE = \(\frac{e^{2}}{8 \pi \varepsilon_{0} r}\) …………(4)
The electrostatic potential energy of hydrogen atom
\(\frac{e^{2}}{8 \pi \varepsilon_{0} r}\)
u = \(\frac{-e^{2}}{4 \pi \varepsilon_{0} r}\) ………..(5)
∴ The total energy E of the electron in a hydrogen atom
E = K.E + Potential energy (U)

The total energy of the electron is negative. This implies that the electron is bound to the nucleus.
If E is positive, the electron will escape from the nucleus.

Atomic Spectra
There are two types of spectra

1. Emission spectra
2. Absorption spectra

1. Emission spectra:
When an atomic gas or vapor is excited, the emitted radiation has a spectrum which contains certain wavelength only. A spectrum of this kind is termed as emission line spectrum. It consists of bright lines on a dark background.

Absorption spectra:
When white light passed through a gas, the transmitted light has spectrum contain certain wavelength only. A spectrum of this kind is termed as absorption line spectrum. It consists of dark lines on a bright background.

1. Spectral series:

The frequencies of the light emitted by a particular element exhibit some regular pattern. Hydrogen is the simplest atom and therefore, has the simplest spectrum, the spacing between lines of the hydrogen spectrum decreases in a regular way. Each of these sets is called a spectral series.

The first such series was observed by a Johann Jakob Balmer in the visible region of the hydrogen spectrum. This series is called Balmer series. Balmer found a simple empirical formula for the observed wavelengths.

where λ is the wavelength, R is a constant called the Rydberg constant, and n may have integral values 3, 4, 5, etc. The value of R is 1.097 × 10⁷m^-1. This equation is also called Balmer formula.

Other series of spectra for hydrogen were discovered. These are known, as Lyman, Paschen, Brackett, and Pfund series. These are represented by the formulae:
Lyman series:

Balmer series:

Paschen series:
This series is in the infrared region. For this series the electron must jump from higher orbit to the third orbit.

Bracket series:
This series is the infrared region, for this the electron must jump from higher energy level to fourth orbit.

P-fund series:
This series is in the infrared region.

Bohr Model Of Hydrogen Atom
Limitations of Rutherford model:
1. Circular motion is an accelerated motion, an accelerated charge emit radiations. So that electron should emit radiation. Due to this emission of radiation, the energy of the electron decreases. Thus the atom becomes unstable.

2. There is no restriction for the radius of the orbit. So that electron can emit radiations of any frequency.

Bohr postulates:
Bohr combined classical and early quantum concepts and gave his theory in the form of three postulates.

• Electrons revolve round the positively charged nucleus in circular orbits.
• The electron which remains in a privileged path cannot radiate its energy.
• The orbital angular momentum of the electron is an integral multiple of h/π.
• Emission or Absorption of energy takes place when an electron jumps from one orbit to another.

Radius of the hydrogen atom:
Consider an electron of charge ‘e’ and mass m revolving round the positively charged nucleus in circular orbit of radius ‘r’. The force of attraction between the nucleus and the electron is

This force provides the centripetal force for the orbiting electron

According to Bohr’s second postulate, we can write
Angular momentum, mvr \(=\frac{n h}{2 \pi}\).
ie. v = \(\frac{n h}{2 \pi m r}\) _____(4)
Substituting this value of ‘v’ in equation (2), we get

Energy of the hydrogen atom:
The K.E. of revolving electron is
K.E\(=\frac{1}{2}\) mv² ______(6)
Substituting the value of equation (3) in eq.(6), we get
K.E = \(\frac{1}{2} \frac{e^{2}}{4 \pi \varepsilon_{0} r}\) ______(7)
The potential energy of the electron,
P.E = \(\frac{-e^{2}}{4 \pi \varepsilon_{0} r}\) _______(8)
ie. The Total energy of the hydrogen atom is,
T.E = Ke + PE

Substituting the value of equation (5) in equation (9) we get

1. Energy levels
Ground state (E₁):
Ground state is the lowest energy state, in which the electron revolving in the orbit of smallest radius.
For ground state n = 1
∴ Energy of hydrogen atom E₁ = \(\frac{-13.6}{n^{2}}\) = -13.6 ev.

Excited State (E₂):
When hydrogen atom receives energy, the electrons may raise to higher energy levels. Then atom is said to be in excited state.

First Excited state:
For first excited state n = 2
∴ Energy of first excited state E₂ = \(\frac{-13.6}{2^{2}}\) = -3.04ev
Similarly energy of second excited state
E₃ = \(\frac{-13.6}{3^{2}}\) = -1.51ev

Energy difference between E₁ and E₂ of H atom:
The energy required to exist an electron in hydrogen atom to its first existed state.
∆E = E₂ – E₁ = ^–3.4 – ^–13.6 = 10.2eV.

Ionization energy:
Ionization energy is the minimum energy required to free the electron from the ground state of atom. (ie. n = 1 to n = ∞)
The ionization of energy of hydrogen atom = 13.6 ev

2. Energy level diagram of hydrogen atom:

Note:
An electron can have any total energy above E = 0ev. In such situations electron is free. Thus there is a continuum of energy states above E = 0ev.

The Line Spectra Of The Hydrogen Atom
According to the third postulate of Bohr’s model, when an atom makes a transition from higher energy state (n_i) to lower energy state (n_f), photon of energy hv_if is emitted.
ie. hν_if = E_ni – E_nf

De Broglie’s Explanation Of Bohr’s Second Postulate Of Quantization
Louis de Broglie argued that the electron in its circular orbit, behalf as a particle wave. Particle waves can produce standing waves under resonant conditions.
The condition to get standing wave,
2πr_n = nλ
n = 1, 2, 3……..
The quantized electron orbits and energy states are due to the wave nature of the electron.

DeBroglie’s Proof for Bohr’s second postulate:
According to De Broglie, the electron in a circuit orbit is a particle wave. The particle wave can produce standing waves under resonant conditions. The condition for resonance for an electron moving in n^th circular orbit of radius r_n,
2πr_n = nλ______(1)
n = 1, 2, 3………
If the speed of electron is much less than the speed of light, wave length

Note:
The quantized electron orbits and energy states are due to the wave nature of the electron.

Limitations of Bohr atom model:

1. The Bohr model is applicable to hydrogenic atoms. It cannot be extended to many electron atoms such as helium
2. The model is unable to explain the relative intensities of the frequencies in the spectrum.
3. Bohr model could not explain fine structure of spectral lines.
4. Bohr theory could not give a satisfactory explanation for circular orbit.

Students can Download Chapter 11 Dual Nature of Radiation and Matter Notes, Plus Two Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Notes Chapter 11 Dual Nature of Radiation and Matter

Introduction
The discovery of cathode rays by Rontgen and discovery of electrons by JJ Thomson were important milestones in the study of atomic structure.

Electron Emission
We know that metals have free electrons. The free electrons cannot normally escape out of the metal surface. If an electron attempts to come out of the metal, the metal surface acquires a positive charge. This positive surface held electrons inside the metal surface.

Work function:
When we give energy to electron in a metal, it can come out of metal. This minimum energy required by an electron to escape from the metal surface is called the work function of the metal. It is generally denoted by Φ₀(hν₀) and measured in eV (electron volt).

Electron volt:
One electron volt is the energy gained by an electron when it has been accelerated by a potential difference of 1 volt
1 eV = 1.602 × 10^-19J.
This unit of energy is commonly used in atomic and nuclear physics.

Different types of electron emission:
The minimum energy required for the electron emission from the metal surface can be supplied by any one of the following methods.

(i) Thermionic emission:
Electrons can come out of metal surface, if heat energy is given to metal.

(ii) Field emission:
By applying a very strong electric field (of the order of 10⁸ Vm^-1) to a metal, electrons can be pulled out of the metal.

(iii) Photoelectric emission:
When light (of suitable frequency) incident on a metal surface, electrons are emitted from the metal surface. These electrons are called photoelectrons. This phenomena is called photo electric effect.

Photoelectric Effect
1. Hertz’s observations:
The phenomenon of photoelectric emission was discovered by Heinrich Hertz in 1887, Heinrich Hertz observed that when light falls on a metal surface, electrons escape from the metal surface.

2. Hallwachs’ and Lenard’s observations:
Wilhelm Hallwachs and Philipp Lenard investigated the phenomenon of photoelectric emission in detail. The experimental set up consist of two metal plates (cathode and anode) inside a evacuated glass tube as shown in figure.

They observed that current flpws in the circuit when emitter plate (C) was illuminated by UV radiation. It means that when light incident on a metal plate electrons are emitted. These electrons move towards the anode and results in current flow.

They also observed that, when a negatively charged zinc plate is illuminated by UV light, it becomes chargeless. He also observed that uncharged Zn plate becomes positively charged when it is illuminated with UV light.

From these observations they concluded that the particles emitted carry negative charge.

Threshold frequency:
The minimum frequency (ν₀) required to produce photo electric effect is called the threshold frequency. It depends on the nature of material.

Experimental Study Of Photoelectric Effect
The experimental setup:

The experimental arrangement consists of two zinc plates enclosed in a quartz bulb. The plates are connected to a battery through a micro ammeter. When ultraviolet light is incident on the cathode plate, the micrometer indicates a current in the circuit.

When the anode is made negative (with respect to cathode) the current decreases and at a certain voltage (V₀), current is completely stopped. This voltage V₀ is called stopping potential. At this stage,
\(\frac{1}{2}\) mV_max² = eV₀
where v_max is the maximum kinetic energy of photo electrons.

1. Effect of intensity of light on photocurrent Experiment:
In this experiment the collector A is maintained at a positive potential. The frequency of the incident radiation and the accelerating potential are kept at fixed.

Then change the intensity of light and measure photoelectric current in each time. Draw a graph between photo current and intensity of light. We get a graph as shown in figure.

Observations:
This graph shows that photocurrent increases linearly with intensity of incident light.

Conclusion:
The photocurrent is directly proportional to the number of photoelectrons emitted per second. This implies that the number off Photoelectrons emitted per second is directly proportional to the intensity of incident radiation.

2. Effect of potential on photoelectric current Experiment:
Keep the plate A at positive accelerating potential. Then illuminate the plate C with light (of fixed frequency v and fixed intensity I₁). Then vary the positive potential of plate A gradually and measure the resulting photocurrent each time.

When the photo current reaches maximum, the polarity of plates are reversed and thus apply a negative potential (retarding potential) to plate A.

Again photocurrent is measured by varying the retarding potential till photocurrent reaches zero. The experiment is repeated for higher intensity I₂ and I₃ keeping the frequency fixed.

Observations:
As accelerating potential increases photo current increases. At a particular anode potential photocurrent reaches maximum. Further increase in accelerating potential does not increase photo current.

When we apply negative potential to A, photo electrons get retarded and hence photocurrent decreases. At particular retarding potential photocurrent becomes zero. This potential is called cut off or stopping potential.

The graph of anode potential with photo current:

The saturation current is found to be large at higher intensity (because photo current is directly proportional to intensity). But stopping potential is same for different intensity at fixed frequency, (ie. for a given frequency of incident radiation stopping potential is independent of its intensity).

Note:
a. The maximum value of photo current is called saturation current (Isat).

b. The retarding anode potential at which photo current reaches zero is called stopping potential (V₀).

When retarding potential is applied, only most energetic electrons can reach collector plate A. At stopping potential no electrons reach plate A, ie stopping potential is sufficient to repel the electron with maximum kinetic energy

c. The stopping potential or maximum value of KE depends only on frequency of incident light, not on its intensity. Hence stopping potential is same for different intensity at constant frequency.

d. At zero anode potential, photocurrent is not zero, ie photo electric effect takes place even if anode potential is not applied.

3. Effect of frequency of incident radiation on stopping potential:
Experiment:
In this experiment, we adjust the intensity of light at various frequencies (say ν₁, ν₂ and ν₃ such that ν₁ < ν₂ < ν₃) and study the variation of photocurrent with collector plate potential.

Observations:

For frequencies ν₁, ν₂ and ν₃ (ν₁ < ν₂ < ν₃) τηε stopping potential are found to be V₀₃ > V₀₂ > V₀₁. It means that stopping potential varies linearly with incident frequency fora given photosensitive material.

The graph of stopping potential with frequency:

The graph shows that

• The stopping potential V₀ varies linearly with the frequency of incident radiation for a given photosensitive material,
• There exists a certain minimum cutoff frequency ν₀ for which the stopping potential is zero.

These observations have two implications:

• The maximum kinetic energy of the photoelectrons varies linearly with the frequency of incident radiation, but is independent of its intensity.
• Fora frequency ν of incident radiation, lower than the cutoff frequency ν₀, no photoelectric emission is possible even if the intensity is large.
• For a frequency ν₀, no photoelectric emission is possible even if the intensity is large. This minimum, cutoff frequency ν₀, is called the threshold frequency. It is different for different metals.

Summary of the experimental features and observations:
Laws of photoelectric emission:

1. For a given frequency of radiation, number of photoelectrons emitted is proportional to the intensity of incident radiation.
2. The kinetic energy of photoelectrons depends on the frequency of incident light but it is independent of the light intensity.
3. Photoelectric effect does not occur if the frequency is below a certain value. The minimum frequency (ν₀) required to produce photo electric effect is called the threshold frequency.
4. Photoelectric effect is an instantaneous phenomenon.

Photoelectric Effect And Wave Theory Of Light
Wage theory of light is not used to explain photo electric effect. Why?
Reasons
1. According to wave theory, when intensity of incident wave increases, the KE of electron must be increased. This is pgainst the experimental observation of photoelectric effect.

2. According to wave theory, absorption of energy by electron takes place continuously. A large number of electrons absorb energy from the wave at a time.

Hence energy received by a single electron will be small. Hence it takes hours to eject an electron from a metal surface. This delay in photoemission is against the experimental observation.

Einstein’s Photoelectric Equation
Energy quantum of radiation:
Einstein explained photoelectric effect based on quantum theory. According to quantum theory, light contain photons having energy hν, when a photon of energy hr incidents on a metal surface, electrons are liberated.

A small portion of the photon energy is used for work function (Φ) and remaining energy is appeared as K.E of the electron.

By law of conservation of energy, we can write,
Photon energy = work function + K.E of electrons
hν = Φ + \(\frac{1}{2}\) mv²
\(\frac{1}{2}\)mv² = hν – Φ______(1)
If threshold frequency ν₀ is incident, we can take K.E = 0
So eq(1) can be written as
0 = hν₀ – Φ
i.e. work function Φ = hν₀______(2)
Substituting eq(2) in eq(1) we get
\(\frac{1}{2}\)mv² = hν – hν₀
\(\frac{1}{2}\)mv² = h(ν – ν₀)______(3)
This is Einstein’s Photoelectric equation.
But we know ν = c/λ and ν₀ = c/λ₀
Substituting these values in eq(3) we get,

Discussion (explanation of photo electric effect on the basis of Einstein’s photo electric equation):
1. If the intensity of the incident light increases, more number of photons interact with electrons and more number of electrons are emitted. Thus the electric current increases with the intensity of the incident light.

2. For a given metal, Φ₀(hν₀) is constant. Hence from 1/2mv² = hν – hν₀, we can understand that KE depends on ‘V’ (incident frequency).

3. From this equation 1/2mv² = hν – hν₀. we can understand that photoemission is not possible, if ν < ν_0.

4. According to quantum theory, a photon interacts only with a single electron (no sharing of energy takes place) so that there is no time delay in photoelectric emission.

Particle Nature Of Light: The Photon:
The photon picture of electromagnetic radiation is as follows:

1. In interaction of radiation with matter, radiation behaves as if it is made up of particles called photons.
2. Each photon has energy E and momentum ρ.
3. Photon energy is independent of intensity of radiation.
4. Photons are electrically neutral and are not deflected by electric and magnetic fields.
5. In a photon-particle collision the total energy and total momentum are conserved.

Wave Nature Of Matter
In 1924, the French physicist Louis Victorde Broglie put forward the hypothesis, that moving particles of matter should display wavelike properties under suitable conditions.

