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KENDRIYAVIDYALAYASANGATHAN
RAIPURREGION
REGIONAL OFFICE, RAIPUR
STUDY cum SUPPORT MATERIAL
PHYSICS
CLASS XII
SESSION 2016-17
SYLLABUS-2016-17
(THEORY)
One Paper Time: 3 hrs.
Max Marks: 70
UNIT
NAME OF CHAPTER
UnitI
Electrostatics
UnitII
Current Electricity
UnitIII
Magnetic Effect of Current and Magnetism
UnitIV
Electromagnetic Induction and Alternating Current
MARKS
15
16
UnitV
Electromagnetic Waves
UnitVI
Optics
UnitVII
Dual Nature of Matter
UnitVIII
Atoms and Nuclei
Unit IX
Electronic Devices
UnitX
Communication Systems
17
10
12
TOTAL
70
Unit I: Electrostatics
Electric Charges; Conservation of charge, Coulomb's law-force between two point charges,
forcesbetween multiple charges; superposition principle and continuous charge distribution.
Electric field, electric field due to a point charge, electric field lines, electric dipole, electric field dueto a
dipole, torque on a dipole in uniform electric field.
Electric flux, statement of Gauss's theorem and its applications to find field due to infinitely longstraight
wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell(field inside and
outside).
Electric potential, potential difference, electric potential due to a point charge, a dipole and systemof
charges; equipotential surfaces, electrical potential energy of a system of two point charges and
ofelectric dipole in an electrostatic field.
Conductors and insulators, free charges and bound charges inside a conductor. Dielectrics andelectric
polarization, capacitors and capacitance, combination of capacitors in series and in parallel,capacitance
of a parallel plate capacitor with and without dielectric medium between the plates,energy stored in a
capacitor.
Unit II: Current Electricity
Electric current, flow of electric charges in a metallic conductor, drift velocity, mobility and theirrelation
with electric current; Ohm's law, electrical resistance, V-I characteristics (linear and nonlinear),electrical
energy and power, electrical resistivity and conductivity. Carbon resistors, color-code for carbon
resistors; series and parallel combinations of resistors; temperature dependence of resistance.
Internal resistance of a cell, potential difference and emf of a cell, combination of cells in series andin
parallel. Kirchhoff's laws and simple applications. Wheatstone bridge, meter bridge.
Potentiometer - principle and its applications to measure potential difference and for comparingEMF of
two cells; measurement of internal resistance of a cell.
Unit III: Magnetic Effects of Current and Magnetism
Concept of magnetic field, Oersted's experiment.
Biot - Savart law and its application to current carrying circular loop.
Ampere's law and its applications to infinitely long straight wire. Straight and toroidal solenoids,
force on a moving charge in uniform magnetic and electric fields. Cyclotron.
Force on a current-carrying conductor in a uniform magnetic field. Force between two parallelcurrentcarrying conductors-definition of ampere. Torque experienced by a current loop in uniformmagnetic
field; moving coil galvanometer-its current sensitivity and conversion to ammeter andvoltmeter.
Current loop as a magnetic dipole and its magnetic dipole moment.Magnetic dipole moment of
arevolving electron. Magnetic field intensity due to a magnetic dipole (bar magnet) along its axis
andperpendicular to its axis. Torque on a magnetic dipole (bar magnet) in a uniform magnetic field;
barmagnet as an equivalent solenoid, magnetic field lines; Earth's magnetic field and magnetic elements.
Para-, dia- and ferro - magnetic substances, with examples. Electromagnets and factors affecting
theirstrengths. Permanent magnets.
Unit IV: Electromagnetic Induction and Alternating Currents
Electromagnetic induction; Faraday's laws, induced EMF and current; Lenz's Law, Eddy currents.Self
and mutual induction.
Alternating currents, peak and RMS value of alternating current/voltage; reactance and impedance;LC
oscillations (qualitative treatment only), LCR series circuit, resonance; power in AC circuits,wattless
current.
AC generator and transformer.
Unit V: Electromagnetic waves
Need for displacement current, Electromagnetic waves and their characteristics (qualitative ideasonly).
Transverse nature of electromagnetic waves.
Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, X-rays, gammarays)
including elementary facts about their uses.
Unit VI: Optics
Reflection of light, spherical mirrors, mirror formula. Refraction of light, total internal reflectionand its
applications, optical fibres, refraction at spherical surfaces, lenses, thin lens formula,lensmaker's
formula. Magnification, power of a lens, combination of thin lenses in contact,combination of a lens and
a mirror.Refraction and dispersion of light through a prism.
Scattering of light - blue color of sky and reddish appearance of the sun at sunrise and sunset.
Optical instruments : Microscopes and astronomical telescopes (reflecting and refracting) andtheir
magnifying powers.
Wave optics: Wave front and Huygens’s principle, reflection and refraction of plane wave at a
planesurface using wave fronts. Proof of laws of reflection and refraction using Huygens’s
principle.Interference, Young's double slit experiment and expression for fringe width, coherent sources
andsustained interference of light. Diffraction due to a single slit, width of central maximum.Resolving
power of microscopes and astronomical telescope. Polarization, plane polarized light, Brewster'slaw,
uses of plane polarized light and Polaroids.
Unit VII: Dual Nature of Matter and Radiation
Dual nature of radiation. Photoelectric effect, Hertz and Lenard's observations; Einstein'sphotoelectric
equation-particle nature of light.
Matter waves-wave nature of particles, de Broglie relation. Davisson-Germer experiment(experimental
details should be omitted; only conclusion should be explained).
Unit VIII: Atoms and Nuclei
Alpha-particle scattering experiment; Rutherford's model of atom; Bohr model, energy levels,hydrogen
spectrum.
Composition and size of nucleus, Radioactivity, alpha, beta and gamma particles/rays and
theirproperties; radioactive decay law.
Mass-energy relation, mass defect; binding energy per nucleon and its variation with mass
number;nuclear fission, nuclear fusion.
Unit IX: Electronic Device
Energy bands in solids (Qualitative ideas only) conductor, insulator and semiconductor;
semiconductor diode - I-V characteristics in forward and reverse bias, diode as a rectifier; IVcharacteristics of LED, photodiode, solar cell, and Zener diode; Zener diode as a voltage regulator.
Junction transistor, transistor action, characteristics of a transistor, transistor as an amplifier
(commonemitter configuration). Logic gates (OR, AND, NOT, NAND and NOR).
Unit X: Communication Systems
Elements of a communication system (block diagram only); bandwidth of signals (speech, TV anddigital
data); bandwidth of transmission medium. Propagation of electromagnetic waves in theatmosphere, sky
and space wave propagation. Need for modulation. Production and detection of anamplitude-modulated
wave.
Marking Pattern in CBSE Board
TYPE OF
QUESTION
Very Short Answer
Question
Short Answer
Question-1
Short Answer
Question –II
Value Based
Questions
Long Answer
Question
GRAND TOTAL
MARKS ON
EACH
QUESTION
1
NO OF
QUESTIONs
TOTAL
MARKS
5
5
2
5
10
3
12
36
4
1
4
5
3
15
26
70
UNIT I
ELECTROSTATICS
WEIGHTAGE:8m
1. Charge: It is something possessed by material objects that makes it possible for them to exert
electrical force and to respond to electrical force.
2. Properties of charges:
(a) Quantisation of charge: It is property by virtue of which all free charges are integral
multiple of a basic unit of charge of an electron.
𝒒 = ±𝒏𝒆 where e=1.6x10-19
(b) Additive nature of charge: It is property by virtue of which total charge of a system is
obtained by simply adding algebraically all charges present any where on the system.
𝑞 = 𝑞1 + 𝑞2 + 𝑞3−−−−−− + 𝑞𝑛
(c) Conservation of charge: It is property by virtue of which total charge of an isolated system
always remains constant.
3. Coulombs law: The force of interaction between two point charges is directly proportional to the
product of charges and inversely proportional to square of distance between them.
𝑭 ∝ 𝒒𝟏 𝒒𝟐 and𝑭 ∝
𝟏
𝒓𝟐
𝒒𝟏 𝒒𝟐
𝒓𝟐
Where k is a constant which depends on system of measurement and nature of medium.
𝑭=𝒌
𝒌=
𝟏
𝟒𝝅∈𝟎
= 9 × 109 Nm2/kg2
4. Unit of charge: SI unit of charge is one coulomb which is that charge which when placed at a
distance of 1m from an equal charge and similar charge in vacuum would repel it by a force of
9x109newtons.
CGS unit is 1statcoulomb or 1 electrostatic unit
1coulomb = 3x109 stat coulomb
5. Electric field: Due to a given charge is the place space around a given charge in which force of
attraction or repulsion due to the charge can be experienced by any other charge.
6. Electric field intensity: At any point is the strength of field at that point. It is defined as the
force experienced by unit positive charge placed at that point.
⃗E =
⃗F
q0
𝐹
𝑞0 →0 𝑞0
𝐸⃗ = lim
7. Electric field intensity due to a point charge:
𝒒
⃗⃗⃗𝑬 = 𝒌
𝒓𝟐
8. Unit of electric field intensity: The SI unit of electric field is newton per coulomb.
9. Electric field intensity due to multiple charges:
Electric field intensity at a point due to a group of charges is equal to the vector sum of the
electric field intensity due to individual charges at the same point.
⃗⃗⃗
𝐸 = 𝐸⃗1 + 𝐸⃗2 … … … + 𝐸⃗𝑁
𝑛
𝐸⃗ = 𝑘 ∑
𝑖=1
𝑞𝑖
𝑟̂
𝑟𝑖2 𝑖
10. Electric field lines:It is the path straight or curved in electric field, such that tangent at any point
of it gives direction of electric field at that point.
Properties of electric field lines:
1. Electric field lines are discontinuous curves. They start from positive charge and end at
negative charge.
2. Tangent to electric field line at any point gives direction of electric field at that point.
3. No two lines of force can intersect each other because at the point of intersection , there will
be two possible direction of electric field which is not possible. Hence the lines do not cross
each other.
4. The electric field lines are always normal to the surface of conductor.
5. The electric field lines contract longitudinally, on account of attraction between unlike
charges.
6. The electric field lines exert a lateral pressure on account of repulsion between like charges.
11. Electric dipole: It is a system of equal and opposite charges separated by a small distance.
12. Dipole moment: It is given by product of magnitude of either charge and distance between the
two charges.
𝑝 = 𝑞(2𝑎)
The 𝐝𝐢𝐫𝐞𝐜𝐭𝐢𝐨𝐧 of dipole moment is from is from positive to negative charge
13. Field intensity on axial line of dipole;
The axial line of a dipole is the line passing through the positive and negative charges of the electric
dipole.
Electric field at P (EB) due to +q
Electric field at P due to -q (EA)
Net field at P is given by
Simplifying, we get
As a special case :
14. Field intensity at a point on the equatorial line of dipole:
Let P be a point Consider a point P on the equatorial line.
The resultant intensity is the vector sum of the intensities along
PA and PB. EA and EB can be resolved into vertical and
horizontal components. The vertical components of EASinθ and EBSinθ cancel each other as they are
equal and oppositely directed. It is the horizontal components which add up to give the resultant
field.
E = 2EA cos 
As 2qa = p
As a special case,
15. Torque on a dipole in uniform electric field:
Force on +q charge=qE along direction of E
Force on –q charge =qE opposite to E
Fnet=qE-qE =0
The forces are equal in magnitude, opposite in direction
acting at different points, therefore they form a couple
which rotates the dipole.
Torque𝜏 = 𝐹 × 𝑝𝑒𝑟𝑝. 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝜏 = 𝐹 × 𝑑𝑠𝑖𝑛𝜃 = 𝑞𝐸 × 𝑑𝑠𝑖𝑛𝜃 = (𝑞𝑑)𝐸𝑠𝑖𝑛𝜃
[𝜏 = 𝑝𝐸𝑠𝑖𝑛𝜃 𝑂𝑟 𝜏⃗⃗ = 𝑝 × 𝐸⃗ ]
16. Electric flux: It is represented by electric field passing normally through a given surface. SI unit
of flux is newton m2/coulomb. It is a scalar quantity.
∆∅ = ⃗𝑬. ⃗⃗⃗⃗⃗⃗
∆𝑺 = 𝑬∆𝑺 𝐜𝐨𝐬 𝜽
17. Gauss’s Law: ‘Electric flux over a closed surface is 1/ε0 times the charge enclosed by it.’
∅=
𝑞
∈0
18. Electric field due to a an infinite long uniformly charged wire:
Gaussian surface is a cylinder with wire as axis, radius r and length l
The field is radial everywhere and hence the electric flux crosses only
through the curved surface of the cylinder.
If E is the electric field intensity at P, then the electric flux through
the Gaussian surface is
∅ = 𝐸 × 2𝜋𝑟𝑙
According to gauss theorem electric flux is
∅=
𝑞
∈0
=
𝜆𝑙
∈0
Hence𝐸 × 2𝜋𝑟𝑙 =
𝜆𝑙
∈0
[∴ 𝑬 =
𝝀
]
𝟐 𝝅 ∈𝟎 𝒓
19. Electric field due to a uniformly charged spherical shell:
Let R be the radius of uniformly charged shell with charge
density′𝜎′.
Case (i)r>R
At points outside the sphere the electric field is radial every where
because of spherical symmetry.
Total electric flux∅ = 𝐸 × 4𝜋𝑟 2
According to gauss theorem electric flux is
∅=
∈0
𝑞
𝐸 × 4𝜋𝑟 2 = ∈
0
hence
[𝐸 =
𝑞
𝑞
4𝜋∈0 𝑟 2
]Electric field due to charged shell is same as that due to a point charge q placed at
the centre of shell
Case (ii)r=R
When point P lies on the surface of the shell or sphere, r = R
𝑞
hence𝐸 × 4𝜋𝑅 2 = ∈
0
[𝐸 =
𝑞
𝜎
=
]
4𝜋 ∈0 𝑅 2 ∈0
Case (iii)r<R
The gaussian surface does not enclose any charge, (charge resides on the surface of the shell)
0
𝐸 × 4𝜋𝑟 2 = ∈
0
hence [E=0]
20. Electric field at a distance r when
r › R𝐸 = 𝑘
𝑞
𝑟2
r=R
𝐸=𝑘
r‹ R
E=0
𝑞
𝑅2
=
𝜎
∈0
21. Electric field due to a thin infinite plane sheet of charge: Let σ be the surface charge density
on the sheet. E.F is independent of the distance from the plane sheet.
𝐸=
𝜎
2 ∈0
22. Electric field due to two thin parallel sheet of charge: Electric field between the plates is
𝜎
∈0
and in the region on either side of the plates 𝐸 = 0
𝐸=
23. Electrostatic potential difference: P.D between two points in electric field is defined as the
amount of work done to move a test charge without acceleration from one point to another. SI
unit of PD is volt.∆𝑽 =
𝑾𝑨𝑩
𝒒
24. Electrostatic potential: Electrostatic potential at any point in electric field is the amount of work
done in moving a unit positive charge from infinity to the point.
𝑽=
𝑾∞𝑩
𝒒
=𝒌
𝒒
𝒓
Potential is a scalar quantity measured in volts.
25. Electrostatic potential at any point due to a dipole: Potential at a distance r from the centre of
dipole at an angle θ with the axis of dipole is
𝑉=k
At a point on the axis of dipole θ=0
𝑉=𝑘
p cos θ
r 2 − a2
𝑟2
𝑝
− 𝑎2
At a point on the equatorial line of dipole θ=90
𝑉 = 0 𝑎𝑠 𝑐𝑜𝑠90 = 0
26. Equipotential surface: It is the surface at every point of which the potential is same.
27. Properties of equipotential surface:
1. No work is done in moving a charge from one point of equipotential surface to the other
2. For any charge configuration, equipotential surface through a point is normal to the electric
field at that point.
3. Where electric field is large the distance between electric field is small and vice versa.
28. Potential energy of system of charges: It is defined as the amount of work done in bringing the
various charges to their respective positions from infinitely large mutual separations.
29. Expression for potential energy for a system of charges:
n
n
qi qj
1
𝑼 = k∑∑
2
rij
j=1
i=1
i≠j
30. Electrostatics of a conductors:
1. Electric field inside a conductor is zero
2. The interior of a conductor can have no excess charge in static situations.
3. Electric field just outside the conductor is normal to the surface of the conductor.
4. Electrostatic potential is constant throughout the volume of the conductor and has the same
value as on its surface.
𝜎
5. Electric field at the surface of conductor is 𝐸 =
∈0
31. Relation between electric potential and electric intensity:
dV
E=−
dr
32. Electrical capacitance: It is ability to store charge. It is numerically the charge required to raise
the potential by unity.
𝑄
𝐶=
𝑉
SI unit of capacity is Farad
1𝐶𝑜𝑢𝑙𝑜𝑚𝑏
1𝐹𝑎𝑟𝑎𝑑 =
1 𝑉𝑜𝑙𝑡
𝐶 = [𝑀−1 𝐿−2 𝑇 4 𝐴2 ]
33. Capacity of isolated spherical conductor: Let R be the radius of spherical conductor.
C = 4π ∈0 R
34. Capacity of a parallel plate capacitor: Let the surface charge density on the plates be σ
Such that 𝜎 =
𝑄
𝐴
Electric field between the plates is given by
𝐸=
𝜎
𝜎
𝜎
+
=
2 ∈0 2 ∈0 ∈0
Potential difference between the plates is V=Ed
V=
Capacity of a capacitor 𝐶 =
𝑄
𝑉
=
𝜎𝐴
𝜎𝑑/∈0
=
𝜎
𝑑
∈0
∈0 𝐴
𝑑
[𝐶 =
∈0 𝐴
]
𝑑
Capacity of a parallel plate capacitor with dielectric:
Let the surface charge density on the plates be σ
Such that 𝜎 =
𝑄
𝐴
Electric field between the plates is given by
⃗⃗⃗⃗
𝐸𝑂 =
𝜎
∈0
𝑎𝑛𝑑 ⃗⃗⃗
𝐸𝑖 =
𝜎
𝑘∈0
where E0 is electric field in air and Ei is electric field in dielectric.
Potential difference between the plates is given by
𝑉 = ⃗⃗⃗⃗⃗
𝐸𝑂 (𝑑 − 𝑡) + ⃗⃗⃗⃗
𝐸𝑖 𝑡 =
𝜎
𝜎
𝜎
𝑡
(𝑑 − 𝑡) +
𝑡 = (𝑑 − 𝑡 + )
∈0
∈0
𝑘 ∈0
𝑘
Capacity of a capacitor 𝐶 =
[𝐶 =
∈0 𝐴
1
𝑑 − 𝑡(1 − 𝑘 )
𝑄
𝑉
=
𝜎𝐴
𝜎
𝑡
(𝑑−𝑡+𝑘)
∈0
=
∈0 𝐴
𝑡
(𝑑−𝑡+𝑘)
]
∈ 𝐴
If d=t then[ 𝐶 = 𝑘 𝑑0 ]
𝑘=
𝐶𝑚
𝐶0
35. Grouping of capacitors:
Parallel combination:
𝑪 = 𝑪𝟏 + 𝑪𝟐 ± − − − − − − +𝑪𝒏
Series combination:
𝟏
𝟏
𝟏
𝟏
=
+
+ ⋯……..+
𝐂 𝐂𝟏 𝐂𝟐
𝐂𝐧
36. Energy stored in capacitor:
Consider a parallel plate capacitor of capacity C. Let at any instant the charge on the capacitor be
Q’. Then potential difference between the plates will be
Suppose the charge on the plates increases by d Q’. The work done will be
𝑑𝑊 = 𝑉 ′ 𝑑𝑄′ =
𝑄′
𝐶
𝑑𝑄′
′
𝑄𝑄
𝑄2
′
The total work done is 𝑊 = ∫0 𝐶 𝑑𝑄 = [2𝐶 ]
This work done is stored as electrical potential energy.
[∴ 𝑈 =
𝑄2 1 2 1
= 𝐶𝑉 = 𝐶𝑉]
2𝐶 2
2
Energy density of parallel plate capacitor:
Capacity of a parallel plate capacitor is
𝐶=
∈0 𝐴
𝑑
1 ∈0 𝐴𝑑2 𝐸 2
1∈ 𝐴
𝑈 = 2 𝑑0 𝑉 2 = 2
[𝒖 =
𝑼
𝑨𝒅
𝑑
1
= 2 ∈0 (𝐴𝑑)𝐸 2
𝟏
= ∈𝟎 𝑬 𝟐 ]
𝟐
Energy is stored in the dielectric medium between the plates of capacitor
1
1 𝑄2 1
𝑈 = 𝐶𝑉 2 =
= QV
2
2 𝐶
2
When a dielectric is inserted between the plates of capacitor and the battery remains connected
1
1
𝑈 = (𝑘𝐶)𝑉 2 = 𝑈 = 𝑘 𝐶𝑉 2 = 𝑘𝑈0
2
2
Total energy is additive in series and parallel combination.
𝑈 = 𝑈1 + 𝑈2 + 𝑈3
S0ME IMPORTANT QUESTIONS AND ANSWERS
1. Define dipole moment of an electric dipole. Is it a scalar or a vector?
Electric dipole moment of an electric dipole is equal to the product of either charge and
distance between the two charges. 𝒑 = 𝒒 × 𝟐𝒂 where p is dipole moment.
It is a scalar quantity.
2. Why must electrostatic field be normal to the surface at every point of a charged conduction?
The component of electric field along the tangent to the surface of the conductor must
be zero.
𝑬 𝐜𝐨𝐬 𝜽 = 𝑶 where θ is angle between and tangent to the surface.
𝑬 ≠ 𝟎, ∴ 𝐜𝐨𝐬 𝜽 = 𝟎 𝑶𝒐𝒓 𝜽 = 𝟗𝟎
hence E is perpendicular to the surface
3. Why do the electric field lines not form closed loop?
No electric field exist from negative to positive charge , hence electric field lines do not
form closed loop.
4. In which orientation a dipole placed in a uniform electric field is in a) Stable,
b)UnstableEquilibrium?
(a) For stable equilibrium the angle between p and E must be 00
(b) For unstable equilibrium the angle between p and E must be 1800
5. Two point charges having equal charge are separated by 1m distance experience a force of
8N. What will be the force if they are held in water at the same distance? (Given kwater = 80)
𝐤=
𝐅𝐚
𝐅𝐚
𝟖
𝟏
∴ 𝐅𝐦 =
=
=
𝐅𝐦
𝐤 𝟖𝟎 𝟏𝟎
6. A dipole, of dipole movement p is present in a uniform electric field E. Write the value of
angle between p and E for which the torque experienced by the dipole is minimum.
𝝉 = 𝒑𝑬 𝐬𝐢𝐧 𝜽 for the torque to be minimum 𝒑𝑬 𝐬𝐢𝐧 𝜽 = 𝟎 ∴ 𝐬𝐢𝐧 𝜽 = 𝑶 𝒐𝒓 𝜽 = 𝟎
7. A charge q is placed at the centre of a cube of side l. What is the flux passing through each
face of the cube?
𝒒
According to gauss theorem electric flux linked with a closed surface is∅ =
∈𝟎
The flux is symmetrically distributed through all the six faces ∴ ∅ =
𝟏 𝒒
𝟔 ∈𝟎
8. Figure shows three pouint charges +2q, -q and +3q. What is the flux through the closed surface
S?
Electric flux through the surface S
∅=
∑𝒒
∈𝟎
=
+𝟐𝒒−𝒒
∈𝟎
=
𝒒
∈𝟎
9. If the radius of Gaussian surface is halved, how will the flux through the Gaussian surface
change?
Even if the radius of the surface is halved, the charge enclosed by the surface does not
change hence the flux remains constant.
10. A hollow metal sphere of radius 5 cms is charged such that potential on its surface is 5 V.
What is the potential at the centre of the sphere?
In side a hollow sphere potential is constant and same as that on its surface.
Hence 𝑽𝒊 = 𝑽𝑺 = 𝟓 𝑽
11. Name a physical quantity whose SI unit is J/C. Is it a scalar or a vector quantity?
J/C is unit of electric potential.It is a scalar quantity.
12. What is the work done to move a test charge q through a distance of 1 cm along the
equatorial axis of dipole?
Potential at any point on the equatorial line is 0. Hence work done W = q∆V =0 as ∆V=0
13. The following graph shows variation of charge Q with voltage V for two capacitors K and L.
In which capacitor is more energy stored ?
Q
L
K
V
The slope of straight line represents capacitances. Therefore capacity of L will be more.
𝟏
Energy stored in a capacitor𝑼 = 𝟐 𝑪𝑽𝟐
14. Draw a plot showing variation of
a) Electric field E and
b) Electric potential V with distance r due to a point
charge Q.
15. Calculate amount of work done in turning an electric dipole
ofdipole moment 3x10-8 C-m from position of unstable
equilibrium to the position of stable equilibrium in a
uniform electric field of intensity 103 N/C
For unstable equilibrium θ=1800and for stable equilibrium
θ=00Required work done
𝑾 = 𝒑 𝑬 (𝐜𝐨𝐬 𝜽𝟏 − 𝐜𝐨𝐬 𝜽𝟐 )
𝑾 = 𝟑 × 𝟏𝟎
−𝟖
× 𝟏𝟎
𝟑 (𝐜𝐨𝐬
+F
I
1/r2
−𝟓
𝟏𝟖𝟎 − 𝐜𝐨𝐬 𝟎) = −𝟔 × 𝟏𝟎 𝑱
2
-F
16.
Plot a graph showing the variation of Coulomb’s force (F) versus 1/r2 where r is the distance
between the two charges of each pair of charge(1µC,2µC) and (1µC,-3µC).
For given pair of charge 𝑭 ∝
17.
𝟏
𝒓𝟐
Magnitude of q1q2 is higher and negative in second case
Two point charges 4µC and -2µC are separated by a distance of
1m. At what point on the line joining the two charges is the electric
potential zero.
Let the potential be zero at a point P at a distance x from the
charge 4µC.
At P V1+V2=0
𝒒𝟏
𝒒𝟐
𝒌 +𝒌
=𝟎
𝒓𝟏
𝒓𝟐
𝒌
A
4µC
B
P
𝟒 × 𝟏𝟎−𝟔
𝟏 × 𝟏𝟎−𝟔
−𝒌
=𝟎
𝒙
𝟏−𝒙
𝟒
𝟏
=
𝒐𝒓 𝟐(𝟏 − 𝒙) = 𝒙
𝒙 𝟏−𝒙
𝟑
𝟐 = 𝟑𝒙 𝒐𝒓 𝒙 = 𝒎
𝟐
𝟑
Potential is zero at a distance of 𝟐 𝒎 from 4µC charge
SOME QUESTIONS FOR PRACTISE
LEVEL-I
1. What is the charge acquired by a body when 1 million electrons are transferred to it?
2. An attractive force of 5N is acting between two charges of +2.0 μC & -2.0 μC placed at some
distance. If the charges are mutually touched and placed again at the same distance, what will be
the new force between them?
3. A charge of +3.0 x 10-6 C is 0.25 m away from a charge of -6.0 x 10-6C.
a. What is the force on the 3.0 x 10-6 C charge?
b. What is the force on the -6.0 x 10-6 C charge?
4. An electric dipole consist of a positive and a negative charge of 4µC each placed at a distance of
5mm. Calculate dipole moment.
5. Three capacitors of capacitances 2µF, 3µF and 4µF are connected in parallel. What is the
equivalent capacitance of the combination? Determine charge on each capacitor, if the
combination is connected to 100V supply?
6. An electric dipole with dipole moment 4x10-9C-m is aligned at 300 with direction of electric field
of magnitude 5x104N/C. Calculate the magnitude of the torque acting on the dipole.
7. A point charge of 2µC is at the centre of cubic Gaussian surface 9.0 cm in edge. What is the net
electric flux through the surface?
8. What is the amount of work done in moving a 200nC charge between two points 5 cm apart on
an equipotential surface?
9. How much work must be done to charge a 24 μF capacitor, when the potential difference
between the plates is 500 V?
10.What is the equivalent capacity of the network given below?
1µC
LEVEL II
1. What is the work done in moving a charge of 100μC through a distance of 1cm along the
equatorial line of dipole?
2. The given graph shows that variation of charge q versus potential difference V for two capacitors
C1 and C2. The two capacitors have same plate separation but the plate area of C2 is double than
that of C1. Which of the lines in the graph correspond to C1 and C2 and why?
– 4 µC are separated by a distance of 1
m in air. At what point on the line joining the charges is the electric potential zero?
4. Two charges +5µC and +20µC are placed 15 cm apart. At what point on the line joining the two
charges is the electric field zero?
5. Two charges +16µC and −9µC are placed 8 cm apart. At what point on the line joining the two
charges is the electric field zero?
6. A 600 pF capacitor is charged by a 200 V supply. It is then disconnected and from the supply and
is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the
process.
7. Keeping the voltage of the charging source constant, what will be the percentage change in the
energy stored in a parallel plate capacitor if the separation between its plates were to be
decreased by 10%.
8. Four charges are placed at the vertices of a square of side d as shown in the
figure.(i) Find the work done to put together this arrangement. (ii) A charge
q0is brought to the center E of the square, the four charges being held fixed
at its corners. How much extra work is needed to do this?
3. Two point charges 5µC and
9. If S1 and S2 are two hollow spheres enclosing charges Q and 2Q respectively
as shown in the figure
(i) What is the ratio of the electric flux through S1 and S2?
(ii) How will the flux through the sphere S1 change, if a medium of dielectric
constant 5 is filled in the space inside S1.
10. A charge of 24μC is given to a hollow sphere of radius 0.2m. Find the
potential
(i) at the surface of the sphere, and
(ii) at a distance of 0.1 cm from the centre of the sphere.
LEVEL III
1. A slab of material of dielectric constant has the same area as the plates of a parallel plate
capacitor but has a thickness 3d / 4, where d is the separation of the plates. How is the
capacitance changed when the slab is inserted between the plates?
2. A parallel plate capacitor with air between the plates has a capacitance of 8µF. What will be the
capacitance if the distance between the plates is doubled and the space between them is filled
with a substance of dielectric constant K=6?
3. Two dipoles, made from charges ±q and ±Q, respectively, have equal dipolemoments. Give the
(i) ratio between the ‘separations’ of these two pairs ofcharges (ii) angle between the dipole axis
of these two dipoles.
4. The capacitors C1, and C2, having plates of area A each, are connected in series, as shown.
Compare the capacitance of this combination with the capacitor C3, again having plates of area
A each, but ‘made up’ as shown in the figure.
5. A point charge +10μC is at a distance 5cm directly above the centre of a
square of side 10cm as shown in fig. What is the magnitude of flux
throughthe square?
6. Two identical charges ,Q each are kept at a distance r from each other. A third
charge q is placed on the line joining the two charges such that all the three
charges are in equilibrium. What is magnitude, sign and position of the charge q?
7. ABCD is a square of side 5m. Charges of +50C, -50C and +50C are placed at A,C and D respectively .
Find the magnitude of resultant electric field at B.
8. A cube with each side a is kept in electric field given by E = Cx as shown in the
figure where C is a positive dimensional constant. Find (i) The electric flux
through the cube, and
(ii) The net charge inside the cube.
10.
Two parallel plate capacitor X and Y have same area of plates and same
separation between them. X has air between the plates whereas Y has a
dielectric of constant k=4
(i) Calculate capacitance of each capacitor if equivalent capacitance is4 μF.
(ii) Calculate potential difference between the plates of X and Y.
(iii) What is the ratio of electrostatic energy stored in X and Y.
UNIT: I ELECTROSTATICS
ANSWERS
LEVEL I
-13
1. Q = Ne 1.6 x10 C
2. F=0
3. FAB = FBA=2.736N
4. P=2x10-8 C-m
5.
6. 10-4Nm
7. 2,26x105Nm2/C
8. W=0
9. W=3J
10. C=15µF
LEVEL II
1.
2.
3.
4.
5.
6.
7.
0
A
5
𝑚 from 5µC charge
9
5 cm from 5 µC charge
24cm from -9µCcharge
6x10-6 J
11.11%
8.
𝑞2
4𝜋∈0
(4 − √2) , 0
𝑄
9. 1: 3, ∅ = 5∈
0
10. (i) 1.08x106V (ii) 1.08x106V
LEVEL III
4𝑘
1.
𝐶
𝑘+3 0
2.
3.
4.
5.
24µF
q a=Q A or a/A=Q/qθ = 0
C3= Ceq
1.88x105Nm2/C
6.
200
3
𝑝𝐹,100 V, 50V, 50V, 200V,10-8C,10-8C,10-8C,2x10-8 C
7. Q/4, Positive, r/2
8. 2.7x1010N/C
9. a3C N-m2/C, a3C𝜖0 Coulombs.
10. Cx=5μF Cy= 20μF
CURRENT ELECTRICITY
The flow of charge through a conductor is called electric current.
𝑑𝑄
I= 𝑑𝑡
It is scalar quantity and its SI unit is Ampere
OHMS LAW
It states that current flowing through a conductor is directly proportional to the potential difference
across the ends of the conductor provided physical conditions like temperature and pressure remains
constant.
V∝I
V=IR
RESISTANCE
It is the obstruction to the flow of current.
Resistance of a conductor is directly proportional to length and inversely proportional to its area of
cross-section.
𝑙
𝑙
𝑅∝ =𝜌
𝑎
𝑎
Where 𝜌 is the resistivity of the material of the conductor
It is defined resistance per unit length per unit area of cross section.
SI unit is ohm m.
It depends on nature of material and temperature’
DRIFT VELOCITY
It is the average velocity with which electrons move through the conductor in presence of external
electric field.
In absence of electric field the elwctrons are in random motion nd the average thermal velocity is =0
𝑒𝐸
F= 𝑚𝑎 = 𝑒𝐸 𝑜𝑟 𝑎 = 𝑚
From I equation of motion
𝑣 = 𝑢 + 𝑎𝑡
𝑣𝑎𝑣𝑔 = 𝑢𝑎𝑣𝑔 + 𝑎𝑡𝑎𝑣𝑔
𝑣 = 𝑎𝜏
𝑒𝐸
𝑒𝑉
𝑣𝑑 =
𝜏=
𝜏
𝑚
𝑚𝑙
𝜏 – Relaxation time – it is the average time between 2 successive collisions.
CURRENT IN TERMS OF DRIFT VELOCITY
n – no.of elecrtrrons per unit volume
𝑎𝑙- volume of the conductor
𝑛𝑎𝑙 – total no. of electrons
𝑒𝑛𝑎𝑙 – total charge
𝑞 𝑛𝑒𝑎𝑙
𝐼= =
= 𝑛𝑒𝑎𝑣𝑑
𝑡
𝑡
CURRENT DENSITY & RESISTIVITY
𝐼
𝑉
𝑉 𝐸
=
= =
𝑎 𝑅𝑎 𝜌𝑙 𝜌
𝐼
𝑛𝑒 2 𝐸
𝑗 = = 𝑛𝑒𝑣𝑑 =
𝜏
𝑎
𝑚
𝑗=
𝜌=
Comparing above equations
𝑚
𝑛𝑒 2 𝜏
VARIATION OF RESISTIVITY WITH TEMPERATURE
CONDUCTORS
Resistivity increases with increase in temperature
ALLOYS
Variation of resistivity is very less hence the alloys are used in making standard resistance.
SEMICONDUCTORS
Resistivity decreases with increase in temperature.
COMBINATION OF RESISTANCE
SERIES COMBINATION
𝑽 = 𝑽𝟏 + 𝑽𝟐 + 𝑽𝟑
Using Ohm’s law𝑰𝑹 = 𝑰𝑹𝟏 + 𝑰𝑹𝟐 + 𝑰𝑹𝟑
𝑹 = 𝑹𝟏 + 𝑹 𝟐 + 𝑹𝟑
PARALLEL COMBINATION
𝑰 = 𝑰𝟏 + 𝑰𝟐 + 𝑰𝟑
𝑉
𝑉
𝑅
𝑅1
Using Ohm’s law =
+
𝑉
𝑅2
+
𝑉
𝑅3
COMBINATION OF CELLS
SERIES COMBINATION
𝑽 = 𝑽𝟏 + 𝑽𝟐 + 𝑽𝟑
𝑬 − 𝑰𝒓 = (𝑬𝟏 − 𝑰𝒓𝟏 ) + (𝑬𝟐 − 𝑰𝒓𝟐 )
𝑬 = 𝑬𝟏 + 𝑬𝟐 𝒂𝒏𝒅 𝒓 = 𝒓𝟏 + 𝒓𝟐
or
1
𝑅
=
1
𝑅1
+
1
𝑅2
+
1
𝑅3
PARALLEL COMBINATION
𝑰 = 𝑰𝟏 + 𝑰 𝟐 + 𝑰𝟑
𝑬 − 𝑽 𝑬𝟏 − 𝑽 𝑬𝟐 − 𝑽
=
+
𝒓
𝒓𝟏
𝒓𝟐
On solving
𝑬=
𝑬 𝟏 𝒓𝟐 + 𝑬 𝟏 𝒓𝟐
𝒓𝟏 𝒓𝟐
𝒂𝒏𝒅 𝒓 =
𝒓𝟏 + 𝒓𝟐
𝒓𝟏 + 𝒓𝟐
COLOUR CODE FOR CARBON RESISTORS:
The first two rings from the end give the first two significant
figures of resistance in ohm.
The third ring indicates the decimal multiplier.
The last ring indicates the tolerance in per cent about the
indicated value
AB x 10C ±D %ohm
Eg.
Letter
Colour
Number
Colour
Tolerance
B
Black
0
Gold
5%
B
Brown
1
Silver
10%
R
Red
2
No colour
20%
O
Orange
3
Y
Yellow
4
G
Green
5
B
Blue
6
V
Violet
7
G
Grey
8
W
9
KIRCHHOFF’S LAWS:
I Law or Current Law or Junction Rule: (∑I=0 )
The algebraic sum of electric currents meeting at a junction in any electrical network is always zero.
I1 - IW
2 - I3 + I4 - I5 = 0
hite
II Law or Voltage Law or Loop Rule:(∑∆V=0)
The algebraic sum of all the potential drops and emf’s along any closed
path in an electrical network is always zero.
Loop ABCA:
- E1 + I1.R1 + (I1 + I2).R2 = 0
Loop ACDA:
- (I1 + I2).R2 - I2.R3 + E2 = 0
Sign Conventions:
 The emf is taken negative when we traverse from positive to negative terminal of the cell through
the electrolyte.
 The emf is taken positive when we traverse from negative to positive terminal of the cell through
the electrolyte.
 The potential falls along the direction of current in a current path and it rises along the direction
opposite to the current path.
 The potential fall is taken negative.
