Plus Two Zoology Notes Chapter 2 Reproductive Health

Students can Download Chapter 2 Reproductive Health Notes, Plus Two Zoology Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Zoology Notes Chapter 2 Reproductive Health

Reproductive health – problems and strategies
The Govt of India has started family planning (1951)and ‘Reproductive and Child Health Care (RCH)
programmes’ that helps
1. To create awareness among people about various reproduction related aspects.

2. Introduction of sex education in schools to avoid myths and misconceptions about sex-related aspects.

3. The information about reproductive organs, adolescence and related changes, safe and hygienic sexual practices, sexually transmitted diseases (STD), AIDS, etc., would help adolescent age group to lead a reproductively healthy life.

4. The information of available birth control options, care of pregnant mothers, post-natal care of the mother and child, importance of breastfeeding, equal opportunities for the male and the female child, etc., are the important components build up a socially responsible and healthy society.

5. The ban on amniocentesis for sex-determination helps to legally check increasing female foeticides, massive child immunisation, etc., are some programmes.

Plus Two Zoology Notes Chapter 2 Reproductive Health

Population Explosion And Birth Control
According to the census report of 2001
Plus Two Zoology Notes Chapter 2 Reproductive Health 1
Population increased due to decreased

  1. Maternal mortality rate (MMR)
  2. Infant mortality rate (IMR)

According to the 2001 census report, the population growth rate was 1.7 per cent.
Marriageable age raised to 18 years for females and 21 years for males. Government was taken measures to check this population growth rate by the contraceptive methods.

An ideal contraceptive should be user-friendly, easily available, effective and reversible with no or least side-effects.. Natural/Traditional, Barrier, lUDs, Oral contraceptives, Injectables, Implants and Surgical methods.

Natural Methods:
Periodic abstinence – Here the couples avoid or abstain from coitus from day 10 to 17 of the menstrual cycle.

Withdrawal or coitus interruptus is another method in which the male partner withdraws his penis from the vagina just before ejaculation so as to avoid insemination.

Plus Two Zoology Notes Chapter 2 Reproductive Health

Lactational amenorrhea This method is based on the fact that ovulation and therefore the cycle do not occur during the period of intense lactation following parturition. So chances of conception are almost nil.

Barrier Method:
Condoms (Nirodh’) are barriers made of thin rubber/ latex sheath that are used to cover the penis in the male or Wagina and cervix in the female, just before coitus. This can prevent conception.
Plus Two Zoology Notes Chapter 2 Reproductive Health 2
Use of condoms have additional benefit of protecting the user from STDs and AIDS. Both the male and the female condoms are disposable.

Diaphragms, cervical caps and vaults are also barriers made of rubber that are inserted into the female reproductive tract to cover the cervix during coitus. So it helps to prevent conception by blocking the entry of sperms through the cervix.
Plus Two Zoology Notes Chapter 2 Reproductive Health 3

Plus Two Zoology Notes Chapter 2 Reproductive Health

Intra Uterine Devices (lUDs):
IUDs are ideal contraceptives for the females. These devices are inserted by doctors or expert nurses in the uterus through vagina.

Two types are copper releasing lUDs (CuT, Cu7, Multiload 375) and the hormone releasing lUDs (Progestasert, LNG-20)

IUDs increase phagocytosis of sperms within the uterus and the Cu ions released suppress sperm motility and the fertilising capacity of sperms.

The hormone releasing lUDs make the uterus unsuitable for implantation It is one of most widely accepted methods of contraception in India.

Oral contraceptives(pills) contain either progestogens or progestogen-estrogen combinations.

Pills have to be taken daily for a period of 21 days starting within the first five days of menstrual cycle. After a gap of 7 days (during which menstruation occurs) it has to be repeated in the same pattern to prevent conception.

It helps to inhibit ovulation and implantation as well as to prevent entry of sperms, eg-Saheli.

Saheli-a new oral contraceptive for the females-was developed by scientists at Central Drug Research Institute (CDRI) in Lucknow.

Plus Two Zoology Notes Chapter 2 Reproductive Health

Administration of progestogens or progestogen-estrogen combinations or lUDs within 72 hours of coitus have been found to be very effective as emergency contraceptives to avoid possible pregnancy due to rape or casual unprotected intercourse.

Surgical Methods (Sterilisation):
Plus Two Zoology Notes Chapter 2 Reproductive Health 4
In male it is called ‘vasectomy’ a small part of the vas deferens is removed or tied up through a small incision on the scrotum.

In female, ‘tubectomy’ a small part of the fallopian tube is removed or tied up through a small incision in the abdomen or through vagina.

Plus Two Zoology Notes Chapter 2 Reproductive Health

Medical Termination Of Pregnancy (MTP)
Government of India legalised MTP in 1971 to avoid its misuse. So it helps to check indiscriminate and illegal female foeticides which are reported to be high in India.

It is important to avoid unwanted pregnancies either due to casual unprotected intercourse or failure of the contraceptive used during coitus or rapes.

MTPs are also essential in the cases where continuation of the pregnancy could be harmful or even fatal either to the mother or to the foetus or both.

MTPs are considered relatively safe during the first trimester, i.e., upto 12 weeks of pregnancy.

Sexually Transmitted Diseases (STDs)
Diseases or infections which are transmitted through sexual inter course are called sexually transmitted diseases (STD) or venereal diseases (VD) or reproductive tract infections (RTI).
Common STDs.

Gonorrhoea
Syphilis
Genital herpes
Chlamydiasis
Genital warts
Trichomoniasis
Hepatitis-B
AIDS

Early symptoms of most of these are minor and include itching, fluid discharge, slight pain, swellings, etc., in the genital region.

Plus Two Zoology Notes Chapter 2 Reproductive Health

Hepatitis-B and HIV are transmitted by sharing of injection needles, surgicaljnstruments, etc. with infected persons and transfusion of blood, from an infected mother to the foetus.

All diseases are completely curable except hepatitis-B, genital herpes, and HIV infections.
The important steps to control STDs are given below.

(i) Avoid sex with unknown partners / multiple partners.
(ii) Always use condoms during coitus.
(iii) In case of doubt, go to a qualified doctor for early detection and get complete treatment if diagnosed with disease.

Infertility
In the case of infertility, corrections are not possible, the couples could be assisted to have children through certain special techniques commonly known as assisted reproductive technologies (ART).

In vitro fertilisation (IVF—fertilisation outside the body in almost similar conditions as that in the body) followed by embryo transfer (ET).
In test tube baby programme Ova from the wife/donor (female) and sperms from the husband/donor (male) are collected and are induced to form zygote under simulated conditions in the laboratory.
The zygote or early embryo (with upto 8 blastomeres) is transferred into the fallopian tube (ZIFT – Zygote intra fallopian transfer) and embryos with more than 8 blastomeres, into the uterus (IUT – intra uterine transfer).
Transfer of an ovum collected from a donor into the fallopian tube (GIFT – gamete intra fallopian transfer) of another female who cannot produce one.

Plus Two Zoology Notes Chapter 2 Reproductive Health

Intra cytoplasmic sperm injection (ICS) sperm is directly injected into the ovum. It is the procedure to form an embryo in the laboratory.
Infertility cases either due to inability of the male partner to inseminate the female or due to very low sperm counts in the ejaculates, could be corrected by artificial insemination (Al) technique.
In this technique, the semen collected either from the husband or a healthy donor is artificially introduced either into the vagina or into the uterus (lUI-Intra uterine insemination) of the female.

