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KENDRIYA VIDYALAYA SANGATHAN
REGIONAL OFFICE, RAIPUR
STUDY cum SUPPORT MATERIAL
CLASS XII
PHYSICS
2015 – 2016
PATRON
Smt. P.B. Usha
DEPUTY COMMISSIONER
KVS RO RAIPUR
COORDINATOR
Sh. R.N. Sendhil Kumar
PRINCIPAL
K V No.2, BOLANGIR
PREPARED BY:
1. Mr. Y.K. Tiwari, PGT(Phy), K.V. Durg
2. Mr. S.K. Mishra, PGT(Phy), K.V. No.2, Raipur
3. Mr. V. K. Verma, PGT (Phy), K.V. No.4, Korba
4. Mr. B.R. Gajpal , PGT (Phy) , K.V. Bilaspur
5. Mr. H. S. Tripathi, PGT (Phy), K.V. Mahasamund
6. Mrs. S Khirbat, PGT (Phy), K.V. BMY Bhilai
7. Mr. Pankaj Kumar Sinha, PGT (Phy), K.V. Dantewada
CLASS XII SYLLABUS-2015-16
(THEORY)
One Paper Time: 3 hrs.
Max Marks: 70
NAME OF CHAPTER
MARKS
UnitI
Electrostatics
15
UnitII
Current Electricity
UnitIII
Magnetic Effect of Current and Magnetism
UnitIV
Electromagnetic Induction and Alternating
Current
UnitV
Electromagnetic Waves
UnitVI
Optics
UnitVII
Dual Nature of Matter
UnitVIII
Atoms and Nuclei
Unit IX
Electronic Devices
UnitX
Communication Systems
TOTAL
16
17
10
12
70
Unit I: Electrostatics
Electric Charges; Conservation of charge, Coulomb's law-force between two point charges,
forces between multiple charges; superposition principle and continuous charge distribution.
Electric field, electric field due to a point charge, electric field lines, electric dipole, electric
field due to 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
long straight 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
system of charges; equipotential surfaces, electrical potential energy of a system of two point
charges and of electric dipole in an electrostatic field.
Conductors and insulators, free charges and bound charges inside a conductor. Dielectrics and
electric polarisation, 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
their relation with electric current; Ohm's law, electrical resistance, V-I characteristics (linear
and nonlinear),electrical energy and power, electrical resistivity and conductivity. Carbon
resistors, colourcode for carbon resistors; series and parallel combinations of resistors;
temperature dependence ofresistance.
Internal resistance of a cell, potential difference and emf of a cell, combination of cells in series
and in parallel. Kirchhoff's laws and simple applications. Wheatstone bridge, metre bridge.
Potentiometer - principle and its applications to measure potential difference and for comparing
EMF 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 parallel
current-carrying conductors-definition of ampere. Torque experienced by a current loop in
uniform magnetic field; moving coil galvanometer-its current sensitivity and conversion to
ammeter and voltmeter.
Current loop as a magnetic dipole and its magnetic dipole moment. Magnetic dipole moment of
a revolving electron. Magnetic field intensity due to a magnetic dipole (bar magnet) along its
axis and perpendicular to its axis. Torque on a magnetic dipole (bar magnet) in a uniform
magnetic field; bar magnet 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
their strengths. 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
ideas only). 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 25
Reflection of light, spherical mirrors, mirror formula. Refraction of light, total internal
reflection and its applications, optical fibres, refraction at spherical surfaces, lenses, thin lens
formula, lens maker'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 colour of sky and reddish apprearance of the sun at sunrise and sunset.
Optical instruments : Microscopes and astronomical telescopes (reflecting and refracting) and
their magnifying powers.
Wave optics: Wave front and Huygen's principle, reflection and refraction of plane wave at a
plane surface using wave fronts. Proof of laws of reflection and refraction using Huygen's
principle. Interference, Young's double slit experiment and expression for fringe width,
coherent sources and sustained interference of light. Diffraction due to a single slit, width of
central maximum. Resolving
power of microscopes and astronomical telescope. Polarisation, plane polarised light, Brewster's
law, uses of plane polarised light and Polaroids.