 The potential rise is taken positive.
 Note: The path can be traversed in clockwise or anticlockwise direction of the loop.
WHEAT STONE BRIDGE:
If no current flows through galvanometer ( VB=VD)
𝑷
𝑸
=
𝑹
𝑺
In the loop ABDA
I1P + IgG – (I – I1)R = 0________________(1)
In the loop BCDA
(I1-Ig)Q – (I-I1+Ig)S - IgG = 0____________(2)
Solving equation (1) & (2)
𝑷
𝑸
=
𝑹
𝑺
METRE BRIDGE:
It works on the principle of Wheatstone bridge.
𝑷 𝑹
=
𝑸 𝑺
𝒍𝟏
𝑹
𝟏𝟎𝟎 − 𝒍𝟏
=
𝒐𝒓 𝑺 = 𝑹
𝟏𝟎𝟎 − 𝒍𝟏
𝑺
𝒍𝟏
POTENTIOMETER:
It is a device used to measure emf of a cell.
PRINCIPLE: The potential drop or fall of potential across any portion of the wire is directly proportional
to the of the wire provided the wire is of uniform area and current flowing through it is constant.
𝐿
V=IR = 𝐼𝜌 𝐴 = kL
Where, k – potential gradient
𝑽 𝝆𝑰
𝑘= =
𝑳
𝑨
NOTE:-For better Sensitivity of Potentiometer k should be less.
COMPARISON OF EMF’S USING POTENTIOMETER:
𝑬𝟏 𝑳𝟏
=
𝑬𝟐 𝑳𝟐
INTERNAL RESISTANCE OF A CELL
𝑬
𝒍𝟏
𝒓 = 𝑹 ( − 𝑰) = 𝑹( − 𝟏)
𝑽
𝒍𝟐
CURRENT ELECTRICITY PREVIOUS YEAR FOR PRACTICE
1.Two wires of equal length, one of copper and the other of manganin have the same resistance. Which
wire is thicker?ANS- Manganin
2. Define drift velocity and relaxation time. Derive the expression for resistivity in terms of number
density of electrons and relaxation time.
3.A cell of emf ‘E’ and internal resistance ‘r’ is connected across a variable resistor ‘R’. Plot a graph
showing variation of terminal voltage ‘V’ of the cell versus the current ‘I’. Using the plot, show how the
emf of the cell and its internal resistance can be determined.
4.Answer the following :
(a) Why are the connections between the resistors in a meter bridge made of thick copper strips ?
(b) Why is it generally preferred to obtain the balance point in the middle of the meter bridge wire ?
(c) Which material is used for the meter bridge wire and why ?
5. Two identical cells, each of emf E, having negligible internal resistance, are connected in parallel
witheach other across an external resistance R. What is the current through this resistance?
6.Write principle of potentiometer. Draw the circuit diagram to compare emf of two primary cells. Write
the formula used.
7. Calculate the value of the resistance R in the circuit shown in the figure so that the current in the
circuit is
0.2 A. What would be the potential difference
between points B and E?
ANS- VBE=1V
8. In the meter bridge experiment, balance point
was observed at J with AJ = l.
(i) The values of R and X were doubled and then
interchanged. What would be the new position of
balance point?
(ii) If the galvanometer and battery are interchanged at the balance position, how will the balance point
get affected?
10. Two conducting wires X and Y of same diameter but different materials are joined in series across a
battery. If the number density of electrons in X is twice that in Y, find the ratio of drift velocity of
electrons in the two wires.
11. State Kirchhoff’s rules. Use these rules to write the expressions for the currents I1, I 2 and I 3 in the
circuit diagram shown.
ANS- I1=2/13,I2=7/13,I3=9/13
12. Calculate the current drawn from the battery in the
given network.
ANS- I=2A
13. The following graph shows the variation of terminal potential difference V, across a combination of
three cells in series to a resistor, versus the current, i:
(i)
Calculate the emf of each cell.
(ii)
For what current i will the power dissipation of the circuit be maximum?
ANS- emf of each cell is 2V and I=1A
14. Two wires X, Y have the same resistivity, but their cross-sectional areas are in the ratio 2 : 3 and
lengths in the ratio 1 : 2. They are first connected in series and then in parallel to a d.c. source.
Find out the ratio of the drift speeds of the electrons in the two wires for the two cases.
ANS (a) Vdx/vdy=3/2
(b)vdx/vdy=2/1
MAGNETIC EFFECTS OF CURRENT & MAGNETISM
1 MARK QUESTIONS
1.
A vector needs three quantities for its specification.name three independent quantitiesneeded to
completely specify the earth’s magnetic field at appoint on the earth’s surface?
Ans. These are(i) angle of declination
(ii) angle of dip or inclination
(iii)horizontal component of the earth’s magnetic field.
2.
Why should a voltmeter have high resistance?
Ans. A voltmeter is always connected in parallel. When connected in parallel, it should drawleast
current, otherwise the potential difference which it has to measure will change.
3.
What is the direction of the force acting on a charge particle q, moving with a velocity , will
cange.v in a uniform magnetic field B
Ans: Force, to both velocity v and magnetic field B.
2.
Magnetic field lines can be entirely confined within the core of a toroid, but not within a
straight solenoid. Why?
Ans: Magnetic field lines can be entirely confined within the core of a toroid because toroidhas no
ends. A solenoid is open ended and the field lines inside it which is parallel to thelength of the
solenoid, cannot form closed curved inside the solenoid.
3.
An electron does not suffer any deflection while passing through a region of uniform
magnetic field. What is the direction of the magnetic field?
Ans: Magnetic field is parallel or antiparallel to velocity of electron i.e., angle between v and B is 0° or
180°.
4.
A beam of a particles projected along +x-axis, experiences a force due to a magnetic field along
the +y-axis. What is the direction of the magnetic field?
Ans:
5.
Ans:
6.
Ans:
7.
Ans:
9.
By Fleming’s left hand rule magnetic field must be along negative Z-axis
What is the characteristic property of a diamagnetic material?
These are the substances in which feeble magnetism is produced in a direction opposite to the
applied magnetic field. These substances are repelled by a strong magnet. These substances have
small negative values of susceptibility and positive low value of relative permeability.
The permeability of a magnetic material is 0.9983. Name the type of magnetic materials it
represents.
As permeability < 1, so magnetic material is diamagnetic.
Where on the surface of Earth is the angle of dip zero?
Angle of dip is zero at equator of earth’s surface.
A narrow beam of protons and deuterons, each having the same momentum, enters a region of
uniform magnetic field directed perpendicular to their direction of momentum. What would be
the ratio of the circular paths described by them?
Ans.
As r=mv/qB
i.e.,
=> r∝ 1/q
so, rp:rd =1:1
10.
Mention the two characteristic properties of the material suitable for making core of a
transformer.
Ans: Two characteristic properties: (i) Low hysteresis loss (ii) Low coercivity.
11.
An electron is moving along positive x axis in the presence of uniform magnetic field along
positive y axis. What is the direction of the force acting on it?
Ans:
12.
negative z direction.
Why should the spring or suspension wire in a moving coil galvanometer have low
torsional constant?
Ans: Sensitivity of a moving coil galvanometer is inversely proportional to the torsional
constant.
13.
Steel is preferred for making permanent magnets whereas soft iron is preferred for
making electromagnets .Give one reason.
Ans: steel-- high retentivity, high coercivity
Soft iron-- high permeability and low retentivity.
14.
Where on the surface of the earth is the vertical component of earth’s magnetic fieldzero?
Ans: At equator.
TWO MARKS QUESTIONS
1.
Define magnetic susceptibility of a material. Name two elements, one having positive
susceptibility
and the other having negative susceptibility. What does negative susceptibility signify ?
Ans: Magnetic susceptibility: It is defined as the intensity of magnetisation per unit magnetising
field,
It has no unit.
Iron has positive susceptibility while copper has negative susceptibility.
Negative susceptibility of a substance signifies that the substance will be repelled by a strong
magnet or opposite feeble magnetism induced in the substance.
2.
Define the term magnetic dipole moment of a current loop. Write the expression for the
magnetic moment when an electron revolves at a speed ‘v’, around an orbit of radius ‘ r’ in
hydrogen atom.
Ans: Magnetic moment of a current loop:
M = NIA
i.e., magnetic moment of a current loop is the product of number of turns, current flowingin the
loop and area of loop. Its direction is perpendicular to the plane of the loop.
Magnetic moment of Revolving Electron, M=evr/2
3.
Define current sensitivity and voltage sensitivity of a galvanometer. Increasing the
current sensitivity may not necessarily increase the voltage sensitivity of a galvanometer.
Justify.
Ans: Current sensitivity :It is defined as the deflection of coil per unit current flowing in it.Current
Sensitivity, S=NAB/ k
Voltage sensitivity :It is defined on the deflection of coil per unit potential
differenceacrossends.
Voltage Sensitivty, SV=NAB/GC
where G is resistance of
galvanometer.
Justification: When number of turnsNis doubled, then the current sensitivity (µN)
isdoubled; but at the same time, the resistance of galvanometer coil (G) will also be doubled,
so voltage sensitivity S will remain unchanged; hence inreasing current sensitivity does not
necessarily increase the voltage sensitivity.
4.
A wire of length L is bent round in the form of a coil having N turns of same radius. If a steady
current I flows through it in a clockwise direction, find the magnitude and direction of the
magnetic field produced at its centre.
𝐿
Ans:𝐿 = 𝑁 × 2𝜋𝑟 => 𝑟 = 2𝜋𝑁
𝜇0 𝑁𝐼 𝜇0 𝜋𝑁 2 𝐼
𝐵=
=
2𝑟
𝐿
5.
A point charge is moving with a constant velocity perpendicular to a uniform
magnetic field as shown in the figure. What should be the magnitude and
direction of the electric field so that the particle moves undeviated along the
same path?
Ans: Magnitude of electric field is vB and its direction is along positive Y-axis.
6.
(i) Write two characteristics of a material used for making permanent magnets. (ii)
Why is core of an electromagnet made of ferromagnetic materials?
Ans: (i) For permanent magnet the material must have high retentivity and high coercivity
(e.g.,steel).
(ii) Ferromagnetic material has high retentivity, so when current is passed in ferromagnetic
material it gains sufficient magnesium immediately on passing a current through it.
7.
Draw magnetic field lines when a (i) diamagnetic, (ii) paramagnetic substance is placed in an
external magnetic field. Which magnetic property distinguishes this behaviour of the field
lines due to the two substances?
Ans:
8.
Ans:
The magnetic susceptibility of diamagnetic substance is small and negative but that of
paramagnetic substance is small and positive.
Deduce the expression for the magnetic dipole moment of an electron orbiting around the central
nucleus.
Consider an electron revolving around a nucleus (N) in circular path of radius r with
Area of current loop (electron orbit), A = p r2
Magnetic moment due to orbital motion, M= IA=evr/2
9.
A circular coil of ‘N’ turns and diameter ‘d’ carries a current ‘I’. It is unwound and rewound to
make another coil of diameter ‘2d’, current ‘I’ remaining the same. Calculate the ratio of the
magnetic moments of the new coil and the original coil.
Ans: Magnetic moment (M) = NIA=NIπd2/4 = NIπR2
Length of wire remains same, so NB = NA/2,
R=2R
On solving
MA/MB= 2/1
10.
Which one of the two, an ammeter or a milliammeter, has a higher resistance and why?
Ans: As the shunt resistance is connected in parallel with the galvanometer, so the milliammeter will
have a higher resistance than the ammeter.
11.
The following figure shows the variation of intensity of magnetization I versus the
applied magnetic field intensity H for two magnetic materials A and B.
(1) Identify the materials A and B
(2) Draw the variation of susceptibility with temperature for B.
Ans: 1) A is paramagnetic material.
2) B is diamagnetic material.
THREE MARKS QUESTIONS
Q.1Distinguish between diamagnetic, paramagnetic and ferromagnetic substances.
Ans.
Property
Effects of magnets
Susceptibility(Xm)
Relative permeability
value(µr)
Permeability value(µ)
Effect of temperature
Examples
Diamagnetic
They
are
feebly
repelled by magnets.
-1=< Xm<0
0=<µr <1
µ< µ0
Independent
temperature.
Paramagnetic
They
are
feebly
attracted by magnets.
0<Xm<ε
1<µr<1+ε
Ferromagnetic
They
are
strongly
attracted by magnets.
Xm>1000
µr>1000
µ> µ0
of Susceptibility
is
inversely proportional
to temperature.
µ>> µ0
Susceptibility decreases
with temperature in a
complex manner.
Bi ,Pb,Cu
Hints: for distinguishing them, use the formula µr=1+Xm.
Q 2. Discuss the motion of a charged particle in a uniform magnetic field with initial velocity
(i) parallel to the field, (ii) perpendicular to the magnetic field and (iii) at an arbitrary angle
with the field direction.
Ans. When a charged particle having charge q and velocity v enter a magnetic field B it
experiences a force
F = qv B sinθ
The direction of this force is perpendicular to both v and B.
Following three cases are possible:
1. When the initial velocity is parallel to the magnetic field:
Here θ=0 , So F = qvB sin0 = 0.
Thus the parallel magnetic field does not exert any force on the
moving charged particle. The charged particle will continue to move
along the line of force.
2. When the initial velocity is perpendicular to the magnetic field:
The particle will move along circular path.
Centripetal force = magnetic force
mv2/r = qvB
or r = mv / qB
3. When the initial velocity makes an arbitrary angle with the field direction:
In this case particle will move along helical path with velocity vCosθ
along the direction of magnetic field
m(vSinθ)2/r = qvSinθB
r = mvSinθ / qB
Q.3 .
State the factors on which the force acting on a charge moving in a
magnetic field depends. Write the expression for this force. When is
this force minimum and maximum?
Ans: Factors on which the force acting on a charge moving in a magnetic field depends are
1. Magnitude of charge’q’
2. Intensity of magnetic field ‘B’
3. Velocity of the particle’v’
4. Angle between ‘v’ and ‘B’ – θ
F = qvBSinθ
For Maximum force θ = 900 then F = qvB
For Minimum force θ = 00 then F = 0
Q4. State Biot-Savart law for magnetic field produced at a point due to a small current element. And
express it in vector form. Derive the expression for magnetic field at the centre of current carrying
circular loop.
Ans According to Biot Savart Law the Magnetic field dB due to small current
element is
1. directly proportional to the current I
2. directly proportional to the current dl
3. directly proportional to the current Sinθ
4.directly proportional to the current 1/r2
Hence dB α
𝐼𝑑𝑙𝑆𝑖𝑛𝜃
𝑟2
or dB = k
𝐼𝑑𝑙𝑆𝑖𝑛𝜃
dB =
𝑟2
𝜇0 𝐼𝑑𝑙𝑆𝑖𝑛𝜃
4𝜋
𝑟2
Vector form of Biot Savart Law
⃗⃗⃗⃗
𝜇0 𝐼𝑑𝑙 ×𝑟
⃗⃗⃗⃗⃗
𝑑𝐵=
3
4𝜋
According to Biot Savart Law dB =
θ= 90 hence dB =
0
Total field B=
𝜇0 𝐼𝑑𝑙
𝑟
𝜇0 𝐼𝑑𝑙𝑆𝑖𝑛𝜃
4𝜋
𝑟2
4𝜋 𝑟 2
𝜇0 𝐼2𝜋𝑟
𝜇0 𝐼
4𝜋 𝑟 2
=
2 𝑟
Q. 5. Derive a mathematical expression for the force acting on a current carrying straight conductor kept
in a magnetic field. Under what conditions is this force (i) zero and (ii) maximum?
Ans. Let ‘n’ be the no. of electrons per unit volume of the conductor.
Total no. of electrons = nAl
Charge q= enAl
Force F =qvBSinθ = enAlvBSinθ = IlBSinθ
⃗
𝐹 = 𝐼𝑙 × 𝐵
For Maximum force θ = 900 then F = IlB
For Minimum force θ = 00 then F = 0
Q. 6. Define current sensitivity and voltage sensitivity of a galvanometer. State the factors on which the
sensitivity of a moving coil galvanometer depends.
Ans.
Current sensitivity: It is the deflection produced in the galvanometer when a unit
current flows through it.
Voltage sensitivity: It is the deflection produced in the galvanometer when a unit
potential difference is applied across its ends.
Voltage sensitivity, Vs=α/V=NBA/k RG
Q.7. State ampere’s circuital law connecting the line integral of B over a closed path to the net current
crossing the area bounded by the path. Use the law to derive the formula for the magnetic field
due to an infinitely long straight current carrying wire.
Q. 8. A long solenoid with closely wound turns has n turns per
unit of its length. A steady current I flow through this
solenoid. Use Ampere’s circuital law to obtain an
expression for the magnetic field at a point on its axis and
close to its mid point.
Ans. The magnetic field along the axis of solenoid is uniform
and outside the solenoid the magnetic field is zero.
Q. 9. How will you convert a galvanometer into an ammeter of range 0 - I
amperes? What is the effective resistance of an ammeter?
Toconvert a galvanometer into an ammeter a small resistance shunt
‘S’ is connected in parallel with the galvanometer.
Ig G= (I - Ig)S Or S = IgG/((I - Ig))
Effective resistance
R= SG/(S+G)
Q. 10. How can a galvanometer be converted into a voltmeter to read a maximum potential
difference V? Discuss with related mathematical expression.
Ans.
A voltmeter is connected in parallel with a circuit element. So it must draw a very small
current, otherwise the voltage to be measured would decrease. Toconvert a galvanometer into
a voltmeter large resistance ‘R’ is connected in series with the galvanometer.
Let RG be the resistance of galvanometer and Ig be the current with
which galvanometer gives full scale deflection. To measure a maximum
potential difference V, a high resistance R is connected in series with it.
Total resistance of the device = R + RG
Therefore by Ohm’s law
Q11. How will you select materials for making permanent magnets, electromagnets and cores of
transformers?
Ans.
A. Permanent magnets- The material used for making permanent magnets must have the
following characteristics:
1. High retentivity 2. High coercivity 3.High permeability.
B. Electromagnets- The material used for making cores of electromagnets must have the
following characteristics:
1. High initial permeability
2. Low retentivity
C. Transformer cores- The material used for making cores of transformers must havethe
following characteristics:
1. High initial permeability
2. Low hysteresis loss
3. Low resistivity
Q12. Define the terms, (i) Magnetisation, (ii) Relative permeability and (iii) magnetic
susceptibility. Give their S I unit, if any
Ans. (i)Magnetisation or intensity of magnetization - It is the magnetic moment developed per unit
volume of a material when placed in a magnetising field. It is a vector quantity.
It’s SI unit is Am-1.
(ii)Relative permeability - It is the ratio of permeability of the medium to
thepermeability of free space.
It is unit less quantity.
(iii)Magnetic susceptibility - Magnetic susceptibility measures the ability of a substanceto take
up magnetisation when placed in a magnetic field. It is the ratio of the intensity of magnetisafion
M to the magnetising field inteitsity H.
As magnetic susceptibility is the ratio of two quantities having the same units, so it has no
unit.
FIVE MARKS QUESTIONS
Q.1(a)Using Biot-Savart law, deduce an expression for the magnetic field on the axis of a circular
current loop. Hence obtain the expression for the magnetic field at the centre of the loop.
(b)Two circular coils of equal radius and carrying equal current are placed such that their centers
coincide and their axis are perpendicular to each other. Find the net magnetic field at the common center.
Ans. Consider a circular loop of wire of radius a and carrying current I, as shown in figure.
Let the plane of the loop be perpendicular to the plane of paper. We wish to find field at an axial point P
at a distance r from the centre C.
Consider a current element dl at the top of the loop. It has an outward coming current.
𝜇0 𝐼𝑑𝑙𝑆𝑖𝑛𝜃
According to Biot Savart Law ⃗⃗⃗⃗⃗
𝑑𝐵=
θ= 900 hence ⃗⃗⃗⃗⃗
𝑑𝐵=
𝜇0 𝐼𝑑𝑙
4𝜋
𝑟2
4𝜋 𝑟 2
The magnetic field is resolved into its components, the cos component being equal and opposite
cancel out and total field is given by sum of sin components
𝜇0 𝐼
(b) Magnetic field at the centre B1 = B2 =
2 𝑟
B2 = B12 + B22
B=√2𝐵
√2𝜇0 𝐼
B=
2 𝑟
Q. 2. With the help of a labeled diagram, explain the principle,
construction, theory and working of a cyclotron.
Ans. It is a device used to accelerate charged particles like protons, deuterons, α- particles, etc., to
very high energies.
Principle: A charged particle can be accelerated to very high energies by making it passthrough
alternating electric field and strong magnetic field perpendicular to each other, by making it
cross the same electric field time and again.
Q3. Derive an expression for the torque on a rectangular coil of area A, carrying a current I and placed in
a magnetic field B. The angle between the direction of B and vector perpendicular to the plane of
the coil is θ.
Ans.
Consider a rectangular coil PQRS suspended in a uniform magnetic field B , with its axis
perpendicular to the field.
Let I be the current flowing through the coil PQRS, a and b be the sides of the coil
PQRS,
A =ab= area of the coil and is the angle between the direction ofBand normal to the plane of
the coil.
According to Fleming’s left hand rule, the magnetic forces on sides PS and QR are equal,
opposite and collinear (along the axis of the loop),so their resultant is zero.
The side PQ experiences a normal inward force equal to IbB while the side RS experiences an
equal normal outward force. These two forces form a couple which exerts a torque given by
𝜏 = Force x perpendicular
distance
= IbBx asinθ = IBAsinθ
If the rectangular loop has N turns, the torque increases N times i.e., 𝜏 =
NIBAsinθ
But NIA = m, the magnetic moment of the loop, so𝜏 =mBsinθ
In vector notation torque is represented as
⃗⃗ × ⃗𝑩
⃗
⃗ = ⃗𝑴
𝝉
The direction of the torque t is such that it rotates the loop clockwise about the axis of
suspension.
Q. 4. With the help of a neat and labeled diagram, explain the underlying principle, construction and
working of a moving coil galvanometer. What is the function of (i) uniform radial field (ii) soft
iron core in such a device?
Ans. A galvanometer is a device to detect current in a circuit, the magnitude of which depends on the
strength of current.
Construction: A pivoted-type galvanometer consists of a rectangular coil of fineinsulated copper wire
wound on a light aluminium frame. The motion of the coil is controlled by a pair of hair springs of
phosphor-bronze. The springs provide the restoring torque. A light aluminium pointer attached to the
coil measures its deflection on a suitable scale.
The coil is placed symmetrically between the concave poles of a permanent horse-shoe magnet.
There is a cylindrical soft iron core which not only makes the field radial but also increases the
strength of the magnetic field.
Theory and working: As the field is radial, the plane of the coil always remains parallel
to the field B . When a current flows through the coil, a torque acts on it. It is
𝜏 = Force x perpendicular distance = NIbB x a sin 90 = NIB (ab) = NIBA
Here ϴ =90 , because the normal to the plane of coil remains perpendicular to the field
in all positions.
The torque 𝜏 deflects the coil through an angle θ. A restoring torque is set up in the coil due to the
elasticity of the springs such that
𝜏 = 𝑘𝛼
Where k is the torsion constant of the springs i.e., torque required to produce unit angular twist.
In equilibrium position,
Restoring torque = Deflecting torque
kα = NIBA
α = NIBA/k
Thus the deflection produced in the galvanometer coil is proportional to the current flowing
through it.
Functions:
(i) A uniform magnetic field provides a linear current scale.
(ii) A soft iron core makes the field radial. It also increases the strength of the magnetic field and
hence increases the sensitivity of the galvanometer.
Q5.
Derive a mathematical expression for the force per unit length acting on each of the two straight
parallel metallic conductors carrying current in the same direction and kept near each other.
Hence define an ampere. Why do such current carrying conductors attract
each other?
Ans.
AB and CD are two parallel conductors carrying currents I1 and I2
respectively separated by distance‘r’. The currents in the same direction force
is attractive force
currents in opposite directions, force is repulsive.
The magnetic field produced by current I1 at any point on wire AB is
B1= µ0I1/2πr
According to Flemings Left Hand Rule the field acts perpendicular to the wire
CD and points into the plane of paper.
The force on current carrying wire RS.
F2=µ0I1I2l/2πr
Force per unit length,
F2 / l=µ0I1I2/2πr
Similarly force on AB due to magnetic field produced by CD
is
F1/ l=µ0I1I2/2πr
According to Fleming’s left hand rule, this force acts at right angles to CD, towards AB in the
plane of the paper. Similarly, an equal and opposite force is exerted on the wire AB by the field
of wire CD. Thus when the currents in the two wires are in the same direction, the forces
between them are attractive.
Definition of ampere:
One ampere is that value of steady current, which on flowing in each of the two parallel
infinitely long conductors of negligible cross-section placed in vacuum at a distance of 1 m
from each other, produces between them a force of 2 X 10-7 newton per meter of their length.
VALUE BASED QUESTION MAGNETIC EFFECTS OF ELECTRIC CURRENT AND MAGNETISM
1. Mr Narasimham , a 65 year old person often complained of neck pain. One day his grandson Akash,
suggested that magnetic therapy is very effective in reducing such pains. He said that the permanent
magnet/electromagnet ,used in the device will help to produce Joule’s heating effects in the blood
stream, which helps the blood flow better.He immediately contacted his friend in Chennai, who was
running Magnetic Therapy Clinic.
a.What two values did Akash exhibit towards his grandfather? Mentionany two.
Ans.Responsible behaviour, concern and awareness
b. What is the SI unit of magnetic induction and define it?
Ans. Tesla (defn)
2. Ms Rajyam joined a PG course in Nanotechnology lab in IIT Chennai. Thefirst day, when she went to
the lab, she met Mr. Antonio, the labassistant.He greeted her and advised her not to touch the wires which
weresuspended from the roof at every part of the lab as they were from highvoltage lines. He also told her
not to bring any of the two wires closer to eachother during any experimental applications. He helped her
in understandingabout the precautions that has to be taken in the lab.
a.What value did Mr. Antonio exhibit towards Ms. Rajyam? Mention any two (Responsible behaviour,
sensitivity, concern for others and alerting thepeople)
b.Why two high voltage power transmission lines should not be close to eachother?
c.Give an expression for the magnetic force that acts between the wires?
3.
In the birthday party of Bharat, a class 7student, his parents gave bigslinkys to all his friends as return
gifts. The next day, during the physicsclass Mr Mohan, the teacher explained them about the production
ofmagnetic fields using current carrying coil and also said that they can makepermanent magnets, using
such coils by passing high currents throughthem. That night Suman, a friend of Bharat, asked his father
about thecoils, and their shape. His father asked him to bring the slinky, that hisfriend gave and explained
the uses of toroid and solenoid.
a.What value did Suman’s father exhibit towards his son?
(Responsibility, makes his child to understand the concepts and to generateinterest in the subjects)
b.What is the difference in the fields produced by the solenoid and Toroid?
The magnetic field lines in a toroid is concentric circles whereas in solenoidit is straight within the turns.
4. Ms Anita found that her son could not hear properly. Thespecialist prescribed hearing aid for her son.
Hearing aids consist ofelectromagnets in the loudspeakers used in the device.
a.What two values does Ms Anita exhibit towards her son and students?Mention any two.
(caring attitude, sensitive towards society, concern for others
b.What is an electromagnet? In what way its hysteresis curve is differentfrom that used for permanent
magnets?
Ans. Electromagnet- temporary magnet.Hysteresis curve has small are,small coercivity, small retentivity.
5. Ms Gomathy wife of Mr Varadan complained about the non availability ofgas cylinders and explained to
him to look out for alternate methods forcooking.Mr Varadan bought an induction stove to overcome the
fuel problem. Thenext day Gomathy used her copper bottom cooker and kept it on the
induction stove. But even after using it for half an hour she found that thecooker was not hot and food not
cooked. As she was not aware of themethod to use the induction stove, she asked her elder daughter
Tanya,studying first year engineering about it. She told her, that some vessels cannot be used on this
stove. She took the instruction manual and explained toher mother, that the stove works on magnetic
induction, and copper being adia magnetic material, will not respond to it.
a.What values did Mr varadan and Tanya exhibit towards Ms Gomathy?
Mention any two (awareness, concern for conservation of energy and fossil fuels, sharing the knowledge)
b.Give few examples of diamagnetic materials and explain how theirsuceptiblity varies with temperature?
Ans.Susceptibility is independent of temperature as they have no permanentdipoles.
6. Hari and Rama class X students, were assigned a project based onmagnetism. In their project work, they
had calculated the value of earth’smagnetic field. When they submitted their project for verification.Mr
Satish, their physics teacher, corrected the mistakes. He alsosuggested few books which could be of use
to them.
a.What values did Mr Satish exhibit towards his students? Mention anyTwo.
Ans.(Honesty, helpfulness, responsible behaviour towards students, concern forthe student to create
interest in the subject)
b.Mention the three magnetic elements required to calculate the value ofearth and draw a neat diagram to
explain them.
Ans. Magnetic declination, magnetic inclination and horizontal componentof earth’s magnetic field.
7. Mr Sairam the chief development officer, in southern railway went on anofficial tour to attend a seminar
on fast moving trains. He met his friend ontosaki in Tokyo after he finished his seminar there. His friend
explainedto Sairam, how Japanese people are concentrating on energy conservationand saving of fossil
fuels using Maglev trains. Mr sairam travelled fromTokyo to Osaka in maglev train and found that sound
is less, traveliing issmooth and understood in what way we are lagging behind Japanese inmass
transporting systems.
a. What values did Mr sairam found from Ontosaki? Mention any two.
(awareness about new technology, concern for energy conservation, decreaseof noise pollution and air
pollution i.e, concern for environment)
b.What is Meisner’s effect?
Ans.When a superconduc tor is cooled in a magnetic field below its criticaltemperature the magnetic field lines
are expelled showing diamagneticproperty. This is called Meissner effect.
8. Ms Ramani a house wife aged 42 years complained of stomach ache oneday. Her husband Mr Srinivas
took her to a nearby hospital. The doctorobserved her and found something wrong near her liver and
suspectedmalignancy. There after checking her MRI scan, a team of doctors advisedher to go through
Carbon radio therapy which is very safe. They said usingcyclotron, high speed ions can be generated that
directly attach thecancerous tissues and destroy them.
1. What values did Mr srinivas and the doctor have exhibited? Mention anytwo.
Ans. Concern for others, helpfulness, presence of mind, responsible citizen.
2.What are the role played by Electric field and magnetic fielding Cyclotron?Ans. The charged particles
are accelerated by the electric field with themagnetic field bringing them again and again to the electric
field thatis the region between the Dees.
ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENT
Gist of the lesson (including basic concepts and important formulae)
1. Magnetic Flux:- The no. of magnetic field lines passing through an area inside a magnetic field
region is known as magnetic flux passing through that area.
The magnetic flux through a plane surface placed inside a uniform magnetic field is given by
 