Plus Two Maths Notes Chapter 12 Linear Programming

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

Kerala Plus Two Maths Notes Chapter 12 Linear Programming

Introduction
A special class of optmisation problems such as finding maximum profit, minimum cost, or minimum use of resources, etc, is Linear Programming Problems. In this chapter we study some linear programming problems and their solutions graphically.

Plus Two Maths Notes Chapter 12 Linear Programming

A. Basic Concepts
A linear Programming Problem is one that is concerned with finding the optimal value (maximum or minimum) of a linear function of several variables (called the objective function) subject to the conditions that the variables are non-negative and satisfy a set of linear inequalities (called linear constraints). Variables are sometimes called decision variables and are non-negative.
A few important LPP are;

  • Diet Problem.
  • Manufacturing Problem.
  • Transportation Problem.

1. The common region determined by the constraints including the non-negative constraints x ≥ 0, y ≥ 0 of a LPP is called the feasible region.

2. Points within and on the boundary of the feasible region represents feasible solution of the constraints. Any point outside the feasible region is an infeasible solution.

3. Any point in the feasible region that gives the optimal value (maximum or minimum) of the objective function is called an optimal solution.

Plus Two Maths Notes Chapter 12 Linear Programming

I. Corner point Method

  1. Find the feasible region of the LPP and determine its corner points (vertices).
  2. Evaluate the objective function Z = ax + by at each corner points. Let M and m be the maxjmum and minimum values at these points.
  3. If the feasible region is bounded, M, and m respectively are the maximum and minimum values of the objective function.
  4. If the feasible region is unbounded, then
    • M is the maximum value of the objective function, if the open half-plane determined by ax + by > M has no points in common with the feasible region. Otherwise the objective function no maximum value.
    • m is the minimum value of the objective function, if the open half-plane determined by ax + by < m has no point in common with the feasible region. Otherwise, the objective function has no minimum value.

Plus Two Physics Notes Chapter 3 Current Electricity

Students can Download Chapter 3 Current Electricity 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 3 Current Electricity

Introduction
In the present chapter, we shall study some of the basic laws concerning steady electric currents.

Electric Current
Rate of flow of electric charge is called electric current. or
Plus Two Physics Notes Chapter 3 Current Electricity - 1

Electric Currents In Conductors

  • Conductors: Free electrons are found in conductors. The electric current in conductors is due to the flow of electrons.
  • Electrolytes: The current in electrolyte is due to flow of ions.
  • Semiconductor: The current in semiconductor is due to flow of both holes and electrons.

Plus Two Physics Notes Chapter 3 Current Electricity

Ohm’s Law
At constant temperature, the current through a conductor is directly proportional to the potential difference between its ends.
V α I (or)
Plus Two Physics Notes Chapter 3 Current Electricity - 2
where R is constant, called resistance of materials.

1. Resistance of a material:
Factors Affecting Resistance of Resistor:
For a given material resistance is directly proportional to the length and inversely proportional to the area of cross-section.
R ∝ \(\frac{\mathrm{L}}{\mathrm{A}}\)
Plus Two Physics Notes Chapter 3 Current Electricity - 3
where ρ is the constant of proportionality and is called resistivity of material.

Resistivity (coefficient of specific resistance) of a Material:
The resistance per unit length for unit area of cross-section will be a constant and this constant is known as the resistivity of the material. The resistivity or coefficient of specific resistance is defined as the resistance offered by a resistor of unit length and unit area of cross-section.
Plus Two Physics Notes Chapter 3 Current Electricity - 4
Resistivity is a scalar quantity and its unit is Ω-m.

Conductance and conductivity:
The reciprocal of resistance is called conductance and the reciprocal of resistivity is called conductivity. The SI unit of conductance is seimen and that for conductivity is seimen per meter. The unit of conductance can also be expressed as Ω-1.

Plus Two Physics Notes Chapter 3 Current Electricity

Current density: Current per unit area is called current density
current density j = \(\frac{I}{A}\)
Vector form
Plus Two Physics Notes Chapter 3 Current Electricity - 5

Mathematical expression of Ohm’s law in terms of j and E:
Considers conductor of length i. Let V’ be the potential difference between the two ends of a conductor.
According to ohms law
We know V= IR
Plus Two Physics Notes Chapter 3 Current Electricity - 6
This potential difference produces an electric field E in the conductor. The p.d. across the conductor also can be written as
V = El _____(2)
Comparing (1) and (2), we get
El = jρl
E = jρ
The above relation can be written in vector form as
Plus Two Physics Notes Chapter 3 Current Electricity - 7
where σ is called conductivity of the material.

Plus Two Physics Notes Chapter 3 Current Electricity

Drift Of Electrons And The Origin Of Resistivity
Random thermal motion of electrons in a metal:
Every metal has a large number of free electrons. Which are in a state of random motion within the conductor. The average thermal speed of the free electrons in random motion is of the order of 105m/s.

Does random thermal motion produce any current? The directions of thermal motion are so randomly distributed. Hence the average thermal velocity of the electrons is zero. Hence current due to thermal motion is zero.

(a) Drift Velocity (vd):
The average velocity acquired by an electron under the applied electric field is called drift velocity.

Explanation: When a voltage is applied across a conductor, an electric filed is developed. Due to this electric field electrons are accelerated. But while moving they collide with atoms, lose their energy and are slowed down. This acceleration and collision are repeated through the motion. Hence electrons move with a constant average velocity. This constant average velocity is called drift velocity.

(b) Relaxation time (τ):
Relaxation time is the average of the time between two successive collisions of the free electrons with atoms.

(c) Expression for drift velocity:
Let V be the potential difference across the ends of a conductor. This potential difference makes an electric field E. Under the influence of electric field E, each free electron experiences a Coulomb force.
F = -eE
or ma = -eE
a = \(\frac{-e E}{m}\) _____(1)
Due to this acceleration, the free electron acquires an additional velocity. A metal contains a large number of electrons.
For first electron, additional velocity acquired in a time τ,
v1 = u1 + aτ1
where u1 is the thermal velocity and τ is the relaxation time.
Similarly the net velocity of second, third,……electron

Plus Two Physics Notes Chapter 3 Current Electricity
v2 = u2 + aτ2
v3 = u3 + aτ3
vn = un + aτn
∴ Average velocity of all the ‘n’ electrons will be
Plus Two Physics Notes Chapter 3 Current Electricity - 8
Vav = 0 + aτ (∴ average thermal velocity of electron is zero)
where τ = \(\frac{\tau_{1}+\tau_{2}+\ldots \ldots \ldots+\tau_{n}}{n}\)
where Vav is the average velocity of electron under an external field. This average velocity is called drift velocity.
ie. drift velocity Vd = aτ _____(2)
Plus Two Physics Notes Chapter 3 Current Electricity - 9

(d) Relation between electric current and drift speed:
Consider a conductor of cross-sectional area A. Let n be the number of electrons per unit volume. When a voltage is applied across a conductor, an electric filed is developed. Let vd be the drift velocity of electron due to this field.
Total volume passed in unit time = Avd
Total number of electrons in this volume = Avdn
Total charge flowing in unit time=Avdne
But charge flowing per unit time is called current I ie. current, I = Avdne
I = neAvd
Now current density J can be written as
J = nevd (J= I/A)
Deduction of Ohm’s law: (Vector form)
We know current density
J = nvde
Plus Two Physics Notes Chapter 3 Current Electricity - 10

Plus Two Physics Notes Chapter 3 Current Electricity

1. Mobility: Mobility is defined as the magnitude of the drift velocity per unit electric field.
Plus Two Physics Notes Chapter 3 Current Electricity - 11

Limitations Of Ohm’s Law
Certajn materials do not obey Ohm’s law. The deviations of Ohm’s law are of the following types.
1. V stops to be proportional to I.
Plus Two Physics Notes Chapter 3 Current Electricity - 12
Metal shows this type behavior. When current through metal becomes large, more heat is produced. Hence resistance of metal increases. Due to increase in resistance the V-I graph becomes nonlinear. This nonlinear variation is shown by solid line in the above graph.