The waves associated with material particles are known as matter waves or de-Broglie’s waves. de-Broglie wave is seen with microscopic particles like proton, electron, and neutron, etc. The wave length of matter waves is called de-Broglie wave length.
De-Broglie wave length,

h – Plank’s constant, m – mass of the particle, v – velocity of the particle.

1. Wavelength of matter waves:
The energy of photon E = hν _____(1)
If photon is considered as a particle of mass ‘m’, the energy of photon can be written as
E = mc² _____(2)
From eq(1) and eq (2) we get
hν = mc²
m = \(\frac{\mathrm{hv}}{\mathrm{c}^{2}}\) ________(3)
Momentum of the electron can be written as
P = mass × velocity ______(4)
Substituting eq (3) in eq(4) ,we get

The wave length of electron wave:
If electron of mass ‘m’ and charge ‘e’ is accelerated through a p.d of V volt, the de-Broglie wavelength can be written as

2. Uncertainty Principle:
According to the principle, it is not possible to measure both the position and momentum of an electron (or any other particle) at the same time exactly.

If (∆x) is the uncertainty in position and (∆p) is the uncertainties in momemtum, the product uncertainties is given by
∆x.∆p =\(\frac{h}{2 \pi}\)

The above equation allows the possibility that if ∆x is zero; then ∆p must be infinite in order that the product is nonzero. Similarly, if ∆p is zero, ∆x must be infinite.

The wave packet description of an electron:

The above wave packet description of matter wave corresponds to an uncertainty in position (∆x) and an uncertainty in momentum (∆p).

Wave packet description for ∆p = 0:

The above wavepacket description of matter wave corresponds to a definite momentum of an electron extends all over space. In this case, ∆p = 0 and
∆x → ∞

Davisson Germer Experiment

Aim: To confirm the wave nature of electron.
Experimental setup:
The Davisson and Germer Experiment consists of filament ‘F’, which is connected to a low tension battery. The Anode Plate (A) is used to accelerate the beam of electrons. A high voltage is applied in between A and C. ’N’ is a nickel crystal. D is an electron detector. It can be rotated on a circular scale. Detector produces current according to the intensity of incident beam.

Working:
The electron beam is produced by passing current through filament F. The electron beam is accelerated by applying a voltage in between A (anode) and C. The accelerated electron beam is made to fall on the nickel crystal.

The nickel crystal scatters the electron beam to different angles. The crystal is fixed at an angle of Φ = 50° to the incident beam.

The detector current for different values of the accelerating potential ‘V’ is measured. A graph between detector current and voltage (accelerating) is plotted. The shape of the graph is shown in figure.

Analysis of graph:

The graph shows that the detector current increases with accelerating voltage and attains maximum value at 54V and then decreases. The maximum value of current at 54 V is due to the constructive interference of scattered waves from nickel crystal (from different planes of crystal). Thus wave nature of electron is established.

Experimental wavelength of electron:
The wave length of the electron can be found from the formula
2d sinθ = nλ ______(1)
From the figure, we get
θ + Φ + θ = 180°
2θ = 180 – Φ, 2θ = 180 – 50°
θ = 65°
for n = 1

equation (1) becomes
λ = 2dsinθ_____(2)
for Ni crystal, d = 0.91 A°
Substituting this in eq. (2), we get
wavelength λ = 1.65 A°
Theoretical wave length of electron:
The accelerating voltage is 54 V
Energy of electron E = 54 × 1.6 × 10¹⁹J
∴ Momentum of electron P = \(\sqrt{2 \mathrm{mE}}\)

= 39.65 × 10^-25 Kg ms^-1
∴ De-Broglie wavelength λ = \(\frac{h}{P}\)

Discussion:
The experimentally measured wavelength is found in agreement with de-Broglie wave length. Thus wave nature of electron is confirmed.

Students can Download Chapter 10 Wave Optic Notes, Plus Two Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Notes Chapter 10 Wave Optic

Introduction
In 1678, the Dutch physicist Christian Huygens put forward the wave theory of light. We will discuss in this chapter.

Wavefront:
The wavefront is defined as the locus of all points which have the same phase of vibration. The rays of light are normal to the wavefront. Wavefront can be divided into 3.

1. Spherical wavefront
2. Cylindrical wavefront
3. Plane wavefront.

1. Spherical Wavefront:

The wavefront originating from a point source is spherical wavefront.

2. Cylindrical Wavefront:

If the source is linear, the wavefront is cylindrical.

3. Plane wavefront:
If the source is at infinity, we get plane wavefront.

Huygen’s Principle
According to Huygen’s principle

1. Every point in a wavefront acts as a source of secondary wavelets.
2. The secondary wavelets travel with the same velocity as the original value.
3. The envelope of all these secondary wavelets gives a new wavefront.

Refraction And Reflection Of Plane Waves Using Hygens Principle
1. Refraction of a plane wave. (To prove Snell’s law):
AB is the incident wavefront and c₁ is the velocity of the wavefront in the first medium. CD is the refracted wavefront and c₂ is the velocity of the wavefront in the second medium. AC is a plane separating the two media.

The time taken for the ray to travel from P to R is

O is an arbitrary point. Hence AO is a variable. But the time to travel a wavefront from AB to CD is constant. In order to satisfy this condition, the term containing AO in eq.(2) should be zero.

where ¹n₂ is the refractive index of the second medium w.r.t. the first. This is the law of refraction.

2. Reflection of plane wave by a plane surface:

AB is the incident wavefront and CD is the reflected wavefront, ‘i’ is the angle of incidence and ‘r’ is the angle of reflection. Let c₁ be the velocity of light in the medium. Let PO be the incident ray and OQ be the reflected ray.
The time taken for the ray to travel from P to Q is

O is an arbitrary point. Hence AO is a variable. But the time to travel for a wave front from AB to CD is a constant. So eq.(2) should be independent of AO. i.e., the term containing AO in eq.(2) should be zero. AO
∴ \(\frac{A O}{C_{1}}\)(sin i – sin r) = 0
sin i – sin r= 0
sin i = sin r
i = r
This is the law of reflection.
Behavior of wave frond as they undergo refraction or reflection.

a. Wave frond through the prism:

Consider a plane wave passing through a thin prism. The speed of light waves is less in glass. Hence the lower portion of the incoming wave frond will get delayed. So outgoing wavefrond will be tilted as shown in the figure.

b. Wave frond through a thin convex lens:

Consider a plane wave passing through a thin convex lens. The central part of the incident plane wave travels through the thickest portion of lens.

Hence central part get delayed. As a result the emerging wavefrond has a depression at the centre. Therefore the wave front becomes spherical and converges to a point F.

c. Plane wave incident on a concave mirror:

A plane wave is incident on a concave mirror and on reflection we have spherical wave converging to the focul point F.

3. The Doppler Effect:
There is an apparent change in the frequency of light when the source or observer moves with respect to one another. This phenomenon is known as Doppler effect in light.

When the source moves away from the observer the wavelength as measured by the source will be larger. The increase in wavelength due to Doppler effect is called as red shift.

When waves are received from a source moving towards the observer, there is an apparent decrease in wavelength, this is referred to as blue shift.

Mathematical expression for Doppler shift:
The Doppler shift can be expressed as

V_radial is the component of source velocity along the line joining the observer to the source.

Coherent And Incoherent Addition Of Waves
Super position principle:
According to superposition principle, the resultant displacement produced by a number of waves at a particular point in the medium is the vector sum of the displacements produced by each of the waves.

Coherent sources:
Two sources are said to be coherent, if the phase difference between the displacements produced by each of the waves does not change with time.

Incoherent sources:
Two sources are said to be coherent, if the phase difference between the displacements produced by each of the waves changes with time.

Constructive interference:
Consider two light waves meet together at a point. If we get maximum displacement at the point of meeting, we call it as constructive interference.

Destructive interference:
Consider two lightwaves meet together at a point. If we get minimum displacement at the point of meeting, we call it as destructive interference.

Mathematical condition for Constructive interference and Destructive interference:

Consider two sources S₁ and S₂. Let P be point in the region of s₁ and s₂. The displacement produced by the source s₁ at P.
y₁ = a cos ωt
Similarly, the displacement produced by the source s₂ at P
y₂ = a cos (ωt + Φ)
Where Φ is the phase difference between the displacements produced by s₁ and s₂
The resultant displacement at P,
Y = y₁ + y₂
= a cos ωt + a cos (ωt + Φ)
= a (cos ωt + cos (ωt + Φ))

Therefore total intensity at P,

Constructive interference:
If we take phase difference Φ = 0, ±2π, ±4π……., we get maximum intensity (4I₀) at P. This is the mathematical condition for constructive interference. The condition for constructive interference can be written in the form of path difference between two waves.

Where n = 0, 1, 2, 3……..

Destructive interference:
If we take phase difference Φ = ±π, ±3π, ±5π………., we get zero intensity at P. This is the mathematical condition for destructive interference. The condition for destructive interference can be written in the form of path difference between two waves.

Where n = 0, 1, 2, 3……..

Interference Of Light Waves And Youngs Double Slit Experiment
Young’s double-slit experiment:

The experiment consists of a slit ‘S’. A monochromatic light illuminates this slit. S₁ and S₂ are two slits in front of the slit ‘S’. A screen is placed at a suitable distance from S₁ and S₂. Light from S₁ and S₂ falls on the screen. On the screen interference bands can be seen.

Explanation:
If crests (ortroughs) from S₁ and S₂ meet at certain points on the screen, the interference of these points will be constructive and we get bright bands on the screen.

At certain points on the screen, crest and trough meet together. Destructive interference takes place at those points. So we get dark bands.

Expression for band width:

S₁ and S₂ are two coherent sources having wave length λ. Let ‘d’ be the distance between two coherent sources. A screen is placed at a distance D from sources. ‘O’ is a point on the screen equidistant from S₁ and S₂.
Hence the path difference, S₁O – S₂O = 0
So at ‘O’ maximum brightness is obtained.
Let ‘P’ be the position of n^th bright band at a distance x_n from O. Draw S₁A and S₂B as shown in figure. From the right angle ∆S₁AP
we get, S₁P² = S₁A² + AP²
S₁P² = D² + (X_n – d/2)² = D² + X_n² – X_nd + \(\frac{d^{4}}{4}\)
Similarly from ∆S₂BP we get,
S₂P² = S₂B² + BP²
S₂P² = D² + (X_n + d/2)²

S₂P² – S₁P² = 2x_nd
(S₂P + S₁P)(S₂P – S₁P) = 2x_nd
But S₁P ≈ S₂P ≈ D
∴ 2D(S₂P – S₁P) = 2x_nd
i.e., path difference S₂P – S₁P = \(\frac{x_{n} d}{D}\) ____(1)
But we know constructive interference takes place at P, So we can take
(S₂P – S₁P) = nλ
Hence eq(1) can be written as

Let x_n+1 be the distance of (n+1)^th bright band from centre o, then we can write

This is the width of the bright band. It is the same for the dark band also.

Diffraction
The bending of light round the comers of the obstacles is called diffraction of light.

1. The single slit diffraction:

Consider a single slit AC having length ‘a’. A screen is placed at suitable distance from slit. B is midpoint of slit, A straight line through B (perpendicular to the plane of slit), meets the screen at O. AD is perpendicular CP.

Calculation of path difference:
Consider a point P on the screen having a angle θ with normal AE. The path difference between the rays (coming from the bottom and top of the slit) reaching at P,
CP – AP = CD
(CP – AP) = a sin θ
path difference, (CP – AP) = a θ______(1)
[for small θ. sin θ ≈ θ]

(I) Position of maximum intensity:
Consider the point ‘O’, the path difference between the rays (coming from AB and BC) reaching at O is zero. Hence constructive interference takes place at ‘O’. Thus maximum intensity is obtained. This point is called central maximum or the principal maximum.

(II) Position of secondary minima:
Let P be a point on the screen such that the path difference between the rays AP and CP be λ.
ie, CP – AP = λ______(2)
Substituting eq (1) in eq (2) we get
θ = λ
(or) θ = \(\frac{\lambda}{a}\)______(3)
Let the slit AC be imagined to be split into two equal halves AB and BC. For every point in AB, there is a corresponding point in BC such hat the distance between the points are equal to a/2 Consider two points K and L such that, KL = a/2. There fore, the path difference between the rays (coming form K and L) at P is,.
LP – KP = \(\frac{a}{2}\)θ_______(4)
Substituting (3) in (4) we get

This means that the rays (coming from K and L) reaching at P are out of phase and cancel each other. Hence the intensity at P becomes zero.
In otherwards, at angle θ = \(\frac{\lambda}{\mathrm{a}}\)
The intensity becomes zero.
Similarly on the lower half of the screen, the intensity is zero for which θ = – \(\frac{\lambda}{\mathrm{a}}\)
The general equation for zero intensity can be written as
θ = \(\pm \frac{n \lambda}{a}\)
Where n = 1, 2, 3,…
For first minima n = 1, and second minima n = 2.

(III) Position of Secondary maxima:
Let P be a point on the screen, such that
CP – AP = \(\frac{3}{2}\)λ
From eq (1),we know (CP – AP) = aθ
Therefore aθ = \(\frac{3}{2}\)λ
The wave front AC can be divided into three equal parts.

The rays from first and second parts will cancel each other and the rays from third part will reach at P. Hence the point P becomes bright.

Similarly the next maximum occurs at θ = \(\frac{5}{2}\)\(\frac{λ}{a}\)
The general equation for maximum can be written
\(\theta=\pm \frac{(2 n+1) \lambda}{2 a}\)

1. (a) Intensity Distribution on the screen of diffraction pattern:

(b) Comparison between interference and diffraction bands:
Interference:

• Interference is due to superposition of waves coming from two wavefronts.
• Interference bands are of equal width.
• Minimum intensity regions are perfectly dark.
• All the bright bands are of equal intensity.

Diffraction:

• Diffraction is due to the superposition of waves coming from different parts of the same wave front.
• Diffraction bands are of unequal width.
• Minimum intensity regions are not perfectly dark.
• All bright bands are not of the same intensity.

2. Seeing The Single Slit Diffraction Pattern:

Take two razor blades and an electric bulb. Hold the two blades as shown in the figure. Observe the glowing bulb through the slit. A diffraction pattern can be seen.

3. Resolving Power Of Optical Instruments:
Resolving power of optical instrument:
The ability of an optical instrument to form distinctly separate images of the two closely placed objects is called is resolving power.

Explanation:

The image of a point object formed by a ideal lens is a point only. But because of diffraction effect, instead of point image, we get a diffraction pattern. Diffraction pattern consists of a bright central circular region surrounded by concentric dark and light rings.

Let us discuss three cases; when we observe two point object through a lens.

1. Unresolved:
If central maxima of two diffraction pattern are overlapped, the image is unresolved. This image can’t be viewed clearly.

2. Just resolved:
If central maxima of two diffraction pattern are just separated, the image is just resolved. In this case image is just distinqushed.

3. Resolved:
If central maxima of two diffraction pattern are separated, the image is resolved. This image can be viewed clearly.

Limit of resolving power of optical instrument:
The minimum distance of separation between two points so that they are just resolved by the optical instrument is known as its limit of resolution. Resolving power is also defined as reciprocal of limit of resolution.

1. Telescope and resolving power:

Telescope consist of two convex lenses called eyepiece and objective .The light falling on objective lens undergoes for diffraction. Hence a diffraction pattern of bright and dark rings is produced around central bright region as shown in figure.
The radius of central bright region,

This radius can be written in terms of angular width,
∆θ ≈ \(\frac{0.61 \lambda}{\mathrm{a}}\)
Where a is the radius and f – focal length of objective lens. λ is the wave length of light used.

This angular width of central bright region is related to resolving power of telescope. When angular width of spot increases, resolving power decreases.

The limit of resolution of telescope, ∆θ ≈ \(\frac{0.61 \lambda}{\mathrm{a}}\)
This equation shows that telescope will have better resolving power if ‘a’ is large and λ is small.

2. Microscope and resolving power:

In microscope the object (microscopic size) is placed slightly beyond f (focal length of objective lens). When the separation between two points in a microscopic specimen is comparable to the wavelength λ of light, the diffraction effect become important.