Unit VII: Dual Nature of Matter and Radiation
Dual nature of radiation. Photoelectric effect, Hertz and Lenard's observations; Einstein's
photoelectric 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 their
properties; 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; I-V
characteristics 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
(common emitter 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
and digital data); bandwidth of transmission medium. Propagation of electromagnetic waves in
the atmosphere, sky and space wave propagation. Need for modulation. Production and
detection of an amplitude-modulated wave. Basic ideas about internet, mobile telephony and
global positioning system (GPS)
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
ELECTROSTATICS
Frictional Electricity:
Column I (+ve Charge)
Column II (-ve Charge)
Glass
Silk
Wool, Flannel
Amber, Ebonite, Rubber, Plastic
Ebonite
Polythene
Dry hair
Comb
Coulomb’s Law:
q1 q2
F=k
In vacuum,
r2
1
4πε0
k=
In medium, k =
1
where ε is the absolute electric permittivity of the
dielectric medium
4πε
The dielectric constant or relative permittivity or specific inductive capacity or
dielectric coefficient is given by
K = εr =
In
F=
In medium, F =
1
q1 q2
4πε0
r2
1
4πε0εr

ε0
q1 q2
r2
ε0 = 8.8542 x 10-12 C2 N-1 m-2
1
4πε0
= 8.9875 x 109 N m2 C-2
or
1
4πε0
= 9 x 109 N m2 C-2
Coulomb’s Law in Vector Form:
+ q1
In vacuum, for q1 q2 >
F12 =
F21 =
1
q1 q2
4πε0
r2
1
q1 q2
4πε0
r2
+ q2
r12
r
F12
r21
F21
q1q2 > 0
r12
- q1
- q2
r12
r
F12
F21
q1q2 > 0
In vacuum, for q1 q2 <
F12 =
1
q1 q2
4πε0
r2
r12 &
F21 =
1
q1 q2
4πε0
r2
+ q1
r21
r12
- q2
F12
F21
r
(in all the cases)
F12 = - F21
F12 =
1
4πε0
q1 q2
3
r
r12
&
q1q2 < 0
F21 =
1
4πε0
q1 q2
r21
3
r
Note: The cube term of the distance is simply because of vector form.
Otherwise the law is ‘Inverse Square Law’ only.
Dielectric Constant or Relative Permittivity:
ε
ε0
K = εr =
Fv
K = εr =
Fm
Linear Charge Density:
λ=
dq
d
l Density:
Surface Charge
σ=
dq
dS
Volume Charge Density:
dq
ρ=
d‫ז‬
Electric Field:
+q
+ q0
-q
+ q0
F
F
q – Source charge, q0 – Test charge, F – Force & E - Field
E=
F
Lt
∆q → 0 ∆q
E=
or
Density of electric lines of force:
∆N
α E
∆A
Electric field due to a point charge:
1
q
4πε0
r2
E (r) =
r
F
q0
or
E=
1
q
4πε0
r2
r
Y
F
+ q0
P (x,y,z)
r
+q
X
O
Z
E
r2
0
Electric Field in terms of co-ordinates:
E (r) =
q
1
4πε0
Superposition Principle:
2
2
2
(x +y +z )
3/2
( x i + y j +zk )
F14
- q5
+ q1
+ q2
F15
F12
F13
- q3
+ q4
F12
F1
F15
F13
F14
F1 = F12 + F13 + F14 + F15
Fa (ra) =
1
4πε0
N
∑ qa qb
b=1
b≠a
ra - rb
│ra - rb │3
Dipole and its Moment:
p
p = (qx 2L)
i
-q
+q
2L
Electric Field due to a dipole:
i) At a point on the axial line:
EP = EB - EA
A
B
O
-q
l
│EP │ =
EP =
EB
P
+
q
p
l
EA
x
1
2px
4πε0
(x2 – l2)2
1
2px
4πε
(x2 – l2)2
If l << x, then
EP ≈
i
2p
4πε0
x3
The direction of electric field intensity at a point on the axial
line due to a dipole is always along the direction of the dipole
moment.
ii) At a point on the equatorial line:
EB
EB
θ
EQ
Q
θ
EB cos θ θ
EQ
EA
EA cos θ θ
y
A
θ
θ
O
-q
B
EA
EB sin θ
Q
EA sin θ
+q
p
l
l
EQ
=
1
q
p
4πε0 ( x2 + l2 )3/2
If l << y, then
EQ ≈
p
4πε0 y3
The direction of electric field intensity at a point on the equatorial line due to a dipole is parallel and
opposite to the direction of the dipole moment.