2. Diode shows this type behavior. We get different values of current for same negative and positive voltages.
Plus Two Physics Notes Chapter 3 Current Electricity - 13

Plus Two Physics Notes Chapter 3 Current Electricity

3. This type behavior is shown by materials like GaAs. there is more than one value of V for the same current I.
Plus Two Physics Notes Chapter 3 Current Electricity - 14
Note: The materials which donot obey ohms law are mainly used in electronics.

Resistivity Of Various Materials
The materials are classified as conductors, semi conductors, and insulators according to their resistivities. Commercially produced resistors are of two types.

  1. Wire bound resistors
  2. Carbon resistors

1. Wire bound resistor:
Wire wound resistors are made by winding the wires of an alloy.
Eg: Manganin, Constantan, Nichrome.

2. Carbon resistors:
Resistors in the higher range are made mostly from carbon. Carbon resistors are compact. Carbon resistors are small in size. Hence their values are given using a colourcode.

(i) Colourcode of resistors:
The resistance value of commercially available resistors are usually indicated by certain standard colour coding.

The resistors have a set of coloured rings on it. Their significance is indicated in the table The first two bands from the end indicated the first two significant digits and the third band indicates the decimal multiplier. The last metallic band indicates the tolerance.
Value of colours:
Plus Two Physics Notes Chapter 3 Current Electricity - 15

Plus Two Physics Notes Chapter 3 Current Electricity
Illustration:
Plus Two Physics Notes Chapter 3 Current Electricity - 16
The colour code indicated in the given sample is Red, Red, Red with a silver ring at the right end. Then the value of given resistance is 22 × 102 ±10%.

Temperature Dependence Of Resistivity:
The resistivity of a material is found to be dependent on the temperature. The resistivity of a metallic conductor is approximately given by,
ρT = ρo[1 + α(T – To)]
where ρT is the resistivity at a temperature T and ρo is the resistivity at temperature To. α is called the temperature coefficient of resistivity.
Variation of resistivity in metals:
Plus Two Physics Notes Chapter 3 Current Electricity - 17
The temperature coefficient (α) of metal is positive. Which means resistivity of metal increases with temperature. The variation of resistivity of copper is as shown in above figure.
Eg: Silver, copper, nichrome, etc.
Variation of resistivity in semi conductor:
Plus Two Physics Notes Chapter 3 Current Electricity - 18
The temperature coefficient (α)of semiconductor is negative. Which means that resistivity decreases with increase in temperature. The variation of resistivity with temperature for a semiconductor is shown in above figure.
Eg: Carbon,Germanium,silicon
Variation of resistivity in standard resistors:
Plus Two Physics Notes Chapter 3 Current Electricity - 19

Plus Two Physics Notes Chapter 3 Current Electricity
standard resistors, the variation of resistivity will be very little with temperatures. The variation of resistivity with temperature for standard resistors is show above.
Eg: Manganin and constantan.
Explanation for the variation of resistivity:
The resistivity of a material is given by
Plus Two Physics Notes Chapter 3 Current Electricity - 20
The above equation shows that, resistivity depends inversely on number density and relaxation time τ.
Metals:
Number density in metal does not change with temperature. But average speed of electrons increases. Hence frequency of collision increases. The increase in frequency of collision decreases the relaxation time τ. Hence the resistivity of metal increases with temperature.

Insulators and semiconductors:
For insulators and semiconductors, the number density n increases with temperature. Hence resistivity decreases with temperature.

Electrical Energy, Power
Plus Two Physics Notes Chapter 3 Current Electricity - 21
Consider two points A and B in a conductor. Let VA and VB be the potentials at A and B respectively. The potential At A is greater than that B and difference in potential is V.

If ∆Q charge flows from A to B in time ∆t. The potential energy of charge will be decreased. The decrease in potential energy due to charge flow from A to B,
= VA∆Q – VB∆Q
= (VA – VB)∆Q
= V∆Q (VA – VB = V)
This decrease in PE appeared as KE of flowing charges. But we know, the kinetic energy of charge carriers do not increase due to the collisions with atoms. During collisions, the kinetic energy gained by the charge carriers is shared with the atoms.

Hence the atoms vibrate more vigorously ie. The conductor heats up. According to conservation of energy, heat developed in between A and B in time ∆t
∆H = decrease in potential energy
∆H = V∆Q
∆H = VI∆t (∵ ∆Q=I∆t)
\(\frac{\Delta \mathrm{H}}{\Delta \mathrm{t}}\) = VI
Rate of workdone is power, ie

Plus Two Physics Notes Chapter 3 Current Electricity
Plus Two Physics Notes Chapter 3 Current Electricity - 22
using Ohm’s law V = IR
Plus Two Physics Notes Chapter 3 Current Electricity - 23
It is this power which heats up the conductor.
Power transmission:
The electric power from the electric power station is transmitted with high voltage. When voltage increases, the current decreases. Hence heat loss decreases very much.

Combination Of Resistors – Series And Parallel Combination
Resistors in series:
Consider three resistors R1, R2 and R3 connected in series and a pd of V is applied across it.
Plus Two Physics Notes Chapter 3 Current Electricity - 24
In the circuit shown above the rate of flow of charge through each resistor will be same i.e. in series combination current through each resistor will be the same. However, the pd across each resistor are different and can be obtained using ohms law.
pd across the first resistor V1 = I R1
pd across the second resistor V2, = I R2
pd across the third resistor V3 = I R3
If V is the effective potential drop and R is the effective resistance then effective pd across the combination is V = IR
Total pd across the combination = the sum pd across each resistor, V = V1 + V2 + V3
Substituting the values of pds we get IR = IR1 + IR2 + IR3
Eliminating I from all the terms on both sides we get
Plus Two Physics Notes Chapter 3 Current Electricity - 25
Thus the effective resistance of series combination of a number of resistors is equal to the sum of resistances of individual resistors.