Where nsinβ is called numerical aperture, n is the refractive index of liquid used in microscope, β is the half angle of the cone of light from the microscopic object with objective lens.
The limit of resolution of microscope d_min = \(\frac{1.22 f \lambda}{2 n \sin \beta}\)
This equation also can be written as d_min = \(\frac{1.22 \lambda}{2 \tan \beta}\)

Note: Telescope is used to resolve objects at far distance but microscope is used to produce magnification of near objects.

4. The Validity Of Ray Optics:
Fresnel distance is the distance beyond which the diffraction properties becomes significant, (ie. the ray optics is converted into wave optics).
Fresnel distance, z_F = \(\frac{\mathrm{a}^{2}}{\lambda}\)
Where ‘a’ is the size of the aperture
For distances much smaller than z_F, the spreading due to diffraction is smaller compared to the size of the beam. It becomes comparable when the distance is approximately z_F. For distances much greater than z_F, the spreading due to diffraction dominates over that due to ray optics.

Polarisation

Consider a long string that is held horizontally, the other end of which is assumed to be fixed. If we move the end of the string up and down in a periodic manner, a wave will propagate in the +xdirection (see above figure). Such a wave can be described by the following equation
y(x,t) = a sin (kx – ωt)
where ‘a’ represent the amplitude and k = 2π/λ represents the wavelength associated with the wave.

Since the displacement (which is along the y-direction) is at right angles to the direction of propagation of the wave, this wave is known as a transverse wave.

Also, since the displacement is in the/direction, it is often called to as a y-polarised wave. Since each point on the string moves on a straight line, the wave is also called to as a linearly polarised wave.

The string always remains confined to the x-y plane and therefore it is also called to as a plane polarised wave.

In a similar manner we can consider the vibration of the string in the x-z plane generating a z-polarised wave whose displacement will be given by
z(x,t) = a sin (kx – ωt)

Unpolorised wave:
If the plane of vibration of the string is changed randomly in very short intervals of time, then it is known as an unpolarized wave.

(a) Polarization property of light:
When light passes through certain crystals like tourmaline, the vibrations of electric field vector are restricted. This property exhibited by light is known as polarization.

Note:

1. Polarization is the property of light which reveals that light is a transverse wave.
2. A sound wave can’t be polarized because sound wave is a longitudinal wave.

Polarizer and analyzer:
When an unpolarized light passes through a tourmaline crystal T₁, the light coming out of T₁ is plane polarized.

In order to check the polarization, another tourmaline crystal T₂ is kept parallel to T₁.

When we look through T₂ we get maximum intensity. Then T₂ is rotated through 90°. If no light is coming, we can say that light from T₁ is plane polarized.

Polarizer: The crystal which produces polarized light is known as polarizer.

Analyzer: The crystal which is used to check weather the light is polarized or not is called the analyzer or detector.

Law of Malus: This law states that when a beam of plane polarized light is incident on an analyzer, the intensity (I) of the emergent light is directly proportional to the square of the cosine of the angle (θ) between the polarizing directions of the polarizer and the analyzer.

I = I_m cos²θ
where I_m is the maximum intensity.

1. Polarisation By Scattering:

The nunpolarized light incident on a dust particle in atmosphere, it is absorbed by electrons in the dust particle. The electrons in the dust particle reradiate light in all directions. This phenomenon is called scattering.

Explanation:
Let a beam of unpolarized light be incident on a dust particle along x-axis. The electrons in the dust particle absorb light and behave as a oscillating dipole. This dipole emit light in all directions.

When an observer observe this particle along y-axis, the observer can receive light from the electron vibrating in z-axis. This light is linearly polarised in z-direction (its plane of polarisation is yz).

This polarised light is represented by dots in the picture. This explains the polarisation of scattered light from the sky.

2. Polarization By Reflection:
At a particular angle of incidence on a medium, the reflected lights is fully polarized. This angle is known as polarizing angle or Brewster’s angle. At polarizing angle, the reflected and refracted rays are mutually perpendicular.

Brewster’s law:
Brewster’s law states that the tangent of the polarizing angle is equal to the refractive index of the material of the reflector.

Let ‘Q ’ be the polarizing angle and ‘n’ be the refractive index of the medium then,
tan θ = n
At polarizing angle, r + θ =90°.

Proof:
Consider an unpolarized light coming from air and is incident on a medium having refractive index n. Let θ be the angle of incidence, Φ be the angle of reflection and ‘r’ be the angle of refraction.
Using snells law, we can write
n = \(=\frac{\sin \theta}{\sin r}\) ______(1)
At the polarizing angle reflected and refracted light are mutually perpendicular
ie. Φ – 90 + r = 180°
∴ r = 90 – Φ______(2)
Substituting eq (2) in eq(1), we get

But we know
Angle of incidence (θ) = angle of reflection(Φ)
∴ n = \(\frac{\sin \theta}{\cos \theta}\)
n = tanθ

Students can Download Chapter 9 Ray Optics and Optical Instruments Notes, Plus Two Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments

Introduction
In this chapter, we consider the phenomena of reflection, refraction and dispersion of light, using the ray picture of light.

Reflection Of Light Byspherical Mirrors
Laws of reflection:

1. According to the first law of reflection, the angle of reflection equals the angle of incidence.
2. According to the second law of reflection, the incident ray, reflected ray and the normal to the point of incidence all lie in the same plane.

1. Sign convention:

• According to this convention, all distances are measured from the pole of the mirror or the optical centre of the lens.
• The distances measured in the same direction as the incident light are taken as positive and
  those measured in the direction opposite to the direction of incident light are taken as negative.
• The heights measured upwards are taken as positive. The heights measured downwards are taken as negative.

2. Focal length of spherical mirrors:
Reflection of light: Spherical mirrors are of two types.

• Concave mirror
• Convex mirror

Principal focus of a concave mirror:

A narrow parallel beam of light, parallel and close to the principal axis, after reflection converges to a fixed point on the principal axis is called principal focus of concave mirror.
Principal focus of a convex mirror:

A narrow parallel beam of light, parallel and close to the principal axis, after reflection appears to diverge from a point on the principal axis is called principal focus of convex mirror.
Relation connecting focal length and radius of curvature:

Consider a ray AB parallel to principal axis incident on a concave mirror at point B and is reflected along BF. The line CB is normal to the mirror as shown in the figure.
Let θ be angle of incidence and reflection.
Draw BD ⊥ CP,
In right angled ΔBCD,
Tanθ = \(\frac{B D}{C D}\) _____(1)
In right angled ΔBFD,
Tan2θ = \(\frac{B D}{F D}\) _____(2)
Dividing (1)and(2)
\(\frac{\tan 2 \theta}{\tan \theta}=\frac{C D}{F D}\) ____(3)
If θ is very small, then tanθ ≈ θ and tan2θ ≈ 2θ
The point B lies very close to P. Hence CD ≈ CP and FD ≈ FP From (3) we get

3. The mirror equation:

Let points P, F, C be pole, focus, and centre of curvature of a concave mirror. Object AB is placed on the principal axis. A ray from AB incident at E and then reflected through F. Another ray of light from B incident at pole P and then reflected. These two rays meet at M. The ray of light from point B is passed through C. Draw EN perpendicular to the principal axis.
ΔIMF and ΔENF are similar.
ie. \(\frac{I M}{N E}=\frac{I F}{N F}\) _____(1)
but IF = PI – PF and NF = PF (since aperture is small)
hence eq. (1) can be written as

[∵ NE = AB)
ΔABP and ΔIMP are similar

From eq.(2) and eq.(3), we get

applying sign convention we get
PI = -v
PF = -F
PA = -u
Substituting these values in eq.(4) we get

This is called mirror formula or mirror equation.
Linear magnification:
Linear magnification is defined as the ratio of the height of the image to the height of the object.

Consider an object AB having height h_o, which produces an image IM having height h_i
In the figure, ΔABP and ΔIMP are equal. ie.

Applying sign convention
PI = -V, PA = -u, h_i = -ve and h_o = +ve
We get

But we know \(\frac{h_{i}}{h_{0}}\) = m (magnification) ie.

This formulae is true fora concave mirror and convex mirror.
Relation connecting v, f, and m
We have
\(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\)
Multiplying throughout by ‘v’, we get
\(\frac{v}{u}+\frac{v}{v}=\frac{v}{f}\)
But m = -v/u
ie. -m + 1 = \(\frac{v}{f}\)
m = 1 – \(\frac{v}{f}\)

Relation connecting u, f and m
We know
\(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\)
Multiplying throughout by ‘u’ we get

Refraction
The phenomenon of bending of light when it travels from one medium to another is known as refraction.
Light from rarer to denser medium:

When light travels from a rarer medium to a denser medium, it deviates towards the normal.
Light from denser to rarer medium:

When light travels from a denser medium to a rarer medium, it deviates away from the normal.

Laws of refraction:
First law:
The incident ray, the refracted ray, and the normal at the point of incidence are all in the same plane.

Second law (Snell’s law):
The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media and for the given colour of light used. This constant is known as the refractive index of second medium w.r. t. the first medium.

Explanation:
If ‘i’ is the angle of incidence in the first medium and ‘r’ is the angle of refraction in the second medium, then by Snell’s law,

Where ₁n₂ is the refractive index of the second medium with respect to the first medium. If the first medium is air, then sini/sinr is known as absolute refractive index of the second medium.
ie, \(\frac{\sin i}{\sin r}=n\)
where ‘n’ is the refractive index of the second medium.

Some examples of refraction:
(a) Apparent depth:

When an object (in a denser medium) is viewed from a rarer medium, it seems to be raised towards the surface. This is called apparent depth.

(b) Twinkling of stars:
Twinkling of stars is due to the refraction of star light at different layers of the atmosphere. Due to this refraction the star at S appears at S¹. But the density of the layer continuously changes. So, the apparent position continuously changes. Thus the star appears to be twinkling.

(c) Apparent shift in the position of the sun at sunrise and sunset:

Sun is visible before sunrise and after sunset because of atmospheric refraction. The density of atmospheric air decreases as we go up. So the rays coming from the sun deviates towards the normal. So the sun at ‘S’ appears to come from ‘S¹’. Thus an observer on earth can see the sun before sunrise and after sunset.

Total Internal Reflection

When a ray of light passes from a denser to rarer medium, after refraction the ray bends away from the normal. If the angle of incidence increases, the angle of refraction increases. When the angle of refraction is 90°, the corresponding angle of incidence is called the critical angle.

If we increases the angle of incidence beyond the critical angle, the ray is totally reflected back to the same medium. This phenomenon is called total internal reflection.

Relation between critical angle and refractive index
Refractive index,

where ‘C’ is the critical angle.
A demonstration for total internal reflection
Demonstration – 1:

Take a soap solution in a beaker. Now direct the laser beam from one side of the beaker such that it strikes the upper surface of water obliquely. Adjust the direction of laser beam until the beam is totally reflected back to water.

Demonstration – 2:

Take a soap solution in a long test tube and shine the laser light from top, as shown in above figure. Adjust the direction of the laser beam such that it is totally internally reflected. This is similar to what happens in optical fibres.
Condition for total internal reflection:

1. Light should travel from denser medium to rarer medium.
2. Angle of incidence in the denser medium should be greater than the critical angle.

Relative critical angle:
Critical angle of a medium A with respect to a rarer medium B is represented as _BC_A. _BC_A is related to the refractive index _Bn_A as

Some Effects And Applications Of Total Internal Reflection
(a) Brilliance of diamond:
Refractive index of diamond is high (n = 2.42) and the critical angle is small (C = 24.41°). More over the faces of the diamond are cut in such a way that a ray of light entering the crystal undergoes multiple total reflections. This multiple reflected light come out through one or two faces. So these faces appear glittering.

(b) Mirage:
On hot summer days the layer of air in contact with the sand becomes hot and rare. The upper layers are comparatively cooler and denser. When light rays travel from denser to rarer, they undergo total internal reflection. Thus image of the distant object is seen inverted. This phenomenon is Known as mirage.

(c) Looming (superior mirage):
Due to the mist and fog in cold countries, distant ship cannot be seen clearly. But due to the total internal reflection, the image of the ship appears hanging in air. This illusion is known as looming.

(d)Total reflection prisms:

A right-angled prism is called a total reflecting prisms. Total reflecting prisms are based on the principle of total internal reflection. With the help of these prisms, the direction of the incident ray can be changed. The refractive index for glass is 1.5 and its critical angle is 42°. When a ray of light makes an angle of incident more than 42° (within the glass) the ray undergoes total internal reflection.

1. Optical fibres:

Optical fibres consist of a number of long fibres made of glass or quartz (n = 1.7). They are coated with a layer of a material of lower refractive index (1.5). When light incident on the optical fibre at angle greater than the critical angle, it undergoes total internal reflection. Due to this total internal reflection, a ray of light can travel through a twisted path.
Uses:

• Used as a light pipe in medical and optical diagnosis.
• It can be used for optical signal transmissions.
• Used to carry telephone, television and computer signals as pulses of light.
• Used for the transmission and reception of electrical signals which are converted into light signals.

Refraction At Spherical Surfaces And By Lenses
Spherical lenses:
There are two types of lenses

• convex lenses and
• concave lenses.

Principal axis:
A straight line passing through the two centers of curvature is called the principal axis of the lens.

Principal focus (F):
A narrow beam of parallel rays, parallel and close to the principal axis, after refraction, converges to a point on the principal axis in the case of a convex lens or appears to diverge from a point on the axis in the case of a concave lens. This fixed point is called the principal focus of the lens.

Focal length:
It is distance between the optic centre and the principal focus.

1. Refraction at a spherical surface:

Consider a convex surface XY, which separates two media having refractive indices n₁ and n₂. Let C be the centre of curvature and P be the pole. Let an object is placed at ‘O’, at a distance ‘u’ from the pole. I is the real image of the object at a distance V from the surface. OA is the incident ray at angle ‘i’ and Al is the refracted ray at an angle ‘r’. OP is the ray incident normally. So it passes without any deviation. From snell’s law,

r₁ = n₂ _____(1)
From the Δ OAC, exterior angle = sum of the interior opposite angles
i.e., i = α + θ ______(2)
Similarly, from ΔIAC,
a = α + β
r = α – β ______(3)
Substituting the values of eq(2) and eq(3)in eqn.(1) we get,
n₁(α + θ) = n₂(α – β)
n₁α + n₁θ = n₂α – n₂β
n₁θ + n₂β = n₂α – n₁α
n₁θ + n₂β = (n₂ – n₁)α _______(4)
From OAP, we can write,

From IAP, β = \(\frac{\mathrm{AP}}{\mathrm{PI}}\), From CAP, α = \(\frac{\mathrm{AP}}{\mathrm{PC}}\)
Substituting θ, β and α in equation (4) we get,

According to New Cartesian sign convection, we can write,
OP = -u, PI = +v and PC = R
Substituting these values, we get

Case -1: If the first medium is air, n₁ = 1, and n₂ = n,

2. Refraction by a lens:
Lens Maker’s Formula (for a thin lens):
Consider a thin lens of refractive index n₂ formed by the spherical surfaces ABC and ADC. Let the lens is kept in a medium of refractive index n₁ Let an object ‘O’ is placed in the medium of refractive index n₁ Hence the incident ray OM is in the medium of refractive index n₁ and the refracted ray MN is in the medium of refractive index n₂.

The spherical surface ABC (radius of curvature R₁) forms the image at I₁. Let ‘u’ be the object distance and ‘v₁‘ be the image distance.
Then we can write,

This image I₁ will act as the virtual object for the surface ADC and forms the image at v.
Then we can write,

Adding eq (1) and eq (2) we get

Dividing throughout by n1, we get

if the lens is kept in air, \(\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}\) = n
So the above equation can be written as,

From the definition of the lens, we can take, when u = 8, f = v
Substituting these values in the eq (3), we get

This is lens maker’s formula

For convex lens,
f = +ve, R₁ = +ve, R₂ = – ve

For concave lens,
f = -ve, R₁ = -ve, R₂ = +ve

Lens formula

Linear magnification: If h_o is the height of the object and h_i is the height of the image, then linear magnification

3. Power of a lens:
Power of a lens is the reciprocal of focal length expressed in meter.

Unit of power is dioptre (D).