ϕ = B  A  BA cos
The magnetic flus passing any surface placed inside a magnetic field (uniform of non-uniform) is
given by


ϕ =  B  dA
2.
3.
4.
5.
6.
The SI unit of magnetic flux is Tm2 or Wb. It is a Scalar quantity.
The dimension of magnetic flux is [ML2T-2A-1].
Electromagnetic Induction:- “Whenever there is change in magnetic flux linked with a conductor or
conducting coil, an emf is induced in the conductor or coil. The emf induced lasts so long as the
magnetic flux linked with conductor or coil changes. This phenomenon is called EMI.”
Faraday’s Law of Electromagnetic Induction:- “The magnitude of the induced emfin a circuit is
proportional to the time rate of change of magnetic flux through the circuit.”
d

dt
Lenz’s Law:-“ The polarity of induced emf is such that it tends to tends to produce a current which
opposes the change in magnetic flux that produces it.”It is based of law of conservation of energy.
Mathematically, Faraday’s & Lenz’s laws are combined in the following expression
d

dt
Induced current and induced charge:- If a coil is closed and has resistance R, then current induced in
the coil,

N d
i 
ampere,
R
R dt
and the induced charge,
N
Total flux linkange
q = i∆t =   =
R
Resistance
Motional emf:(a) The emf induced in a straight conductor moving inside a uniform magnetic field with a velocity
perpendicular to its length as well as the magnetic field induction
ɛ = Bvl
(b) The emf induced in a disc rotating inside a uniform magnetic field directed parallel to the axis of
rotation of disc/ the emf induced in a straight conductor rotating about its one end inside a uniform
magnetic field directed parallel to the axis of rotation
ɛ=
Bvl Bwl 2

2
2
7. Eddy Currents:- “When bulk pieces of conductors are subjected to changing magnetic flux, induced
currents are produced in them due to electromagnetic induction. The flow pattern of these currents
resemble swirling water, so these are known as eddy currents or whirlpool currents.”
 Eddy currents are undesirable in many devices such as transformers, electric motors etc. since
they heat up the core and dissipate electrical energy in the form of heat.