Plus Two Physics Notes Chapter 3 Current Electricity

1. Resistors in parallel:
Consider three resistors R1, R2 and R3 connected in parallel across a pd of V volt. Since all the resistors are connected across same terminals, pd across all the resistors are equal.
Plus Two Physics Notes Chapter 3 Current Electricity - 26
As the value of resistors are different current will be different in each resistor and is given by Ohm’s law
Current through the first resistor
I1 = \(\frac{V}{R_{1}}\)
Current through the second resistor
I2 = \(\frac{V}{R_{2}}\)
Current through the third resistor
I3 = \(\frac{V}{R_{3}}\)
Total current through the combination is
I = \(\frac{V}{R}\), where R is the effective resistance of parallel combination.
Total current through the combination = the sum of current through each resistor
I = I1 + I2 + I3
Substituting the values of current we get
Plus Two Physics Notes Chapter 3 Current Electricity - 27
Eliminating V from all terms on both sides of the equations, we get
Plus Two Physics Notes Chapter 3 Current Electricity - 28
Thus in parallel combination reciprocal of the effective resistance is equal to the sum of reciprocal of individual resistances. The effective resistance in a parallel combination will be smaller than the value of smallest resistance.

Plus Two Physics Notes Chapter 3 Current Electricity

Cells, Emf, Internal Resistance
Electrolytic cell:
Electrolytic cell is a simple device to maintain a steady current in an electric circuit. A cell has two electrodes. They are immersed in an electrolytic solution.

E.M.F:
E.M.F. is the potential difference between the positive and negative electrodes in an open circuit. ie. when no current is flowing through the cell.

Voltage:
Voltage is the potential difference between the positive and negative electrodes, when current is flowing through it.

Internal resistance of cell:
Electrolyte offers a finite resistance to the current flow. This resistance is called internal resistance (r).

Relation between ε, V and internal drop:
Consider a circuit in which cell of emfs is connected to resistance R. Let r be the internal resistance and I be the current following the circuit.

According to ohms law,
current flowing through the circuit
Plus Two Physics Notes Chapter 3 Current Electricity - 29
V is the voltage across the resistor called terminal voltage. Ir is potential difference across internal resistance called internal drop.

Plus Two Physics Notes Chapter 3 Current Electricity

Cells In Series And Parallel
Plus Two Physics Notes Chapter 3 Current Electricity - 30
Consider two cells in series. Let ε1, r1 be the emf and internal resistance of first cell. Similarly ε2, r2 be the emf and internal resistance of second cell. Let I be the current in this circuit.
From the figure, the P.d between A and B
VA – VB = ε1 – Ir1 _____(1)
Similarly P.d between B and C
VB – VC = ε2 – Ir2 ______(2)
Hence, P.d between the terminals A and C
VAC = VA – VC = VA – VB + VB – VC
VAC = [VA – VB] + [VB – VC]
when we substitute eqn. (1) and (2) in the above equation.
VAC = ε1 – Ir1 + ε2 – Ir2 VAC = (ε1 – ε2) – I(r1 + r2)
VAC = εeq – Ireq
where εeq = ε1 + ε2, and req = r<sub1 + r2
The rule of series combination:

  1. The equivalent emf of a series combination of n cells is the sum of their individual emf.
  2. The equivalent internal resistance of a series com-bination of n cells is just the sum of their internal resistances.

Cells in parallel:
Plus Two Physics Notes Chapter 3 Current Electricity - 31
Consider two cells connected in parallel as shown in figure. ε1, r1 be the emf and internal resistance of first cell and ε2, r2 be the emf and internal resistance of second cell. Let I1 and I2 be the current leaving the positive electrodes of the cells.
Total current flowing from the cells is the sum of I1 and I2.
ie. I = I1 + I2 ______(1)
Let VB1 and VB2 be the potential at B1 and B2 respectively. Considering the first cell, P.d between B1 and B2.
Plus Two Physics Notes Chapter 3 Current Electricity - 32
Considering the second cell, P.d between B1 and B2
Plus Two Physics Notes Chapter 3 Current Electricity - 33

Plus Two Physics Notes Chapter 3 Current Electricity
Substituting the values I1 and I2 in eq.(1), we get
Plus Two Physics Notes Chapter 3 Current Electricity - 34
Plus Two Physics Notes Chapter 3 Current Electricity - 35
If we replace the combination by a single cell between B1 and B2, of emf and εeq and internal resistance req, we have
V = εeq – Ireq ______(3)
The eq(2) and eq(3) should be same.
Plus Two Physics Notes Chapter 3 Current Electricity - 36
The above equation can be put in a simpler way.
Plus Two Physics Notes Chapter 3 Current Electricity - 37
If there are n cells of emf ε1, ε2,………..εn and internal
resistance r1, r2………..rn respectively, connected in parallel.
Plus Two Physics Notes Chapter 3 Current Electricity - 38

Kirchoff’s Rules
1. First law (Junction rule): The total current entering the junction is equal to the total current leaving the junction.
Explanation:
Plus Two Physics Notes Chapter 3 Current Electricity - 39

Plus Two Physics Notes Chapter 3 Current Electricity
Consider a junction ‘O’. Let I1 and I2 be the incoming currents and I1, I4 and I5 be the outgoing currents.
According to Kirchoff’s first law,
Plus Two Physics Notes Chapter 3 Current Electricity - 40

2. Second law (loop rule): In any closed circuit the algebraic sum of the product of the current and resistance in each branch of the circuit is equal to the netjemf in that branch.

OR

Total emf in a closed circuit is equal to sum of voltage drops
Plus Two Physics Notes Chapter 3 Current Electricity - 41
Explanation: Consider a circuit consisting of two cells of emf E1 and E2 with resistances R1, R2 and R3 as shown in figure. Current is flowing as shown in figure
Applying the second law to the closed circuit ABCDE1A.
-I3R3 + E1 + -I1R1 = 0
Similarly for the closed loop ABCDE2A.
-I2R2 + -I3R3 + E2 = 0
For the closed loop AE2DE1A
-I1R1 + I2R2 + -E2 + E1 = 0
Note:

  • Voltage drop in the direction of current is taken as negative (and vice versa).
  • emf is taken as positive, if we go -ve to +ve terminal (and vice versa)

Plus Two Physics Notes Chapter 3 Current Electricity

Wheatstone’s Bridge
Four resistances P, Q, R, and S are connected as shown in figure. Voltage ‘V’ is applied in between A and C. Let I1, I2, I3 and I4 be the four currents passing through P, R, Q, and S respectively.
Plus Two Physics Notes Chapter 3 Current Electricity - 42
Working:
The voltage across R
When key is closed, current flows in different branches as shown in figure. Under this situation
The voltage across P, VAB = I1P
The voltage across Q, VBC = I3Q __(1)
The voltage across R, VAD = I2R
The voltage across S, VDC = I4S
The value of R is adjusted to get zero deflection in galvanometer. Underthis condition,
I1 = I3 and I2 = I4 _____(2)
Using Kirchoffs second law in loopABDA and BCDB, weget
VAB = VAD ______(3)
and VBC = VDC _______(4)
Substituting the values from eq(1) into (3) and (4), we get
I1P = I2R ______(5)
and I3Q = I4S _____(6)
Dividing Eq(5) by Eq(6)
Plus Two Physics Notes Chapter 3 Current Electricity - 43
[since I1 = I3 and I2 = I4]
This is called Wheatstone condition.