4. Combination of thin lenses in contact:

Consider two thin convex lenses of focal lengths f₁ and f₂ kept in contact. Let O be an object kept at a distance ‘u’ from the first lens L₁, I₁ is the image formed by the first lens at a distance v₁.
Then from the lens formula, we can write,

This image will act as the virtual object for the second lens and the final image is formed at I (at a distance v). Then

If the two lenses are replaced by a single lens of focal length ‘F’ the image is formed at V. Then we can write,

where P is the power of the combination, P₁ and P₂ are the powers of the individual lenses.

Magnification (combination of lenses):
If m₁, m₂, m₃,…….. are the magnification produced by each lens,
then the net magnification,
m = m₁. m₂. m₃……….
Relation connecting m, u and f:

Relation connecting m,v and f:

Refraction Through A Prism

ABC is a section of a prism. AB and AC are the refracting faces, BC is the base of the prism, ∠A is the angle of prism.
Aray PQ incidents on the face AB at an angle i₁. QR is the refracted ray inside the prism, which makes two angles r₁ and r₂ (inside the prism). RS is the emergent ray at angle i₂.
The angle between the emergent ray and incident ray is the deviation ‘d’.
In the quadrilateral AQMR,
∠Q + ∠R = 180°
[since and N₁M are normal] ie,
∠A + ∠M = 180° ____(1)
In the Δ QMR.
∴ r₁ + r₂ + ∠M = 180° _____(2)
Comparing eq (1) and eq (2)
r₁ + r₂ = ∠A ______(3)
From the Δ QRT,
(i₁ – r₁) + (i₂ – r₂) = d
[since exterior angle equal sum of the opposite interior angles]
(i₁ + i₂) – (r₁ + r₂) = d
but, r₁ + r₂ =A
∴ (i₁ + i₂ ) – A = d
(i₁ + i₂) = d + A _____(4)

It is found that for a particular angle of incidence, the deviation is found to be minimum value ‘D’.
At the minimum deviation position,
i₁ = i₂ = i, r₁ = r₂ = r and d = D
Hence eq (3) can be written as,
r + r= A
or r = \(\frac{A}{2}\) ______(5)
Similarly eq (4) can be written as,
i + i = A + D
i = \(\frac{A+D}{2}\) _____(6)
Let n be the refractive index of the prism, then we can write,
n = \(\frac{\sin i}{\sin r}\) ______(7)
Substituting eq (5) and eq (6) in eq (7),

i – d curve:

It is found that when the angle of incidence increases deviation (d) decreases and reaches a minimum value and then increases. This minimum value of the angle of deviation is called the angle of minimum deviation.

Dispersion By A Prism
Dispersion: The splitting of the white light into its component colours is called dispersion.

The pattern of colour components of light is called the spectrum of light.

Reason for dispersion:
The refractive index is different for different colours. Refractive index for violet is higher than red. This variation of refractive index of medium with the wavelength causes dispersion.

Some Natural Phenomena Due To Sunlight
1. The rainbow:
The rainbow is an example of the dispersion of sunlight by the water drops in the atmosphere. The conditions for observing a rainbow are that the sun should be shining in one part of the sky while it is raining in the opposite part of the sky.
There are two types rainbow

• Primary rainbow
• secondary rainbow.

(i) Primary rainbow:

In a primary rainbow, after refraction at the surface of water droplet, the ray suffers one internal reflection and finally comes out of the drop by forming an inverted spectrum. The maximum deviated light is red (42°) and the least deviated light is violet (40°).

(ii) Secondary rainbow:

secondary rainbow, after refraction at the surface of water droplet, the ray suffers two total internal reflection and finally comes out of the droplet by forming a spectrum. The most deviated light in this spectrum is violet (53°) and the least deviated light is red (50°).

2. Scattering of light:
When sunlight travels through the earth’s atmosphere, it changes its direction by atmospheric particles. This is called scattering. Light of shorter wavelength is scattered much more than light of longer wavelength. Scattering is possible only when size of the particles is comparable to the wavelength of incident light.

Rayleigh’s scattering law:
The intensity of the scattered light from a molecule is inversely proportional to the 4^th power of the wavelength.
ie, \(I \alpha \frac{1}{\lambda^{4}}\)
I – Intensity of Scattering

Blue colour of sky:
According to Rayleigh scattering, scattering is inversely proportional to the fourth power of its wavelength. Hence shorterwavelength is scattered much more than longer wavelength. Thus blue colour is more scattered than the other colours. So sky appears blue.

Whiteness of clouds:
Clouds contain large partides (dust, H₂O), which scatter all colours almost equally. Hence clouds appear white.

Colours of the sunset (or sunrise):
At sunrise and sunset light has to travel a longer distance before reaching the earth. During this time, smaller wavelengths are scattered away. The remaining colours is red. Hence sky appears red in colour.

Optical Instruments
Mirrors, lenses and prisms, periscope, Kaleidoscope, Binoculars, telescopes, microscopes are some examples of optical devices Our eye is one of the most important optical device.

1. The eye:

Human eye consists of an eyeball of size 2.5cm in diameter. The very thin skin in front of the eye is known as cornea. Behind cornea, the empty space is known as aqueous humor. The small wall behind cornea is known as iris. In this iris a small circular opening is there, which is known as pupil. Iris can adjust its tension to vary the size of the pupil.

Behind the iris a muscular membrane is there which is known as ciliary muscle. The focal length of the crystalline lens can be adjusted to see the object any separation by adjusting the tension of ciliary muscles. The backwall of eye is known as retina.

It consists of light sensitive cells known as rods and cones. The rods are sensitive to intensity and cones are sensitive to colour. The signals from retina are transferred to the brain by optic nerves.

The brightest point in the retina is known as yellow spot and the lowest point in the eye (retina) is known as blind spot. The space between the lens and retina is filled by a liquid which is known as vitreous humor.

Defects of Vision
a. Myopia or shortsightedness:

A person suffering from myopia can see only nearby objects but cannot see objects beyond a certain distance clearly. This defect occurs due to

1. Elongation of eyeball
2. Short focal length of eye lens

It can be corrected by using a concave lens of suitable focal length.

b. Hypermetropia or Far sightedness:

A person suffering from this defect can see only distant object clearly but cannot see nearby objects clearly. This defect occurs due to

1. Decrease in the size of the eyeball.
2. Increase in focal length of the eyeball.

This defect can be corrected by using a converging lens (convex).

c. Astigmatism:

A person suffering from astigmatism cannot focus objects in front of the eye clearly. It can be corrected by using a cylindrical lens of suitable focal length.

d. Presbyopia:
It is the farsightedness occurring due to awakening of ciliary muscles. It can be corrected by using a lens of bifocal length.

1. The microscope:
Simple microscope: A simple microscope is a converging lens of small focal length.

Working: The object to be magnified is placed very close to the lens and the eye is positioned close to the lens on the other side. Depending upon the position of object, the position of image is changed.

Case 1:
If the object is placed, one focal length away or less, we get an erect, magnified and virtual image at a distance so that it can be viewed comfortably ie. at 25cm or more. (This 25cm is denoted by the symbol D).

Case 2:
If the object is placed at a distance f (focal length of lens), we get the image at infinity.

Mathematical expression of magnification:
Image at D:
If the image is formed at ‘D’, we can take u = -D. Hence the lens formula can be written as

The image is formed at D, ie. v = -D

This equation is used to find magnification of simple microscope when image at D (D ≈ 25cm).

Image at infinity:

If the object is placed at f, the image forms at infinity. In this case, magnification,

Suppose the object has a height h, the angle subtended is
tanθ₀\(=\frac{h}{D}\), θ₀\(=\frac{h}{D}\)______(2)
where ‘D’ is the comfortable distance of object from the eye (least distinct vision).

When the final image is formed at infinity,
θ_i = \(\frac{h^{1}}{v}\) ______(3)
When h₁ is the height of image and v is the image distance

This equation is used to find magnification of simple microscope when image at infinity.

2. Compound microscope:
Apparatus: A compound microscope consists of two convex lenses, one is called the objective and the other is called eye piece.

The convex lens near to the object is called objective. The lens near to the eye is called eye piece. The two lenses are fixed at the ends of two co-axial tubes. The distance between the tubes can be adjusted.

Working:
The object is placed in between F and 2F of objective lens. The objective lens forms real inverted and magnified image (I₁M₁) on the other side of the lens.

This image will act as object or eyepiece. Thus an enlarged, virtual, and inverted image is formed, (this image can be adjusted to be at the least distance of distinct vision, D).

Magnification: The magnification produced by the compound microscope

Where m₀ & m_e are the magnifying power of objective lens and eyepiece lens.

Eyepiece acts as a simple microscope.
Therefore m_e = 1 + \(\frac{D}{f_{e}}\) _____(2)
m₀ = \(\frac{v_{0}}{u_{0}}\) ______(3)
We know magnification of objective lens
Where v₀ and u₀ are the distance of the image and object from the objective lens.
Substituting (2) and (3) in (1), we get

for compound microscope, u_o » f_o (because the object of is placed very close to the principal focus of the objective) and v_o ≈ L, length of microscope (because the first image is formed very close to the eye piece).

where L is the length of microscope, f₀ is the focal length of objective lens.
Case 1: If the final image is formed at infinity, magnification of eye piece D
m \(=\frac{D}{f_{e}}\)
∴ Total magnification of compound microscope

(3) Telescope: Astronomical telescope is used to observe heavenly bodies.
There are two types of telescopes

1. Refracting and
2. Reflecting Telescope.

(1) Refracting Telescope:
Constructional details

It consists of two convex lenses, one is called objective and other is called eyepiece. These two lenses are fitted at the ends of two coaxial tubes. The distance between the two lenses can be varied.

Working:
The objective lens forms the image (IM) of a distant object at its focus. This image (formed by objective) is adjusted to be focus of the eyepiece.

Magnification:
The magnifying power of a telescope is the ratio of the angle subtended by the image at the eye to the angle subtended by the object at the objective.

(For small values tan α ≈ α)

But IC = f_o (the focal length objective lens) and IC¹ = f_e(the focal length eyepiece lens.)
∴ m = \(\frac{f_{0}}{f_{e}}\)
In this case the length of the telescope tube is (f₀ + f_e).

Case 1: When the image formed by the objective is within the focal length of the eyepiece, Then the final image is formed at the least distant of distinct vision. In this case, magnifying power.

(2) Reflecting Telescope:
Newtonian types reflecting Telescope:
The Newtonian reflector consists of a parabolic mirror made of an alloy of copper and tin. It is fixed atone end of a metal tube.

The parallel rays from a distant stars incident on the mirror M₁. After reflection from the mirror, the ray incident on a plane mirror M₂.

The reflected ray from M₂ enter into eye piece E. The eyepiece forms a magnified, virtual and erect image. Magnifying power of Newton Telescope
m = \(\frac{f_{0}}{f_{e}}\) or m = \(\frac{R}{2 f_{\theta_{g}}}\)
where
f_o — is the focal length of concave mirror
f₂ — is the focal length of eyepiece.
R – Radius of curvature of concave reflector.

Students can Download Chapter 8 Electromagnetic Waves Notes, Plus Two Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Notes Chapter 8 Electromagnetic Waves

Introduction
In this chapter we shall study the basic concepts of electromagnetic waves.

Displacement Current
Amperes circuital law in ac circuit: Consider a capacitor connected to a AC source using conducting wires. AC current can flow through a capacitor. Hence magnetic field is produced around the conducting wire. This magnetic field can be found using amperes circuital law.

Magnetic field at P
Method – 1 (To find magnetic field at P)

Consider a point P, which lies outside and very close to a capacitor as shown in the figure. We can find magnetic field at P using amperes circuital law. In order to find an magnetic field at P, consider a open surface (amperien loop having pot like surface) with a boundary of circle of radius r.
Applying amperes circuital law we get
\(\oint\)B.dI = µ₀i
Where ‘i’ is the current passing through the surface. (This surface lies outside to capacitor)
Integrating we get B.2πr = µ₀i
B = \(\frac{\mu_{0} i}{2 \pi r}\) _____(1)

Method – 2 (To find magnetic field at P)

Consider a open surface (amperien loop having pot like surface) extended to interior of capacitor with a boundary of circle of radius r.
Applying amperes circuital law we get
\(\oint\)B.dI = µ₀0
(since the current passing through the closed surface is zero, surface lies in between the plates)
ie. B = 0 ______(2)

Discussion of method 1 and method 2: Amperian circuital law is independent of size and shape of pot like surface. Hence we expect same value of B in eq(1)and eq(2). But we got different values at the same point P. Hence we can understand that there is a mistake in the amperes circuital law in AC circuits.

Maxwells correction in amperes circuital law:
To solve the above mistake, Maxwell introduced a term in the amperes circuital law. The modified amperes circuital law can be written as
\(\oint\)B.dI = µ₀(i_c + i_d)
Where i_d is called displacement current. Its value is

The above modified amperes circuital is known as Ampere- Maxwell law. This law is applicable for both AC and DC circuits.

Question 1.
Show that conduction current i_c is equal to displacement current i_d
Answer:
The flux passing through the surface in between plates (see figure 2)

Capacitor is connected to ac voltage. Hence the charge on the plate also changes with time. Hence the flux passing through the pot shape surface changes with time.
ie. the flux in between capacitor changes.
The change influx,

This means that the conduction current passing through the conduction wire is converted into displacement current, when it passes in between plates of capacitor.
1. The total current i is the sum of the conduction current and the displacement current
So we have

2. Outside the capacitor plates, we have only conduction current and no displacement current inside the capacitor there is no conduction current and there is only displacement current.

Electromagnetic Waves
1. Sources of Electromagnetic waves:
Question 2.
How are electromagnetic waves produced?
Answer:
Consider a charge oscillating with some frequency (An oscillating charge is an example of accelerating charge). This oscillation produces an oscillating electric and magnetic field in space. The oscillating electric and magnetic fields (EM Wave) propagates through the space. The experimental production of electromagnetic wave was done by Hertz’s experiment in 1887.

2. Nature of electromagnetic waves:
Characteristics of Electromagnetic waves:
(i) Electromagnetic waves propagate in the form of mutually perpendicular magnetic and electric
fields. The direction of propagation of wave is perpendicular to both magnetic and electric field vector.

(ii) Velocity of electromagnetic waves in free space is,

The speed of electromagnetic wave in a material medium is given by

(iii) The ratio of magnitudes of electric and magnetic field vectors in free space is constant

E and B are in same phase

(iv) No medium is required for propagation of transverse wave.

(v) Electromagnetic waves show properties of reflection, refraction, interference, diffraction and polarization.

(vi) Electromagnetic waves have capability to carry energy from one place to another.

Mathematical Expression:

Consider a plane electromagnetic wave travelling along the Z direction. The electric and magnetic fields are perpendicular to the direction of wave motion.
The electric field vector along the Y direction.
E_x = E₀sin(kz – ωt)
and B_Y = B₀sin(kz – ωt)
where E₀ is the amplitude of electric field vector, B₀ is the amplitude of magnetic field vector, ω is the angular frequency and k is related to the wave length λ of the wave,
k = 2π/λ.

Electromagnetic Spectrum
Electromagnetic waves include visible light waves, X-rays, gamma rays, radio waves, microwaves, ultraviolet and infrared waves. The classification is based roughly on how the waves are produced or detected.

1. Radio waves:
Radio waves are produced by the accelerated motion of charges in conducting wires. They are used in radio and television communication systems. They are generally in the frequency range from 500 kHz to about 1000 MHz.

2. Microwaves:
Microwaves (short-wavelength radio waves), with frequencies in the gigahertz (GHz) range, are produced by special vacuum tubes (called klystrons, magnetrons, and Gunn diodes). Due to their short wavelengths, they are suitable for the radar systems used in aircraft navigation. Microwave ovens are domestic application of these waves.

3. Infrared waves:
Infrared waves are produced by hot bodies and molecules. Infrared waves are sometimes referred to as heatwaves. Infrared lamps are used in physical therapy.

Infrared rays are widely used in the remote switches of household electronic systems such as TV, video recorders etc. Infrared radiation also plays an important role in maintaining the earth’s warmth or average temperature through the greenhouse effect.