If the observation point is far away or when the dipole is very short, then the electric field intensity at a
point on the axial line is double the electric field intensity at a point on the equatorial line.
i.e. If l << x and l << y, then EP = 2 EQ
Torque:
p
θ
E
t = pxE
t
Case i: If θ = 0°, then t = 0.
Case ii: If θ = 90°, then t = pE
(maximum value).
Case iii: If θ = 180°, then t = 0.
Work done on a dipole:
W = p E (cosθ1 - cos θ2)
Potential Energy U = - p E cos θ
Note: Potential Energy can be taken zero arbitrarily at any position of the dipole.
Case i: If θ = 0°, then U = - pE
(Stable Equilibrium)
Case ii: If θ = 90°, then U = 0
Case iii: If θ = 180°, then U = pE (Unstable Equilibrium)
Work done:
B
WAB = dW = - E . dl =
A
qq0
4πε0
Work done by electric field in a closed loop is zero.
B
E . dl = 0
A
1
1
[r
B
-
rA
]
Electric Potential:
WAB
q0 =
4πε0
W∞B
q0 =
1
1
q
[r
-
B
rA
]
q
W∞B
V=
q0
=V
4πε0 r
Electric Potential Difference:
VB - VA
WAB
= ∆V =
q0
Electric Potential due to a single point charge:
E
dx +q0
B
+q
Q
r
x
q
V =
4πε0 r
Electric Potential due to group of point charges:
P
q0E
∞
q1
r1
qn
rn
q2
+1 C
r2
P
r3
r4
q4
V=
q3
1
n
qi
i=1
│ r - ri│
∑
4πε0
( in terms of position
vector )
Electric Potential due to an Electric Dipole:
i) At a point on the axial line:
A
B
-q
p
O
+q
l
l
x
VP =
1
p
4πε0 (x2 – l2)
ii) At a point on the equatorial line:
+1 C
P
Q
y
A
θ
θ
-q
p
O
+q
l
l
B
q
VQ = 0
Electrostatic Potential Energy:
i) Electrostatic Potential Energy
of a Two Charges System:
U =
1
q1q2
4πε0
│ r2 - r1
│
Y
A (q1)
r1
B (q2)
r2
or
U=
O
1
q1q2
4πε0
r12
Electric Flux:
dΦ = E . dS = E dS cos θ
r2 - r1
Z
X
Φ=
E . dS = E S cos θ = E . S
S
Special Cases:
• For 0° < θ < 90°, Φ is
positive.
• For θ = 90°, Φ is
zero.
• For 90° < θ < 180°, Φ
is negative.
Gauss’s Theorem:
1
ΦE =
E . dS =
S
ε0
n
∑ qi
i=1
1. Electric Field Intensity due to an Infinitely Long Straight Charged Wire:
E=
1
4 πε0
2λ
r
The direction of the electric field intensity is radially outward from the positive line charge. For negative
line charge, it will be radially inward.
Note:
The electric field intensity is independent of the size of the Gaussian surface constructed. It depends only
on the distance of point of consideration. i.e. the Gaussian surface should contain the point of
consideration.
2. Electric Field Intensity due to an Infinitely Long, Thin Plane Sheet of Charge:
E=
=
σ
2 ε0
The direction of the electric field intensity is normal to the plane and away from the positive charge
distribution. For negative charge distribution, it will be towards the plane.
Note:
The electric field intensity is independent of the size of the Gaussian surface constructed. It neither
depends on the distance of point of consideration nor the radius of the cylindrical surface.
If the plane sheet is thick, then the charge distribution will be available on both the sides. So, the charge
enclosed within the Gaussian surface will be twice as before. Therefore, the field will be twice.