Plus Two Physics Notes Chapter 3 Current Electricity

Meter Bridge
Uses: Meter Bridge is used to measure unknown resistance.
Principle: It works on the principle of Wheatstone bridge condition (P/Q=R/S).
Plus Two Physics Notes Chapter 3 Current Electricity - 44
Circuit details:
Unknown resistance X’ is connected in between A and B. Known resistance (box) is connected in between B and C. Voltage is applied between A and C. A100cm wire is connected between A and C. Let r be the resistance per unit length. Jockey is connected to ‘B’ through galvanometer.
Working: A suitable resistance R is taken in the box. The position of jockey is adjusted to get zero deflection.
If ‘l’ is the balancing length from A, using Wheatstone’s condition,
Plus Two Physics Notes Chapter 3 Current Electricity - 45
knowing R and l, we can find X (resistance of wire)
Resistivity: Resistivity of unknown resistance (wire) can be found from the formula
\(\rho=\frac{\pi r^{2} X}{l}\)
Where r (the radius of wire) is measured using screw gauge. l (the length of wire) is measured using meter scale
Note: Meter bridge is most sensitive when all the four resistors are of the same order

Potentiometer
(a) Comparison of e.m.f of two cells using potentiometer:
Plus Two Physics Notes Chapter 3 Current Electricity - 46

Plus Two Physics Notes Chapter 3 Current Electricity
Principle: Potential difference between two points of a current carrying conductor (having uniform thickness) is directly proportional to the length of the wire between two points.

Circuit details: A battery (B1), Rheostat and key are connected in between A and B. This circuit is called primary circuit. Positive end of E1 and E2 are connected to A and other ends are connected to a two way key. Jockey is connected to a two key through galvanometer. This circuit is called secondary circuit.

Working and theory: Key in primary circuit is closed and then E1 is put into the circuit and balancing length l1 is found out.
Then, E1 α l1 ______(1)
Similarly, E2 is put into the circuit and balancing length (l2 ) is found out.
Then, E2 α l2 _______(2)
Dividing Eq(1) by Eq(2),
\(\frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}}\) _____(3)

(b) Measurement of internal resistance using potentiometer:
Principle: Potential difference between two points of a current carrying conductor (having uniform thickness) is directly proportional to the length of the wire between two points.

Circuit details: Battery B1, Rheostat and key K1 are connected in between A and B. This circuit is called primary.
Plus Two Physics Notes Chapter 3 Current Electricity - 47
In the secondary circuit a battery E having internal resistance ‘r’ is connected . A resistance box (R) is connected across the battery through a key (K2). Jockey is connected to battery through galvanometer.

Working and theory : The key (K1) in the primary circuit is closed and the key is the secondary (K2) is open. Jockey is moved to get zero deflection in galvanometer. The balancing length l1 (from A) is found out.
Then we can write.
E1 α l1 _____(1)
Key K2 is put in the circuit, corresponding balancing length (l2) is found out. Let V be the applied voltage, then
V1 α l1 _____(2)
‘V’ is the voltage across resistance box.
Current through resistance box
ie, voltage across resistance,
V = \(\frac{E R}{(R+r)}\) ____(3)
Substituting eq (3) in eq (2),
\(\frac{E R}{(R+r)} \alpha l_{2}\) ____(4)
Dividing eq (1) by eq (4),
Plus Two Physics Notes Chapter 3 Current Electricity - 48

Plus Two Physics Notes Chapter 3 Current Electricity

Question 1.
Why is potentiometer superior to voltmeter in measuring the e.m.f of a cell?
Answer:
Voltmeter takes some current while measuring emf. So actual emf is reduced. But potentiometer does not take current at null point and hence measures actual e.m.f. Hence potentiometer is more accurate than voltmeter.

Plus Two Physics Notes Chapter 6 Electromagnetic Induction

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.

Plus Two Physics Notes Chapter 6 Electromagnetic Induction

The Experiments Of Faraday And Henry
Faraday and Henry conducted a series of experiments to develop principles of electro magnetic induction.
Experiment – 1
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 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
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 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
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 3

Plus Two Physics Notes Chapter 6 Electromagnetic Induction
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
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 4
Magnetic flux through a plane of area A placed in uniform magnetic field B can be written as
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 5
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
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 6
If the coil contain N turns, the total induced emf is given by,
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 7

Plus Two Physics Notes Chapter 6 Electromagnetic Induction

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
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 8
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.
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 9
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.
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 10
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
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 11
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
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 12

Plus Two Physics Notes Chapter 6 Electromagnetic Induction

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,
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 13
This current flows through the arm PQ. The arm PQ is placed in a magnetic field. Hence force acting on the arm,
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 14
Substituting eq. (1) in eq. (2)
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 15
If this arm is pulling with a constant velocity v, the power required for motion P = Fv
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 16
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,
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 17
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
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 18
so the above equation can be written as
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 19

Plus Two Physics Notes Chapter 6 Electromagnetic Induction
We also know magnitude of induced emf
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 20
Comparing (1) and (2), we get
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 21

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
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 22
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
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 23

Plus Two Physics Notes Chapter 6 Electromagnetic Induction
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
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 24
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.

Plus Two Physics Notes Chapter 6 Electromagnetic Induction

(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 µr
then M = µrµ0nAN

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
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 25

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
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 26

Plus Two Physics Notes Chapter 6 Electromagnetic Induction
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 µr the L = µ0µrnAN.

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,
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 27

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
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 28

Plus Two Physics Notes Chapter 6 Electromagnetic Induction
Hence the total work done (when the current grows from 0 to I0 is)
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 29
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.
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 30
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:
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 31
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
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 32
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,
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 33

Plus Two Physics Notes Chapter 6 Electromagnetic Induction
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 = \(\frac{V}{R}\)
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 34
(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:
Plus Two Physics Notes Chapter 6 Electromagnetic Induction - 35

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments

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:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 1

  • 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.

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments

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

  • Concave mirror
  • Convex mirror

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 2
Principal focus of a concave mirror:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 3
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:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 4
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:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 5
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
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 6

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments

3. The mirror equation:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 7
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
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 8
[∵ NE = AB)
ΔABP and ΔIMP are similar
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 9
From eq.(2) and eq.(3), we get
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 11

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments
applying sign convention we get
PI = -v
PF = -F
PA = -u
Substituting these values in eq.(4) we get
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 12
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.
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 13
Consider an object AB having height ho, which produces an image IM having height hi
In the figure, ΔABP and ΔIMP are equal. ie.
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 14
Applying sign convention
PI = -V, PA = -u, hi = -ve and ho = +ve
We get
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 15
But we know \(\frac{h_{i}}{h_{0}}\) = m (magnification) ie.
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 16

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments
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 – \(\frac{v}{f}\)
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 17
Relation connecting u, f and m
We know
\(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\)
Multiplying throughout by ‘u’ we get
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 18

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:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 19
When light travels from a rarer medium to a denser medium, it deviates towards the normal.
Light from denser to rarer medium:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 20
When light travels from a denser medium to a rarer medium, it deviates away from the normal.

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments

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,
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 21
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:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 22
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.
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 23

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments

(c) Apparent shift in the position of the sun at sunrise and sunset:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 26
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
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 27
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,
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 28
where ‘C’ is the critical angle.
A demonstration for total internal reflection
Demonstration – 1:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 29
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:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 30

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments
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 BCA. BCA is related to the refractive index BnA as
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 31

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:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 32
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:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 33
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.