Eddy currents are minimised by using laminations of metal to make a metal core. The
laminations are separated by an insulating material like lacquer. The plane of the laminations
must be arranged parallel to the magnetic field, so that they cut across the eddy current paths.
This arrangement reduces the strength of the eddy currents.
8. Applications of eddy currents:(i) Magnetic braking in trains: (ii) Electromagnetic damping:
(iii) Induction furnace:
(iv) Electric power meters:
9. Self-Induction:- “When the current in a coil is changed, a back emf is induced in the coil that
opposes the change in the current. This phenomenon is known as self induction or electrical inertia.”
10. Self-Inductance L :This quantity is the measure of self-induction of a coil. It is a scalar quantity. SI unit of this quantity
is Henry and the dimension is [ML2T-2A-2]. This quantity is also known as ‘coefficient of selfinduction’.
Self-inductance of a coil is defined numerically equal to, “the back emf induced in the coil when the
current flowing through its turns changes at the rate of 1 A/s.” OR “the magnetic flux linked with the
coil when the current flowing through its turns is unity.”
The formula for the self-inductance of any coil is

L   or L 
I
dI
dt
11. Self-inductance of a long solenoid is
L = μrμ0n2Al
12. Mutual-Induction:-“When two coils are placed nearby and the current in one coil is changed, an emf
is induced in the neighbouring coil due to the change in magnetic flux linked with it. This
phenomenon is known as mutual induction.”
13. Mutual-Inductance M12:- the mutual induction between two coils is given mathematically by
quantity ‘Mutual-Inductance’ or ‘coefficient of mutual induction’.
Mutual inductance of two coils is defined as the “Magnetic flux linked with one coil due to the unit
amonut of current flowing in the neighbouring coil”. OR “the back emf induced in one coil due the
unit rate of change of current in the neighbouring coil”.
M 12 
2
I1
or M 12 
2
dI 1
dt
Mutual-Inductance of two coils depends on the linkage of magnetic field lines between them apart
from other factors. If the magnetic fields lines of one coil are completely linked with the
neighbouring coil then they are called perfectly coupled.
For two perfectly coupled co-axial solenoids, mutual-inductance is given by
M12 = μ0n1 n2 πr2l where r is the radius of inner coil.
OR,
M12 = √³L1L2´ where L1 and L2 are the self-inductances of the two coils.
14. The magnetic energy stored in a current carrying solenoid:U = ½ LI2
15. Alternating emf and alternating current:- The emf/ current whose polarity/ direction reverses after a
regular interval of time periods is called an alternating emf/ alternating current.
The alternating emf / alternating current produced by an a.c. generator is a sinusoidally varying
alternating emf/ current.
The instantaneous magnetic flux associated with coil is   NBA cos(t   ) or   NBA sin( t   )
The instantaneous emf and instantaneous current is
e =E0 sin (ɷt+ϕ) or e = E0cos (ɷt+ϕ)
i = I0 sin (ɷt+ϕ) or i = I0cos (ɷt+ϕ)
The peak value or amplitude of the emf/ current is
E0 = NBA and I 0 = NBA / R
16. The average value of alternating emf / alternating current is ‘zero’ for full cycle.
2V
2I 0
17. The average value of alternating emf / alternating current for the ‘half’ cycle is 0 or


18. RMS value / Effective value / Virtual value of the alternating current:- It is the value of alternating
emf/ current that is measured by a.c. metres which are based on heating effect of current.
RMS value of the current is numerically equal to that value of constant (D.C.) current which when flows
through a resistor for a certain time period produces the same amount of heat as is produced by the
alternating current in the same time period for same resistor.
Irms = Io/√2 = 0.707 I0 and Erms = Eo/√2 = 0.707 E0
19. Phase difference between the alternating current and alternating voltage:The potential difference across resistor remains in phase with the alternating current.
The potential difference across inductor leads the alternating current with a phase angle π/2.
The potential difference across capacitor lags behind the alternating current by a phase angle π/2.
(a) Phasor diagram and a wave diagram for a resistor v =Vm sin ɷt1 and i = Im sin ɷt1
(b) Phasor diagram and a wave diagram for a inductor v =Vm sin ɷt1 and i = Im sin (ɷt1-π/2)
(c) Phasor diagram and a wave diagram for a capacitor v =Vm sin ɷt1 and i = Im sin (ɷt1+π/2)
20. L-C-R series circuit:-
21. Impedance and reactance:- “The obstruction offered by the pure inductance or pure capacitance to
the flow of a.c. which is frequency dependent and has the dimension of resistance but is not a source
of power dissipation is called reactance.”
“The obstruction offered by a circuit to the flow of a.c. that comprises of a frequency dependent
component as well as frequency independent component is called impedance of the circuit. It has a
dimension of resistance.”
The SI unit of reactance and impedance is Ω.
Reactance of an Inductor (or Inductive Reactance) XL = ѡL
Reactance of a Capacitor (or Capacitive Reactance) XC = 1/ѡC
Impedance of a series L-C-R circuit Z = √{(XL – XC)2 + R2} and voltage V = √{(VL – VC)2 + VR2}
22. Phasor diagram for a series L-C-R circuit with an a.c. source:-
23. Phase difference between voltage and current for a series L-C-R circuit: X  XC 
 V  VC
When the source frequency f >fr   tan 1  L
 or   tan 1  L
R


 VR



 X  XL 
 V  VL 
When the source frequency f<fr   tan 1  C

 or   tan 1  C
R


 VR 
24. Graphs of Reactance Vs frequency and Impedance Vs frequency for a.c. circuits:-
25. Graph of current Vs frequency for a series L-C-R circuit:-
26. Resonance in a series L-C-R circuit:“The a.c. current flowing in the series L-C-R circuit is maximum for a certain frequency of the a.c.
source when the impedance of the circuit is minimum. This phenomenon is called resonance. And the
corresponding frequency when the impedance is minimum is called the resonance frequency.”
27. The resonance frequency is:At resonance XL = XC
Angular resonance frequency  r 
1
and resonance frequency f r 
1
2 LC
LC
28. Quality factor of series L-C-R circuit:The mathematical factor that is measure the sharpness of resonance of L-C-R circuit is called the
quality factor of the series L-C-R circuit.
The Q-factor of L-C-R circuit is defined as
“the ratio of voltage drop across inductor (or capacitor) to the voltage drop resistor at resonance.”
OR
“the ratio of resonance frequency to the frequency band width of the resonant curve.”
1 L r
L
=
= r
R C 2
R
29. Power dissipation in a.c. circuit:The average power dissipation in a.c. circuit depends upon the phase difference between current and
voltage.
For pure inductive or capacitive circuit, where the phase difference between current and voltage is
π/2, the average power dissipation is zero as for one half of the cycle the electrical energy is
Q=
transformed into magnetic/ electrostatic energy and in the next half cycle the magnetic energy/
electrostatic energy is retransformed into electrical energy.
The power dissipation in an a.c. circuit is <P> = ErmsIrmscosϕ
R
The factor ‘cosϕ’ is called power factor. cos  
Z
The power factor is maximum at resonance and is zero for pure capacitive or pure inductive
circuit.
30. Watt-less current:- the component of current phasor that is at π/2 phase difference with the voltage is
called watt-less current as it is not source of any power dissipation.
The watt-less current is given by expression Irmscos ϕ
31. Transformer:It is a device used for converting low alternating voltage at high current into high voltage at low
current and vice-versa.
Principle: It works on the principle of mutual induction i.e. if two coils are inductively coupled and
when current or magnetic flux is changed through one of the two coils, then induced emf is produced
in the other coil.
•
According to Faradays law, the emf induced in secondary
If the current in the
secondary is very small
The emf induced in primary
As
the
primary
has
negligible resistance
From equation (i) and (ii)
Transformers are of two types:1. Step – up transformer
The transformer having more number of turns in the secondary coil than
primary coil (i.e.NS>NP) and used to convert low voltage at high current to high voltage at low
current.
2. Step – down transformer The transformer having more number of turns in the primary coil than
secondary coil (i.e.NS<NP) and used to convert high voltage at low current to low voltage at high
current.
Transformation Ratio:“The output to input voltage ratio of transformer is equal to the ratio of no. of turns in the secondary
to the primary of transformer. This ratio is called the transformation ratio of the of the transformer.”
vS N S
=
vP N P
v I
P
Efficiency of Transformer:- η = s s  0
v p I p Pi
100% efficient transformer is called an ideal transformer as there is no energy loss in this transformer
For an ideal transformer;a) The windings of transformer should have no resistance.
b) The output of transformer should be in open circuit.
c) There should be no flux leakage.
d) There should be no hysteresis loss or any other loss.
The various sources of energy loss in transformer are:1. Flux losses:- The coupling of the two coils of transformer is not perfect. As a result, whole of the
magnetic flux linked to the primary coil can not be linked to the secondary coil.
2. Copper losses:- Some electrical energy is always converted into heat energy in the resistance of the
copper wire used in the winding of the coil.
3. Iron losses:- The changing magnetic flux leads to production of induced emf in the iron core of the
transformer which also leads to loss of some electric energy in the eddy current produced in the iron
core. It is minimized using laminated iron core. It is prepared by joining two similar iron strips
together after coating with varnish. As a single iron strip is very thin so its resistance becomes large
which leads to production of very small eddy currents in it and in this way only a small amount of
heat is produced in the core.
4. Hysteresis losses:- Due to alternating current flowing through the coil, the iron core is magnetised
and demagnetised again and again. During each cycle of magnetisation and demagnetisation, some
energy is lost due to Hysteresis. It can be minimized by selecting a material for iron core whose area
of hysteresis loop is very small.
5. Humming losses:- Due to passage of AC current, the core of the transformer starts vibrating and
produces humming sound in which some part of the electrical energy is lost in the form of sound.
Uses of transformer:1.
2.
3.
4.
Transformers are used for voltage regulators and stabilized power supplies.
Small transformers are used in radio sets, televisions, telephones, loud speakers etc.
A step-up transformer is used in the production of X-rays.
A step-down transformer is used for obtaining large current for electric welding or in the induction
furnace for melting metals.
VERY SHORT ANSWER QUESTION S (1 marks)
Q1.
An electric lamp, connected in series with a capacitor and an a.c. source is glowing with certain
brightness. How does the brightness of bulb changes on reducing the capacitance?
Ans.
On reducing the capacitance, capacitive reactance increases which reduce brightness of lamp.
Q2.
If the speed of rotation of armature is increased twice how would it affect the (a) maximum
e.m.f produced (b) frequency of the e.m.f?
(e=NBAω ;f=ω/2Π)
A choke coil and a bulb are connected in series to a d.c. source. The bulb shines brightly. How
does the brightness changes when an iron core is inserted in the choke coil?
Ans.
Q3.
Ans.
Brightness of bulb will not change because at steady d.c. , the choke coil has no inductive
reactance.
Q4.
The current in the wire PQ is increasing. In which direction does the induced current flows in the
current loop.
Ans.
Clockwise
Q5.
Give the direction in which the induced current flows in the wire loop, when the magnet moves
towards it as shown in figure.
P
Q
N
S
Ans.
Clockwise when looked from magnet side of the loop.
Q6.
Why a transformer cannot be used to step up d.c. voltage?
Ans.
D.C. cannot produce varying field for secondary winding, therefore induced emf cannot be
produced in it.
Q7.
What is the phase difference between the the voltage across an inductor and a capacitor in an a.c.
circuit.
Ans.
1800
Q8.
Give the phase difference between applied a.c. voltage and current in a LCR circuit at resonance.
Ans.
00 i.e. in phase
Q9.
What is the power consumed in (i) purely inductive and (II) purely capacitive a.c. circuits?
Ans.
Zero
Q10.
What is the power dissipation of an a.c. circuit in which voltage and current are given by :
V = 300 sin (ωt –π/2) and I = 10 sin ωt ?
Ans.
Power in a.c. circuit is ,P = VI cosφ
Here φ = π/2 and cosπ/2 = 0
Thus P = 0
Q11.
In series LCR circuit, when voltage and current are in same phase?
Ans.
At resonance
Q12.
What is the power factor of an LCR series circuit at resonance.
Ans.
Unity.
Q13.
If number of turns of a solenoid is doubled, keeping the other factors constant, how does self
inductance of the solenoid changes?
Ans.
As L α N2thus Lˈα 4 N2
Hence self inductance increases to four times.
Q14.
Ans.
A plot of magnetic flux (φ) versus current is shown in figure for two inductors A & B. which of
the two has larger value of self inductance?
L = φ/I
A
B
Φ
For given I, A has larger value of φ, so A has larger
self inductance.
I
Q15.
Why is a.c. more dangerous than d.c. for same voltage?
Ans.
A.c. of same r.m.s. voltage as that of d.c. will have higher value of maximum voltage given as
Vmax = √2 Vrms
This increases the value of a.c. which makes it more dangerous.
SHORT ANSWER QUESTION (2 or 3 marks)
Q1.
State faraday's laws of electromagnetic induction (EMI).
Ans.
1stlaw: When magnetic flux linking with a coil changes, an e.m.f. is induced in the coil. This
induced e.m.f. lasts so long as the change in magnetic flux continues.
2nd law: the magnitude of the induced e.m.f. produced in a coil is directly proportional to the rate
of change of magnetic flux dφ/dt linked with it.
I.e. I e I = dφ /dt
Q2.
State Lenz’s law. Show that this law follows the principle of conservation energy.
Ans.
Lenz’s law states that induced e.m.f. opposes the cause that produces this e.m.f.
In the arrangement shown in the figure, direction of the induced current is such that it produces
magnetic field which opposes the movement of magnet towards the coil. Deflection of
Galvanometer indicates the presence of electrical energy. Some work has to be done to move the
magnet which results into electrical energy. Electrical energy produced in the coil is basically
due to the mechanical energy applied to move the magnet towards the coil. Hence Lenz’s law
follows from the principle of energy conservation.
Q3.
How are eddy currents produced? Give two applications of eddy currents.
Ans.
Eddy current are circulating currents produced in a metal itself due to EMI when it is placed in
changing magnetic flux in accordance with faradays laws of EMI. Eddy currents are useful in
induction furnaces and dead beat galvanometer.
Non-uniform magnetic field
Eddy current
Metal
Q4.
Define self-inductance. Write its unit. Give expression for self-inductance of a long solenoid
having N turns.
Ans.
It is defined as the induced as the induced e.m.f. produced in the coil through which the rate of
decrease of current is unity.
OR
It is defined as the magnetic flux linked with a coil when unit current flows through it.
Its S.I. unit is henry. Self-inductance of long solenoid is given by L= μ0μrN2A/l .
Q.5
Define mutual inductance. Write its S.I. unit. Give two factors on which the coefficient of mutual
inductance between a pair of coils depends.
Ans. Mutual induction of the two coils or circuits can be defined as the magnetic flux linked with the
secondary coil due to the flow of unit current in the primary coil. Its S.I. unit is henry. It depends
upon number of turns of both the coils, area of cross-section of primary coil and length of coil.
Q.6 Draw a labeled diagram of a step down transformer. Mention two sources of energy loss in a
transformer.
Ans. Losses in a transformer are mainly because of (i) iron loss in the core of the transformer and (ii)
copper loss i.e. I2R loss in windings of the transformer.
Q7. How do R, XL and XC get affected when the frequency of applied AC is doubled?
Ans: a) R remains unaffected
b) XL=2πfL, so doubled
c) XC=1/2πfC, so halved
Q8.
An electric lamp connected in series with a capacitor and an AC source is glowing with certain
brightness. How does the brightness of the lamp change on reducing the capacitance?
Ans: Brightness decreases. (As C decreases, XC increases. Hence Z increases and I decreases.)
Q9.
The peak value of an AC is 5A and its frequency is 60Hz. Find its rms value. How long will the
current take to reach the peak value starting from zero?
Ans: Irms= 3.5A . Time period T=(1/60)s . The current takes one fourth of the time period to reach the
peak value starting from zero. t =T/4 =(1/240)s.
Q10.
Ans:
Q11.
Ans:
Q12
Ans:
When an AC source is connected to a capacitor with a dielectric slab between its plates, will
Therms current increase or decrease or remain constant?
The capacitance increases, decreasing the reactance Xc. Therefore the rms current increases.
In an AC circuit V and I are given by V=100Sin100t volts and I= 100 Sin(100t+π/3)ma
respectively. What is the power dissipated in the circuit?
V0=100V I0=100A Ф= π/3 P=VrmsIrmsCos Ф=2500W
The natural frequency of an LC circuit is 1,25,000 Hz. Then the capacitor C is replaced by
another
capacitor with a dielectric medium k, which decreases the frequency by 25 KHz.
What is the value of k?
υ1=1/2π√LC υ2=1/2π√kLC k=( υ1/ υ 2)2=(1.25)2=1.56.
Q13.Obtain the resonant frequency and Q factor of a series LCR circuit with L= 3H, C= 27μF and
R= 7.4 Ώ. Write two different ways to improve quality factor of a series LCR circuit
Ans: Q=45,ω0=111rad/s
Q14. An ac generator consists of a coil of 50 turns and an area of 2.5m2 rotating at an angular
speed of 60 rad/s in a uniform magnetic field of B= 0.3T between two fixed pole pieces. The
resistance of the circuit including that of the coil is 500Ώ
(i) What is the maximum current drawn from the generator?
(ii)What is the flux through the coil when current is zero?
(iii)What is the flux when current is maximum?
(4.5A, 375Wb, zero)
Q15. For given a.c. circuit, distinguish among resistance, reactance and impedance.
Ans.
S.No. Resistance (R)
Inductive
Reactance (XL)
Capacitive
Reactance (XC)
Impedance (Z)
1.
It is opposition to
the flow of any
type of current.
It opposes the
flow of variable
current.
It opposes direct
current.
It is the total opposition
offered to current (Due to
resistance, inductive
reactance and capacitive
reactance.)
2.
It is independent
of frequency of
source of supply.
It depends
directly on the
frequency of
source.
It depends
inversely on the
frequency of
source.
It depends on the
frequency of the source.
3.
It is given by,
It is given by
It is given by
It is given by,
R = ρ I/a
XL =2 πνL
XC = 1/2 πνC
Z = √[R2+(XL-XC)2]
Long Answer Questions (5 Marks)
Q.1.(a) Derive the phase relation between current and voltage for a series LCR circuit using phasor
diagram.
(b)Obtain the resonance frequency. Draw a plot showing the variation of the peak current (i0) with
frequency of the a.c. source used.
Ans.(a) Across resistance current and voltage are in same phase
Across inductance the voltage leads current by 900
Across capacitance the voltage lags current by 900
Therefore
V
= √{(VL –
VC)2 + VR2}
V=I Z, VR= I R, VL= I XL , VC= I XC
Z = √[R2+(XL-XC)2]
 XL  XC 