σ
E=
ε0
3. Electric Field Intensity due to a Uniformed Charged This Spherical Shell:
i) At a point P outside the shell:
Electric field due to a uniformly charged thin
spherical shell at a point outside the shell is
such as if the whole charge were concentrated
at the centre of the shell.
q
E=
4πε0 r2
σ R2
E=
ε0 r2
ii) At a point A on the surface of the shell:
q
E=
4πε0 R2
σ
E=
ε0
Electric field due to a uniformly
charged thin spherical shell at a
point on the surface of the shell is
maximum.
iii) At a point B inside the shell:
E=0
This property E = 0 inside a cavity is used for electrostatic shielding.
Electric Field Intensity due to a Uniformed Charged This Spherical Shell:
Electrical Capacitance:
q
C=
V
E
E
ma
x
O
R
r
Capacitance of an Isolated Spherical Conductor:
C = 4πε0 r
Capacitance of Parallel Plate Capacitor:
C=
A ε0
d
A ε0 εr
C=
d
(with dielectric medium)
Series Combination of Capacitors:
C
q
1
V
C
q
C
q
2
V
V
V
1/C = 1/C1 + 1/C2 + 1/C3
Parallel Combination of Capacitors:
3
VC q
1
V
V
C
2
C
1
q
2
q
3
3
V
Energy Stored in a Capacitor:
q
q
U=
0
dq
U
1
2
ε0 Ad
q2
2
C
U=
or
C
Energy Density:
C = C1 + C2
1
+ C3
1
U
ε0
=
A
2
o
r
Loss of Energy on Sharing of Charges between the Capacitors in Parallel:
Total charge after sharing = Total charge before sharing
(C1 + C2) V = C1 V1 + C2 V2
C1 V1 + C2 V2
V=
C1 + C2
The total energy before sharing is
1
1
C1 V12
Ui =
2
+
C2 V22
2
The total energy after sharing is
Ui – Uf =
Ui – Uf > 0
C1 C2 (V1 – V2)2
2 (C1 + C2)
or Ui > Uf
Therefore, there is some loss of energy when two charged capacitors are connected together.
The loss of energy appears as heat and the wire connecting the two capacitors may become hot.
Polarization of Dielectrics:
The dielectric constant is given by
K=
E0
E0 - Ep
Capacitance of Parallel Plate Capacitor with Dielectric Slab:
C > C0. i.e. Capacitance increases with introduction of dielectric slab.
Dielectric Constant:
K=
C
C0
ELECTROSTATICS PREVIOUS YEAR QUESTION FOR PRACTICE
1.Why do the electric field lines never cross each other ?
2.A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another
uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined
system to that stored initially in the single capacitor.
ence of a uniform
electric field E.
(b) Consider two hollow concentric spheres, S1 and S2, enclosing charges 2Q and 4Q respectively as shown in
the figure. (i) Find out the ratio of the electric flux through them. (ii) How will the electric flux through the
Deduce the necessary expression.
4. Two charges of magnitudes – 2Q and + Q are located at points (a, 0) and (4a, 0) respectively. What is the
electric flux due to these charges through a sphere of radius ‘3a’ with its centre at the origin?
5. A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate
capacitor but has the thickness d/2, where d is the separation between the plates. Find out the
expression for its capacitance when the slab is inserted between the plates of the capacitor.
6. Draw the equipotential surfaces due to an electric dipole. Locate the points where the potential
due to the dipole is zero.
7. A charge ‘q’ is placed at the centre of a cube of side l. What is the electric flux passing through
each face of the cube?
8. The dipole is aligned parallel to the field. Find the work done in rotating it through the angle
of 180°
9. A test charge ‘q’ is moved without acceleration from A to
along the path from A to B and then
from B to C in electric field E as shown in the figure. (i)
Calculate the potential difference between
A and C. (ii) At which point (of the two) is the electric
potential more and why?
C
10. Plot a graph showing the variation of coulomb force (F) versus 1/r2 , where r is the distance between the two
–
11. Net capacitance of th
connected in parallel? Find the ratio of energy stored in the two configurations if they are both connected to the
same source.