4. Visible rays:
It is the part of the spectrum that is detected by the human eye. It starts from 4 × 10¹⁴ Hz to 7 × 10¹⁴ Hz (ora wavelength range of about 700 – 400 nm).

5. Ultraviolet rays (UV):
It covers wavelengths ranging from about 4 × 10^-7m to 6 × 10^-10m (0.6 nm to 400 nm)). UV radiation is produced by special lamps and very hot bodies. The sun is an important source of ultraviolet light.

UV light in large quantities has harmful effects on humans. Exposure to UV radiation induces the production of more melanin, causing tanning of the skin. UV radiation is absorbed by ordinary glass. Hence, one cannot get tanned or sunburn through glass windows.

Due to its shorter wavelengths, UV radiations can be focussed into very narrow beams for high precision applications such as eye surgery. UV lamps are used to kill germs in water purifiers.

6. X-rays:
It covers wavelengths from about 10^-8m to 10^-13m (4nm – 10nm). One common way to generate X-rays is to bombard a metal target by high energy electrons. X-rays are used as a diagnostic tool in medicine and as a treatment for certain forms of cancer.

7. Gamma rays:
They lie in the upper-frequency range of the electromagnetic spectrum and have wavelengths of^rom about 10^-10m to less than 10^-14m. This high-frequency radiation is produced in nuclear reactions and also emitted by radioactive nuclei. They are used in medicine to destroy cancer cells.

Different Types Of Electromagnetic Waves:

Students can Download Chapter 7 Alternating Current Notes, Plus Two Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Notes Chapter 7 Alternating Current

Alternating Current
AC current is commonly used in homes and offices. The main reason for preferring ac voltage over dc voltage is that ac voltages can be easily converted from one voltage to the other and can be transmitted over long distances. In this chapterwe will deal the properties of ac and its flowthrough different devices (inductor, capacitor, etc).

Ac Voltage Applied To A Resistor

Consider a circuit containing a resistance ‘R’ connected to an alternating voltage.
Let the applied voltage be
V = V_o sinωt ______(1)
According to Ohm’s law, the current at any instant can be written as
I = \(\frac{V_{0} \sin \omega t}{R}\)
Where I₀ = V_o/R is the peak value of current. Comparing eq(1) and eq(2), we can understand that the current and voltage are in same phase.
Graphical variation of current and voltage:

R.M.S value (or Virtual value, effective value) of current and voltage:
The mean value of emf and current for one cycle is zero. Hence to measure ac, the root mean square (rms) values are considered.

The r.m.s value or virtual value of an AC is the square root of the mean of the squares of the instantaneous value of current taken over a complete cycle.
I_rms = \(\frac{I_{0}}{\sqrt{2}}\) and V_rms = \(\frac{V_{0}}{\sqrt{2}}\)
where I₀ – maximum current, V₀ – maximum voltage, (r.m.s.- root mean square).

Power dissipated in the resistor:
The average power consumed in one complete cycle,

Substituting current and voltage, We get

Representation Of Ac Current And Voltage By Rotating Vectors – Phasors
To represent the phase relation between current and voltage, phasors are used. Aphasoris a vector which rotates about the origin with an angular speed ω. The vertical components of phasors of V and I represent instantaneous value of V and I at a time t (see figure). The length of phasors give maximum amplitudes of V and I.

Phasor diagram of v and i for the circuit containing resistor only

The figure(a) represent the voltage and current phasors and their relationship at time t₁. Fig (b) shows the graphical variation of V and I.

Ac Voltage Applied To An Inductor

Consider a circuit containing an inductor of inductance ‘L’ connected to an alternating voltage.
Let the applied voltage be
V = V_o sinωt _____(1)
Due to the flow of alternating current through coil, an emf, \(\frac{d I}{d t}\) is produced in the coil. This induced emf is equal and opposite to the applied emf (in the case of ideal inductor)

Integrating, we get

Where I_o = \(\frac{V_{0}}{L \omega}\),
The term Lω is called inductive reactance. Comparing eq(1) and eq(2), we can understand that, the current lags behind the voltage by an angle 90°.
Graphical variation of current and voltage:

Phasor diagram:

Inductive reactance X_L:
The resistance offered by an inductor to a.c. flow is called inductive reactance.
Inductive reactance

Power Consumed by an Inductor Carrying AC:
The instantaneous value of voltage and current in a pure inductor is
V = V_o sinωt
I = I_o cosωt
The average power consumed per cycle.

The above expression indicates that the average power or net energy consumed by an inductor carrying ac for a full cycle is zero.

Ac Voltage Applied To A Capacitor

Consider a circuit containing a capacitor of capacitance ‘C’ connected to alternating voltage.
Let the applied voltage be V = V_o sinωt _____(1)
The instantaneous current through capacitor

Substituting eq.(1) in eq.(2), we get

\(\frac{1}{\mathrm{C} \omega}\) is called capacitative reactance
Comparing eq(1) and eq(3), we can understand that, the current leads the voltage by an angle 90°
Graphical variation of current and voltage:

Phaser diagram:

Capacitative Reactance X_c:
The resistance offered by a capacitor to ac flow is called Capacitative reactance
Capacitative reactance

1. Power consumed by a capacitor carrying current:
The instantaneous value of voltage and current in a pure inductor is
V = V_o sinωt
I = I_o cosωt
The average power consumed per cycle.

The above expression indicates that the average power or net energy consumed by a capacitor carrying ac for a full cycle is zero.

Ac Voltage Applied To A Series Lcr Circuit

Consider a circuit containing an inductance L, resistance R and capacitance C connected in series. An alternating voltage V = V_o sinωt is applied to the circuit.
Phasor Diagram:

Let V_R be the voltage across R. This voltage is represented by a vector OA (since I and V_R are in same direction). Let V_L be the voltage across L. This voltage is represented by a vector OB (since the voltage V_L leads the current by angle 90°).

Similarly, let V_c be the voltage across C. This voltage is represented by a vector OC (since the voltage V_c lags the current by angle 90°).

The phase difference between V_L and V_c is Φ(ie. they are in opposite directions). So the magnitude of net voltage across the reactance is (V_L – V_c). This is represented by a vector OD in phasor diagram.

The final voltage in the circuit is the vector sum of V_R and (V_L – V_c). The final voltage is represented by diagonal OE.

1. Impedances of LCR circuit:
From the right angled triangle OAE,
Final voltage, V = \(\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{c}}\right)^{2}}\)

Where Z is called impedance of LCR circuit

Phase Difference: Let Φ be the phase difference between final voltage V and current I
From fig (2), we can write

Expression for current:
The eq(2) shows that there is a phase difference between current and voltage. The instantaneous current lags the voltage by an angle (Φ).
If V = V_o sinωt is the applied voltage, the current at any instant can be written as
I = I_o sin(ωt – Φ) _____(3)
Where I_o is the peak value of current. It’s value can be written as

2. Analytical solution:
If we apply V = V_msinωt to an LCR circuit, we can write
V_L + V_R + V_C = V_m sinωt

Substituting these values in eq.(1), we get

The above equation (2) is like the equation for a forced, damped oscillator. Hence we can take the solution of above equation as
q = q_m sin(ωt + θ)

Substituting these values in eq.(2) we get

Multiplying and dividing by Z = \(\sqrt{R^{2}+\left(X_{0}-X_{L}\right)^{2}}\), we have

Substituting these values in eq.(4), we get
q_mωz[cosΦcos(ωt + θ) + sinΦsin(ωt + θ)] = V_msinωt
q_mωz cos(ωt + θ – Φ) = V_m sinωt ______(5)
(∴ cos(A – B) = cosA cosB + sinA sinB)
Comparing the two sides of the eq.(5) we get
V_m = q_mωz = i_mz
where i_m = q_mω
and cos(ωt + θ – Φ) = sinωt
sin (ωt + θ – Φ + π/2) = sinωt (∵ sin(θ + π/2) = cosθ)
ωt + θ – Φ + π/2 = ωt
(θ – Φ) = -π/2
Therefore, the current in the circuit is

Thus, the analytical solution for the amplitude and phase of the current in the circuit agrees with that obtained by the technique of phasors.

3. Resonance:
When ωL = \(\frac{1}{\omega C}\), the impedance of the LCR circuit becomes minimum. Hence current becomes maximum. This phenomena is called resonance.

The frequency of the applied signal at which the impedance of LCR circuit is minimum and current becomes maximum is called resonance frequency.
Expression for resonance frequency
Resonance occurs at ωL = \(\frac{1}{\omega C}\)

Impedance at resonance: Resonance occurs at ωL = \(\frac{1}{\omega C}\). Substituting this condition in eq(1), in section 7.6, we get
Impedance, Z = R
Current at resonance:
substituting ωL = \(\frac{1}{\omega C}\) in eq(2) in section 7.6 we get,
TanΦ = 0, or Φ = 0
Substituting this value in eq(3) in section 7.6, we get
current I= I_o sin ωt
Where I_o = V_o/R
Graphical variation of current with ω in LCR circuit:

Variation of current through LCR circuit with angular frequency co for two cases (1) R= 100Ω and (2) R=200Ω, is shown in the graph
Note: It is important to note that resonance phenomenon is exhibited by a circuit only if both L and C are present in the circuit. Only then the voltages across L and C cancel each other (both being out of phase).

The current amplitude ( V_m/R) is the total source voltage appearing across R. This means that we cannot have resonance in a RL or RC circuit.

4. Sharpness of resonance:

The figure shows the variation of current i with © in a LCR circuit.
Bandwidth: At ω₀, the current in the LCR circuit is maximum. Suppose we choose a value of ω for which the current amplitude is \(\frac{1}{\sqrt{2}}\) times its maximum value.

We can see that there are two values of ω(ω₁ and ω₂) and forwhich current is \(\frac{i_{m}}{\sqrt{2}}\).

The difference ω₂ – ω₁ is called bandwidth.
If we take ω₁ = ω₀ – ∆ω and ω₂ = ω₀ + ∆ω
We get bandwidth, ω₂ – ω₁ = 2∆ω.

Expression for bandwidth and sharpness of resonance:
We know that the current in the LCR circuit

We know that the current in the LCR circuit becomes \(\frac{i_{m}}{\sqrt{2}}\) at ω₂ = ω₀ + ∆ω. Substituting this m in eq(1), we get

But ω₂ = ω₀ + ∆ω, substituting this above equation.

Sharpness of resonance: The quantity \(\left(\frac{\omega_{0}}{2 \Delta \omega}\right)\) is
called sharpness of resonance.
From eq.(4),weget

When bandwidth increases, the sharpness of resonance decreases, ie. the tuning of the circuit will not be good.

Quality Factor (Q): The ratio \(\frac{\omega_{0} L}{R}\) is called the quality factor. When R is low or L is large, the quality factor becomes large. Lange quality factor means that the circuit is more selective.

Power in AC circuit: The power factor
Power in AC circuit with LC and R: In ac circuit the Voltage vary continuously.
∴ The average power in the circuit for one full cycle of period,


(since sin 2A = 2sinA CosA)
The mean value of sin²ωt over a complete cycle is 1/2 and the mean value of sin2ωt over a complete cycle is zero.

True power = Apparent power × power factor
The term P_av called true power. V_rms × I_rms is called the apparent power and cosΦ is called power factor.
powerfactor = \(\frac{\text { True power }}{\text { apparent power }}\)
Power factor is defined as the ratio of true power to apparent power.

Case – 1 (In purely resistive circuit)
In this case, current and voltage are in same phase. Hence Φ = 0
∴ P_av = V_rms I_rmsCosO
True power, P_av = V_rms I_rms

Case – 2 (In a purely inductive and purely capacitative circuit (no resistance)). In this case, the angle between voltage and current is 90°.
∴ P_av = V_rms I_rmsCos 90
True power, P_av = 0
Which means that, the power consumed by the circuit is zero. The current in such a circuit (purely inductive and purely capacitive) doesn’t do any work. A current that does not do any work is called wattles or idle current.

Lc Oscillations

A capacitor can store electrical energy. An inductor can store magnetic energy. When a charged capacitor is connected to an inductor, the electrical energy( of capacitor) transfers to magnetic energy (of inductor) and vise versa. Thus energy oscillates back and forth between capacitor and inductor. This is called L. C. Oscillations.
Expression for frequency:
Applying Kirchoff’s second rule, we get

Transformers
Principle: It works on the principle of mutual induction.
Construction:

A transformer consists of two insulated coils wound over a core. The coil, to which energy is given is called primary and that from which energy is taken is called secondary.

Working and mathematical expression :
Let V₁ N₁ be the voltage and number of turns in the primary. Similarly, let V₂, N₂ be voltage and number of turns in the secondary.

When AC is passed, a change in magnetic flux is produced in the primary. This magnetic flux passes through secondary coil.

If Φ₁ and Φ₂ are the magnetic flux of primary and secondary, we can write Φ₁ α N₁ and Φ₂ α N₂.
Dividing Φ₁ and Φ₂
\(\frac{\phi_{1}}{\phi_{2}}=\frac{N_{1}}{N_{2}}\)
[since Φ is proportional to number of turns] or \(\phi_{1}=\frac{\mathrm{N}_{1}}{\mathrm{N}_{2}} \phi_{2}\)
Taking differentiation on both sides we get

Step up Transformer:
If the output voltage is greater than input voltage, the transformer is called step up transformer. In a step up transformer N₂ > N₁ and V₂ > V₁.

Step down transformer:
If the output voltage is less than the input voltage, then the transformer is called step down transformer. In a step down transformer N₂ < N₁ and V₂ < V₁.

Efficiency of a transformer:
The efficiency of a transformer is defined as the ratio of output power to input power.

For an ideal transformer, efficiency = 1
i.e, V₁I₁ = V₂I₂

1. Power losses in a transformer
(i) Joule loss or Copper loss:
When current passes through a coil heat is produced. This energy loss is called Joule loss. It can be minimized by using thick wires.

(ii) Eddy current loss: This can be minimized by using laminated cores. Laminated core increases the resistance of the coil. Thus eddy current decreases.

(iii) Hysteresis loss: When the iron core undergoes cycles of magnetization, energy is lost. This loss is called hysteresis loss. This is minimized by using soft iron core.

(iv) Magnetic flux loss:
The total flux linked with the coil may not pass through secondary coil. This loss is called magnetic flux loss. This loss can be minimized by closely winding the wires.

Students can Download Chapter 6 Electromagnetic Induction Notes, Plus Two Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Notes Chapter 6 Electromagnetic Induction

Introduction
In this chapter we are going to discuss the laws governing electromagnetic induction; how energy can be stored in a coil, generation of ac, the relation between voltage and current in various circuit components and finally the working of transformer.

The Experiments Of Faraday And Henry
Faraday and Henry conducted a series of experiments to develop principles of electro magnetic induction.
Experiment – 1

Connect a coil to a galvanometer G as shown in the figure. When the north pole of a bar magnet is pushed towards the coil, galvanometer shows deflection. The deflection indicates that a current is produced in the coil.

The galvanometer does not show any deflection when the magnet is held stationary. When the magnet is pulled away from the coil, the galvanometer shows deflection in the opposite direction.

Conclusion of experiment 1:
The relative motion between magnet and coil produces an electric current in the first coil.
Experiment 2

Connect a coil C₁ to a galvanometer G. Take another coil C₂ and connect it with a battery. A steady current in the coil produces a steady magnetic field. When the coil C₂ is moved towards the coil C₁, the galvanometer shows a deflection. This deflection indicates that the electric current is induced in the coil G.

When the coil C₂ is moved away, the galvanometer shows a deflection in the opposite direction. When the coil C₂ is kept fixed, no deflection is produced in the coil C₁.
Conclusion of Experiment – 2:
The relative motion between two coils induces an electric current in the first coil.
Experiment – 3

In this experiment coil C₁ is connected to galvanometer G. The second coil C₂ is connected to a battery through a key K.

When the key K is pressed, the galvanometer shows a deflection. If the key is held pressed continuously, there is no deflection in the galvanometer. When the key is released, a momentary deflection is observed again, (but in opposite direction).

Conclusion of experiment – 3:
The change in current in second coil induces a current in the first coil.