R


  tan 1 
I = V /Z
(b) Clearly I will be maximum when Z is minimum.
i.e. for electrical resonance
(XL - XC) = 0
Or XL = XC
i.e. ωL = 1/ωC i.e. ω = 1/√(LC), where
ω is the angular frequency of the circuit.
i.e. νm= 1/2π√(LC),
Q.2.
Ans.
Draw a schematic diagram of a step-up transformer. Explain its
working
principle. Deduce the expression for the secondary to primary voltage in terms of the number of
turns in the two coils. In an ideal transformer, how is this ratio related to the currents in the two
coils?
For figure refer to NCERT text book part 1 fig. 7.20. page no. 260.
Principle : It is based on the principle of mutual induction. It is a phenomena of inducing e.m.f.
in coil due to rate of change of current in near by coil.
Primary e.m.f.
Ep= -Npdφ/dt = VP ( Resistance of Primary is very small)
Secondary e.m.f.
Es = -Nsdφ/dt = VS (Resistance of Secondary is very high)
Thus
Es/Ep = Vs/Vp = Ns/Np
In ideal transformer, input power = output power
i.e.ipVp = isVs i.e. Us/Up=Ns/Np=is/ip
NUMERICAL PROBLEMS
Q1.
Ans.
The electric mains in a house are marked 220 V, 50 Hz. Write down the equation for
instantaneous voltage.
Given Vrms= 220 V and ν = 50 Hz
V0 = √2 Vrms
= 1.414 × 220 = 311 V
And ω = 2πν = 2× 3.14 × 50 = 314 rad-1
Equation for instantaneous voltage, e = 311 sin 314 t
Q2. An alternating voltage E = 200 sin 300 t is applied across a series combination of R = 10 Ω and an
Inductor of 800 mH. Calculate (i) impedance of circuit (ii) peak value of current in circuit (iii)
powerfactor of the circuit.
Ans.(i) Impedance, Z = √(R2 + ω2 L2 )
Here ω = 300 rad-1
Z = √[102 + (300 × 0.8)2]
On solving Z = 240.2 Ω
(ii) peak value of current,
I0 = E0/Z = 200/240.2 = 0.83 A
(iii) Power factor, cosφ = R/Z = 10/ 240.2 = 0.042
Q3. A series LCR circuit with L = 5 H, C = 80 μF, R = 40 Ω is connected to a variable frequency source
of 230 V (i) determine resonance frequency of the circuit (ii) obtain impedance of circuit and
amplitude of current at resonance.
Ans.Given :Vrms = 230 V
(i) Resonance frequency, ω = 1/√(LC) = 1/ √(5 × 80 ×10-6) = 50 rad-1
(ii) At resonance, impedance ,Z = R = 40 Ω
And Amplitude, I0 = V0/Z = (230 ×√2)/ 40
{ V0 = Vrms√2}
= 8.13 A
Q4.The output voltage of an ideal transformer connected to a 240 V a.c. mains is 24 V. When this
transformer is used to light a bulb with rating 24 V, 24 W. calculate current in primary coil of the
circuit.
Ans.
Since V1/V2 = I2/I1
I1 = I2× V2/ V1
I2 = W/V2 = 24/24 = 1A
Thus
I1 = 1 × 24/240 = 0.1 A
Q.5
Prove that the average power over a complete cycle of a.c. through an ideal inductor is zero.
Ans.
For inductor current lags voltage by 900
Average power over a full cycle, PL = average of PL
Average of sin 2ωt for full cycle =
<P>= <VI>=<VOSinωt IO Cosωt>
1
1
<P>= 2 𝑉0IO<2Sin ωt Cosωt>=2 𝑉0 IO<Sin2 ωt > = 0 ( because <Sin2 ωt > = O)
Previous year questions
1 mark questions
1. State Lenz’s law. A metallic rod horizontally along east-west direction is allowed to fall under
gravity. Will there be an emf induced at its ends? Justify your answer.
2. Predict the direction of induced current in metal rings 1 and 2 when current I in the wire is
steadily decreasing?
1
I
3.
4.
5.
6.
7.
2
How does the mutual inductance of a pair of coils change when (i) distance between the coils is
increased and (ii) number of turns in the coils is increased?
How can the self- inductance of a given coil having N number of turns area of cross-section A
and length l be increased?
Define self-inductance of a coil. Write its SI unit.
Mention any two useful applications of eddy current
A graph of magnetic flux (ø) versus current (I) is shown in the figure
for two inductors A and B. Which of the two has larger value of selfinductance?
8. In the given figure a bar magnet is quickly moved towards a conducting loop having a capacitor.
Predict the polarity of the plates A and B of the capacitor.
S
N
2 marks questions
9. A metallic rod of length L is rotated with angular frequency of ω with one end hinged at the
centre and the other end at the circumference of a circular metallic ring of radius L about an axis
passing through the centre and perpendicular to the plane of the ring. A constant and uniform
magnetic field B parallel to the axis is presents everywhere. Deduce the expression for the emf
between the centre and the metallic ring.
10. A current is induced in coil C1 due to the motion of current carrying coil C 2. (i) Write any two
ways by which a large deflection can be obtained in the galvanometer G. (ii) Suggest an
alternative device to demonstrate the induced current in place of a galvanometer.
11. Predict the polarity of the capacitor in the situation described by adjoining figure. Explain the
reason too.
S
N
S
N
12. Two identical loops one of copper and the other of aluminium are rotated with the same angular
speed in the same magnetic field. Compare (i) the induced emf and (ii) The current produced in
the two coils. Justify your answer.
13. (i) When primary coil P is moved towards secondary coil S (as shown in figure below) the
galvanometer shows momentary deflection. What can be done to have larger deflection in the
galvanometer with the same battery?(ii) State the related law.
S
P
14. A coil B is connected to low voltage bulb B and placed near another coil A as shown in the
figure. Give reasons to explain the following observations (i) The bulb B lights (ii) Bulb gets
dimmer if the coil Q is moved towards left.
15. A conducting rod of length l is moved in a magnetic field of magnitude B with velocity v such
that the arrangement is mutually perpendicular. Prove that the emf induced in the rod is
|𝐸| = Blv.
16. A rectangular coil of area A having number of turns N is rotated at f revolutions per second in a
uniform magnetic field B the field being perpendicular to the coil. Prove that the maximum emf
induced in the coil is 2πfNBA.
17. An alternating voltage given by V = 70 sin 100 πt is connected across a pure resistor 25 Ω. Find
(i) the frequency of the source (ii) the rms current through the resistor.
18. i) The graphs (I) and (II) represent the variation of the opposition offered by the circuit element
to the flow of alternating current with frequency of the applied emf corresponding to each graph.
(ii) Write the expression for the impedance
Offered by the series combination of the above two elements connected across the AC source.
Which will be ahead in phase in this circuit voltage or current?
O
Frequency
(I)
O
Frequency
(II)
3 marks questions
19. (i) State faraday’s law of electromagnetic induction. (ii) A jet plane is travelling towards west at
a speed of 1800 km/h. what is the voltage difference developed between the ends of the wing
having a span of 25m if the earth’s magnetic field at the location has magnitude of 5x10-4 T and
the dip angle is 30o?
20. (i) State the law that gives the polarity of the induced emf. (ii) A 1.5 μF capacitor is connected to
220 V , 50 Hz source. Find the capacitive reactance and the rms current.
21. A coil of number of turns N area A is rotated at a constant speed ω in a uniform magnetic field B
and connected to a resistor R. Deduce expressions for (i) maximum emf induced in the coil (ii)
power dissipation in the coil.
22. An AC voltage V = Vo sin ωt is applied across a pure inductor L. Obtain an expression for the
current I in the circuit and hence obtain the (i) inductive reactance of the circuit and (ii) the phase
of the current flowing with respect to the applied voltage.
23. An AC voltage V = Vo sin ωt is applied across a pure capacitor C. Obtain an expression for the
current I in the circuit and hence obtain the (i) capacitive reactance of the circuit and (ii) the
phase of the current flowing with respect to the applied voltage.
5 marks questions
24. State Faraday’s law of electromagnetic induction. Figure shows a rectangular conductor PQRS in
which the conductor PQ is free to move in a uniform magnetic field B perpendicular to the plane
of paper. The field extends from x=0 to x=b and is zero for x > b. Assume that only the arm PQ
possesses resistance r. When the arm PQ is pulled outward from x = 0 to x = 2b and is then
moved backward to x = 0 with constant speed v obtain the expression for the flux and the
induced emf. Sketch the variation of these quantities with distance 0 ≤ x ≤ 2b.
(All
India
2010)
● S ●
●
●
●
●
●
P
●
●
●
●
●
●
●
Q●
● R●
●
●
X=0
x=b
x = 2b
25. (i)State Lenz’s law. Give one example to illustrate this law. The Lenz’s law is a consequence of
the principle of conservation of energy. Justify this statement. (ii) Deduce an expression for the
mutual induction of two long coaxial solenoids but having different radii and different number of
turns.
26. (i) What are eddy currents? Write their two applications. (ii) Figure shows a rectangular
conducting loop PQR in which arm RS of length l is movable. The loop is kept in a uniform
magnetic field B directed downward perpendicular to the plane of loop. The arm RS is moved
with a uniform speed v. Deduce an expression for (a) the emf induced across the arm RS (b) the
external force required to move the arm and (c) the power dissipated as heat.
ELECTROMAGNETIC WAVES
1. Concept of displacement current
Displacement current is that current which appears in a region in which the electric field (and hence
electric flux) is changing with time.
Note- We have ID = ε0 dΦ/dt = ε0d(EA) /dt = ε0d[(q/ ε0A)A] /dt = dq /dt = I
2. Modified Ampere’s circuital Law
∮B.dl = µ0 (I + ε0 dΦ/dt )
3. Learn only one order either increasing wavelength or frequency .other calculated by c= ν λ
,velocity of light c=3x108m/s
S.No Name
Frequency Wavelengt Production
Uses
.
h
Range
(Hz)
Range
1.
Gamma
rays
1019 – 1023
10-11 to 1014
m
Emitted by radioactive
nuclei,
Produced in nuclear reaction
In medicine, to
destroy cancer
cells.
2.
X – rays
1016 - 1020
10-8 to 1012
m
3.
Ultraviolet
rays
1015 – 1017
(4 × 10-7 to Produced by special lamps &
6 × 10-10) m very hot bodies (sun).
For eye surgery,
to kill germs in
water purifiers.
4.
Visible
rays
4 × 1014 –
7 × 1014
700 – 400
nm
Jumping of electrons in
higher orbits
Provide us
information about
the world.
5.
Infrared
rays (heat
waves)
1012 - 1014
1mm - 700
nm
Produced by hot bodies and
molecules.
Infrared detectors
used in earth
satellite, used in
green house to
keep plants
warm.
6.
microwave
s
1010 - 1012
0.1 – 1 mm
Produced by special vacuum
tubes (klystrons, gun diode &
magnetrons)
Microwave oven,
for radar system in
aircraft navigation.
7.
Radio
waves
10 - 109
> 0.1 m
Produced by accelerated
motion of charges in
conducting wires.
In radio &
television
communication
system, in cellular
phones to transmit
voice
communication.
Generated by bombarding a
metal target by high energy
electron
Question Answer related to this topic
Used as diagnostic
tool in medicine,
to study crystal
structures
Q.1
What is the ratio of speed of infrared rays and ultra violet rays in
vacuum? (1 mark)
1:1
Q.2
Write the following radiations in ascending order in respect of
their frequencies X-ray, microwaves, radio waves. (1 mark)
radio wave, microwave , X-ray.
Q.3
Name the electromagnetic radiation to which waves of wavelength
in the range 10-2 m belong. Give one use of this part of Em
spectrum. (1 mark)
Infrared, used in remote control.
Q.4
Which part of the electromagnetic spectrum has the largest
penetrating power?
(1 mark)
Gama rays
Q.5
Give a reason to show that microwaves are better carrier of signal
for long range transmission.
(1 mark)
As high frequency wave reduces size of transmitting antenna.
Q.6
Name the EM waves used for studying crystal structure of
solids.
(1 mark)
X-rays
Q.7
Name the EM waves used for treatment of cancer tumors.
mark)
(1
Y-rays
Q.8
Name the electromagnetic radiation used for viewing objects
through haze and fog. (1 mark)
I.R.
Q.9
Identify the electromagnetic radiations as given : frequency =
1020 HZ. (1 mark)
X – rays
Q.10
Write the expression for the velocity of e.m. waves in terms of
permittivity and permeability. (1 mark)
Q.11
Sketch a schematic diagram depicting electric and magnetic fields
for an electromagnetic wave propagating along z-direction.
(2
marks)
Q.12
The oscillating magnetic field in a plane electromagnetic wave is given by B Y=8 x 106 sin(20 x 1011t + 300πx)T. Calculate the wavelength of e.m. wave. Write down the
expression for oscillating electric field. (2 marks)
As B = Bo sin(wt + bx)
BY = 8 x 10-6 sin(2 x 1011t + 300πx)
ω = 2 x 1011rad/s and
K = 300π = 2π/λ
λ = 2π/300 = 1/150m = 0.006m
EZ = E0 sin(wt + kx)
Where E0 = CBO = 3 x 108 x 8 x 10-6 = 2400N/C
EZ = 2400 sin(2 x 1011t + 300πx)
Q.13
The oscillating electric field of an electromagnetic wave is given by E Y=30 sin(2 x
1011t + 300πx) Vm-1. Find the dirn of propagation of wave and write down the
expression for magnetic field ?
(2 marks)
EY = 30 sin(21 x 1011t + 300πx)
Comparing with EY = E0 sin(wt + kx) dirn of propagation is –x direction &
BZ = B0 sin( 2 x 1011t + 300πx)
Where B0 = E0/C = 10-7 T
Q.14
Find the wave length of e.m. waves of frequency 5 x 1019 Hz. Give its two
applications. (2 marks)
Gamma rays, use (i) To detect flaws in metal castings
(ii) Medicinal uses
Q.15
Mention four properties of em waves.
(2 marks)
(i) They are produced by accelerating or oscillating charge.
(ii) They do not require material medium for propagation.
(iii) They follow the law of super position.
(iv) They propagate with speed of light in vacuum irrespective of their wavelength.
Q.16
Why are infrared radiations referred to as heat waves also? Name the radiations
which are next to these radiations in spectrum having (i) shorter wavelength and (ii)
longer wavelength.
(2 marks)
I.R. waves get produced by molecules of hot bodies.
(i) Visible
(ii) Microwaves.
Q.17
From the following, identify the e.m. waves having (i) maximum (ii) Minimum
frequency.
(2 marks)
(a) Radio waves (b)Gamma rays (c)Visible light (d)Microwaves
(e)U.V.rays (f)I.R.rays
Maximum frequency – Gamma rays
Minimum frequency – Radio waves
Q.18
Electromagnetic radiations with wavelength
(i)
λ1 are
used
to
kill
germs
in
water
(ii)
λ2 are
used
in
T.V
communication
(iii) λ2 plays an important role in maintaining the earth’s warmth.
purifiers
system
Name the part of e.m. spectrum to which these radiation belong. Arrange these
radiation in decreasing order of wavelength. (2 marks)
(i) U.V. (ii) Radio waves (iii) I.R radiation
Radio waves I.R, U.V.
Q.19
Draw a sketch of a plane e.m. wave propagating along x axis. Depict clearly the
directions of electric and magnetic field varying sinusoidally. (2 marks)
Q.20
Arrange the following electromagnetic radiation in ascending order of their
frequencies.
(a)Microwaves (b)Radio waves (c)X-rays (d) Gamma rays
Write two uses any one of this. (2 marks)
(R M I V U X Y)
Radio wave , micro wave , x rays, gamma rays
Uses – X-rays (i) radiography (ii) Crystal structure
Q.21
The magnitude of magnetic field in a plane e.m. wave is given as BX=0, BY=2 x107 sin(0.5 x 103 x + 1.5 x 1011t)T.
(3 marks)
(a)
Determine
the
wavelength
(b) Write an expression for the electric field.
BY = BO sin (1.5 x 1011t + 0.5 x 103x)T
BO = 2 x 10-7 T,
ω = 2πν = 1.5 x 1011
ω = 1.5 x 1011
K = 0.5 x 103
⇒
and
frequency
of
wave.
EX = 0 , EY = 0, EZ = CBO sin (1.5 x 1011t + 0.5 x 103x)
EZ = 60 sin (1.5x 1011t + 0.5 x 103x)
Q.22
Identify the following e.m. radiation as per wave length given below. Write one
application
of
each.
(3marks)
-3
-3
(i)10 nm
(ii)10 m
(iii)1nm
(i) Infrared – used in remotes
(ii) Microwave – used in Radar
(iii) U.V. rays – In water purifier.
Q.23
Give two uses each of (i) radio waves (ii) Microwaves.
(3 marks)
Radio waves – used in (a) Cellular phones communication
(b) T.V. communication
Microwaves – used in (a) RADAR
(b) Microwave ovens
Q.24
Identify the following electromagnetic radiations as given below Write one application
of each ? (3 marks)
(a) 1nm
(b)10-12m
(c)10-8m
(a) U.V rays – In water purifier.
(b) γ rays – medicinal use
(c) X rays – studying crystal structure
Q.25
Name the constituent radiation of e.m. spectrum which
(i) is used in satellite communication
(ii) is used for studying crystal structure
(iii) is similar to radiation emitted during decay of nuclei
(iv) has wavelength b/w 390nm and 770nm
(v) is absorbed from sunlight by ozone layer
(vi) produces intense heating effect
(i) Radio wave
(ii) X – ray
(iii) Γ – ray
(iv) Visible
(v) U.V.ray
(vi) I.R.rays
Q.26
Name the following constituent radiation of electromagnetic radiation which is
suitable for
(a) Radar
system
used
in
aircraft
navigation
(b) Treatment of cancer tumors
Write two application of each of them. (3 marks)
(a) Microwaves – Uses in automobiles as speed determination microwave ovens.
(b)
γ-rays
uses
(i)
To
detect
flaws
in
metal
(ii) Sterilization
castings
Q.27
A plane electromagnetic wave of frequency 25MHz travels in free space along the xdirection at a particular point in space and time
time. (3 marks)
Q.28
. Determine the
at this
Write the order of frequency range and one use each of the following e.m. radiation (i)
Micro waves (ii) U.V rays (iii) Gamma rays ? (3 marks)
(i) 1010Hz use :- Micro wave ovens
(ii) 1015Hz use :- In water purifier
(iii) 1022Hz use :- Medicinal use
Q.29
The electric field of plane an wave is given Ex=0 , Ey=0, Ez=0.5cos{2π x 108(t-4c)}
(i) What is dirn of propagation of wave?
(ii) Compute the components of magnetic fields.
(3 marks)
(i) dirn = xdirn/x-axis
(ii) Bx = 0 , By = Bo cos{2π x 108(t-x/c)} T
Bz = 0
Where Bo = 0.5 x 3 x 108 T
Q.30 Value based questions
1. A woman and her daughter of class XII in KV were in the kitchen, preparing a feast for visitors using
the new microwave oven purchased last evening. Suddenly, the daughter noticed sparks inside the oven
and unplugs the connection after switching it off. She found that inside the microwave oven a metallic
container had been kept to cook vegetable. She informs her mother that no metallic object must be used
while cooking in microwave oven and explains the reasons for the same.
a) What attitude of the daughter inspires you?
b) Give another use of a microwave oven
c) The frequency of microwave is 3 x 1011 Hz. Calculate its wavelength.
ANS:a) Presence of mind, Knowledge of subject;
b) Used in telecommunication; &
c) λ= ʋ /c = (3 x 1011))/ (3 x 108). = 10-3.m
2. Two persons were playing and one of them got injured as he fell from the top of a tree. He was taken
to the Hospital where, he was admitted for a fracture after X-ray. The boy’s mother, who on hearing the
above incident rushed to the hospital, was worried and restless; the other boy consoled her and the
doctor who, overheard them, informed her about the latest developments in the Medicine field and
assured a speedy recovery of her son.
a) How would you rate the qualities of the boy who consoled his friend’s mother?
b) How X-ray is produced?
c) Mention the fields where X-ray is used.
(ANS: caring for others, giving timely assistance, positive attitude; b) & c) Refer NCERT Text book)
3. Gopal visits his friend Naresh. In his house Naresh was playing with his kid sister and in spite of the
broad day light, Gopal notices the tube light burning and advises him to save electricity; He also claims
that the heat inside the room increases which would lead to global warming.
a) What are the values associated with the decision saving electricity?
b) Which electromagnetic wave is responsible for increase in the average temperature of the earth? Give
other applications of the electromagnetic wave.
ANS: a) concern for society/nation, awareness about global warming,
c) Infrared waves; applications- to treat muscular strain, solar water heaters & cookers)
4. Two friends were passing through the market. They saw two welders using the welding machine. One
welder was using the goggles face mask with window in order to protect his face. The other one was
welding with naked eyes. They went to the welder who was not using face mask and explained him the
advantages of using goggles and face masks. Next day, the welder bought asset of goggles and began to
do his work fearlessly.
a. What values were displaced by two friends?
b. Why do welders wear glass goggles or face masks with glass windows while carrying out welding?
Ans- a. Knowledge, creating awareness and social responsibilities.
b.Welder wear special glass goggles or face mask with glass windows to protect their eyes from large
amount of harmful UV radiation produced by welding arc.
5. Many people like to watch CID programme on a TV channel. In this programme, a murder mystery is
shown. This murder mystery is solved by CID team. Each member of the team works with full
dedication. They collect information and evidences from all possible sources and then tend to lead the
correct conclusion. Sometimes they also use ultraviolet rays in the forensic laboratory. Some people get
surprised to know the advantages of ultraviolet rays because they only aware of the fact that ultraviolet
rays coming from the sun produce harmful effects.
a. What values were displayed by the members of CID team?
b. What is the use of ultraviolet rays in forensic laboratory?
Ans. (a). Team spirit, sense of responsibilities and,(b). UV rays are used in the detection of forged
documents, finger prints, etc
RAY OPTICS AND OPTICAL INSTRUMENTS
GIST OF THE LESSON
Light is a form of energy that gives sense of vision to our eyes.
Light waves are electromagnetic waves, whose nature is transverse. The speed of light in vacuum is 3 x
108mls but it is different in different media.
The speed and wavelength of light change when it travels from one medium to another but its
frequency remains unchanged.
Reflection of Light
The ray of light is turned back into the same medium on striking a highly polished surface such as a
mirror, this phenomenon is called reflection of light.
Laws of Reflection
There are two laws of reflection.
(i) The angle of incidence (i) is always equal to the angle of reflection (r).
(ii) The incident ray, the reflected ray and the normal at the pointof incidence all three lie in the same
plane.
Different properties of image formed by plane mirror
 Size of image = Size of object
 Magnification = Unity
 Distance of image = Distance of object
 A plane mirror may form a virtual image.
 A man may see his full image in a mirror of half height of man.
 To see complete wall behind himself a person requires a plane mirror of at least one
third the height of wall. It should be noted that person is standing in the middle of the
room.
Sign Convention for Spherical Mirrors
1. All distances are measured from the pole of the mirror.
2. Distances measured in the direction of incident light rays are taken as positive.
3. Distances measured in opposite direction to the incident light rays are taken as negative.
4. Distances measured above the principal axis are positive.
5. Distances measured below the principal axis are negative.
Focal length of spherical mirrors
Rays parallel to principal axis after reflection from a concave/ convex mirror meet at a point or appear to
diverge from a point on principal axis called focus.
Focal Length The distance between the pole and focus is called focal length (f).
Relation between focal length and radius of curvature is given by
f = r/2
Image formation by a spherical mirror:
(i) The ray from the point which is parallel to the principal axis. The reflected ray goes through the
focus of the mirror.
(ii) The ray passing through the centre of curvature of a concave mirror or appearing to pass through it
for a convex mirror. The reflected ray simply retraces the path.
(iii) The ray passing through (or directed towards) the focus of the concave mirror or appearing to pass
through (or directed towards) the focus of a convex mirror. The reflected ray is parallel to the principal
axis.
(iv) The ray incident at any angle at the pole. The reflected
ray follows laws of reflection.
The mirror equation
Mirror formula is the relationship between object distance
(u), image distance (v) and focal length.
1 1 1
= +
f v u
Linear Magnification
The ratio of height of image (I) formed by a mirror to the height of the object (O) is called
linear magnification (m).
Linear magnification (m) = I/O = -v/u
Refraction of Light
The deviation of light rays from its path when it travels
from one transparent medium to another transparent
medium is called refraction of light.
Snell experimentally obtained the following laws of
refraction:
(i)
The incident ray, the refracted ray and the
normal to the interface at the point of incidence,
all lie in the same plane.
(ii)
The ratio of the sine of the angle of incidence to
the sine of angle of refraction is constant.n21 =
sin i
sin r
where n21 is a constant, called the refractive index of the second medium with respect to the first medium
.
If n21 is the refractive index of medium 2 with
respect to medium 1 and n12 the refractive index
of medium 1 with respect to medium 2,then it
should be clear that
1
n21 =
n12
Refractive Index: The ratio of speed of light in
c
vacuum (c) to the speed of light in any medium (u) is called refractive index of the medium . n21 = v
The refraction of light through the atmosphere is responsible for many interesting phenomena.
 The sun is visible a little before the actual sunrise and until a little after the actual sunset due to
refraction of light through the atmosphere.
 Refraction by the atmosphere makes the sun oval.
 Twinkling of stars is caused by the passing of light through different layers of a turbulent
atmosphere.
Critical Angle
The angle of incidence in a denser medium for which the angle of
refraction in rarer medium becomes 90°. is called critical angle
(C).
Critical angle for glass = 42°
Critical angle for water = 48°
Critical angle increases with temperature
Total Internal Reflection (TIR)
When a light ray travelling from a denser medium towards a rarer
medium is incident at the interface at an angle of incidence greater
than critical angle, then light rays reflected back in to the denser
medium. This phenomenon is called TIR.
Conditions necessary for TIR
Total internal reflection occurs
 if angle of incidence in denser medium exceeds critical angle.
 a ray of light enters from a denser medium to a rarer medium.
Relation between critical angle and refractive index
1
n12 = sin C.
Total internal reflection in nature and its technological applications
MIRAGE
(a) A tree is seen by an observer at its place when the air above the ground is at uniform temperature,
(b) When the layers of air close to the ground have varying temperature with hottest layers near the
ground, light from a distant tree may undergo total internal reflection, and the apparent image of the tree
may create an illusion to the observer that the tree is near a pool of water.
Prisms designed to bend rays by90º and 180º or to invert image without changing its size make use
of total internal reflection
Optical fibres
Optical fibres are fabricated with high quality composite glass/quartz fibres. Each fibre consists of a core
and cladding. The refractive index of the material of the core is higher than that of the cladding.
Optical fibres are extensively used for transmitting and receiving electrical signals which converted to
light by suitable transducers.
Optical fibres can also be used for transmission of optical signals. For example, these are used as a ‘light
pipe’ to facilitate visual examination of internal organ like esophagus, stomach and intestines.
It is available in decorative lamp with fine plastic fibres with their free ends forming a fountain like
structure.
REFRACTION AT SPHERICAL SURFACES:
n2 n1 n2 − n1
− =
v
u
R
Refraction through a lens
1
n2
1
1
[ = ( − 1) ( − )]
f
n1
R1 R 2
Magnification(m) produced by a lens is defined, as the ratio of the size of the image to that of the
object.
When we apply the sign convention, we see that, for erect (and virtual) image formed by a convex or
concave lens, m is positive, while for an inverted (and real) image, m is negative.
Linear magnification (m) = I/O = v/u
Power of a lens
The power P of a lens is defined as the tangent of the angle by which it
converges or diverges a beam of light falling at unit distant from the optical
centre.
1
tan δ =
f
1
P=
f
The SI unit for power of a lens is dioptre (D): 1D = 1m–1. The power of
a lens of focal length of 1 metre is one dioptre. Power of a lens is positive for a converging lens and
negative for a diverging lens
Combination of thin lenses in contact:
Two lenses A and B of focal length f1 and f2 placed in contact with each other.
The effective focal length f of their combination is given by
1 1 1
= +
f f1 f2
The effective power P of their combination is given by
P = P1 + P2
The total magnification m of the combination is
m = m1 × m2
If lens of refractive index n2 is placed in a medium of refractive index n1.
A convex lens will behave as convex if n2>n1
A concave lens will behave as concave if n2>n1
A convex lens will behave as concave if n2 < n1
A concave lens will behave as convex if n2 < n1
A convex lens will behave as plane glass sheet if
n2 = n1( no refraction will occur)
A concave lens will behave as plane glass sheet
if n2 = n1( no refraction will occur)
Refraction through prism
Prism is uniform transparent medium bounded between two refracting
surfaces, inclined at an angle.
Angle of Deviation
The angle sub tended between the direction of incident light ray and
emergent light ray from a prism is called angle of deviation (δ).
Prism Formula
A+δm
⌈n21 =
sin (
2
A
sin 2
)
⌉
DISPERSION BY A PRISM
When a narrow beam of sunlight, usually called white light, is
incident on a glass prism, the emergent light is seen to be consisting
of several colours violet, indigo, blue, green, yellow, orange and
red (given by the acronym VIBGYOR). The red light bends the
least, while the violet light bends the most. The phenomenon of
splitting of light into its component colours is known as
dispersion. The pattern of colour components of light is called the
spectrum of light.
 Red light is at the long wavelength end (~700 nm) while the violet light is at the short
wavelength end (~ 400 nm).
 Dispersion takes place because the refractive index of medium for different wavelengths
(colours) is different.
 Red light travels faster than violet light in a glass prism.
Scattering of light
Blue colour of sky:
 As sunlight travels through the earth’s atmosphere, it gets scattered (changes its direction) by the
atmospheric particles.
 Light of shorter wavelengths is scattered much more than light of longer wavelengths.
 The amount of scattering is inversely proportional to the fourth power of the wavelength. This is
known as Rayleigh scattering.
Hence, the bluishcolour predominates in a clear sky, since blue has a shorter wavelengththan red and
is scattered much more strongly.
Clouds are generally white
 Clouds have droplets of water with a >> λ
 large scattering objects (for example, raindrops, large dust or ice) all wavelengths are scattered
nearly equally.
 Hence clouds are generally white
Reddish appearance of the sun and full moon near the horizon.
 At sunset or sunrise, the sun’s rays have to pass through a larger distance in the atmosphere
 Most of the blue and other shorter wavelengths are removed by scattering.
 The least scattered light reaching our eyes, therefore, the sun looks reddish.
Optical instruments
Simple Microscope
It is used for observing magnified images of objects. It is consists of a converging lens of small focal
length.
Magnifying Power
(i) When final image is formed at least
distance of distinct vision (d) then 𝑚 =
1+𝑑
𝑓
where, f= focal length of the lens.
(ii) When final image is formed at infinity, then M
= d/f
Compound Microscope
It is a combination of two convex lenses called objective lens and eye piece separated by a distance.
Both lenses are of small focal lengths but fo< fe, where fo and fe are focal lengths of objective lens and
eye piece respectively.
Astronomical(refracting telescope) Telescope
The telescope is used to provide angular magnification of distant objects. It also has an objective and an
eyepiece. But here, the objective has a large focal length and a much larger aperture than the eyepiece.
Magnifying power
When final image is formed at infinity,
𝑚=
𝑓𝑜
𝑓𝑒
Length of telescope tube is 𝐿 = 𝑓0 + 𝑓𝑒
Aberration of Lenses
The image formed by the lens suffer from following two drawbacks
(i)
Spherical Aberration Aberration of the lens due to which the rays passes through the lens
are not focussed at a single and the image of a point object placed on the axis is blurred called
spherical aberration.
(ii)
Chromatic AberrationImage of a white object formed by lens is usually coloured
andblurred. This defect of the image produced by lens is called chromatic aberration.
Reflecting telescope
Telescopes with mirror objectives are called reflecting telescopes.
They have several advantages. First, there is no chromatic aberration in a mirror. Second, if a parabolic
reflecting surface is chosen, spherical aberration is also removed.
WAVE OPTICS
Wavefront:
A wave front is the locus of points having the same phase of oscillation. Rays are the lines perpendicular
to the wavefront, which show the direction of propagation of energy. The time taken for light to travel
from one wavefront to another is the same along any ray.
Huygens’ Principle.
According to Huygens’
(a) Each point on the given wave front (called primary wave front) acts as a fresh source of new
disturbance, called secondary wavelet, which travels in all directions with the velocity of light in the
medium
(b) A surface touching these secondary wavelets, tangentially in the forward direction at any instant
gives the new wavefront at that instant. This is called secondary wave front.
Doppler effect
Doppler effect is the shift in frequency of light when there is a relative motion between the source and
the observer. The effect can be used to measure the speed of an approaching or receding object.
𝛥𝜈 𝑣𝑟
=
𝜈0
𝑐
Coherent and Incoherent Addition of Waves.
Two sources are coherent if they have the same frequency and a stable phase difference.
Two sodium lamps illuminating two pinholes we will not observe any interference fringes. This is
because of the fact that the light wave emitted from an ordinary source (like a sodium lamp) undergoes
abrupt phase changes in times of the order of 10–10 seconds. Thus the light waves coming out from two
independent sources of light will not have any fixed phase relationship and would be incoherent,
The interference term averaged over many cycles is zero if
(a) The sources have different frequencies; or
(b) The sources have the same frequency but no stable phase difference.
Diffraction
Diffraction refers to light spreading out from narrow holes and slits, and bending around corners and
obstacles. Different parts of the wavefront at the slit act as secondary sources: diffraction pattern is the
result of interference of waves from these sources.
Diffraction is a general characteristic exhibited by all types of waves, be it sound waves, light waves,
water waves or matter waves. Since the wavelength of light is much smaller than the dimensions of most
obstacles; we do not encounter diffraction effects of light in everyday observations.
Resolving Power
The ability of an optical instrument to produce separate and clear images of two nearby objects,
is called its resolving power
Limit of Resolution
The minimum distance between two nearby objects which can be just resolved by the
instrument, is called its limit of resolution (d).
Resolving power of a microscope = 1/d = 2 n sin θ / λ
Where, d = limit of resolution, λ = wavelength of light used.
μ = refractive index of the medium between the objects and objective lens and θ = half of the cone
angle.
Resolving power of a telescope = 1/dθ = d/1.22 λ
where, dθ = limit of resolution, A = wavelength of light used and d =
diameter of aperture of objective
Polarisation:
The phenomenon of restricting the vibrations of light in a particular direction, perpendicular to direction
of wave motion is called polarisation
The plane ABCD in which vibrations are present is called plane of vibration and plane EFGH whichis
perpendicular to plane of vibration is called plane of polarization.
Law of Malus: When a beam of completely plane polarized is incident on an analyser, the resultant
intensity of light transmitted through the analyser is given by
𝐼 = 𝐼0 𝑐𝑜𝑠 2 θ
where θ is the angle between plane of transmission of analyser and polarizer.
Polarisation by scattering:
Polarisation of the blue scattered light from the sky.The incident sunlight is unpolarised (dots and
arrows). A typical molecule is shown. It scatters light by 90º polarised normal to the plane of the paper
(dots only).
Polarisation by reflection:
When unpolarised light is incident on the boundary between two transparent media, the reflected light is
polarised with its electric vector perpendicular to the plane of incidence when the refracted and reflected
rays make a right angle with each other.
The angle of incidence in this case is called Brewster’s angle and is denoted by ip. We can see that ip is
related to the refractive index of the denser medium. Since we have ip+r = π/2, we get
𝑠𝑖𝑛 𝑖
from Snell’s law
𝑛21 =
𝑠𝑖𝑛 𝑟
if i=ip
r = 90-ip
𝑛21 =
𝑠𝑖𝑛 𝑖𝑝
𝑠𝑖𝑛(90 − 𝑖𝑝 )
=
𝑠𝑖𝑛𝑖𝑝
𝑐𝑜𝑠𝑖𝑝
𝑛21 = tan 𝑖𝑝 (BREWESTER’S LAW)
IMPORTANTDERIVATIONS
Mirror formula.
The figure shows an object AB at a distance u from the pole of a
concave mirror. The image A1B1 is formed at a distance v from the
mirror. The position of the image is obtained by drawing a ray
diagram.
But ED = AB
From equations (1) and (2)
If D is very close to P then EF = PF
But PC = R, PB = u, PB1 = v, PF = f
By sign convention
PC = -R, PB = -u, PF = -f and PB1 = -v
Equation (3) can be written as
R= 2 f
Dividing equation (4) throughout by uvf we get
Relation between critical angle and refracive index of the medium.
Ray of light is travelling from
denser(water) to rarer (air)medium.
According to Snells law
𝑠𝑖𝑛 𝑖
𝑛12 =
𝑠𝑖𝑛 𝑟
If angle of incidence i =C
Then r= 900
𝑠𝑖𝑛 𝑐
1
𝑛12 =
𝑜𝑟 ⌊𝑛21 =
⌋
𝑠𝑖𝑛 90
𝑠𝑖𝑛 𝑐
Refraction at spherical surface
O : Point object, OP = u = Object
Distance
P : Pole, C : centre of curvature, PC
= R = Radius of curvature I : Point
Image, PI = v = Image distance
i, r, α, β and γ are small enough for
sin i= i= tan i to hold .
n1 and n2 refractive indices of rarer
& denser media (with respect to
vacuum or air as incidence
medium) ;
𝑛2
𝑛1
𝑛2 𝑠𝑖𝑛 𝑖 𝑖
=
=
=
𝑛1 𝑠𝑖𝑛 𝑟 𝑟
𝑛21 =
𝑛21
𝑛1 𝑖 = 𝑛2 𝑟
In triangle MOC 𝑖 = 𝛼 + 𝛾
In triangle MCI 𝛾 = 𝑟 + 𝛽
𝑛1 (𝛼 + 𝛾) = 𝑛2 (𝛾 − 𝛽) _______________________(1)
𝑀𝑃
𝑀𝑃
𝑀𝑃
Where 𝛼 = 𝑃𝑂 , 𝛽 = 𝑃𝐼 and 𝛾 = 𝑃𝐶
Substituting values of α, β and γ in eq. (1)
𝑀𝑃
𝑀𝑃
𝑀𝑃
𝑀𝑃
𝑛1 ( 𝑃𝑂 + 𝑃𝐶 ) = 𝑛2 ( 𝑃𝐶 − 𝑃𝐼 )_____________________(2)
Where PO = - u, PI = -v and PC = R
Substituting these values in eq.(2)
𝑛2 𝑛1 𝑛2 − 𝑛1
⌈ −
=
⌉
𝑣
𝑢
𝑅
Lens Makers formula for a thin convex lens
Consider a convex lens (or concave lens) of absolute
refractive index n 2 to be placed in a rarer medium of
absolute refractive index n1.
Considering the refraction of a point object on the surface
the image is formed at I1 which is at a distance
𝑣 ′ and object distance is u.
CI1= P1I1 = 𝑣 ′ (as the lens is thin)
CC1 = P1C1 = R1
CO = P1O = u
It follows from the refraction due to convex spherical surface XP1Y
𝑛2
𝑣′
−
𝑛1
𝑢
=
𝑛2 −𝑛1
𝑅1
____________________________(1)
The refracted ray from A suffers a second refraction on the surface XP2Y and emerges along BI.
Here the object distance is 𝑣 ′ and image distance is 𝑣Therefore I is the final real image of O.
CI1= P2I1 = 𝑣 ′ (as the lens is thin)
CC2 = P2C2 = R2
CI = P2I = 𝑣
𝑛1
𝑣
𝑛
− 𝑣2′ =
𝑛1 −𝑛2
𝑅2
______________________________(2)
Adding (1) & (2)
𝑛1
𝑣
−
𝑛1
𝑢
1
1
= (𝑛2 − 𝑛1 ) (𝑅 − 𝑅 ) ____________________(3)
1
1
1
2
1
But 𝑓 = 𝑣 − 𝑢
1
𝑛
1
1
Hence[𝑓 = (𝑛2 − 1) (𝑅 − 𝑅 )]
1
1
2
XP1Y,
Refraction through prism
The passage of light incident from air into a glass prism is deviated due to refraction occurring twice, once at
the boundary separating air-class and next at the boundary separating glass-air as shown in fig
𝑠𝑖𝑛 𝑖
According to Snells Law 𝑛21 = 𝑠𝑖𝑛 𝑟_________________(1)
In cyclic quadrilateral ALOM
A + LOM = 180º ___________________________(2)
From the triangle LOM
r1 + r + LOM = 180º _____________________(3)
𝐴
Hence A = 2r𝑜𝑟 ⌈ 𝑟 = 2 ⌉ __________________(4)
Comparing these two equations, we get
𝑟1 + 𝑟2 = 𝐴
In minimum deviation position
r1 = r = r
𝐴
𝑜𝑟 ⌈ 𝑟 = 2 ⌉________________________________(5)
The total deviation δ is the sum of deviations at the two faces,
δ = (i – r1 ) + (e – r ) that is,
δ =i + e – A
In minimum deviation position 𝛿 = 𝛿𝑚 & i = e
Hence 𝛿𝑚 = 2i –A
𝐴+𝛿𝑚
𝑖=(
2
)_________________________________(6)
Substituting values of i and r in eq. (1)
𝐴+𝛿𝑚
⌈𝑛21 =
𝑠𝑖𝑛 (
2
𝐴
𝑠𝑖𝑛 2
)
⌉
HUYGENS PRINCIPLE
 Each point of the wave front is the source of a secondary disturbance and the wavelets emanating from
these points spread out in all direction with the speed of wave.
 These wavelets are referred to as secondary wavelets.
 The common tangent to the secondary wavelets gives new position of wave front at a later time.
Reflection on the basis of Wave Theory
Let t be the time in which the incident wave front reach from Q to P’ in the same time the reflected wave
front starting from P reaches Q’
QP’=PQ’=v t
In triangle PQP’ and triangle PQ’P’