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments

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

  • convex lenses and
  • concave lenses.

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 34
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:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 35
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,
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 36
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,
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 37

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments
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,
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 38
According to New Cartesian sign convection, we can write,
OP = -u, PI = +v and PC = R
Substituting these values, we get
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 39
Case -1: If the first medium is air, n1 = 1, and n2 = n,
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 40

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.
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 41
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,
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 42
This image I1 will act as the virtual object for the surface ADC and forms the image at v.
Then we can write,
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 43
Adding eq (1) and eq (2) we get
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 44

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments
Dividing throughout by n1, we get
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 45
if the lens is kept in air, \(\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}\) = n
So the above equation can be written as,
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 46
From the definition of the lens, we can take, when u = 8, f = v
Substituting these values in the eq (3), we get
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 47
This is lens maker’s formula
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 48
For convex lens,
f = +ve, R1 = +ve, R2 = – ve
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 49
For concave lens,
f = -ve, R1 = -ve, R2 = +ve
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 50
Lens formula
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 51
Linear magnification: If ho is the height of the object and hi is the height of the image, then linear magnification
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 52

3. Power of a lens:
Power of a lens is the reciprocal of focal length expressed in meter.
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 53

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments
Unit of power is dioptre (D).

4. Combination of thin lenses in contact:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 54
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,
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 55
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
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 56
If the two lenses are replaced by a single lens of focal length ‘F’ the image is formed at V. Then we can write,
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 57
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:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 58

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments
Relation connecting m,v and f:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 59
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 60

Refraction Through A Prism
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 61
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)

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments
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),
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 62
i – d curve:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 63
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.

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments

Dispersion By A Prism
Dispersion: The splitting of the white light into its component colours is called dispersion.
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 64
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:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 65
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:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 66
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.

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments

Rayleigh’s scattering law:
The intensity of the scattered light from a molecule is inversely proportional to the 4th 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:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 67
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:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 68

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments
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:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 69
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:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 70
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.

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments

1. The microscope:
Simple microscope: A simple microscope is a converging lens of small focal length.
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 71
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
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 72
The image is formed at D, ie. v = -D
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 73
This equation is used to find magnification of simple microscope when image at D (D ≈ 25cm).

Image at infinity:
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 78

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments
If the object is placed at f, the image forms at infinity. In this case, magnification,
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 79
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).
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 80
When the final image is formed at infinity,
θi = \(\frac{h^{1}}{v}\) ______(3)
When h1 is the height of image and v is the image distance
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 81
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.
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 82

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments
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
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 83
Where m0 & me are the magnifying power of objective lens and eyepiece lens.
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 84
Eyepiece acts as a simple microscope.
Therefore me = 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
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 85
for compound microscope, uo » fo (because the object of is placed very close to the principal focus of the objective) and vo ≈ L, length of microscope (because the first image is formed very close to the eye piece).
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 86
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
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 87

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments

(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
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 88
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.
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 89

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments
(For small values tan α ≈ α)
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 90
But IC = fo (the focal length objective lens) and IC1 = fe(the focal length eyepiece lens.)
∴ m = \(\frac{f_{0}}{f_{e}}\)
In this case the length of the telescope tube is (f0 + fe).

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.
Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments - 91

(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.

Plus Two Physics Notes Chapter 9 Ray Optics and Optical Instruments

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
fo — is the focal length of concave mirror
f2 — is the focal length of eyepiece.
R – Radius of curvature of concave reflector.

Plus Two Physics Notes Chapter 8 Electromagnetic Waves

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)
Plus Two Physics Notes Chapter 8 Electromagnetic Waves 1

Plus Two Physics Notes Chapter 8 Electromagnetic Waves
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)
Plus Two Physics Notes Chapter 8 Electromagnetic Waves 2
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(ic + id)
Where id is called displacement current. Its value is
Plus Two Physics Notes Chapter 8 Electromagnetic Waves 3

Plus Two Physics Notes Chapter 8 Electromagnetic Waves
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 ic is equal to displacement current id
Answer:
The flux passing through the surface in between plates (see figure 2)
Plus Two Physics Notes Chapter 8 Electromagnetic Waves 4
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,
Plus Two Physics Notes Chapter 8 Electromagnetic Waves 5

Plus Two Physics Notes Chapter 8 Electromagnetic Waves
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
Plus Two Physics Notes Chapter 8 Electromagnetic Waves 6

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,
Plus Two Physics Notes Chapter 8 Electromagnetic Waves 7
The speed of electromagnetic wave in a material medium is given by
Plus Two Physics Notes Chapter 8 Electromagnetic Waves 8

Plus Two Physics Notes Chapter 8 Electromagnetic Waves

(iii) The ratio of magnitudes of electric and magnetic field vectors in free space is constant
Plus Two Physics Notes Chapter 8 Electromagnetic Waves 9
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:
Plus Two Physics Notes Chapter 8 Electromagnetic Waves 10
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.
Ex = E0sin(kz – ωt)
and BY = 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.

Plus Two Physics Notes Chapter 8 Electromagnetic Waves

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.

Plus Two Physics Notes Chapter 8 Electromagnetic Waves

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.
Plus Two Physics Notes Chapter 8 Electromagnetic Waves 11

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.

Plus Two Physics Notes Chapter 8 Electromagnetic Waves

Different Types Of Electromagnetic Waves:
Plus Two Physics Notes Chapter 8 Electromagnetic Waves 12

Plus Two Physics Notes Chapter 7 Alternating Current

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).

Plus Two Physics Notes Chapter 7 Alternating Current

Ac Voltage Applied To A Resistor
Plus Two Physics Notes Chapter 7 Alternating Current - 1
Consider a circuit containing a resistance ‘R’ connected to an alternating voltage.
Let the applied voltage be
V = Vo 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 = Vo/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:
Plus Two Physics Notes Chapter 7 Alternating Current - 2
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.

Plus Two Physics Notes Chapter 7 Alternating Current

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.
Irms = \(\frac{I_{0}}{\sqrt{2}}\) and Vrms = \(\frac{V_{0}}{\sqrt{2}}\)
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,
Plus Two Physics Notes Chapter 7 Alternating Current - 3
Substituting current and voltage, We get
Plus Two Physics Notes Chapter 7 Alternating Current - 4

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
Plus Two Physics Notes Chapter 7 Alternating Current - 5
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.

Plus Two Physics Notes Chapter 7 Alternating Current

Ac Voltage Applied To An Inductor
Plus Two Physics Notes Chapter 7 Alternating Current - 6
Consider a circuit containing an inductor of inductance ‘L’ connected to an alternating voltage.
Let the applied voltage be
V = Vo 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)
Plus Two Physics Notes Chapter 7 Alternating Current - 7
Integrating, we get
Plus Two Physics Notes Chapter 7 Alternating Current - 8
Where Io = \(\frac{V_{0}}{L \omega}\),
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:
Plus Two Physics Notes Chapter 7 Alternating Current - 9
Phasor diagram:
Plus Two Physics Notes Chapter 7 Alternating Current - 10
Inductive reactance XL:
The resistance offered by an inductor to a.c. flow is called inductive reactance.
Inductive reactance
Plus Two Physics Notes Chapter 7 Alternating Current - 11
Power Consumed by an Inductor Carrying AC:
The instantaneous value of voltage and current in a pure inductor is
V = Vo sinωt
I = Io cosωt
The average power consumed per cycle.
Plus Two Physics Notes Chapter 7 Alternating Current - 12
The above expression indicates that the average power or net energy consumed by an inductor carrying ac for a full cycle is zero.