Magnetic Flux

Magnetic flux through a plane of area A placed in uniform magnetic field B can be written as

Φ = BAcosθ

Faraday’S Law Of Induction
Faraday’s law of electromagnetic induction states that the magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.
Mathematically, the induced emf is given by

If the coil contain N turns, the total induced emf is given by,

Lenz’S Law And Conservation Of Energy
Lenz’s law:
Lenz’s law states that the direction emf (or current) is such that it opposes the change in magnetic flux which produces it,
Mathematically the Lenz’s law can be written as

The negative sign represents the effect of Lenz’s law. The magnitude of the induced emf is given by the Faraday’s law. But Lenz’s law gives the direction induced emf.
Lenz’s law is an accordance with the law of conservation of energy.

When the north pole of the magnet is moved towards the coil, the side of the coil facing north pole becomes north as shown in above figure. (Current is produced in the coil and flows in anticlockwise direction).

So work has to be done to move a magnet against this repulsion. This work is converted into electrical energy.

When the north pole of the magnet is moved away from the coil, the end of the coil facing the north pole acquires south polarity. So work has to be done to overcome the attraction. This work is converted into electrical energy. This electrical energy is dissipated as heat produced by the induced current.

Motional Electromotive Force

Consider a rectangular frame MSRN in which the conductor PQ is free to move as shown in figure. The straight conductor PQ is moved towards the left with a constant velocity v perpendicular to a uniform magnetic field B. PQRS forms a closed circuit enclosing an area that change as PQ moves. Let the length RQ = x and RS = I.
The magnetic flux Φ linked with loop PQRS will be BIx.
Since x is changing with time the rate of change of flux Φ will induce an e.m.f. given by

Energy Consideration: A Quantitative Study
Let ‘r’ be the resistance of arm PQ. Consider the resistance of arm QR, RS, and SP as zero. When the arm is moved,
The current produced in the loop,

This current flows through the arm PQ. The arm PQ is placed in a magnetic field. Hence force acting on the arm,

Substituting eq. (1) in eq. (2)

If this arm is pulling with a constant velocity v, the power required for motion P = Fv

The external agent that does this work is mechanical. Where does this mechanical energy go?
This energy is dissipated as heat. The power dissipated by Joule law,

From eq. (3) and (4), we can understand that the workdone to pull the conductor is converted into heat energy in conductor.
Relation between induced charge and magnetic flux:
We know

so the above equation can be written as

We also know magnitude of induced emf

Comparing (1) and (2), we get

Eddy Currents
Whenever the magnetic flux linked with a metal block changes, induced currents are produced. The induced currents flow in a closed paths. Such currents are called eddy currents.
Experiment to Demonstrate Eddy Currents
Experiment -1

Allow a rectangular metal sheet to oscillate in between the pole pieces of strong magnet as shown in figure. When the plate oscillates, the magnetic flux associated with the plate changes. This will induce eddy currents in the plate. Due to this eddy current the rectangular metal sheet comes to rest quickly.
Experiment – 2

Make rectangular slots on the copper plate. These slots will reduce area of plate. Allow this copper plate to oscillate in between magnets. Due to this decrease in area, the eddy current is also decreased. Hence the plate swings more freely.

Some important applications of Eddy Currents:
1. Magnetic braking in trains:
Strong electromagnets are situated above the rails. When the electromagnets are activated, eddy currents induced in the rails. This eddy current will oppose the motion of the train.

2. Electromagnetic damping:
Certain galvanometers have a core of metallic material. When the coil oscillates, the eddy currents are generated in the core. This eddy current opposes the motion and brings the coil to rest quickly.

3. Induction furnace:
Induction furnace can be used to melt metals. A high frequency alternating current is passed through a coil. The metal to be melted is placed in side the coil. The eddy currents generated in the metals produce heat, that melt it.

4. Electric power meters:
The metal disc in the electric power meter (analogue type) rotates due to the eddy currents. This rotation can be used to measure power consumption.

Inductance
An electric current can be induced in a coil by two methods:

1. Mutual induction
2. Self induction

1. Mutual inductance:
Mutual induction:
The phenomenon of production of an opposing e.m.f. in a circuit due to the change in current or magnetic flux linked with a neighboring circuit is called mutual induction.
Explanation

Consider two coils P and S. P is connected to a battery and key. S is connected to a galvanometer. When the key is pressed, a change in magnetic flux is produced in the primary.

This flux is passed through S. So an e.m.f. is produced in the ‘S’. Thus we get a deflection in galvanometer. similarly, when key is opened, the galvanometer shows a deflection in the opposite direction.

Mutual inductance or coefficient of mutual induction:
The flux linked with the secondary coil is directly proportional to the current in the primary
i.e. Φ α I Or Φ = MI
Where M is called coefficient of mutual induction or mutual inductance.
If I = 1, M = Φ
Hence mutual inductance of two coils is numerically equal to the magnetic flux linked with one coil, when unit current flows through the other.

(i) Mutual inductance of two coils:
Expression for mutual inductance:
Consider a solenoid (air core) of cross sectional area A and number of turns per unit length n. Another coil of total number of turns N is closely wound over the first coil. Let I be the current flow through the primary. Flux density of the first coil B= µ₀nI
Flux linked with second coil, Φ = BAN
Φ = µ₀nIAN _____(1)
But we know Φ = MI _______(2)
From eq(1) and eq(2), we get
∴ MI = µ₀nIAN
M = µ₀nAN
If the solenoid is covered over core of relative permeability µ_r
then M = µ_rµ₀nAN

Relation between induced e.m.f. and coefficient of mutual inductance:
Relation between induced e.m.f and mutual inductance
We know induced e.m.f

2. Self inductance:
Self-induction
The phenomenon of production of an induced e.m.f in a circuit when the current through it changes is known as self- induction.
Explanation

Consider a coil connected to a battery and a key. When key is pressed, current is increased from zero to maximum value. This varying current produces a changing magnetic flux around the coil. The coil is situated in this changing flux, so that an e.m.f. is produced in the coil.

This induced e.m.f. is produced in the coil. This induced e.m.f is opposite to applied e.m.f (E). Hence this induced e.m.f is called back e.m.f.

Similarly, when key is released, a back e.m.f is produced which opposes the decay of current in the circuit.

Thus both the growth and decay of currents in a circuit is opposed by the back e.m.f. This phenomenon is called self – induction.

Mathematical expression for self inductance :
Consider a solenoid (air core) of length /, number of turns N and area cross section A. let ‘n’ be the no. of turns per unit length (n = N/l)
The magnetic flux linked with the solenoid,
Φ = BAN
Φ = µ₀nIAN (since B = µ₀nI)
but Φ = LI
∴ LI = µ₀nIAN
L = µ₀nAN
If solenoid contains a core of relative permeability µ_r the L = µ₀µ_rnAN.

Definition of self inductance:
We know NΦ = LI
If I = 1, we get L = NΦ
Self inductance (or) coefficient of self induction may be defined as the flux linked with a coil, when a unit current is flowing through it.
Note: Physically, the self inductance plays the role of inertia.
Relation between induced emf and coefficient of self induction:
When the current through a coil is varied, a back emf produced in the coil. Using Lens law, emf can be written as,

4. Energy stored in an inductor:
When the current in the coil is switched on, a back emf (ε = -Ldt/dt) is produced. This back emf opposes the growth of current. Hence work should be done, against this e.m.f.
Let the current at any instant be ‘I’ and induced emf
E = \(\frac{-\mathrm{d} \phi}{\mathrm{dt}}\)
i.e., work done, dw= EIdt

Hence the total work done (when the current grows from 0 to I₀ is)

This work is stored as potential energy.
V = \(\frac{1}{2}\)LI²

Ac Generator
An ac generator works on electromagnetic induction. AC generator converts mechanical energy into electrical energy.

The structure of an ac generator is shown in the above figure. It consists of a coil. This coil is known as armature coil. This coil is placed in between magnets. As the coil rotates, the magnetic flux through the coil changes. Hence an e.m.f. is induced in the coil.

1. Expression for induced emf:

Take the area of coil as A and magnetic field produced by the magnet as B. Let the coil be rotating about an axis with an angular velocity ω.

Let θ be the angle made by the areal vector with the magnetic field B. The magnetic flux linked with the coil can be written as

Φ = BA cosθ
Φ = BA cosωt [since θ = ωt)
If there are N turns
Φ = NBA cosωt
∴ The induced e.m.f. in the coil,

Let ε₀ = NAB ω,
then s = ε₀ sin ωt.

Expression for current:
When this emf is applied to an external circuit .alternating current is produced. The current at any instant is given by
I = \(\frac{V}{R}\)

(V = ε₀ sin ω)
I = I₀ sin ωt
Where I₀ = ε₀/R, it gives maximum value of current. The direction of current is changed periodically and hence the current is called alternating current.

Variation of AC voltage with time:

Students can Download Chapter 5 Magnetism and Matter Notes, Plus Two Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Notes Chapter 5 Magnetism and Matter

Introduction
The word magnet is derived from the name of an island in Greece called magnesia where magnetic ore deposits were found.
Properties of a magnet

1. When a bar magnet is freely suspended, it points in the north-south direction.
2. There is a repulsive force when north poles (or south poles) are brought close together.
3. We cannot isolate the north or south pole of a magnet.
4. It is possible to make magnets out of iron and its alloys.

Note: The earth behaves as a magnet with the magnetic field pointing approximately from the geographic south to the north.

The Bar Magnet
A magnet has two poles. One pole is North pole and the other South pole.
Magnetic poles:
These two points near the ends of a magnet at which the power of attraction of the magnet is mostly concentrated are called its magnetic poles.
Note: A current carrying solenoid behaves like a bar magnet.

1. The magnetic field lines:
Properties of magnetic field lines

1. The magnetic field lines of a magnet form continuous closed loops.
2. The tangent to the field line at a given point represents the direction of magnetic field at that point.
3. Flux density of magnetic field represents the strength of magnetic field.
4. The magnetic field lines do not intersect

Field lines of bar magnet:

Field lines of current carrying solenoid:

Field lines of electric dipole:

2. Bar magnet as an equivalent solenoid:
Magnetic field along the axis of a solenoid or bar magnet

Consider a solenoid of radius ‘a’ and numberof turns per unit length‘n’. Let 2l be the length and I be the current flowing through the solenoid. Consider a point P at a distance ‘r’ from the centre of solenoid. To find magnetic field at P, we take a circular element of thickness dx at a distance x from the centre of solenoid.
The magnetic field at P due to this small element,

where ndx = N
(total number of turns in a circular element of thickness dx.)
Integrating from x = – l to x = + l, we get

If the point lies at large distance from the solenoid, we can take,
[(r – x)² + a²]^3/2 ≈ r³
r>>a, r>>x
Hence eq.(1) can be written as

Where m is called magnetic moment of the solenoid.

3. The dipole in a uniform magnetic field:
Torque acting on a magnetic dipole:
Consider a magnetic dipole of dipole moment ‘m’ placed in a uniform magnetic field B. If this dipole is rotated to an angle θ, a restoring torque will act on the needle. ie τ = -mBSinθ.
But we know rotational torque τ = la, where I is the moment of inertia of the magnetic dipole and α is the angular acceleration,
lα = -mBSinθ
(Restoring torque and rotational torque are equal in magnitude but opposite in direction)
But α = \(\frac{d^{2} \theta}{d t^{2}}\), for small rotations sinθ ≈ θ.

This equation represents that, the oscillation of this magnetic needle is simple harmonic. When we compare the above equation with standard harmonic.

Potential energy of a magnetic dipole:
The work done in rotating a magnet in a magnetic field is stored in it as its potential energy. If dipole is rotated through an angle (dθ) in a uniform magnetic field B, work done for this rotation,
dw = τdθ
dw = mBsinθdθ
If this magnetic needle is rotated from θ₁ to θ₂, total work done

If dipole is rotated from stable equilibrium (θ₁ = π/2) to θ₂ = 0 we get,
W = -mBcosθ

This work done is stored as magnetic potential energy, ie.

4. The electrostatic analog:
Permanent Magnets And Electromagnets

Substances which retain theirferromagnetic property at room temperature for a longtime, even after the magnetizing field has been removed are called permanent magnets.

The hysteresis curve helps us to select such materials. They should have high retentivity so that the magnet is strong and high coercivity so that the magnetization is not lost by strong magnetic fields. The material should have a wide hysteresis loop. Steel, Alnico, cobalt-steel and nickel are examples.

Electromagnets are usually ferromagnetic materials with low retentivity, low coercivity and high permeability. The hysteresis curve should be narrow so that the energy liberated as heat is small.
The hysteresis curves of both these materials are shown in the above figure.

Magnetism And Gauss’s Law
Gauss’s law in magnetism: The net magnetic flux through any closed surface is zero.

Explanation:

Consider a Gaussian surfaces represented by I and II. Both cases demonstrates that the number of magnetic field lines leaving the surface is balanced by the number of lines entering it. This is true for any closed surface.

The Earth’s Magnetism
Earth’s magnetic field a rise due to electrical currents produced by motion of metallic fluids in the outer core of the earth. This is known as the dynamo effect.

The magnetic field of the earth behaves as magnetic dipole located at the centre of the earth. The axis of the dipole does not coincide with the axis of rotation of the earth. The axis of dipole is titled by 11.3° with axis of rotation of the earth.

The pole near the geographic north pole of the earth is called north magnetic pole (N_m). Likewise, the pole near the geographic south pole is called the south magnetic pole. (S_m).
Note: The north magnetic pole (N_m) behaves like the south pole of a bar magnet (inside the earth). Similarly, the south magnetic pole (S_m) behaves like the north pole of a bar magnet.

(i) Magnetic declination and dip:
The elements of earth’s magnetic field:
The earth’s magnetic field at a place can be completely specified in terms of three quantities. They are

1. Declination
2. Dip
3. Horizontal intensity

Magnetic meridian:
Magnetic meridian at a place is the vertical plane passing through the earth’s magnetic poles.
Geographic meridian:
Geographic meridian at a place is the vertical plane passing through the geographic poles.

1. Magnetic Declination (I):

Declination at a place is the angle between the geographic meridian and magnetic meridian at that place.

2. Dip or Inclination (θ):

The angle between the earth’s magnetic field and the horizontal component of the earth’s magnetic field at a place is called dip. Dip angle changes from place to place. On the equator, the dip is zero and at the poles, the dip is 90°.

3. Horizontal Intensity B_h:
The horizontal intensity at a place is the horizontal components of the earths field.
Relation between Dip, Horizontal intensity and Earth’s magnetic field:

Let B be the Earth’s magnetic field and θ be the angle of dip. Let B_h be the horizontal intensity and Bvthe vertical intensity of the earth’s magnetic field. Then from figure ,we get
B_h = B cos θ
The vertical component, B_v = B sin θ
∴ Tanθ = \(\frac{B_{v}}{B_{h}}\)
and resultant field, B = \(\sqrt{\mathbf{B}_{\mathrm{h}}^{2}+\mathbf{B}_{\mathrm{v}}^{2}}\)

Magnetization And Magnetic Intensity
The magnetic properties of a substance can be studied by defining some parameter such as

1. Intensity of magnetization (M)
2. Magnetic intensity vector (H)
3. Susceptibility
4. Permeability

1. Intensity of magnetisation (M):
It is defined as the magnetic moment per unit volume. It is the measure of the extent to which a specimen is magnetized.

2. Magnetic Intensity Vector (Magnetising field):
It is defined as the magnetic field which produces an induced magnetism in a magnetic substance. If H is the magnetising field and B the induced magnetic field in the material.
ie. H = \(\frac{B}{\mu}\)
where µ is the constant called the magnetic permeability of the medium.

3. Magnetic susceptibility (χ):
Magnetic susceptibility of a specimen is the ratio of its magnetization to the magnetising field,
ie. χ = \(\frac{M}{H}\)

4. Magnetic permeability (µ):
It is the ratio of magnetic field inside a specimen to the magnetising field.
ie. µ = \(\frac{B}{H}\)
µ = µ₀µ_r
µ₀ – Permeability of free space
µ_r – Relative permeability of a medium.