Angle PQP’= Angle PQ’P’= 90
PP’ is the common side
QP’=PQ’=v t
Triangle PQP’≈ Triangle PQ’P’
AngleQPP’ = AngleQ’PP’
i = r
Refraction on the basis of wave theory
Let t be the time in which the incident wave front reach from Q to P’ in the same time the refracted wave
front starting from P reaches Q’ such that QP’=C1t and PQ’=C2t
In triangle PQP’
𝑠𝑖𝑛 𝑖 =
In triangle
𝑄𝑃′ 𝐶1 𝑡
=
_________(1)
𝑃𝑃′ 𝑃𝑃′
𝑠𝑖𝑛 𝑟 =
𝑃𝑄 ′ 𝐶12 𝑡
=
__________(2)
𝑃𝑃′ 𝑃𝑃′
Hence from eq. (1) and (2)
[
𝑠𝑖𝑛 𝑖 𝐶1 𝑛2
=
=
= 𝑛21 ]
𝑠𝑖𝑛 𝑟 𝐶2 𝑛1
INTERFERENCE OF LIGHT
CONDITION FORCONSTRUCTIVE AND DESTRUCTIVE INTERFERENCE OF LIGHT
Let the waves from two coherent source of light be represented by
𝑦1 = 𝑎 𝑐𝑜𝑠 𝜔𝑡__________________(1)
𝑦2 = 𝑎 𝑐𝑜𝑠(𝜔𝑡 + ∅)______________(2)
According to principal of superposition, resultant displacement is given by
y = y1 + y2
= a[𝑐𝑜𝑠 𝜔𝑡 + 𝑐𝑜𝑠 (𝜔𝑡 + ∅]
= 2 a cos∅ cos (ωt + ∅/2)
The amplitude of the resultant displacement is 2a cos (∅/2) and therefore the intensity at that point will
be
𝐼 = 4𝑎2 𝑐𝑜𝑠 2 ∅ /2) = 4I0
∅ /2)_________________(3)
Where ∅ = 0, ±2𝜋, ±4𝜋 … … ..corresponds to constructive interference leading to maximum intensity.
Hence condition for constructive interference ∅ = 2𝑛𝜋_____________(4)
Where ∅ = ±𝜋, ±3𝜋 … … ..corresponds to destructive interference leading to minimum intensity.
(2𝑛+1)𝜋
Hence condition for destructive interference ∅ =
_________(5)
2
Young’s double slit experiment and fringe width
Light from the two coherent sources (S1&S2) superimpose to produce interference pattern on the screen placed
at a distance D from the plane of the slit.
The intensity of light reaching at P depends on path difference between the waves reaching at the point.
The path difference between the waves reaching at P = 𝑆2 𝑃 − 𝑆1 𝑃
𝐷 2
𝐷 2
(𝑆2 𝑃)2 − (𝑆1 𝑃)2 = 𝐷2 + (𝑦 + ) − 𝐷2 − (𝑦 − ) = 2𝑦𝑑
2
2
𝑆2 𝑃 − 𝑆1 𝑃 =
2𝑦𝑑
2𝑦𝑑
=
𝑆2 𝑃 + 𝑆1 𝑃
2𝐷
path difference ∆𝑥 =
𝒚=
𝒏𝝀𝑫
𝒅
𝐷
= 𝑛𝜆 for constructive interference
𝒘𝒉𝒆𝒓𝒆 𝒏 = 𝟎, 𝟏, 𝟐, 𝟑 … …position of bright band
path difference ∆𝑥 =
𝒚=
𝑦𝑑
(𝟐𝒏+𝟏)𝝀𝑫
𝟐𝒅
𝑦𝑑
𝐷
=
(2𝑛+1)𝜆
2
for destructive interference
𝒘𝒉𝒆𝒓𝒆 𝒏 = 𝟏, 𝟐, 𝟑 … …position of dark band
Distance between two consecutive dark bands or two consecutive bright bands is called fringe width (β).
𝜷 = 𝒚𝒏 − 𝒚𝒏−𝟏 =
𝝀𝑫
𝒅
Diffraction at single slit
The phenomenon of bending of light around the corners of an obstacles or aperture.
Condition for Diffraction – The size of obstacle or aperture should be of the order of the wavelength of
thelight.
Condition for secondary Minima
asinθn = nλ
a -Size of aperture, n –order of minima, λ –wavelength of light used.
Condition for secondary Maxima –
asinθn = (2n +1)λ/2
a -Size of aperture, n –order of minima, λ –wavelength of light used.
Width of fringes = β = Dλ/a
Width of central Bright fringe = 2 β = 2 Dλ/a
QUESTIONS
1Mark Questions
1. Write the relation between angle of incidence I, angle of prism A and angle of minimum deviation 𝛿𝑚
for a glass prism.
Ans: 2i = A + 𝛿𝑚
2. A concave mirror, of aperture 4cm, has a point object placed on its principal axis at a distance of 10cm
from the mirror. The image, formed by the mirror, is not likely to be a sharp image. State the likely
reason for the same.
Ans: The incident rays are not likely to be paraxial.
3. How does power of lens vary when incident red light is replaced by blue light?
Ans: Wavelength decreases hence power increases
4. An object is held at the principal focus of a concave lens of focal length f. Where is the image formed?
Ans: That is image will be formed between optical centre and focus of lens; towards the side of the object.
5. What is the geometrical shape of the wavefront when a plane wave passes through a convex lens?
Ans: The wavefront is spherical of decreasing radius.
6. What is the angle between the plane of polariser and analyser, in order that the intensity of transmitted
through analyser reduces to half?
Ans: 450
7. A diverging lens of focal length ‘F’ is cut into two identical parts each forming a plano-concave lens.
What is the focal length of each part?
Ans: Focal length of each half part will be twice the focal length of initial diverging lens.
8. How the angular separation of interference fringes in Young’s double slit experiment change when the
distance between the slits and screen is doubled?
Ans: Angular separation between fringes, θ = λ/d where λ = wavelength, d = separation between coherent
sources. So, θ is independent of distance between the slits and screen. So angular separation (θ ) will
remain unchanged.
9. Two thin lenses of power +6 D and – 2 D are in contact. What is the focal length of the combination?
Ans: Net power of lens combination P = P1 + P2 = + 6 D - 2 D = + 4 D
∴ Focal length, f = 1/P = ¼ m = 25 cm
10. How does the angular separation between fringes in single-slit diffraction experiment change when the
distance of separation between the slit and screen is doubled?
Ans: Angular separation is θ = β / D = λ / d
Since θ is independent of D, angular separation would remain same.
12. How does the fringe width, in Young’s double-slit experiment, change when the distance of separation
between the slits and screen is doubled?
Ans: The fringe width is, β = D λ / d
If D (distance between slits and screen) is doubled, then fringe width will be doubled.
13. When light travels from a rarer to a denser medium, the speed decreases. Does this decrease in speed
imply a decrease in the energy carried by the light wave? Justify your answer.
Ans: No; when light travels from a rarer to denser medium, its frequency remains unchanged.
According to quantum theory, the energy of a light beam depends on frequency and not on speed.
14. For the same value of angle incidence, the angles of refraction in three media A, B and C are 15°, 25°
and 35° respectively. In which medium would the velocity of light be minimum?
Ans: From Snell's law, n = sin i/sin r = c/v
For given i, v α sin r ; r is minimum in medium A, so velocity of light is minimum in medium A.
15. How does resolving power of telescope change if the incident yellow light is replaced by blue light?
Ans: Resolving power = D / 1.22 λ
𝝀𝒃 < 𝝀𝒚 Hence RP decreases
16. When monochromatic light is incident on a surface separating two media, the reflected and refracted
light both have the samefrequency as the incident frequency. Explain why?
Ans: Reflection and refraction arise through interaction of incident light with the atomic constituents of
matter. Atoms may be viewed as oscillators, which take up the frequency incident light. Thus, the
frequency of scattered light equals the frequency of incident light.
17. Unpolarised light is incident on a plane glass surface. What should be the angle of incidence so that the
reflected and refracted rays are perpendicular to each other?
Ans:For i + r to be equal to 𝜋/2, we should have tan iB
This gives iB = 57°. This is the Brewster’s angle for air to glass interface.
18. In a single-slit diffraction experiment, the width of the slit is made double the original width. How does
this affect the size and intensity of the central diffraction band?
Ans: In single slit diffraction experiment fringe width is, β = 2Dλ / d
If d is doubled, the width of central maxima is halved. Thus size of central maxima is reduced to half.
Intensity of diffraction pattern varies square of slit width. So, when the slit gets double, it makes the
intensity four times.
19. Under what condition does a convex lens of glass having certain refractive index, acts as a plane glass
sheet?
Ans: When refractive index of lens is equal to refractive index of liquid.
20. You are given following three lenses. Which two lens you will use to make objective and eyepiece of an
astronomical telescope?
LENS
POWER
APERTURE
L1
3D
8cm
L2
6D
1cm
L3
10D
1cm
Ans:
L1 as objective.
L3 as eyepiece
21.
The line AB in the ray diagram represents a lens. State whether the lens is convex or concave?
Ans: AB represents concave lens.
22. A convex lens is placed in contact with a plane mirror. A point object at a distance of 20 cm on the axis
of this combination has its image coinciding with itself. What is the focal length of the lens?
Ans: Since image coincides with object, it implies that ray must be falling normally on the plane mirror.
This implies that the ray after passing through lens becomes parallel. So, object must be at the focus of
lens. So, focal length of lens = 20 cm.
23.‘Two independent monochromatic sources of light cannot produce a sustained interference pattern’. Give
reason.
Ans:In independent monochromatic sources phase difference changes at a rate of 108 Hz. Hence, the
interference pattern obtained also fluctuates with 108 Hz and therefore, it is not sustainable as result of
persistence of vision.
Q24. A concave lens of refractive index 1.5 is immersed in a medium of refractive index 1.65. What is the
nature of the lens?
Ans: Since  for lens. <for surrounding. It behaves like converging lens.
2 Mark Questions
1. An object AB is kept in front of a concave mirror as shown in the fig.
(i)
Draw ray diagram showing image formation by the concave
mirror
(ii)
How will the position and intensity of the image be affected if the
lower half of the mirror’s reflecting surface is painted black?
Ans: (i) Image formed will be inverted diminished between C and F.
(iii)
No change in position of image and its intensity will get reduced.
2. Draw a labeled ray diagram to show the image formation in a refracting type astronomical telescope.
Why should the diameter of the objective of a telescope be large?
Ans: For large light gathering power and higher resolution, the
diameter of the objective should be large.
3. Define resolving power of a compound microscope. How does
resolving power of a compound microscope change when
the
(i) Refractive index of the medium between the object and objective lens increases?
(ii) Wavelength of the radiation used is increased?
Ans: Resolving power of a microscope is defined as the reciprocal of the minimum separation of two
points seen distinctly.
Resolving power = 2 n sinθ / 1.22 λ 232
(i) Increase in the refractive index (n) of the medium increases resolving power because RP α n
(ii) On increasing the wavelength of the radiation, resolving power decreases because RP α 1/λ
4. Define resolving power of a telescope. How does it get affected on
(i) Increasing the aperture of the objective lens?
(ii) Increasing the focal length of the objective lens?
Ans: Resolving power of a telescope is defined as the reciprocal of the smallest angular separation
between two distant objects.
Resolving power = D / 1.22 λ where is aperture of the objective lens
(i) Resolving power increases on increasing the aperture of the objective lens, since RP α D.
(ii) Resolving power does not get affected on increasing the focal length of objective lens, since RP is
independent of focal length.
5. How will the angular separation and visibility of fringes in Young’s double slit experiment change when
(i) screen is moved away from the plane of the slits, and
(ii) width of the source slit is increased?
Ans. (i) Angular separation = β / D = λ/d
It is independent of D;
β = D λ/d increases, so visibility of fringes increases.
(ii) Remains unchanged but fringes becomes less and less sharp; so visibility of fringes decreases. If the
condition s/S =λ/d is not satisfied, the interference pattern disappears.
6. In single slit diffraction pattern a slit of width‘d’ is illuminated with red light of wavelength 650nm. For
what value of‘d’ will
(i)
The first minima fall at an angle of diffraction of 300.
(ii)
The first maxima fall at an angle of diffraction of 300.
Ans. For first minima 𝑑𝑠𝑖𝑛𝜃 = 𝜆
Substituting value of 𝜃and λ , d = 1300nm
For first maxima 𝑑𝑠𝑖𝑛𝜃 = 𝜆
Substituting value of 𝜃and λ , d = 1950nm
7. Two convex lens of same focal length but of aperture A1 and A2 (A2<A1), are used as objective lenses
in two astronomical telescopes having identical lenses. What is the ratio of their resolving power?
Which telescope will you prefer and why?
𝐴
Ans R=1.22𝜆
𝑅1 𝐴1
=
𝑅2 𝐴2
A1 because it has greater resolving power
8. A convex lens of focal length 10 cm is placed co-axially 5cm away from a concave lens of focal length
10cm. If an object is placed 30cm in front of the convex lens, find the position of the final image formed
by the combined system.
AnsFor convex lens u1=-30cm, f1= +10cm
1
1
1
= −
𝑓1 𝑣1 𝑢1
V1=15cm
U2= 15 - 5=10cm,f2=-10cm Hence v2= ∞ (on the left of second lens)
9. (i) State the principle on which the working of an optical fiber is based.
(ii) What are the necessary conditions for this phenomenon to occur?
Ans: (i) The working of optical fiber is based on total internal reflection.
Statement: When a light ray goes from denser to rarer medium at an angle greater than critical angle,
the ray is totally reflected in first (denser) medium. This phenomenon is called total internal reflection.
(ii) Conditions:
(a) Ray of light must go from denser medium to rarer medium.
(b) Angle of incidence must be greater than critical angle (i. e., i > C).
10. Draw a labelled ray diagram of a reflecting telescope. Mention its two advantages over the refracting
telescope.
Ans: Ray Diagram
Advantages:
(i) It is free from chromatic and spherical aberrations.
(ii) Its resolving power is greater than refracting telescope due to larger aperture of mirror.
11. Write down the conditions to obtain the sustained interference fringe pattern of light. What is the effect
on the interference fringes in Young’s double slit experiment, when monochromatic source is replaced
by a source of white light?
Ans: Conditions for sustained interference
(i)
The two sources of light must be coherent to emit light of constant phase difference.
(ii)
The amplitude of electric field vector of interfering wave should be equal to have greater
contrast between intensity of constructive and destructive interference.
When monochromatic light is replaced by white light, then coloured fringe pattern is obtained on the
screen.
12. State briefly two features which can distinguish the characteristic features of an interference pattern
from those observed in diffraction pattern.
Ans:
S.No Interference Pattern
Diffraction Pattern
1.
All the bright bands are of same
Bright bands are not of same intensity.
intensity.
2.
Intensity of minima is very small or zero. The intensity of minima is never zero. There
There is a good contrast between bright
is poor contrast between bright and dark
and dark bands.
bands.
13. Draw a ray diagram of compound microscope. Write the expression for its magnifying power.
𝑣
1+𝐷
Ans: 𝑚 = 𝑢𝑜 ( 𝑓 )
𝑜
𝑒
5 Mark Questions
Q1.
(i)Draw a ray diagram to show refraction of a ray of monochromatic light passing through a glass prism.
Deduce the expression for the refractive index of glass in terms of angle of prism and angle of minimum
deviation.
(ii)Explain briefly how the phenomenon of total internal reflection is used in fibre optics.
Q2.
Trace the rays of light showing the formation of an image due to a point object placed on the axis of a
spherical surface separating the two media of refractive indices n1 and n2 .Establish the relation between
the distance of the object, the distance of image and the radius of curvature from the central point of
spherical surface. Hence, derive the expression of the lens maker’s formula.
Q3.
(i) Draw a ray diagram for formation of image of a point object by a thin double convex lens having
radii of curvatures R1 and R2 and hence, derive lens maker’s formula.
(ii)Define power of a lens and give its SI units. If a convex lens of length 50 cm is placed in contact
coaxially with a concave lens of focal length 20 cm, what is the power of the combination?
Q4.
(i) Use Huygens geometrical construction to show how a plane wave front at t=0 propagates and
produces a wave front at a later time.
(ii) Verify, using Huygens principle, snell’s law of refraction of a plane wave propagating from a denser
to a rarer medium.
(iii) When monochromatic light is incident on a surface separation two media, the reflected and refracted
light both have the same frequency. Explain why?
Q5.
(i) What are coherent sources? Why they are necessary for observing sustained coherent sources
obtained in the Young’s double slit experiment?
(ii) Show that the superposition of the waves originating from the two coherent sources,S1 and S2 having
displament,Y1=a cos𝜔t and Y2 = a cos(𝜔t+Ø)at a point produce a resultant intensity,
Hence, write the conditions for the appearance of dark and bright fringes.
Q7.
In a Young’s double slit experiment,
(i)
Deduce the conditions for constructive and destructive interference. Hence, write the expression
for the distance between two consecutive bright or dark fringes.
(ii)
What change in the interferences pattern do you observe, if the two slits, S1and S2 are taken as
point sources?
(iii)
Plot a graph of the intensity distribution vs. path difference in this experiment. Compare this with
the intensity distribution of fringes due to difference do you observe?
(i) How does an unpolarised light incident on a Polaroid gets polarized?
Q8.
Describe briefly, with the help of a necessary diagram, the polarization of light by reflection from a
transparent medium.
(ii) Two Polaroid’s, A and B are kept in crossed positions. How a third Polaroid, C should be placed
between them so that the intensity of polarized light transmitted by Polaroid reduce to 1/8th of the
intensity of unpolarised light incident on A?
Q9.
(i) Obtain the conditions for the bright and dark fringes in diffraction pattern due to a single narrow slit
illuminated by monochromatic source.
Explain Cleary why the secondary maxima go on becoming weaker with increasing order?
(ii)When the width of the slit is made double, how would this affect the size and intensity of the central
diffraction band? Justify your answer.
Q10.
(i) Draw the ray diagram for the formation of image of an object by a convex mirror and use it
(along with the sign convention) to derive the mirror formula.
(ii)Use the mirror formula to show that for an object, kept between the pole and focus of a concave
mirror, the image appears to be formed behind the mirror.
Value Based Questions With Answers (Unit – Optics)
1. Shweta’s grandmother often complains of headache. Shweta asked her to visit an eye specialist for a
check up, but she refused saying that her eye sight is O.K. Some other day, her grandmother asked
Shweta to thread a needle. Shweta understood her problem and took her to the eye specialist who
prescribed her spectacles of suitable power.
Read the above passage and answer the following questions:
What could Shweta make out?
Can you guess the nature of lens prescribed?
What values are displayed by Shweta?
Answer:
a) Shweta could make out that her grandmother is suffering from hypermetropia have difficulty in
viewing nearby objects.
b) Yes, the eye specialist must have prescribed a convex lens of suitable power.
c) Shweta has displayed concern for the health of her grandmother in particular and senior citizens in
general. Like kids, elderly people must be provided care with love.
2. Mona and Anushka are friends, both studying in class 12. Mona is in Science stream and Anushka is in Arts
stream. Both of them go to market to purchase sunglasses. Anushka feels that any coloured glasses with
fancy look are good enough. Mona tells her to look for UV protection glasses, Polaroid glasses and photo
sensitive glasses.
Read the above passage and answer the following questions:
a) What are UV protection glasses, Polaroid glasses and photo sensitive glasses?
b) What values are displayed by Mona?
Answer:
a) UV protection glasses are those which filter ultra-violet rays that are harmful to our eyes. Polaroid
glasses help in reducing the glare. Photo-sensitive glasses get darker in strong day light. They protect
our eyes from strong sunlight especially at noon.
b) Mona has displayed concern for her friend. She has put to use the knowledge she acquired in her
science classes. Mugging up things for examinations is of no use. What we are taught in class room
must be used in practice.
3. The rays of light falling on a convex lens in a direction parallel to principal axis of the lens, get refracted
through the lens and meet actually at a single point F on the principal axis of the lens. This point is called
principal focus of the lens.
Read the above passage and answer the following questions:
a) Is principal focus of a convex lens, a real point? Is the same true for a concave lens?
b) A distinct image of a distant tree is obtained on a screen held at 40cm from a convex lens. What is its
focal length?
c) Our teachers and parents advise us to stay focused. What does it imply?
Answer:
a) Yes, the principal focus of a convex lens is a real point. This is because rays refracted through
convex lens meet actually at this point.
b) F=distance of screen from the lens = 40cm
c) It implies that we concentrate all out energies/efforts at a single point/problem so that we can resolve
the same easily. Staying focused means that we do not divert our energies and attention to several things
at a time. This would lead us nowhere. Thus, the secret of success is to stay focussed.
4. The formula governing reflection of light from a spherical mirror is :
1 1 1
= +
𝑓 𝑣 𝑢
This is known as mirror formula and is applicable equally to concave mirror and convex mirror. The
linear magnification of the mirror is given by (m) = I/O = -v/u
Read the above passage and answer the following questions:
a) An object is held at a distance of 30cm in front of a concave mirror of radius of curvature 40cm.
Calculate distance of the image from the object? What is linear magnification of the mirror?
b) The object is moved to a distance of 40cm in front of the mirror. How is focal length of mirror
affected?
c) What values of life do you learn from the mirror formula?
Answer:
a) Here, u = - 30cm, R = - 40cm, v ?
Substituting in mirror formula
v = 60-30=30cm, on the same side as object.
Magnification, m = -2 (negative sign for inverted image)
b) Focal length (f) of mirror remains unaffected. On changing u; v changes and not f.
c) Mirror formula reveals that f depends only on R, and not on ‘u’ pr ‘v’. In fact on changing ‘u’, v
changes, but’ f’ remains constant.
In day to day life, ‘u’ corresponds to a situation that arises and ‘v’ corresponds to our responses to the
situation. We are like a mirror. Our nature/curvature determines our focal length. The mirror formula
implies that our nature is not affected by the situation that comes up. Response to a particular situation
will depend on our nature.
5. During summer vacation Ravi and Rohit decided to go for a 3-D film (movie). They have heard about
this film through their friends. They were asked buy special glasses to view the film. Before they go for
a movie, they approached their Physics teacher to know about these glasses. Physics teacher explained
when two polarizers are kept perpendicular to each other (crossed polarizers) the left eye sees only the
image from the left end of the projector and the right eye sees only the image from the right lens. The
two images have the approximate perspectives that the left and right eyes would see in reality the brain
combine the images to produce a realistic 3-D effect.
Read the above passage and answer the following questions:
(a)What qualities do these boys possess?
(b)What do you mean by Polarization?
(c) Mention the other applications of polarization.
Answer:
a) Curiosity to learn, approaching the teacher to learn new things, inquisitiveness.
b) The phenomenon of restricting the oscillations of a light wave (electric vector) in a particular
direction is called polarization of light.
c)
In Sun glasses, Liquid Crystal Displays, CD players etc.
************************************
DUAL NATURE OF MATTER AND RADIATION
IMPORTANT CONCEPTS
Work function-The minimum amount of energy required by an electron to just escape from the metal surface is
known as work function of the metal.
One Electron Volt (1eV)-It is the kinetic energy gained by an electron when it is accelerated through a
potential difference of 1 volt.
1eV = 1.6X10-19J
Photon-According to Planck's quantum theory of radiation, an electromagnetic wave travels in the form of
discrete packets of energy called quanta. One quantum of light radiation is called a photon.
The main features of photons are as follows:(i) A photon travels with the speed of light.
(ii) The rest mass of a photon is zero i.e., a photon cannot exist at rest.
(iii) Energy of a photon, E = hν
(iv) Momentum of a photon, p =mc
Photoelectric Effect: -The phenomenon of emission of electrons from a metallic surface when light of
appropriate frequency (above threshold frequency) is incident on it, known as photoelectric effect.
Photoelectric Effect experimental setup-
Factors which Effect Photoelectric Effect:(i) Effect of potential on photoelectric current: It is can be shown in fig. potential v/s photoelectric current.
(ii) Effect of intensity of incident radiations on photoelectric current: For frequency of radiations as constant.
(iii) Effect of frequency of the incident radiations on stopping potential: For constant intensity.
The stopping potential Vo depends on(i) The frequency of incident light and (ii) the nature of
emitter material. For a given frequency of incident light, the stopping potential is independent
of its intensity.
eVo =(1/2)mv2max=Kmax
From this graph between frequency ν, stopping potential Plank's constant (h) can be determined
Intercept on x- axis- Threshold frequency
Intercept on y-axis- Work function/ electron
Slope= tan𝜃 =
ℎ
Intercept on x- axis- Threshold frequency
Intercept on y-axis- Work function
Slope= tan𝜃 = ℎ
𝑒
De-Broglie HypothesisAccording to de Broglie, every moving particle is associated with a wave which controls the particle in every
respect. The wave associated with a particle is called matter wave or de Broglie wave.
λ =h/p = h/mv
This is known as de-Broglie equation.
de-Broglie wavelength of an electron of kinetic energy Kℎ
ℎ
1.227
𝜆=
=
=
𝑛𝑚
√2𝑚𝐾 √2𝑚𝑒𝑉
√𝑉
Davisson and Germer experimentThis experiment proves the existence of de-Broglie waves.It establishes the wave nature of electron particle.
Theory-A sharp diffraction is observed in the electron distribution at an accelerating voltage of 54 V and
scattering angle 50°. The maximum of intensity obtained in a particular direction is due to constructive
interference of electrons scattered from different layers of the regularly spaced atoms of the crystal.
Questions/Answers related to this topic
Q.1
The frequency of incident radiation is greater than threshold frequency 0 in a photocell
.How will the stopping potential vary if frequency is increased. (1 mark)
Q.2
If the intensity of the incident radiation in a photocell is increased. How does the stopping
potential vary? (1 mark)
No effect.
Q.3
Two metals P and Q have work function 2ev and 4ev respectively. Which of the two
metal have smaller threshold wavelength.
(1 mark)
Q.4
The stopping potential in an experiment in Photoelectric effect is 1.5 eV. What is the
maximum kinetic energy of the photoelectron? (1 mark)
Kmax = 1.5 eV
Q.5
In an experiment on photoelectric effect, the following graph were obtained. Name the
characteristics of incident radiation that was kept constant. (1 mark)
frequency
Q.6
State Einstein’s Photoelectric equation.
E=h =h
Q.7
0
(1 mark)
+ Kmax
How does the maximum kinetic energy of photoelectrons vary with work function of
metal? (1 mark)
Decreases.
Q.8
With what purpose was famous Davisson - Germer experiment with electrons per
formed? (1 mark)
To proves the existence of de-Broglie waves
Q.9
An electron and a proton have same De broglie wavelength associated with them. How
are their Kinetic energies related to each other?
Q.10
Ultra violet light of wavelength 2271 Ao is incident on two photo sensitive material
having work function W1 and W2 (W1 >> W2). In which case will the kinetic energy of
emitted electrons be greater? (1 mark)
For W2 metal
.11
Two lines A and B in the plot given below show the variation of De Broglie wavelength λ versus
1/ . Where V is the accelerating potential difference for two particles carrying the same charge
.Which
one
represents
a
particles
of
smaller
mass.
(2
marks)
Q.12
Derive an expression for the de Broglie Wavelength of an electron moving under potential
difference of V volt. (2 marks)
Let an electron be accelerated by applying a potential difference V volt. Then
W=Qv=
Q.13
A particle of mass M at decays into two particle of masses m1 and m2having velocity v1 and
v2 respectively .Find the ratio of de Broglie wavelength of two particles. (2 marks)
Q.14
The wavelength λ of a photon and De Broglie wavelength is of an electron have the same value
.Show that energy of photon is 2 λ mc/n times the energy of the electron. (2 marks)
Q.15
For a photosensitive surface threshold wavelength is λ0 .Does photo emission occur if wavelength
(λ) of incident radiation is more than λ0 , less than λ0. Justify your answer. (2 marks)
(i) No, as work function
Energy if incident radiation Decreases.
(ii) Yes, Energy of incident radiation increases.
Q.16
When a monochromatic yellow colored light beam is incident an a given photosensitive surface,
photo
electrons
are
not
ejected
to
green
colored
monochromatic
beam.
What will happen if the same surface is exposed to (i) violet and (ii) red colored monochromatic
beam of light.
(2 marks)
λv <λy , λR > λy
(i) For violet, color, Photo emission will take place as Energy increases.
(ii) For Red color, No emission of electrons.
Q.17
A Source of light is placed at a distance of 50 c.m from a photocell and cut –off potential is found
to be Vo .If the distance b/w source and photocell is made 25 c.m. What will be new cut –off
potential? (2 marks)
Same, as intensity increases and stopping potential remains same.
Q.18
Show the graphical variation of stopping potential with the frequency of incident radiation .How
do we determine the Planck’s constant using Graph. (2 marks)
Q.19
Explain the effect of increase of intensity and potential difference on photoelectrons kinetic
energy.
Due to increase in Intensity , No effect on kinetic energy of Photo electrons as well as on
Potential Difference. As due to increase in Intensity , there is only an increase in the number of
Photons per unit area , and not the energy incident.
Q.20
Calculate the number of photons emitted per second by transmitter of 10 KW power; radio waves
of frequency 6×105 Hz. (2 marks)
Q.22
The following graph shows the variation of stopping potential Vo with the frequency of the
incident radiation for two Photosensitive metals X and Y.
(i) Explain which metal has smaller threshold wavelength .
(ii) Explain giving reason, which metal emits photoelectrons having smaller kinetic energy.
marks)
(3
Y as
Y as K = h
- Φo
Work function of Y will be more as compared to X.
Q.23
A proton and an alpha particle are accelerated through the same potential .Which one them has
Higher De Broglie wave length. (3 marks)
as mα =4mp and qα =2qp
Q.24
Show the graphical variation of photocurrent with intensity of incident radiation at constant
potential difference b/w electrons and the graphical variation of photocurrent of incident
radiation. (3 marks)
Q.25
State the laws of Photoelectric effect. Explain it on the basis of Einstein equation.
(3 marks)
Laws
:
(i) It is an instantaneous process
(ii) No Photo emission takes place below threshold frequency of material, no matter how
intense the incident beam.
(iii) The maximum photo current (saturation current ) does not depends upon stopping potential
or frequency but depends on intensity of incident radiation.
(iv) Stopping potential is independent on intensity of incident radiation.
Q.26
Draw a schematic diagram of the experimental arrangement used by Davisson &Germer to
establish wave nature of electrons .Explain briefly how the De Broglie relation was verified
experimentally. (3 marks)
Q.27
An electromagnetic wave of wave length λ is incident on a photosensitive surface of negligible
work function .if photoelectrons emitted from this surface have the De Broglie
wavelength λ1 them show that
.
(3 marks)
E = h = Φo + Kmax
Q.28
X-rays of wavelength λ fall on the photosensitive surface emitting electrons. Assuming that the
work function of the surface can be neglected, Show that De Broglie wave length of electrons
emitted will be
.
(5 marks)
As E= h = Φo + K
Q.29
IF the frequency of incident radiation and photocell is doubled for same intensity, what charges
will you observe in (3 marks)
(i) Kinetic energy of photo electrons
(ii) Photoelectric current
(iii) Stopping potential.
(i) Kinetic energy will be increased
(ii) No effect.
(iii) Will increase.
Q.30
Sketch a graph b/w of incident radiation and stopping potential for a given photosensitive
material. What information can be drawn from the value of intercept on the potential axis?
marks)
E = h = Φo + eVo
Q.21
Radition of frequency 1015 Hz are in incident on two photosensitive surfaces A and B.
Following observations are recorded:
Surface A: No photoemission takes place.
Surface B: photoemission takes place but photoelectrons hace zero energy.
Explain the above observations on the basis of Einstein’s photo electric equation.
For ‘A’, Energy incident is less than work function.
For ‘B’, Energy incident is equal to work function of metal.
(3 marks)
(5
UNIT—VIII- ATOMS & NUCLEI
1. Rutherford’s 𝜶-Particle scattering experiment (Geiger – Marsden experiment)
Scattering of 𝛼-particles by heavy nuclei is in accordance with coulomb’s law. Rutherford observed that
number of 𝛼-particles scattered is given by
1
N ∝ 𝑠𝑖𝑛4 𝜃
⁄2
2. Distance of closest approach : Estimation of size of nucleus
1 𝑍𝑒𝑋 2𝑒
𝑟0 = 4𝜋𝜀 1 2
0
2
m𝑣
3. Bohr’s atomic model
Radius of orbit 𝑟 =
v =
2𝜋𝑍𝑒 2
(4𝜋𝜀0 )𝑐ℎ
(4𝜋𝜀0 )𝑛2 ℎ2
4𝜋 2 𝑚𝑍𝑒 2
𝑐
𝑐
X𝑛= 𝛼𝑛
Frequency
Where 𝛼 =
2𝜋𝑍𝑒 2
(4𝜋𝜀0 )𝑐ℎ
v=
1
= 137
2𝜋𝑍𝑒 2
(4𝜋𝜀0 )𝑛ℎ
is called fine structure constant
4. Energy of electron
𝒎𝒁𝟐 𝒆𝟒 𝟏
En = − 𝟖𝜺𝟐 𝒉𝟐 (𝒏𝟐 )
𝟎
Rydberg constant.
𝟏𝟑.𝟔
En = − 𝒏𝟐 eV
Formula –
En = −
𝒁𝟐 𝑹𝒄𝒉
𝑚𝑒 4
R = 8𝜀2 𝑐ℎ3 = 1.097 X 107 m-1 and is called
𝒏𝟐
0
1
1
1
2
𝜈̅ = 𝑅 [𝑛 2 − 𝑛 2 ]
where 𝜈̅ is called wave number.
K.E. = - ( Total Energy )
P.E.= - 2 K.E.
5. Spectral Series of Hydrogen Atom
6. Energy level diagram for hydrogen atom
We know that for hydrogen atom, energy of an electron in nth orbit is given by
13.6
En = − 𝑛2 eV
7. Atomic Mass Unit (amu)
1
One atomic mass unit is defined as 12th of the actual mass of c-12 atom.
1
1
1 u = 12 X mass of C-12 atom = 12 X 1.992678 X 10-26 kg = 1.66 X 10-27 kg.
8. Electron Volt (eV)
It is the energy acquired by an electron when it is accelerated through a potential difference of 1 volt.
1 eV = 1.6 X 10-19 J & 1 MeV = 1.6 X 10-13 J
9. Relation Between amu & MeVEinstein ‘s Mass Energy Equivalence Relation is E =mc2
1amu =1u =931 MeV
10. Nuclear Density (𝝆) =2.3 X 1017 Kg/m3 obviously, nuclear density is independent of mass number A.
11. Properties of nuclear Forces
(i) Nuclear forces are very short range attractive forces.
(ii) Nuclear forces are charge independent.
(iii) Nuclear forces are non-central forces.
(iv) Nuclear forces do not obey inverse square law.
12. Nuclear force as a separation between two nucleons
13. Potential energy of a pair of nucleons as a separation between two nucleons
14. Nucleus consist of protons and neutrons.
Nucleus of protons in a nucleus zXA is Z and number of neutrons ,N =A-Z
15. Radius of Nucleus :- R= R0A1/3 where R0 = 1.2 x 10-15m
16. Mass Defect (∆𝒎)
∆𝒎 = [Z mp + (A – Z) Mn ] − MN
17. Packing fraction (P.F.)
It is defined as the mass defect per nucleon.
∆𝑚
i,e,
P.F. = 𝐴
Nucleus is stable if P.F.>1 & unstable if P.F.< 1
18. Binding Energy (B.E.)The binding energy of a nucleus may be defined as the energy required to break
up a nucleus in to its constituent protons and neutrons and to separate them to such a large distance that
they may not interact with each other. It is equivalent energy of mass defect.
i,e,
B.E. = ∆𝑚 X c2
⇨
B.E. = [{Z mp + (A – Z) Mn} −MN ] x c2
𝐵.𝐸.
19. Binding Energy per nucleonB.E. per nucleon = 𝐴
20. Einstein ‘s Mass Energy Equivalence Relation is E =mc2 1amu =1u =931 MeV
21. Rutherford –Soddy formula :(i) Number of atoms un-decayed after time t
N=N0e-λt
(ii)
N/N0 =[1/2]n
Where n = t\T is number of half lives.
22. Relation between half –life (T) mean life (𝝉 ) and disintegration constant (λ ) is
𝜏 =1/λ and T = 0.693𝜏 = 0.693/λ
23. Displacement Laws:
(i)
For α -particle
zXA
z-2YA-4 + 2He4
(ii)
For β - particle
zXA
A
z+1Y
+
-1β
0
+ ν
(iii) For gamma– ray
( zXA)*(Excited State)zXA (Ground state) + γ
24. In nuclear fission a heavy nucleus break into lighter nuclei .Nearly 0.1 % mass is converted into
energy .In each fission of 92 U 235 with slow neutron 200 MeV energy is released
25. In nuclear fusion two lighter nuclei combine to form a heavy and 0.7 % mass is converted into energy
1 MARK QUESTION
Qn1.What is the ratio of the radii of orbits corresponding to first excited state and ground state in hydrogen
atom?
Ans:- r2 / r1 = (n2 /n1 )2 = (2/1) 2
= 4: 1
Qn2. Two nuclei have mass numbers in the ratio 1: 8. Find the ratio of their nuclear radii and nuclear densities.
Ans;- R1 / R2 = ( A1/ A2) 1/3 = ( 1/ 8 ) 1/3 = 1 / 2 , d1/ d2 = 1:1 Density does not depends on mass (same)
Qn3. What is the ground state energy of electron in case of 3Li 7 ?
Ans:- E n = - 13.6 Z 2 / n2eV
Putting Z= 3 , n = 2
E n = - 30.4 eV
Qn4.Find first excitation energy and excitation potential of hydrogen atom.
Ans:- E = E2 – E1 = -3.4 – ( - 13.6 ) eV
= 10.2 eV
and Potential = 10.2 Volt
Qn5.Find ionisation energy and ionisation potential of hydrogen atom.
Ans:- E n = - 13.6 Z 2 / n2eV , put Z =1 , n=1
E n = - 13.6 eV
Hence ionisation energy = + 13.6 eV, ionisation potential = 13.6 V
Qn6. Tritium has half-life of 12.5 years against β decay. What fraction of the sample will remain undecayed
after 25 years ?
Ans:- N/ N0 = (1/2 ) t/ T
= (1/2 )25/12.5
= 1/4
2 MARKS QUESTION
Qn1. With the help of an example explain how the neutron to proton ratio changes during α – decay of nucleus.
Ans:- 92U238 →92Th234+2α4
238−92 146
N to P ratio before α-decay= 92 = 92 =1.59
N to P ratio after α-decay=
234−90 144
90
= 90 =1.60
146 144
<
92
90
This show that the N to P ratio increases during α-decay of a nucleus
Qn2. A radioactive isotope has half-life of 5 years after how much time is its activity reduces to 3.125% of its
original activity?
𝑅
Ans:- We know that 𝑅0= (
1 n𝑅
) 𝑅0=3.125/100 = 1/ 32 = (1/2)5
2
n=5 and n=t/T or t= n x T = 5x5=25years.
Qn3. A heavy nucleus X of mass number 240 and binding energy per nucleon 7.6 MeV is split into two
fragments Y and Z of mass numbers 110 and 130. The binding energy of nucleons in Y and Z is 8.5 MeV per
nucleon. Calculate the energy Q released per fission in MeV.
Ans:- Total energy of nucleus X = 240 × 7.6 = 1824 MeV
Total energy of nucleus Y = 110 × 8.5 = 935 MeV
Total energy of nucleus Z = 130 × 8.5 = 1105 MeV
Therefore, energy released from fission, Q = 935 + 1105 − 1824 = 216 MeV
Qn4. The ground state energy of hydrogen atom is -13.6eV.What is the K.E& P.E of the electron in this state?
Ans:- K.E= - ( Total Energy ) =13.6 eV,
P.E=-2K.E=-27.2 eV
Qn5.At a given instant there are 25% un-decayed radioactive nuclei in a sample. After 10 seconds the number
of un-decayed nuclei reduces to 12.5 %.calculate the i) mean life of the nuclei ii) the time in which the number
of the un-decayed nuclei will further reduce to 6.25 % of the reduced number.
Ans:- T =10s, λ=.0693/ T ,
τ=1/λ= 1.44 T = 14.43 sec
N=1/16(N0/8) →t= n x T =4x10=40sec
Qn 6.A radioactive nucleus ‘A’ decays as given below:
Β
α
A
A1
A2
If the mass number & atomic number of A1 are 180 & 73 respectively, find the mass number & atomic number
of A &A2
Ans:- For A— 180 & 72,
For A2—176 & 71
Q7. What is the shortest wavelength present in the Paschen series of hydrogen spectrum?
(R = 1.097 x 10 7 m-1 )
1
1
Ans:- 1/λ = R ⌊𝑛2 − 𝑛2 ⌋
1
put n1=3, n2=∞
, λ=9/R=8204Ǻ
2
Qn8. Calculate the frequency of the photon which can excite an electron to -3.4 eV from
-13.6 eV.
Ans:- E = E2 – E1 , = -3.4 – ( - 13.6 ) eV ,
= 10.2 eV = 10.2 x 1.6 x 10 -19 J
E = h ν = 10.2 x 1.6 x 10 -19 J , ν = 10.2 x 1.6 x 10 -19 / 6.6 x 10 -34
,
= 2.5x1015Hz
Q9. The energy levels of an atom are as shown below. a) Which of them will result in the transition of a photon
of wavelength 275 nm? b) Which transition corresponds to the emission of radiation maximum wavelength?
Ans:- (a) E=hc/λ = 6.6 x 10 -34 x 3x 10 8 / 275 x10 -9 x 1.6 x10 -19 , =4.5eV , transition B
(b) Eα 1/λ
transition A provides minimum energy of 2 eV , Hence maximum wavelength
3 MARKS QUESTION
Qn1. Draw the graph showing the variation of binding energy per nucleon with the mass number.What are the
main inferences from the graph? Explain with the help of this plot the release of energy in the processes of
nuclear fission and fusion .
Ans
The variation of binding energy per nucleon versus mass number is shown in figure:-
Inferences from graph
1.
The nuclei having mass number below 20 and above 180 have relatively small binding energy and
hence they are unstable .
2.
The nuclei having mass number 56b and about 56 have maximum binding energy -5.8 MeV and so they
are not stable.
3
Some nuclei have peaks ,e. g 2 He4 , 6C12 ,8O16 ; this indicates that theses are relatively more stable than
their neighbours .
Explanation : - When a heavy nucleus (A ≥ 235 say ) break into two lighter nuclei (nuclear fission ), the binding
energy per nucleon increase i.e nucleons get more tightly bound .This implies that energy would be released
in nuclear fission .
When Two very light nuclei (A ≤ 10) join to from a heavy nucleus ,the binding is energy per nucleon of fused
heavier nuclear more than the binding energy per nucleon of lighter nuclei,so again energy would be released in
nuclear fusion .
Q:2
Define half -life of a radioactive sample .which of the following radiations : α –rays ,β –rays and γ-rays
(i)
(ii)
(iii)
(iv)
Are similar to X-rays
Are easily absorbed by matter
Travel with the greatest speed
Are similar in nature to cathode rays ?
Ans
Half –life : The half –life of a radioactive sample is defined as the time in which the mass of sample is
left one half of the original mass.
(i)
(ii)
(iii)
(iv)
γ-rays are similar to X-rays
α –rays are easily absorbed by matter
γ-rays travel with greatest speed
β –rays are similar to cathode rays .
Q:-3 Define the term ‘ Activity ‘ of a radioactive substance .State its SI unit .Give plot of activity of a
radioactive species versus time.
Ans
The activity of a radioactive elements at any instant is equal to its rate of decay at that instant S.I unit of
activity is Becquerel . (=1 disintegration /second ). The plot is shown is figure .
Q:-4
How does the size of nucleus depend on its mass number ? Hence explain why the density of nuclear
matter in independent of the size of nucleus?
Ans
The radius (size) R of nuclear is related to its mass number (A) as
R=R0 A1/3 where R0 =1.1 x 10- 15m
If m is the average mass of a nucleon ,then mass of nucleus =mA ,where A is mass number .
Volume of nucleus = (4/3) π R3 = (4/3) π R 0 3 A
Density of nucleus = Mass / Volume = 3m / 4 π R 0 3
∴
Nuclear density is independent of mass number .
Q:-5 A radioactive nucleus A undergoes a series of decay according to following scheme:
A
α
A1
β-1
A2
α
γ
A3
A4
The mass number and atomic number of A are 180 and 72 respectively .What are these numbers for A4?
Ans The decay scheme may completely be represented as
180
72A
α
76
70A
β-1
176
71A
α
69A3
172
γ
69A4
172
Clearly ,mass number of A4 is 172 and atomic number is 69.
Q:-6 You are given two nuclides 3X7 and 3Y4
(i)
(ii)
Are they isotopes of the same element ? Why ?
Which one of the two is likely to be more stable?
Ans (i) The two nuclides are isotopes of the same elements because they have the same Z.
(2) The nuclide 3Y4 is more stable because is has less neutron to proton ratio .
Q:- 7 Derive the relation Nt = N0e-λt.
Or
Use basic law of radioactive decay to show that radioactive nuclei follow an exponential decay law?
Or
State the law of radioactive decay .If No is the number of radioactive nuclei at some initial time t 0 ,find out the
relation to determine the number N present at a subsequent time .
Ans Radioactive decay Law:The rate of decay of radioactive nuclei is directly proportional to the number of undecayed nuclei at that time .
𝑑𝑁
𝑑𝑡
= −λ N
Where λ .is the decay constant . Suppose initially the number of atoms in radioactive elements is N0 and the
number of atoms after time t . According to Rutherford and Soddy law .
𝑑𝑁
𝑑𝑡
= −λ N where λ disintegration constant .
𝑑𝑁
𝑁
= −λ dt
Integration loge N = λ t +C
here c is a constant of integration . If N0 is initial number of radioactive nuclei ,then at t = 0,N= N0 ; so
Loge N0 =0 + C⇒C = loge N0
Substituting this equation in (1) ,we get
Loge N– loge N0 =
λt
Loge N/ loge N0 =
λt
N=N0e—λt
Q:8 Define half life of radioactive substance .Establish its relation with the decay constant ?
Or
Define – life of a radioactive sample .Using exponential decay law obtain the formula for the half –life of a
radioactive in terms of its disintegration constant/
Ans Half-life of a radioactive elements is define as the time in which number of radioactive nuclei becomes
half of its initial value
Expression for half time :-time radioactive decay equation is
N = N0e—λt
When
t = T,N=
𝑁0
2
Substituting and solving
Or
e-λt = 1/2
Taking log of both sides
Or
λT = loge 2
𝑇=
0.693
𝜆
****************************************
ELECTRONIC DEVICES
Electronics is a branch of physics which deals with flow of current through inert gases, vacuum or
semiconductors.
Types of metals on the basis of conductivity(i) Conductor :-low resistivity & high conductivity
(ii) Insulator:- high resistivity & low conductivity
(iii)Semiconductor:- resistivity & conductivity lies between them
Energy Bands in Solids:
Valence band(VB):- energy level Completely filled by valence electron
Conduction band(CB):- Energy level either empty or partially filled by valence electron
Forbidden Energy gap :- Energy gap between VB & CB
For Germanium the forbidden energy gap is 0.7ev while it in 1.1ev silicon.
Types of metals on the basis of Energy bands(i) Conductor :-CB & VB are overlap to each other so electron easily available in conduction band.
(ii) Insulator:- Much energy gap between CB & VB so no electron easily available in conduction band.
(iii)Semiconductor:- energy gap between CB & VB are less so electron can jump in conduction band.
(i)
Elemental semiconductors: Si and Ge
(ii)
Compound semiconductors: Examples are: CdS, GaAs, CdSe, InP, etc.
1) Semiconductors are the basic materials used in present solid state deviceslike diode, transistor, ICs,etc.
TYPES OF SEMICONDUCTORS
1) Pure semiconductors are called intrinsic. Semiconductors, ne=nh ie no. ofelectrons is equal to no. of holes.
Holes are electron vacancies with aneffective positive charge.
2) Impure semiconductors are called extrinsic The number of charge carriers can be changed by doping.
Such semiconductors are called extrinsic semiconductors
TYPES OF EXTRINSIC SEMICONDUCTORS
(i)
N-type Semiconductor:- It is obtained by doping Si or Ge with pentavalent atomic(donors) like
As, Sb, P etc, (ne>>nh )
(ii)
P-type:- Semiconductor:- It is obtained by doping Si or Ge with trivalent
atoms(acceptors) like B, Al, In, etc. (nh>>ne )
P-N junction:Arrow shows the direction of conventional current.
Depletion layer: -Formation of p-n junction produces a depletion layer consisting ofimmobile ion cores devoid
of charge carriers with a width of 10-3 mm. This layer is formed due to diffusion of majority carrier across the
junction
Potential barrier:- Potential difference due to negative immobile ions on p-side and positive immobile ions on
n-side is called potential barrier which is produced about 0.7 V for a silicon p-n junction and 0.3Vfor
Germanium p-n junction.
Biasing of Diode :-p-n junction diode join with external battery.
Forward biasing:-When positive terminal of battery join with p-region & negative terminal of battery join
with n-region. Such biasing is called forward biasing (short P-Positive & N-Negative)
Reversed biasing:-When positive terminal of battery join with n-region & negative terminal of battery join
with p-region. Such biasing is called reversed biasing (short P-Negative)
NOTE;-(i) In forward bias ,the barrier is decreased while ,it increases in reversebias.Hence forward current is
more (mA) while it is very small (µA) inreverse bias.
(ii) Diodes can be used for rectifying ac voltage. With the help of a capacitor orsuitable filter ,a dc voltage can
be obtained.
(iii)
(iv)
There are some special purpose diodes. Zener diode is used as a voltage regulator.
p-n junctions have been used to obtain many photonic or optoelectronics
devices. Eg- photodiodes , Solar cells, LED and diode LASER
Identification of important topics /concepts
1) Difference between insulator ,conductor and semiconductor on the basis ofEnergy band diagram.
2) Difference between n-type and p-type semiconductor on the basis ofdoping and energy band diagram.
3) Definition of important terms like depletion layer,forward bias,and reversebias,barrier potential, doping.
4) Graph forward Bias and reverse bias of a p-n junction diode.
5) Diode as rectifier-working and circuit diagram with graph.
6) Use of Zener diode as a voltage regulator
7) CE Amplifier circuit with working and graph.
8) Logic gates-AND,OR,NAND,NOR,NOT, with symbols and truth tables.
9) Some simple digital circuits with combination of gates
IMPORTANT DERIVATIONS COVERING WHOLE UNIT (3 & 5 Marks)
Que1. What is semiconductor diode . How a diode can be made forward and reverse bias. Draw its V-I
characteristic curve .
Ans. A semiconductor diode is basically a p-n junction with metallic contactsprovided at the ends for external
voltage.
Forward bias: In forward bias, the p-type is connected with the positiveterminal and the n-type is connected
with the negative terminal.
Reverse bias : In reverse bias , the p-type is connected with the negativeterminal and the n-type is connected
with the positive terminal.
Que2. What is zener diode. Draw V-I characteristic curve of zener diode. Explain itsuse as an voltage regulator
with circuit diagram.
Ans. It is designed to operate in the reverse breakdown voltage regioncontinuously without being damaged.
A zener diode has unique feature that voltage drop across it , is independentof current through it.
The resistor, RS is connected in series with the zener diode to limit the current flow through the diode
Any increase or decrease in voltage appears across the series resistance RS and t6he voltage across zener diode
remains constant
Que 3.
What is junction transistor. Write its types with symbol. Giving circuitdiagram of p-n-p transistor in CE draw
input & output characteristic curve.
Ans. A junction transistor is a three terminal solid state device obtained bygrowing a thin layer of one type
semiconductor in between two thick layers ofother similar type semiconductor
Transistor are of two types1. n-p-n transistors- it consist of a thin section p-type
semiconductor
sandwiched between two thicker section of n-type
semiconductor.
2. p-n-p transistor-it consist of a thin section of n-type
semiconductor
Que 4What is amplifier? Discuss use of n-p-n transistor as an amplifier with circuitdiagram. What is phase
relation between input & output waveform.
Ans. A device which increases the amplitude of the input signal is calledamplifier.In common emitter amplifier,
input signal to be amplified is applied betweenbase-emitter circuit and the output amplified signal is taken
across the loadresistance in emitter- collector circuit.
There is a phase difference of π between input and output signal.
Que 5. From the diagram shown below identify whether the diode is forward orreverse biased .
Ans. (a) Forward bias (b) Reverse bias.
Que 6. What is meant by rectifier? Discuss working of full wave rectifier with circuitdiagram. Draw its input &
output wave forms.
Ans. Rectifier is a device which convert ac signal to dc.
Working:-When the diode rectifies whole of the AC wave, it is called ‘fullwave rectifier’.During the positive
half cycle of the input ac signal, the diode D1 conducts andcurrent is through BA.During the negative half
cycle, the diode D2 conducts and current is throughBA.
Que 7What is half wave rectifier. Giving circuit diagram & input-outputwaveform explain its working.
Ans. Half wave rectifier is a device which changes half cycle of ac to dc.
Working:- In first half cycle of ac the diode is forward bias & conduct but insecond half cycle the diode is
reverse bias & hence not conduct. Hence it giveshalf dc
Que 8. (a) Draw transfer characteristic curve of Base-biased C-E transistor.
(b) Mention the region where the transistor used as switch & where asAmplifier.
.
(a) Active- Amplifier
((b) Switch- ON Switch- saturation region
OFF-Switch-cut off region
Amplifier – active region
Que 9. You are given two circuits as shown in Fig. Giving truth table identify thelogic operation carried out by
the two circuitsElectronic Devices
Que 10. What is logic gate. Name the basic gates. Give symbol, Boolean expression &truth table for AND gate.
Ans.A logic gate is a digital circuit that follows certain logical relationship betweenthe input and output voltage.
Que11. What is a solar cell? How does it works? Give its one use.
Ans: Solar cell is device for converting solar energy into electricity. It isbasically a p-n junction operating in a
photovoltaic mode without externalbias.
Working: When light photons fall at the junction electron-hole pairs aregenerated. those more in opposite
direction due to junction field. Thesecharges accumulate at the two sides of the junction and photo voltage
isdeveloped.Use: It is used in calculators etc.
SOME IMPORTANT QUESTIONS FROM PREVIOUS YEAR PAPERS
Q1.In a semiconductor the concentration of electrons is 8x1013cm-3 and that of holesis 5x1012cm-3. Is a p-type
or n-type semiconductor?
Ans : As concentration of electrons is more than holes, the given extrinsic semiconductor is n-type.
Q2.The energy gaps in the energy band diagrams of a conductor, semiconductor andinsulator are E1, E2 and
E3. Arrange them in increasing order.
Ans: The energy gap in a conductor is zero, in a semiconductor is ≈ 1eV and in aninsulator is ≥ 3eV. E1=0,
E2=1eV, E3≥3eV . E1 < E2 < E3.
Q3. Find the truth table of following gates
Q6.The current gain (α) of a transister in common base configuration is 0.98. Whatdoes It physically mean?
Ans :The current gain α=0.98 means that 98% of charge carriers of an emitterreach the collector and constitute
the collector current
Q8.Name the gate obtained from the combination of gates shown if figure. Draw the
logic symbol. Give the truth table of the combination.
Ans : The gate is NOR gate the logic symbol is shown in
figTruth table of NOR gate
Q 9.Name the logic gate shown in fig. and write its truth
table.
Ans : . The given logic gate is NAND gate
Truth table of NAND gate
Q10. Show the output waveforms(Y) for the following inputs A and B of(i) OR gate (ii) NAND gate
Ans
:
Q11. For CE transistor amplifier, the audio signal voltage across the collector
resistance of 2 kΩ is 2V. Suppose the currentamplification factor of the transistor is 100,find the input signal
voltage voltage and basecurrent if the base resistance is 1kΩ.
Ans : Given Rc=2kΩ, RB=1k Ω, V0=2V, Input
voltage Vi=?
β = IC/IB=100
V0= ICx Rc=2V
IC=2/ Rc=10-3A
Base current= IB= IC/β=10μA
Base resistance, RB=VBB/IB
Therefore Vi(VBB)= RBx IB = 0.01V
Q12.Two amplifiers are connected one after another in series (cascaded). The firstamplifier has a voltage gain
of 10 and the second has a voltage gain of 20. If theinput signal is 0.01 volt, calculate the output signal.
Ans : Total voltage gain
AV=A1X A2= 200
Voltage gain AV=Output Voltage/ Input Voltage
Output Voltage V0= AVX Vi = 2V
Q13.In a common emitter mode of a transister, the dc current gain is 20, the emittercurent is 7mA.
Calculate (i)base current and (ii)collector curent.
Ans : Given β=20, iE=7mA
(i) β=iC/iB= iE-iB/ iB
iB= ie/(1+ β) =1/3 mA
(ii) iC = iE-iB= 20/3 mA
Q14.A semiconductor has equal electron and hole concentration of 6X108/m3. Ondoping with certain impurity,
electron concentration increases to 9X1012/m3.
(i) Identify the new semiconductor obtained after doping.
(ii) Calculate the new hole concentration.
Ans : (i) The doped semiconductor is n-type.
(ii) ne nh = ni2 hence nh = ni2/ ne=4x104m-3
Q 15. How does a light emitting diode (LED) works? Give two advantages of LED’s over the conventional
incandescent lamps.
Ans: Working of LED :- LED works in forward bias at the junction when majority charge carrier recombine
with minority charge carriers, which grow in number due to diffusion of charges across the junctions,
energy is released in the form of photons.
Advanteges:- (i)Low operational voltage and (ii) Fast on-off switches capability
Q 16. Draw a circuit diagram to slow how a photo diode is viewed. Draw its characteristics curres for three
different illumination intensities.
Ans:
Q 17. Identify the logic gates marked P and Q in the given logic circuit. Write down the output at X for the
inputs
(i)
A = 0, B = 0 and
(ii) A = 1, B = 1.
A
Q
P
B
X
Ans:
P→
NOR gateQ → AND gate
A
0
1
B
0
1
A+B
0
1
𝐴+𝐵
1
0
𝐴+𝐵 B
0
0
Q 18. In half wave rectification, What is the output frequency if the input frequency is 50 Hz. What is the
output frequency of a full wave rectifier for the same input frequency.
Ans: 50 Hz for half wave ,100 Hz for full wave.
COMMUNICATION SYSTEMS
Every communication system has three essential elements-transmitter, medium/channel and receiver. The
block diagram shown in Fig. depicts the general form of a communication system.
•
•
•
•
•
•
•
•
•
•
•
•
•