12. Name the physical quantity whose S.I. unit
13.Show that the electric field at the surface of a charged conductor is given by E=
/ε0 n
surface charge density and n is a unit vector normal to the surface in the outward direction.
figure. Determine (a) equivalent capacitance of the network and (b) charge on each capacitor.
15.Draw 3 equipotential surfaces corresponding to a field that uniformly increases in magnitude but remains
constant along Z-direction. How are these surfaces different from that of a constant electric field along Zdirection?
16. Two identical parallel plate (air) capacitors C1 and C2 have capacitances C each. The between their plates is
now filled with dielectrics as shown. If the two capacitors still have equal capacitance, obtain the relation
between dielectric constants K, K1 and K2 .
17
SOLUTION
1. At the point of intersection two direction of electric field which is not possible.
2.Since charge remains constant U=q2/2C after connection capacitance =2C
Therefore U’/U=1/2 U’ is energy after connection.
3.(b)Φ1/Φ2=2/6=1/3 when dielectric is filled then Φ1’=Φ/εr
4.Φ=-2Q/ε0
5.C=2KCo/K+1 Where C0=Q0/V0
6.
Electric potential is zero at all points in the plane passing through the dipole equator.
7.q/6ε0
8. W
PE
9. (i) VC – VA = 4E
(ii) VC > VA
Because direction of electric field is in decreasing potential.
10.
Since, magnitude of the slope is more for attraction, therefore, attractive force is greater than
repulsive force.
11 When these capacitors are connected in parallel, net capacitance, Cp
C
When these two combinations are connected to same source the potential difference across each
combination is same.
Ratio of energy stored,
Us: Up = 1: 9
12.Potential and scalar quantity.
13.
14. (a) C1, C2 and C3 are in series, their equivalent capacitance C
C
C4 is in parallel with C
Ceq C
C
(b) Charge on capacitor C4 is
q
C4V
15.K=(K1+K2)/2
16.
17.Φ=1.05Nm2C-1., Q=9.27x10-12C
CONCEPT MAP
PREPARED BY RANCHI REGION
CONCEPT MAP
ELECTROSTATICS
CONCEPT MAP
PREPARED BY RANCHI REGION
ELECTROSTATICS
CONCEPT MAP
PREPARED BY RANCHI REGION
ELECTROSTATICS
Whose
reciprocal is
called
Capacitance (C)
Elastance
S.I Unit
Farad(F)
Device giving large
capacity in similar
space
Capacity
depends
upon
Capacitor
Mode of combination
With dielectric
Potential Energy
 W=(1/2) Q xV
 W=(1/2)CV2
 W=(1/2)Q2/C

Common Area
CαA
 Distance C α
(1/d)
 Medium in
between
CαK
Energy Density = ½ so E2
Without dielectric
Series
1/c=1/k1 c1+1/k2 c2
Series
PARALLEL
1/c=1/c1+1/c2
C=c1+c2
ELECTROSTATICS
Parallel C=k1 c1+k2 c2
ccc
c
Spherical
Parallel Plate
C=εoA/d

Outer Sphere Earthed
Cylindrical
C=2πεol/2.3026 ln (b/a)
ab
C=4πεob−a

Inner Sphere Earthed C=4πεob
K= εr = ε/ε0= F0/FM = C/C0 = E0/E = V0/V >1
CURRENT ELECTRICITY
I
a
b
c
d
0
t
Drift Velocity and Current:
vd = - (eE / m) τ
vd = a τ
I = neA vd
Current is directly proportional to drift velocity.
Current density:
J = I / A = nevd
In vector form, I = J . A
Ohm’s Law:
IαV
or V α I or V = R I
I
0
V
Resistance in terms of physical features of the conductor:
m
R = l2
ne τ A
R =ρ
m
l
where ρ =
A
is resistivity or specific
resistance
ne2τ
Relations between vd , ρ, l, E, J and V:
ρ=
E / J = E / nevd
(since, J = I / A = nevd )
(since, E = V / l )
vd = E /(neρ)
When temperature
increases,
vd decreases and ρ
increases.