Relation between permeability and susceptibility:
Let a magnetic material be kept in a solenoid. The specimen gets magnetized by induction. The resultant field inside the specimen is the sum of the field due to the current in the solenoid and the field due to the magnetization of the material.
Resultant field B = Field due to current B₀ + Field due to magnetization B_m.
∴ B = B₀ + B_m
But B_m = µ₀M, B₀ = µ₀H
∴ B = µ₀H + µ₀M

Magnetic Properties Of Materials
Materials can be classified as diamagnetic, paramagnetic or ferromagnetic in terms of the susceptibility χ. A material is diamagnetic if χ is negative, para-if χ is positive and small, and Ferro-if χ is large and positive.

1. Diamagnetism:
Diamagnetic substances are those which have tendency to move from stronger to the weaker part of the external magnetic field.
Diamagnetic material in external magnetic field:

Figure shows a bar of diamagnetic material placed in an external magnetic field. The field lines are repelled and the field inside the material is reduced.

Explanation of diamagnetism:
Electrons in an atom orbit around nucleus. These orbiting electrons produce magnetic field. Hence atom possess magnetic moment.

Diamagnetic substances are the ones in which resultant magnetic moment in an atom is zero. When magnetic field is applied, those electrons having orbital magnetic moment in the same direction slow down and those in the opposite direction speed up.

Thus, the substance develops a net magnetic moment in direction opposite to that of the applied field and hence it repels external magnetic field.

Examples:
Some diamagnetic materials are bismuth, copper, lead, silicon, nitrogen (at STP), water and sodium chloride.

Meissner effect:
The phenomenon of perfect diamagnetism in superconductors is called the Meissner effect.

2. Paramagnetism:
Paramagnetic substances are those which get weakly magnetized in an external magnetic field. They get weakly attracted to a magnet.

Reason for paramagnetism:
The atoms of a paramagnetic material possess a permanent magnetic dipole moment. But these magnetic moments are arranged in all directions.

Due to this random arrangement net magnetic moment becomes zero. But in the presence of an external field B₀, the atomic dipole moment can be made to align in the same direction of B0. Hence paramagnetic material shows magnetism in external magnetic field.
Paramagnetic material in external magnetic field:

Figure shows a bar of paramagnetic material placed in an external field. The field lines gets concentrated inside the material, and the field inside is increased.

Examples:
Some paramagnetic materials are aluminium, sodium, calcium, oxygen (at STP) and copper chloride.

Curie law of magnetism:
Curie law of magnetism states that the magnetisation of a paramagnetic material is inversely proportional to the absolute temperature T.

In the case of paramagnetic materials it can be shown that the magnetic susceptibility at temperature is given by

Where C is a constant called curie’s constant.

3. Ferromagnetism:
Ferromagnetic substances are those which gets strongly magnetized in an external magnetic field. They get strongly attracted to a magnet.
Ferro magnetic materials without external magnetic field:

The atoms in a ferromagnetic material possess a dipole moment. These dipoles align in a common direction over a macroscopic volume called domain. Each domain has a net magnetization. Domains are arranged randomly. Hence net magnetic moment of all domains is zero, so ferromagnetic substance does not show magnetism.
Ferro magnetic materials in external magnetic field:

When we apply an external magnetic field B₀, the domains arranged in the direction of B₀ and grow in size. Thus, in a ferromagnetic material the field lines are highly concentrated.

Question 1.
What happens when the external field is removed?
Answer:
In some ferromagnetic materials the magnetisation persists even if external field is removed. Such materials are called hard ferromagnets. Such materials are used to make permanent magnets.
Eg: Alnico
There is a another class of ferromagnetic materials in which the magnetisation disappears on removal of the external field. Such materials are called soft ferromagnetic materials.
Eg: soft iron

Curie temperature:
The ferromagnetic property depends on temperature. At high temperature, a ferromagriet becomes a paramagnet. The domain structure disintegrates with temperature. This disappearance of magnetisation with temperature is gradual. The temperature of transition from ferromagnetic to paramagnetism is called the Curie temperature T_c.

Variation of B and H in paramagnetic materials:
Figure shows the plot of B versus H. As H is gradually increased from zero, B also increase from zero along OP.

As H increases, more and more magnetic dipoles get aligned in the direction of the field. So M increases and hence B increases.

When all the dipoles get aligned in the direction of the field, the curve becomes almost flat. After this there is no increase of B with H.

When H is gradually decreased from P₁ there is no corresponding decrease in the magnetization. The shifting of domains in the ferromagnetic materials is not completely reversible and some magnetization remains even when H is reduced to zero.

The value of the magnetic field when H is zero is called the remanent field Br (Retentivity). If the current in the solenoid is now reversed so that H is in the opposite direction, the magnetic field B can be gradually brought to zero at the point C. The value of H needed to reduce B to zero is called the coercive force He (Coercivity).

The remaining part of curve is obtained by applying H in reverse direction. From these variations it is clear that B always lags behind H. This phenomenon is known as magnetic hysteresis (hysteresis means to lag behind).

The area enclosed by the hysteresis curve gives the loss of energy in the form of heat during the magnetisation – demagnetization cycle.

Permanent Magnets And Electromagnets

Substances which retain theirferromagnetic property at room temperature for a longtime, even after the magnetizing field has been removed are called permanent magnets.

The hysteresis curve helps us to select such materials. They should have high retentivity so that the magnet is strong and high coercivity so that the magnetization is not lost by strong magnetic fields. The material should have a wide hysteresis loop. Steel, Alnico, cobalt-steel and nickel are examples.

Electromagnets are usually ferromagnetic materials with low retentivity, low coercivity and high permeability. The hysteresis curve should be narrow so that the energy liberated as heat is small.
The hysteresis curves of both these materials are shown in the above figure.

Students can Download Chapter 4 Moving Charges and Magnetism Notes, Plus Two Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism

Introduction; Oersted Experiment
The magnetic effect of current was discovered by Danish Physicist Hans Christians Oersted. He noticed that a current in a straight wire makes a deflection in a magnetic needle.

The deflection increases on increasing current. He also found that reversing the direction of current reverses direction of needle. Oersted concluded that current produces a magnetic field around it.

Magnetic Force
1. Sources and fields:
The static charge is the source of electric field. The source of magnetic field is current or moving charge. Both the electric and magnetic fields are vector fields and both obeys superposition principle.

2. Lorentz Force:
The force experienced by moving charge in electric and magnetic field is called Lorentz force. The Lorentz force experienced by charge ‘q’ moving with velocity ‘v’, is given by

= F_electric + F_magnetic
The features of Lorentz Force:

1. The Lorentz force on positive charge is opposite to that on negative charge because it depends on charge ‘q’.
2. The direction of Lorentz force is perpendicular to velocity and magnetic field. Its direction is given by screw rule or right hand rule.
3. Only moving charge experiences magnetic force. For static charge (v = 0), magnetic force is zero.

Note:

1. A charge particle moving parallel or antiparallel to magnetic field will not experience magnetic force and moves undeviated.
2. The work done by magnetic force is zero. Because magnetic force is always perpendicular to direction of velocity.
3. A charged particle entering perpendicular magnetic field (θ = 90°) will make a circular path.
4. The unit of B is Tesla.

3. Magnetic force on current carrying conductor:
Consider a rod of uniform cross section ‘A’ and length ‘e’. Let ‘n’ be the number of electrons per unit volume (number density). ‘v_d’ be the drift velocity of electrons for steady current ‘I’.
Total number of electrons in the entire volume of rod = nAl
Charge of total electrons = nA l.e
‘e’ is the charge of a single electron.
The Lorentz force on electrons,

(I = neAV_d)

Motion In Magnetic Field
Case I:
The charged particle enters perpendicular to magnetic field.(\(\overrightarrow{\mathrm{V}}\) is perpendicular to \(\overrightarrow{\mathrm{B}}\))
When charged particle moves perpendicular to magnetic field, it experiences a magnetic force of magnitude, qVB and the direction of the force is perpendicular to both \(\overrightarrow{\mathrm{B}}\) and \(\overrightarrow{\mathrm{V}}\). This perpendicular magnetic field act as centripetal force and charged particle follows a circular path.

Mathematical explanation:
Let a charge ‘q’ enters into a perpendicular magnetic field B with velocity V. Let r be the radius of circular path. The centripetal force for charged particle is provided by magnetic force.

Thus radius of circle described by charged particle depends on momentum, charge and magnetic field. If ω is the angular frequency
ω = \(\frac{v}{r}\)
Thus from (1) we get

The frequency ν = \(\frac{q B}{2 \pi m}\)
Thus frequency of revolution of charge is independent of velocity (and hence energy)
The time period T = \(\frac{2 \pi \mathrm{m}}{\mathrm{qB}}\)
(ν = \(\frac{1}{T}\)).

Case II:
The charged particles enters at an angle ‘θ’ with magnetic field.
Since the charged particle enters at an angle ‘θ’ with magnetic field, its velocity will have two components; a component parallel to magnetic field, V_॥ (Vcosθ) and a component perpendicular to the magnetic field, V_⊥(Vsinθ).

The parallel component of velocity remains unaffected by magnetic field and it causes charged particle to move along the field.

The perpendicular component makes the particle to move in circular path. The effect of linear and circular movement produce helical motion.

Pitch and Helix: The distance moved along magnetic field in one rotation is called pitch ‘P’

The radius of circular path of motion is called helix.

Motion In Combined Electric And Magnetic Fields
1. Velocity selector:
A transverse electric and mag¬netic field act as velocity selector. By adjusting value of E and B, it is possible to select charges of particular velocity out of a beam containing charges of different speed.

Explanation:
Consider two mutually perpendicular electric and magnetic fields in a region. A charged particle moving in this region, will experience electric and magnetic force. If net force on charge is zero, then it will move undeflected. The mathematical condition for this undeviation is

The charges with this velocity pass undeflected through the region of crossed fields.

2. Cyclotron
Uses: It is a device used to accelerate particles to high energy.
Principles: Cyclotron is based on two facts

1. An electric field can accelerate a charged particle.
2. A perpendicular magnetic field gives the ion a circular path.

Constructional Details:
Cyclotron consists of two semicircular dees D₁ and D₂, enclosed in a chamber C. This chamber is placed in between two magnets. An alternating voltage is applied in between D₁ and D₂. An ion is kept in a vacuum chamber.

Working:
At certain instant, let D₁ be positive and D₂ be negative. Ion (+ve) will be accelerated towards D₂ and describes a semicircular path (inside it). When the particle reaches the gap, D₁ becomes negative and D₂ becomes positive. So ion is accelerated towards D₁ and undergoes a circular motion with larger radius. This process repeats again and again.

Thus ion comes near the edge of the dee with high K.E. This ion can be directed towards the target by a deflecting plate.

Mathematical expression:
Let V be the velocity of ion, q the charge of the ion and B the magnetic flux density. If the ion moves along a semicircular path of radius ‘r’, then we can write

[Since θ =90°, B is perpendicular to v]
or v = \(\frac{q B r}{m}\) _____(1)
Time taken by the ion to complete a semicircular path.

Eq. (2) shows that time is independent of radius and velocity.

Resonance frequency (cyclotron frequency):
The condition for resonance is half the period of the accelerating potential of the oscillator should be ‘t’. (i.e.,T/2 = t or T = 2t). Hence period of AC
T = 2t

K.E of positive ion

Thus the kinetic energy that can be gained depends on mass of particle charge of particle, magnetic field and radius of cyclotron.
Limitations:

1. As the particle gains extremely high velocit, the mass of particle will be changed from its constant value. This will affect the normal working of cyclotron as frequency depends of mass of particle.
2. Very small particles like electron can not be accelerated using cyclotron. This is because as the mass of electron is very small the cyclotron frequency required becomes extremely high which is practically difficult.
3. Neutron can’t be accelerated

Magnetic Field Due To Current Element; Biot Savart Law

The magnetic field at any point due to an element of current carrying conductor is

1. Directly proportional to the strength of the current (I)
2. Directly proportional to the length of the element (dl)
3. Directly proportional to the sine of the angle (θ) between the element and the line joining the midpoint of the element to the point.
4. Inversely proportional to the square of the distance of the point from the element

The direction of magnetic field is perpendicularto the plane containing d/and rand is given by right hand screw rule.
In the above expression \(\frac{\mu_{0}}{4 \pi}\) is the constant of proportionality and µ₀ is called the permeability of vacuum. Its value is 4π × 10^-7 TmA^-1.
Note: A magnetic field acting perpendicularly in to the plane of the paper is represented by the symbol ⊗ and a magnetic field acting perpendicularly out of the paper is represented by the symbol ⵙ.

Comparison between Biot-Savart Law and Coulomb’s law
Similarities:

1. The two laws are based on inverse square of distance and hence they are long range.
2. Both electrostatic and magnetic fields obey superposition principle.
3. The source of magnetic field is linear; (the current element \(\overrightarrow{\mathrm{ldl}}\)). The source of electrostatic force is also linear; (the electric charge).

Differences:

Magnetic Field On The Axis Of A Circular Current Loop

Consider a circular loop of radius ‘a’ and carrying current ‘I’. Let P be a point on the axis of the coil, at distance x from A and r from ‘O’. Consider a small length dl at A. The magnetic field at ‘p’ due to this small element dl,

\(\mathrm{dB}=\frac{\mu_{0} \mathrm{Idl}}{4 \pi \mathrm{x}^{2}}\) _____(1)
[since sin 90° – 1]
The dB can be resolved into dB cosΦ (along Py) and dB sinΦ (along Px).
Similarly consider a small element at B, which produces a magnetic field ‘dB’ at P. If we resolve this magnetic field we get.
dB sinΦ (along px) and dB cosΦ (along py¹)
dB cosΦ components cancel each other, because they are in opposite direction. So only dB sinΦ components are found at P, so total filed at P is

but from ∆AOP we get, sinΦ = a/x
∴ We get

Point at the centre of the loop: When the point is at the centre of the loop, (r = 0)
Then,

1. Magnetic field at the centre of loop:
The magnetic field at a distance x from centre of loop is given by

The direction of magnetic field due to current carrying circular loop is given by right hand thumb rule.

Thumb Rule: Curl of palm of right hand around circular coil with fingers pointing in the direction of current. Then extended thumb gives the direction of magnetic field.

Note:

1. An anticlockwise current gives a magnetic field out of the coil and a clockwise current gives a magnetic field into the coil.
2. The current carrying loop is equivalent to magnetic dipole of dipole moment m = IA

Ampere’s Circuital Law
According to ampere’s law the line integral of magnetic field along any closed path is equal to µ₀ times the current passing through the surface.

Applications Of Ampere’s Circuital Law
1. Long straight conductor:
Consider a long straight conductor carrying T ampere current. To find magnetic field at ‘P’, we construct a circle of radius r (passing through P).

According to Ampere’s circuital law we can write

[B and dl are parallel]
B∫dl = µ₀I
B2πr = µ₀I
B = \(\frac{\mu_{0} I}{2 \pi r}\)

2. Magnetic field due to long solenoid:

Consider a solenoid having radius T. Let ‘n’ be the number of turns per unit length and I be the current flowing through it.
In order to find the magnetic field (inside the solenoid) consider an Amperian loop PQRS. Let V ‘ be the length and ‘b’ the breadth
Applying Amperes law, we can write

Substituting the above values in eq (1),we get
Bl = µ₀ l_enc ____(2).
But l_enc = n l I
where ‘nl ’ is the total number of turns that carries current I (inside the loop PQRS)
∴ eq (2) can be written as
Bl = µ₀ nIl
B = µ₀nI
If core of solenoid is filled with a medium of relative permittivity µ_r. then
B = µ₀µ_rnl

3. The toroid:
Consider a toroid of average radius ‘r’. Let ‘n’ be the number of turns per unit length. Let I be the current flowing through the toroid. In order to find magnetic field inside the toroid, an camperian loop of radius ‘r’ is considered.

Applying Amperes law to the loop, we can write

Where ‘n2πr ‘ is the total number of turns of the solenoid that carries current I (inside the Amperian loop) Integrating the eq(1) we get
B 2πr = µ₀2πrI
B = µ₀n I
If the core of the solenoid is filled with a medium of relative permeability µ_r then the above equation is modified as

Note: The magnetic field due to toroid is same as that due to solenoid.