TRANSDUCER: Any device that converts one form of energy into another can be termed as
transducer.
SIGNAL: Information converted in electrical form and suitable for transmission is called a signal.
Signals can be either analog or digital.
NOISE: Noise refers to the unwanted signals that tend to disturb the transmission and processing of
message signals in a communication system.
TRANSMITTER: A transmitter processes the incoming message signals so as to make it suitable for
transmission through a channel and subsequent reception.
RECEIVER: A receiver extracts the desired message signals from the received signals at the channel
output.
ATTENUATION: The loss of strength of a signal while propagating through a medium is known as a
attenuation.
AMPLIFICATION: It is the process of increasing the amplitude (and consequently the strength) of a
signal using an electronic circuit called the amplifier. Amplification is necessary to compensate for the
attenuation of signal in communication systems.
RANGE: It is the largest distance between a source and destination up to which the signal is received
with sufficient strength.
BANDWIDTH: Bandwidth refers to the frequency range over which an equipment operates or the
portion of the spectrum occupied by the signal.
MODULATION: The original low frequency message/information signal cannot be transmitted to long
distances. The low frequency message signal is superimposed on a high frequency wave, which acts as a
carrier of the information. This process is known as modulation.
DEMODULATION: The process of retrieval of information from the carrier wave at the receiver is
termed modulation. This is the reverse process of modulation.
REPEATER: A repeater is a combination of a receiver and a transmitter. A repeater picks up the signal
from the transmitter, amplifies and retransmits it to the receiver sometimes with a change in carrier
frequency. Repeaters are used to extend the range of a communication system.
BANDWIDTH OF SIGNALS
In a communication system, the message signal can be voice, music, and picture or computer data. Each
of these signals has different ranges of frequencies.
For speech signals, frequency range 300 Hz to 3100 Hz. Therefore speech signal requires a bandwidth of
2800 Hz (3100 Hz – 300 Hz) for commercial telephonic communication.
To transmit music, an approximate bandwidth of 20 kHz is required because of the high frequencies
produced by the musical instruments.
The audible range of frequencies extends from 20 Hz to 20 kHz.
 Video signals for transmission of pictures require about 4.2 MHz of bandwidth.
 A TV signal contains both voice and picture and is usually allocated 6 MHz of bandwidth for
transmission.
 BANDWIDTH OF TRANSMISSION MEDIUM
 Coaxial cable is a widely used wire medium, which offers a bandwidth of approximately 750 MHz.
 Such cables are normally operated below 18 GHz.
 Communication through free space using radio waves takes place over a very wide range of frequencies:
from a few hundreds of kHz to a few GHz.
 Optical communication using fibers is performed in the frequency range of 1 THz to 1000
THz(microwaves to ultraviolet).
 An optical fiber can offer a transmission bandwidth in excess of100GHz.
MODE OF COMMUNICATION OF ELECTRO MAGNETIC WAVES:
GROUND WAVE PROPAGATION (up to few MHz)
 The antennas should have a size comparable to the wavelength λ of the signal (at least ~ λ/4).
 At longer wavelengths (i.e., at lower frequencies), the antennas have large physical size and they are
located on or very near to the ground.
 The wave glides over the surface of the earth.
 A wave induces current in the ground over which it passes and it is attenuated as a result of absorption
of energy by the earth.
 The attenuation of surface waves increases very rapidly with increase in frequency.
 The maximum range of coverage depends on the transmitted power and frequency (less than a few
MHz)
SKY WAVE PROPAGATION (Frequency range from a few MHz up to 30-40 MHz)
•
Communication can be achieved by ionospheric reflection of radio waves back towards the earth.
•
Ionosphere extends from a height of ~ 65 Km to about 400 Km above the earth’s surface.
•
The ionospheric layer acts as a reflector for a certain range of frequencies (3 to 30 MHz).
•
Electromagnetic waves of frequencies higher than 30 MHz penetrate the ionosphere and escape.
SPACE WAVE PROPAGATION (Frequency greater than 40MHz)
A space wave travels in a straight line from transmitting antenna to the receiving antenna.Space waves are
used for line-of-sight (LOS) communication as well as satellite communication
 A space wave travels in a straight line from transmitting antenna to the receiving antenna.
 At these frequencies, the antennas are relatively smaller and can be placed at heights of many
wavelengths above the ground.
 If the transmitting antenna is at a height hT ,then you can show that the distance to the horizon ddT is
given as
 dt = √2RhT
 the maximum line-of-sight distance dM between the two antennas having heights hT and hR above the
earth is given by
MODULATION
NEED OF MODULATION
Size of the antenna or aerial
 For transmitting a signal, antenna should have a size comparable to the wavelength of the signal (at least
L=λ/4 ) that the antenna properly senses the time variation of the signal.
 For an electromagnetic wave of frequency 20 kHz, the wavelength L is 15 km. Obviously, such a long
antenna is not possible to construct and operate.
 If transmission frequency is high (for example, if ν is 1 MHz, then L is 300 m).
 Hence to decrease the size of antenna modulation is done.
Effective power radiated by an antenna
• A theoretical study of radiation from a linear antenna (length L) shows that the power radiated is
𝑳 𝟐
proportional to (𝝀)
•
•
The power radiated increases with decreasing L, i.e., increasing frequency.
For a good transmission, we need high powers and hence this also points out to the need of using high
frequency transmission.
Mixing up of signals from different transmitters
To avoid mixing up of signals there is a need for translating the original low frequency baseband
message or information signal into high frequency wave before transmission
TYPES OF MODULATION
• There are three types of modulations
 (i) Amplitude modulation (AM),
 (ii) Frequency modulation (FM) and
 (iii) Phase modulation (PM),
•
•
•
•
•
•
AMPLITUDE MODULATED WAVE
Amplitude modulation is the process in which the amplitude of the carrier wave changes in accordance
with the instantaneous value of the message signal.
Let 𝑐(𝑡) = 𝐴𝑐 sin 𝜔𝑐 t
m(𝑡) = 𝐴𝑚 sin 𝜔𝑚 t
𝑐𝑚 (𝑡)= (𝐴𝑐 + 𝐴𝑚 sin 𝜔𝑚 t) sin 𝜔𝑐 t
𝑐𝑚 (𝑡)= 𝐴𝑐 sin 𝜔𝑐 t + 𝜇𝐴𝑐 sin 𝜔𝑐 t sin 𝜔𝑚 t
µ = Am/Ac is the modulation index; in practice,µ is kept ≤1 to
avoid distortion.
𝑐𝑚 (𝑡)=𝐴𝑐 sin 𝜔𝑐 t+
𝜇𝐴𝑐
2
cos(𝜔𝑐 − 𝜔𝑚 )t+
𝜇𝐴𝑐
2
cos(𝜔𝑐 + 𝜔𝑚 )
•
Here (𝜔𝑐 − 𝜔𝑚 ) and (𝜔𝑐 + 𝜔𝑚 ) are respectively called the lower side and upper side frequency.
PRODUCTION OF AM WAVE
INPUT OF SQUARE LAW DEVICE
𝑥(𝑡) = 𝑚(𝑡) + 𝑐(𝑡) = 𝐴𝑚 sin 𝜔𝑚 t + 𝐴𝑐 sin 𝜔𝑐 𝑡
OUTPUT OF SQUARE LAW DEVICe
𝑦(𝑡) = 𝐵𝑥(𝑡) + 𝐶𝑥 2 (𝑡)
• The output of square law device is passed through a band pass filter which rejects dc and the sinusoids
of frequencies ωm , 2ωm and 2 ωc and retains the frequencies ωc , (ωc –ωm) and (ωc + ωm) . The output of
the band pass filter therefore is of the same form as obtained earlier and is therefore an AM wave.
• BLOCK DIAGRAM OF TRANSMITTER