When l increases,
vd decreases.
vd = V /(neρl)
Temperature coefficient of Resistance:
R2 –
Rt –
o α
α R
R1t2 –
r
If R2 < R1, then α
is – ve.
R0 – Resistance at 0°C
Rt – Resistance at t°C
R1 – Resistance at t1°C
R2 – Resistance at t2°C
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.
Eg.
B V B Gold
17 x 100 = 17 ± 5% Ω
C
AB x 10 ± D % ohm
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
G R B Silver
52 x 106 ± 10% Ω
BVB
52 x 100 = 52 ± 20% Ω
B B ROY of Great Britain has Very
Good Wife
Series combination of resistors:
R = R1 + R2 + R3
R1
R3
R2
R is greater than the greatest of all.
Parallel combination of resistors:
R1
1/R =1/R1 + 1/R2 + 1/R3
R2
R is smaller than the smallest of all.
R3
Internal Resistance of a cell:
E r
v
I
I
R
V
E =V+v
= IR + Ir
= I (R + r)
I = E / (R + r)
This relation is called circuit equation.
E
r = ( ----- - 1) R
V
Cells in Series combination:
Er
Er
Er
I
I
R
V
(i) If R << nr, then I = E / r (ii) If nr << R, then I = n (E / R)
Current I =
nE
nr + R
Conclusion: When internal resistance is negligible in
comparison to the external resistance, then the cells are
connected in series to get maximum current.
Cells in Parallel combination:
E r
E r
I
E r
I
R
V
Current I =
nE
nR + r
(i) If R << r/n, then I = n(E / r) (ii) If r/n << R, then I = E / R
Conclusion: When external resistance is negligible in
comparison to the internal resistance, then the cells are
connected in parallel to get maximum current.
KIRCHHOFF’S LAWS:
I Law or Current Law or Junction Rule:
The algebraic sum of electric currents at a junction in any
electrical network is always zero.
I2
I1
I3
O
I5
I1 - I2 - I3 + I4 - I5 = 0
I4
Sign Conventions:
•
•
The incoming currents towards the
junction are taken positive.
The outgoing currents away from the
junction are
negative.
Note: The charges cannot accumulate
at ataken
junction.
The number of
charges that arrive at a junction in a given time must leave in the
same time in accordance with conservation of charges.
II Law or Voltage Law or Loop Rule:
The algebraic sum of all the potential drops and emf’s along any
closed path in an electrical network is always zero.
I1
I1 E1
R1
A
Loop ABCA:
B
I2
I1 + I2
D
I2
E2
R2
- E1 + I1.R1 + (I1 + I2).R2 = 0
I1
R3
Loop ACDA:
C
I2
- (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. Note: The path can be traversed
in clockwise or anticlockwise
direction of the loop.
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.
B
Wheatstone Bridge:
P
Currents through the arms are assumed
by applying Kirchhoff’s Junction Rule.
Applying Kirchhoff’s Loop Rule for:
Q
I1
I1 - I g
Ig
A
G
C
Loop ABDA:
S
R
-I1.P - Ig.G + (I - I1).R = 0
Loop BCDB:
I
II1
I - I1 +
D Ig
- (I1 - Ig).Q + (I - I1 + Ig).S + Ig.G = 0
When Ig = 0, the bridge is said to balanced.
I
E
I
I
By manipulating the above equations, we get
P
R
Q
S
Metre Bridge:
R.B (R)
X
G
A
B
l cm
J 100 - l cm
R
E
Potentiometer:
l
K
X
100 - l
I
+
E
V
A
0
l cm J
100
200
A
+
300
Rh
B
400
K
V
0
l
Vαl
V / l is a constant
Comparison of emf’s using Potentiometer:
E
I
E
A
A
+
0
1
B
K
400
G
l2 J
l
200
Rh
R.
B
+
+
E
100
J
300
A drum is attached to the axle of the wheels of a train. It rotates when the train is moving. To stop the train, a
strong magnetic field is applied to the rotating drum. The eddy currents set up in the drum oppose the
rotation of the drum and the train stops.
CURRENT ELECTRICITY PREVIOUS YEAR FOR PRACTICE
1.Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.0 x 10–7 m2
carrying a current of 1.5 A. Assume the density of conduction electrons to be 9 x 1028 m–3.