Plus Two Physics Notes Chapter 7 Alternating Current

Ac Voltage Applied To A Capacitor
Plus Two Physics Notes Chapter 7 Alternating Current - 13
Consider a circuit containing a capacitor of capacitance ‘C’ connected to alternating voltage.
Let the applied voltage be V = Vo sinωt _____(1)
The instantaneous current through capacitor
Plus Two Physics Notes Chapter 7 Alternating Current - 14
Substituting eq.(1) in eq.(2), we get
Plus Two Physics Notes Chapter 7 Alternating Current - 15
\(\frac{1}{\mathrm{C} \omega}\) 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:
Plus Two Physics Notes Chapter 7 Alternating Current - 16
Phaser diagram:
Plus Two Physics Notes Chapter 7 Alternating Current - 17
Capacitative Reactance Xc:
The resistance offered by a capacitor to ac flow is called Capacitative reactance
Capacitative reactance
Plus Two Physics Notes Chapter 7 Alternating Current - 18

1. Power consumed by a capacitor carrying current:
The instantaneous value of voltage and current in a pure inductor is
V = Vo sinωt
I = Io cosωt
The average power consumed per cycle.
Plus Two Physics Notes Chapter 7 Alternating Current - 19
The above expression indicates that the average power or net energy consumed by a capacitor carrying ac for a full cycle is zero.

Plus Two Physics Notes Chapter 7 Alternating Current

Ac Voltage Applied To A Series Lcr Circuit
Plus Two Physics Notes Chapter 7 Alternating Current - 20
Consider a circuit containing an inductance L, resistance R and capacitance C connected in series. An alternating voltage V = Vo sinωt is applied to the circuit.
Phasor Diagram:
Plus Two Physics Notes Chapter 7 Alternating Current - 21
Let VR be the voltage across R. This voltage is represented by a vector OA (since I and VR are in same direction). Let VL be the voltage across L. This voltage is represented by a vector OB (since the voltage VL leads the current by angle 90°).

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

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

The final voltage in the circuit is the vector sum of VR and (VL – Vc). The final voltage is represented by diagonal OE.

1. Impedances of LCR circuit:
From the right angled triangle OAE,
Final voltage, V = \(\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{c}}\right)^{2}}\)
Plus Two Physics Notes Chapter 7 Alternating Current - 22
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
Plus Two Physics Notes Chapter 7 Alternating Current - 23

Plus Two Physics Notes Chapter 7 Alternating Current
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 = Vo sinωt is the applied voltage, the current at any instant can be written as
I = Io sin(ωt – Φ) _____(3)
Where Io is the peak value of current. It’s value can be written as
Plus Two Physics Notes Chapter 7 Alternating Current - 24

2. Analytical solution:
If we apply V = Vmsinωt to an LCR circuit, we can write
VL + VR + VC = Vm sinωt
Plus Two Physics Notes Chapter 7 Alternating Current - 25
Substituting these values in eq.(1), we get
Plus Two Physics Notes Chapter 7 Alternating Current - 26
The above equation (2) is like the equation for a forced, damped oscillator. Hence we can take the solution of above equation as
q = qm sin(ωt + θ)
Plus Two Physics Notes Chapter 7 Alternating Current - 27
Substituting these values in eq.(2) we get
Plus Two Physics Notes Chapter 7 Alternating Current - 28
Multiplying and dividing by Z = \(\sqrt{R^{2}+\left(X_{0}-X_{L}\right)^{2}}\), we have
Plus Two Physics Notes Chapter 7 Alternating Current - 29

Plus Two Physics Notes Chapter 7 Alternating Current
Substituting these values in eq.(4), we get
qmωz[cosΦcos(ωt + θ) + sinΦsin(ωt + θ)] = Vmsinωt
qmωz cos(ωt + θ – Φ) = Vm sinωt ______(5)
(∴ cos(A – B) = cosA cosB + sinA sinB)
Comparing the two sides of the eq.(5) we get
Vm = qmωz = imz
where im = qmω
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
Plus Two Physics Notes Chapter 7 Alternating Current - 30
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 = \(\frac{1}{\omega C}\)
Plus Two Physics Notes Chapter 7 Alternating Current - 31
Plus Two Physics Notes Chapter 7 Alternating Current - 32

Plus Two Physics Notes Chapter 7 Alternating Current
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= Io sin ωt
Where Io = Vo/R
Graphical variation of current with ω in LCR circuit:
Plus Two Physics Notes Chapter 7 Alternating Current - 33
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 ( Vm/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:
Plus Two Physics Notes Chapter 7 Alternating Current - 34
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.

Plus Two Physics Notes Chapter 7 Alternating Current

Expression for bandwidth and sharpness of resonance:
We know that the current in the LCR circuit
Plus Two Physics Notes Chapter 7 Alternating Current - 35
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
Plus Two Physics Notes Chapter 7 Alternating Current - 36
But ω2 = ω0 + ∆ω, substituting this above equation.
Plus Two Physics Notes Chapter 7 Alternating Current - 37
Plus Two Physics Notes Chapter 7 Alternating Current - 38

Plus Two Physics Notes Chapter 7 Alternating Current
Sharpness of resonance: The quantity \(\left(\frac{\omega_{0}}{2 \Delta \omega}\right)\) is
called sharpness of resonance.
From eq.(4),weget
Plus Two Physics Notes Chapter 7 Alternating Current - 39
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,
Plus Two Physics Notes Chapter 7 Alternating Current - 40
Plus Two Physics Notes Chapter 7 Alternating Current - 41
(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.
Plus Two Physics Notes Chapter 7 Alternating Current - 42
True power = Apparent power × power factor
The term Pav called true power. Vrms × Irms 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.

Plus Two Physics Notes Chapter 7 Alternating Current

Case – 1 (In purely resistive circuit)
In this case, current and voltage are in same phase. Hence Φ = 0
∴ Pav = Vrms IrmsCosO
True power, Pav = Vrms Irms

Case – 2 (In a purely inductive and purely capacitative circuit (no resistance)). In this case, the angle between voltage and current is 90°.
∴ Pav = Vrms IrmsCos 90
True power, Pav = 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
Plus Two Physics Notes Chapter 7 Alternating Current - 43
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
Plus Two Physics Notes Chapter 7 Alternating Current - 44

Plus Two Physics Notes Chapter 7 Alternating Current

Transformers
Principle: It works on the principle of mutual induction.
Construction:
Plus Two Physics Notes Chapter 7 Alternating Current - 45
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
Plus Two Physics Notes Chapter 7 Alternating Current - 46
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.
Plus Two Physics Notes Chapter 7 Alternating Current - 47
For an ideal transformer, efficiency = 1
i.e, V1I1 = V2I2

Plus Two Physics Notes Chapter 7 Alternating Current

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.

Plus Two Physics Notes Chapter 5 Magnetism and Matter

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.