BLOCK DIAGRAM OF RECEIVER
DETECTION OR DEMODULATION
 It is the process of retrieval of message signal from the amplitude modulated wave
1.
2.
3.
4.
5.
6.
7.
8.
SOME IMPORTANT QUESTIONS FOR PRACTISE
Explain ground wave, Sky wave and space wave and ground wave propagation with suitable example.
What is the function of following in communication system?
(a) Reateter (b) Transducer
Define modulation index. Why the amplitude of modulating signal kept less than amplitude of carrier
wave?
Distinguish between point to point and broadcast communication modes with example.
What is meant by demodulation? With the help of block diagram explain the process of demodulation.
For an amplitude modulated wave, the maximum amplitude is found to be 10 V while the minimum
amplitude is 2 V. Determine the value of modulation index. What would be the value of modulation
index if the minimum amplitude is zero volts? Why is modulation index generally kept less than 1.
A carrier wave of frequency 1·5 MHz and amplitude 50 V is modulated by a sinusoidal wave of
frequency 10 kHz producing 50% amplitude modulation. Calculate the amplitude of the AM wave and
frequencies of the side bands produced.
Block diagram of receiver is shown in the fig. Identify X and Y and write its functions
9. Define amplitude modulation and draw amplitude modulated and frequency modulated wave.
10. A transmitting antenna at the top of tower has a height of 36m and the height of receiving antenna is
49m. What is the maximum distance between them for satisfactory communication in the LOS mode?
(radius of earth = 6400Km.)
11. Draw block diagram of simple modulator to produce amplitude modulated wave.
12. A schematic arrangement for transmitting a message signal (20 Hz to 20KHz) is given below
Write two drawbacks from which the arrangement suffers and draw the correct diagram.,