2.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.
3.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 ?
4.A resistance of R Ω draws current from a potentiometer as shown in the figure. The potentiometer has a total
resistance Ro Ω. A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R
when the sliding contact is in the middle of the potentiometer.
5.Two identical cells, each of emf E, having negligible internal resistance, are connected in parallel with
each other across an external resistance R. What is the current through this resistance?
6.Define the current sensitivity of a galvanometer. Write its S.I. unit.
Figure shows two circuits each having a galvanometer and a battery of 3 V.
When the galvanometers in each arrangement do not show any deflection, obtain the ratio R1 / R2
7. Two wires of equal length, one of copper and the other of manganin have the same resistance.
Which wire is thicker?
8.Write principle of potentiometer.
9.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?
10. In the given circuit, assuming point A to be at zero potential, use Kirchhoff’s rules to determine
the potential at point B.
11.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?
12.Two heating elements of resistance R1 and R2 when operated at a constant supply of voltage, V,
consume powers P1 and P2 respectively. Deduce the expressions for the power of their
combination when they are, in turn, connected in (i) series and (ii) parallel across the same voltage
supply.
13. 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.
14.Write any two factors on which internal resistance of a cell depends. The reading on a high resistance
s of the cell are also connected to
the cell.
15. 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.
16. Calculate the current drawn from the battery in the given network.
17. 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.
I(ii) For what current i will the power dissipation of the circuit be maximum ?
18. 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.
SOLUTION
1.
2.
I=neavd
vd=1.041x10-3m/s
Negative slope represents r and intercept represent on v axis represents E.
3 (a)Thick strip and low resistivity of copper have negiligible resistance.
(b)To increase sensitivity
(c) Low thermal coefficient
4.Voltage across is =(2R/Ro+4R) V Ncert-p-123
5.I=E/R
6.Current sensitivity-deflection produced in the galvanometer when unit current flows through it
Unit rad/ampere
R1/R2=3:2
7. Manganin
8. Potential drop across a wire is directly proportional to its length provided the current is constant and wire is
of uniform cross section area.
9..VBE=1V
10.VB=1V
11(a).New balance point l’=100-l (b) No change
12. Ps=P1P2/(P1+P2) Pp=P1+P2
13.In Series current remains constant I=neAvd therefore vdx/vdy=1/2
14. The internal resistance of a cell depends on
(i) distance (l) between electrodes.
(ii) area (A) of immersed part of electrode, and
(iii) nature and concentration of electrolyte.
Internal resistance=1.1 V
15 I1=2/13,I2=7/13,I3=9/13
16.I=2A
17.emf of each cell is 2V and I=1A
18.(a) Vdx/vdy=3/2 (b)vdx/vdy=2/1
CONCEPT MAP
PREPARED BY GUWAHATI REGION
ZIET BHUBANESWAR- PHYSICS WORKSHOP FROM 06.08.2013 TO 08.08.2013
CURRENT ELECTRICITY
C0NCEPT MAP
PREPARED BY GUWAHATI REGION
CURRENT ELECTRICITY
÷E
Is determined
by
÷A
T
T
T
=Temperature
CURRENT ELECTRICITY
MAGNETIC EFFECTS OF CURRENT & MAGNETISM
Gist of the unit
1. Moving electron produces electric and magnetic field both. A Stationary electron produces electric field
only.
2. In a perpendicular field a charge particle transverse a circular path. There will be no change in speed of
particle.
3. The source of magnetic field is a current element or moving charge.
4. The magnetic field strength at a point is zero when the point lies along the direction of current
element.
5. A magnetic field does not cause any force on a stationary charge.
6. The path of a charged particle in a transverse magnetic field is circle.
7. The path of a charged particle entering obliquely in a magnetic field is a helix.
8. The magnetic force only causes a change in direction of a moving charge, it does not impart kinetic energy
to the moving charge. This means that the kinetic energy of a charged particle in uniform magnetic field
remains constant.
9. A cyclotron is used to accelerate charged particles but not used to accelerate electrons.
10.
An ideal ammeter has zero resistance.
11.