Force Between Two Parallel Currents, The Ampere Force Between Two Parallel Conductors

P and Q are two infinitely long conductors placed parallel to each other and separated by a distance r, Let the current through P and Q be l₁ and l₂ respectively.
Magnetic field at a distance ‘r’ from P is

Conductor ‘Q’ is placed in this magnetic field.
If l₂ is the length of the conductor ‘Q’, the Lorentz force on ‘Q’ is

∴ Force per unit length can be written as,

Where f = \(\frac{F}{\ell_{2}}\)
Note:

1. When currents are in the same direction, the force is attractive
2. If the currents are in the opposite direction, the force is repulsive.

Definition of ampere:
An ampere is defined as that constant current which if maintained in two straight parallel conductors of infinite lengths placed one meter apart in vacuum will produce between a force of 2 × 10^-7 Newton per meter length.

Force On A Current Loop, Magnetic Dipole
1. Torque on a rectangular current loop in uniform magnetic filed:

Considers rectangular coil PQRS of N turns which is suspended in a magnetic field, so that it can rotate (about yy 1). Let ‘l’ be the length (PQ) and ‘b’ be the breadth (QR).

When a current l flows in the coil, each side produces a force. The forces on the QR and PS will not produce torque. But the forces on PQ and RS will produce a Torque.
Which can be written as
τ = Force × ⊥ distance _______(1)
But, force = BlI ______(2)
[since θ = 90° ]
And from ∆QTR , we get
⊥ distance (QT) = b sin θ ______(3)
Substituting the vales of eq (2) and eq (3) in eq(1) we get
τ = BIl b sin θ
= BIA sin θ [since lb = A (area)]
τ = IAB sin θ
τ = mB sin θ [since m = IA]

If there are N turns in the coil, then
τ = NIAB sin θ

2. Circular Current loop as a magnetic di pole:
Current loop of any shape act as magnetic dipole.
Current loop acts as magnetic dipole:
The magnetic field due to circular loop of radius R carrying current I at a distance ‘x’ from the centre of loop (on the axis of loop) is given by,

The magnetic field at large distance (x>>R) on axis of loop is

Dividing and multiplying by π

Comparison of magnetic dipole and electric dipole:
The equation (1) is similar to electric field due to electric dipole at a distance ‘x’ from the centre of dipole on its axial line.

Comparing eq(1) and (2), we get 1

m → P
B → E
From this comparison it is clear that a circular current loop acts as a magnetic dipole.

3. The magnetic dipole moment of a revolving electron:
According to Bohr’s model of atom, electrons are revolving around nucleus in its orbit. The electron revolving in its orbit can be considered as circular current loop.

Consider an electron of charge e, revolving around nucleus of charge +ze as shown in figure. The uniform, circular motion of electron constitute current ‘I’. If T is the period of revolution e
I = \(\frac{e}{T}\) _____(1)
If r is the radius of orbit and V s the orbital speed then
T = \(\frac{2 \pi r}{v}\)
Substituting this in (1), we get
I = \(\frac{e v}{2 \pi r}\)
The magnetic moment associated orbiting electron is denoted by µ₁

A = πr², area of orbit
Dividing and multiplying by m_e (Mass of electron)

Applying Quantum Theory, Bohr has proposed that angular momentum of electron can take only discrete values given by,
l = \(\frac{\mathrm{nh}}{2 \pi}\) (Bohr’s quantization condition where n = 1, 2, 3, ……..etc) where h is Plank’s constant. Thus

The orbital magnetic moment of electron is given by

Bohr Magneton: We get the minimum value of magnetic moment, when n = 1 ie

(when n = 1)
Its value is 9.27 × 10^-24 Am². This is called Bohr magneton.
Gyromagnetic Ratio:
The orbital magnetic moment of electron is related to orbital angular momentum ‘l’ as

The ratio of orbital magnetic moment to orbital angular momentum is constant. This constant is called gyromagnetic ratio. Its value is 8.8 × 10¹⁰ c/kg for an electron.

The Moving Coil Galvanometer
It is an instrument used to measure small current.
Principle: A conductor carrying current when placed in a magnetic field experiences a force, (given by Fleming’s left hand rule).
Construction:

A moving coil galvanometer consists of rectangular coil of wire having area ‘A’ and number of turns ‘n’ which is wound on metallic frame and is placed between two magnets. The magnets are concave in shape, which produces radial field.

Working: Let ‘l’ be the current flowing the coil, Then the torque acting on the coil.
τ = NIAB Where A is the area of coil and B is the magnetic field.
This torque produces a rotation on coil, thus fiber is twisted and angle (Φ). Due to this twisting a restoring torque (τ = KΦ) is produced in spring.
Under equilibrium, we can write
Torque on the coil = restoring torque on the spring
or NIAB = kΦ
or Φ = (\(\frac{\mathrm{BAN}}{\mathrm{K}}\))I
The quantity inside the bracket is constant for a galvanometer.
Φ α I
The above equation shows that the deflection depends on current passing through galvanometer.

1. Ammeter and voltmeter:
For measuring large current, the galvanometer can be converted in to ammeter and voltmeter.
Ammeter:
Ammeter is an instrument used to measure current in the circuit.

A galvanometer can be converted into an ammeter by a low resistance (shunt) connected parallel to it.

Theory:
Let G be the resistance of the galvanometer, giving full deflection fora current Ig.

To convert it into an ammeter, a suitable shunt resistance ‘S’ is connected in parallel. In this arrangement Ig current flows through Galvanometer and remaining (I – Ig) current flows through shunt resistance.
Since G and S are parallel
P.d Across G = p.d across
Ig × G = (I – Ig)S

Connecting this shunt resistance across galvanometer we can convert a galvanometer into ammeter.

2. Conversion of galvanometer into voltmeter:
To convert a galvanometer into a voltmeter, a high resistance is connected in series with it.

Theory:
Let Ig be the current flowing through the galvanometer of resistance G. Let R be the high resistance co connected in series with G.

From figure we can write
V = IgR + IgG
V – IgG = IgR

Using this resistance we can covert galvanometer in to voltmeter.

Current sensitivity:
The current sensitivity of galvanometer is the deflection produced by unit current.

The current sensitivity can be increased by increasing number of turns.

The voltage sensitivity:
The voltage sensitivity of galvanometeris the deflection produced by unit voltage.

The increase in number of turns will not change voltage sensitivity.
When number of turns double (N → 2N), the resistance of the wire will be double (ie. R → 2R). Hence the voltage sensitivity does not change.

Students can Download Chapter 8 The d and f Block Elements Notes, Plus Two Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Notes Chapter 8 The d and f Block Elements

The d-block (Transition elements) – elements of the groups 3 – 12 in which the d-orbitals are progressively filled. The f-block (Inner transition elements) – elements in which the 4f and 5f orbitals are progressively filled.

The Transition Elements (d-block):
Their position is in between more electropositive s-block and more electronegative p-block elements.

General Electronic Configuration:
(n -1) d^1-10ns^1-2
The transition metals are classified as,

1. 3d series – 1^st transition series (Sc – Zn)
2. 4d series – 2^nd transition series (Y – Cd)
3. 5d series – 3^rd transition series (La, Hg – Hg)
4. 6d series – 4^th transition series (Ac, Rf – Cn)

Pseudo transition elements – Zn, Cd and Hg are not regarded as transition elements because their orbitals are completely filled in the ground state as well as in their common oxidation states, [(n – 1)d¹⁰ns²]. But they included in transition series due to some similarity to transition metals.

General Properties of Transition Elements:
1. Physical Properties:
High tensile strength, ductility malleability, high thermal and electrical conductivity and metallic lustre, very much hard and low volatile (except Zn, Cd and Hg), high mp (due to interatomic metallic bonding) and bp.

2. Variation in Atomic and Ionic Sizes:
Decreases with increasing atomic number because the new electron enters a d-orbital with low shielding power.

3. Ionisation Enthalpies:
Due to an increase in nuclear charge which accompanies the filling of the inner d-orbitals, there is an increase in ionisation enthalpy along the series.

First ionisation potential/enthalpy of the 5d series are higher than those of the 3d and 4d metals. This is due to lanthanoid contraction caused by poor shielding of the 4f electrons.

4. Oxidation State:
Transition elements shows various oxidation states which aries due to incomplete filling of d-orbital. The elements which give the greatest number of oxidation state occur in or near the middle of the series, e.g. Mn (+2 to +7)

5. Trends in the M²⁺/M Standard Electrode Potential:
The general trend towards less negative E° values across the series is related to the general increase in the sum of the first and second ionisation enthalpies.

6. Trends in Stability of Higher Oxidation State:
In halides, the ability of fluorine to stabilise the highest oxdn. state is due to either higher lattice energy or higher bond enthalpy. The stability of Cu²⁺(aq) rather than Cu⁺(aq) is due to the much more negative Δ_hydH° of Cu²⁺(aq) than Cu⁺(aq).

7. Chemical Reactivity:
Many of them are sufficiently electropositive to dissolve in mineral acids, a few are unaffected by simple acids. The metals of the first series are relatively more reactive and are oxidised by 1M H⁺ (except Cu).

8. Magnetic Properties:
(a) Diamagnetism:
Due to paired electrons, they are weakly repelled by applied magnetic field.

(b) Paramagnetism:
Due to presence of unpaired electrons paramagnetic substances are weakely attracted by applied magnetic field.

(c) Ferromagnetic:
Extreme form of paramagnetism, very strongly attracted by magnetic field. For transition elements, the magnetic moment is determined by the number of unpaired electrons and is calculated by spin-only formula,
\(\mu=\sqrt{n(n+2)}\)
where n is the number of unpaired electrons. The unit is Bohr magneton (BM). e.g.

9. Formation of Coloured Ions:
Most of the transition metal ions are coloured due to d-d transition of electrons. When an electron from a lower energy d orbital is excited to a higher energy d orbital, the energy of excitation corresponds to the frequency of light absorbed. The colour observed corresponds to the complementary colour of the light absorbed,

Example:

1. 1. • Sc³⁺(3d°), Ti⁴⁺ (3d°), Zn²⁺ (3d¹⁰) – Colourless
      • Ti³⁺ (3d¹) – Purple
      • Mn²⁺ (3d⁵) – Pink
      • Fe²⁺ (3d⁶) – Yellow
      • Fe³⁺ (3d⁵) – Green.

10. Formation of Complex Compounds:
They can form a large number of complex compounds due to the comparatively smaller sizes of the metal ions, their high ionic changes and the availability of d- orbital for bond formation.

(a) Catalytic Properties:
It is due to their ability to adopt multiple oxidation states and to form complexes, eg: V₂O₅ (Contact process), Fe (Haber’s process), Ni/Pt/Pd (Hydrogenation of hydrocarbon), TiCl₄ & Al(C₂H₅)₃ (Zeigler – Natta catalyst – polymerisation of ethene and propene).

11. Formation of Interstitial Compounds:
They are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals. They hey are non-stoichiometric and are neither ionic nor covalent.

Characteristics – high m.p, very hard, retain metallic conductivity, chemically inert.

12. Alloy Formation:
Due to similar radii, they form alloys very easily.

Some Important Compounds of Transition Elements:
(a) Potassium Dichromate (K₂Cr₂O₇):
Obtained by the fusion of chromite ore (K₂Cr₂O₄ with Na/K₂CO₃ in pressure of air.

4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂

The yellow solution of sodium chromate is filtered and acidified with H₂SO₄ to give orange sodium di chromate.

2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + H₂O
Na₂Cr₂O₇ is converted into K₂Cr₂O₇ by adding KCl.
Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl

The chromate and dichromate are interconvertible in aqueous solution depending upon P^H of the solution.
2CrO₄^2- + 2H⁺ → Cr₂O₂^2- + H₂O
CrO₇^2- + 2OH^– → 2CrO₄^2- + H₂O

Uses:
K₂/Na₂Cr₂O₇-strong oxidising agents. Na₂Cr₂O₇ used in organic chemistry due to its greater solubility. K₂Cr₂O₇ is used as primary standard in volumetric analysis. Oxidising action in acidic.
solution:
Cr₂O₂^2- + 14H⁺ + 6e^– → 2Cr³⁺ + 7H₂O
e.g. It oxidises l^– to l₂, S^2- to S, Sn²⁺ to Sn⁴⁺ and Fe²⁺ to Fe³⁺

(b) Pottassium Permanganate (KMnO₄):
Preparation:
By the fusion of pyrolusite ore (MnO₂) with KOH and oxidising agent like KNO₃to give dark green K₂MnO₄ which disproportionates in a neutral or acidic solution to give KMnO₄.

2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
3MnO₄^2- + 4H⁺ → 2MnO₄^– + MnO₂ + H₂O

Commercial preparation:
By the electrolytic oxidation of MnO₄^2- ion (Manganate ion).

Laboratory preparation:
By oxidising Mn²⁺ salt using peroxodisulphate.
2Mn²⁺ + 5S₂O₈^2- + 8H₂O → 2MnO₄^– + 10SO₄^2- + 16H^–

Properties:
Dark purple colour, isostructural with KClO₄, on heating decomposes at 513 K
(2KMnO₄ → K₂MnO₄ + MnO₂ + O₂).
It has temperature dependent paramagnetism. Manganate ion – green, paramagnetic (one unpaired electron). Permanganate ion – purple, diamagnetic. The manganate and permanganate ions are tetrahedral.

Acidified permanganate solution oxidises oxalates to CO₂, Fe²⁺ to Fe³⁺, NO₂^– to N03^–, I^– to I₂, S^2- to S, SO₃^2- to SO₄^2-.

Uses:
In analytical chemistry; in organic chemistry as oxidising agent; for bleaching wool, cotton, silk and other textile fibres; fordecolourisation of oils.

The Inner Transition Elements (f-block):
It consist of the two series, lanthanoids (the 14 elements following La) and actinoids (the 14 elements following Ac).

The Lanthanoids:
1. General Electronic Configuration:
(n – 2) f^1-14(n-1)d^0-1ns²

2. Atomic and Ionic Sizes:
There is a regular (steady) decrease in the size of atoms/ions with increase in atomic number as we move across from La to Lu. This slow decrease in size is known as lanthanoid contraction.

(a) Cause of Lanthanoid Contraction:
The 4f electrones constitute inner shells and are ineffective in screening the nuclear charge. Consequently, the attraction of the nucleus for the electrones in the outer most shell increases with increase in atomic number and the electron cloud shrinks. As a result, the size of the lanthanoids decreases.

Consequences:
(a) Similarity of second and third transition series:
The atomic radii of 2^nd row transition series are almost similar to those of third row transition series. Zrand Hf have almost similar radii. This makes it difficult to separate the elements in the pure state.

(b) Variation in the basic strength of hydroxides:
The size of M³⁺ ion decreases and covalent character M-OH increases. OH^– ions are not easily released. Hence the basic strength of oxides and hydroxides decrease from lanthanum to lutetium.

3. Oxidation States:
They display variable oxidation state. The most stable oxidaiton state is +3. They also show +2 and +4 oxidaiton states.

4. General Characteristics:
Due to f-f transition, they form coloured ions. They form carbides when heated with carbon, liberates H₂ from dilute acids, form halides, oxides and hydroxides, form alloys, e.g. Misch metal.

Actinoids:
14 elements from Th to Lr, radio active, most of the elements are man made.

1. Electronic configuration-similar to Lanthanoids, but the last electron is filled in 5f – orbital.

2. Ionic Sizes:
The gradual decrease in the size of the atoms or ions across the series (actinoid contraction). It is greater because of poor shielding by 5f electrons.

3. Oxidation States:
Common +3 oxidation state, also show +4, +5, +6, +7. But +3 and +4 ions tend to hydrolyse.

4. General Characteristics and Comparison with Lanthanoids:
Highly reactive metals. HCl acid attacks all metals, alkalies have no action, magnetic properties are more complex than those of lanthanoids.

Application of d- and f-block Elements:
Iron and steels-most important construction materials, TiO- used in pigment industry, MnO₂ – used in dry battery cells. Battery industry also requires Zn and Ni/Cd. Cu, Ag and Au – coinage metals.

Metals and metal compounds – essential catalysts.
PdCl₂ – used in Wacker Process
AgBr -used in photography.