Plus Two Physics Notes Chapter 5 Magnetism and Matter

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:
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 1
Field lines of current carrying solenoid:
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 2

Plus Two Physics Notes Chapter 5 Magnetism and Matter
Field lines of electric dipole:
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 3

2. Bar magnet as an equivalent solenoid:
Magnetic field along the axis of a solenoid or bar magnet
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 4
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,
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 5
where ndx = N
(total number of turns in a circular element of thickness dx.)
Integrating from x = – l to x = + l, we get
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 6

Plus Two Physics Notes Chapter 5 Magnetism and Matter
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
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 7
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.
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 8

Plus Two Physics Notes Chapter 5 Magnetism and Matter
This equation represents that, the oscillation of this magnetic needle is simple harmonic. When we compare the above equation with standard harmonic.
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 9
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
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 10
If dipole is rotated from stable equilibrium (θ1 = π/2) to θ2 = 0 we get,
W = -mBcosθ
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 11
This work done is stored as magnetic potential energy, ie.
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 12

4. The electrostatic analog:
Permanent Magnets And Electromagnets
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 13

Plus Two Physics Notes Chapter 5 Magnetism and Matter
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.
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 14
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 15

Magnetism And Gauss’s Law
Gauss’s law in magnetism: The net magnetic flux through any closed surface is zero.
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 16
Explanation:
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 17
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.

Plus Two Physics Notes Chapter 5 Magnetism and Matter

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.
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 18
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 (Nm). Likewise, the pole near the geographic south pole is called the south magnetic pole. (Sm).
Note: The north magnetic pole (Nm) behaves like the south pole of a bar magnet (inside the earth). Similarly, the south magnetic pole (Sm) 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):
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 19
Declination at a place is the angle between the geographic meridian and magnetic meridian at that place.

2. Dip or Inclination (θ):
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 20

Plus Two Physics Notes Chapter 5 Magnetism and Matter
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 Bh:
The horizontal intensity at a place is the horizontal components of the earths field.
Relation between Dip, Horizontal intensity and Earth’s magnetic field:
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 21
Let B be the Earth’s magnetic field and θ be the angle of dip. Let Bh be the horizontal intensity and Bvthe vertical intensity of the earth’s magnetic field. Then from figure ,we get
Bh = B cos θ
The vertical component, Bv = 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.
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 22

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 = µ0µr
µ0 – Permeability of free space
µr – Relative permeability of a medium.

Plus Two Physics Notes Chapter 5 Magnetism and Matter

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 Bm.
∴ B = B0 + Bm
But Bm = µ0M, B0 = µ0H
∴ B = µ0H + µ0M
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 23

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:
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 24
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.

Plus Two Physics Notes Chapter 5 Magnetism and Matter

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:
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 25
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
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 26
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:
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 27
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:
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 28

Plus Two Physics Notes Chapter 5 Magnetism and Matter
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 Tc.

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.
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 29
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.

Plus Two Physics Notes Chapter 5 Magnetism and Matter

Permanent Magnets And Electromagnets
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 30
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.
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 31
Plus Two Physics Notes Chapter 5 Magnetism and Matter - 32

Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism

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.

Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism

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
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 1
= Felectric + Fmagnetic
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). ‘vd’ 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,
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 2
(I = neAVd)

Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism

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.
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 3
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.
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 4
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
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 5
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).
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 6

Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism
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’
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 7
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
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 8
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.

Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism

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.
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 9
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
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 10
[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.
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 11
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
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 12
K.E of positive ion
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 13
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 14

Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism
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
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 15
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

Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 16
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:
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 17

Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism

Magnetic Field On The Axis Of A Circular Current Loop
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 18
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,
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 19
\(\mathrm{dB}=\frac{\mu_{0} \mathrm{Idl}}{4 \pi \mathrm{x}^{2}}\) _____(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
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 20
but from ∆AOP we get, sinΦ = a/x
∴ We get
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 21
Point at the centre of the loop: When the point is at the centre of the loop, (r = 0)
Then,
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 22

Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism

1. Magnetic field at the centre of loop:
The magnetic field at a distance x from centre of loop is given by
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 23
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.
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 24
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.
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 25

Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism

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).
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 26
According to Ampere’s circuital law we can write
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 27
[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:
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 28
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
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 29
Substituting the above values in eq (1),we get
Bl = µ0 lenc ____(2).
But lenc = 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 µr. then
B = µ0µrnl

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.
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 30

Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism
Applying Amperes law to the loop, we can write
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 31
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 µr then the above equation is modified as
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 32
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
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 33
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
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 34
Conductor ‘Q’ is placed in this magnetic field.
If l2 is the length of the conductor ‘Q’, the Lorentz force on ‘Q’ is
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 35
∴ Force per unit length can be written as,
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 36
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.

Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism

Force On A Current Loop, Magnetic Dipole
1. Torque on a rectangular current loop in uniform magnetic filed:
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 37
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]
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 38
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,
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 39
The magnetic field at large distance (x>>R) on axis of loop is
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 40
Dividing and multiplying by π
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 41
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.
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 42
Comparing eq(1) and (2), we get 1
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 43

Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism
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.
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 44
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
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 45
A = πr2, area of orbit
Dividing and multiplying by me (Mass of electron)
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 46
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
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 47
The orbital magnetic moment of electron is given by
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 48
Bohr Magneton: We get the minimum value of magnetic moment, when n = 1 ie
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 49
(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
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 50

Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism
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:
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 51
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.
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 52
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
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 53

Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism
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.
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 54
From figure we can write
V = IgR + IgG
V – IgG = IgR
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 55
Using this resistance we can covert galvanometer in to voltmeter.

Current sensitivity:
The current sensitivity of galvanometer is the deflection produced by unit current.
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 56
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.
Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism - 57

Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism
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.

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

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) d1-10ns1-2
The transition metals are classified as,

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

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

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 ΔhydH° 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).

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

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.
Plus Two Chemistry Notes Chapter 8 The d and f Block Elements img 1

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:

      • 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).

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

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

Plus Two Chemistry Notes Chapter 8 The d and f Block Elements
The chromate and dichromate are interconvertible in aqueous solution depending upon PH of the solution.
2CrO42- + 2H+ → Cr2O22- + H2O
CrO72- + 2OH → 2CrO42- + H2O
Plus Two Chemistry Notes Chapter 8 The d and f Block Elements img 2

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:
Cr2O22- + 14H+ + 6e → 2Cr3+ + 7H2O
e.g. It oxidises l to l2, S2- 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
3MnO42- + 4H+ → 2MnO4 + MnO2 + H2O

Commercial preparation:
By the electrolytic oxidation of MnO42- ion (Manganate ion).
Plus Two Chemistry Notes Chapter 8 The d and f Block Elements img 3

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

Laboratory preparation:
By oxidising Mn2+ salt using peroxodisulphate.
2Mn2+ + 5S2O82- + 8H2O → 2MnO4 + 10SO42- + 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.
Plus Two Chemistry Notes Chapter 8 The d and f Block Elements img 4
Acidified permanganate solution oxidises oxalates to CO2, Fe2+ to Fe3+, NO2 to N03, I to I2, S2- to S, SO32- to SO42-.

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) f1-14(n-1)d0-1ns2

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 2nd 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.

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

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.