An ideal voltmeter has infinite resistance.
12. To make scale of galvanometer linear, the magnetic field is made radial by using cylindrical pole pieces.er
13. The nature of magnetic field in a moving coil galvanometer is radial.
14. The value of dip increases from 0° to 90° as we move from earth’s equator to north pole.
Formulae
Biot-Savart law (Magnetic field due to current element)
0 I dl sin
dB
r2
4
Force acting on a charge moving in a magnetic field
B
F = qvBsin
or
F = q( v
)
Magnetic field on the axis of a circular current loop
NIa2
0
B
3 .
2
2
2 (r a ) 2
Magnetic field due
to an infinitely long straight current carrying wire
0I
B
2
Magnetic field at a point on the axis of a solenoid
B
0nI
Motion of a charged particle in a uniform magnetic field
mv
r=
2
T
qB
qB
Torque on a rectangular coil in a uniform magnetic field
= NIBA sin
Force per unit length acting on each of the two straight parallel metallic conductors carrying
current
0 I1
I
F2
2
f
l
2
Deflection in moving coil galvanometer
∝=NIBA/K
Conversion of galvanometer into ammeter and voltmeter
I gG
V
S
R
RG
I I
Ig
g
Elements of Earth’s magnetic field
H BEcos
V BEsin
P.E. of a magnetic dipole in a uniform magnetic field
U
mBcos
Magnetic dipole moment of a revolving electron
M=evr/2=neh/4πme
Magnetising field intensity
H =nI
Concepts for Slow and average Learners
1.
2.
3.
4.
5.
6.
7.
8.
Application of Biot-Savart’s law- current carrying circular loop.
Cyclotron
Moving coil galvanometer
Conversion of galvanometer into voltmeter and ammeter.
Expression for Solenoid and toroid.
Magnetic elements of earth’s surface
Classification of magnetic materials.
Classification of magnetic materials.
ONE MARK QUESTIONS
1.
What is the direction of the force acting on a charge particle q, moving with a velocity v in a uniform
magnetic field B?
Ans: Force, F=q vB sinθ
Obviously, the force on charged particle is perpendicular 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 +yaxis. What is the direction of the magnetic field?
Ans: By Fleming’s left hand rule magnetic field must be along negative Z-axis
5.
Ans:
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.
6. An electron and a proton moving with the same speed enter the same magnetic field region at right angles
to the direction of the field. Show the trajectory followed by the two particles in the magnetic field. Find
the ratio of the radii of the circular paths which the particles may describe.
7.
The permeability of a magnetic material is 0.9983. Name the type of magnetic materials it represents.
Ans: As permeability < 1, so magnetic material is diamagnetic.
8.
Where on the surface of Earth is the angle of dip zero?
Ans: Angle of dip is zero at equator of earth’s surface.
9.
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?
mv
1
Ans: As
r
qB
q
So, rp:rd
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.
Ans:
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?
negative z direction.
12.
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.
Ans:
Define current sensitivity and voltage sensitivity of a galvanometer. Increasing the
current sensitivity may not necessarily increase the voltage sensitivity of a galvanometer. Justify.
Current sensitivity :It is defined as the deflection of coil per unit current flowing in it.Current
Sensitivity, S=NAB/ C
Voltage sensitivity :It is defined on the deflection of coil per unit potential differenceacross its
ends.
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.
Ans:
L
L
N
r
2N
2
NI
0
I
0
B
2r
5.
L
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?
105
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:
The magnetic susceptibility of diamagnetic substance is small and negative but that of paramagnetic
substance is small and positive.
8.
Ans:
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
speedv. The revolving electron is equivalent to electric
current
e
e
ev
I
/
T 2v
2
Area of current loop (electron orbit), A = p r2
Magnetic moment due to orbital motion, M= IA=evr/2
9.
Ans:
A steady current (I1) flows through a long straight wire. Another wire carrying steady current (I2) in the
same direction is kept close and parallel to the first wire. Show with the help of a diagram how the
magnetic field due to the current I1 exerts a magnetic force on the second wire. Write the expression for
this force.
Suppose two long thin straight conductors (or wires) PQ and RS are placed parallel to each other in
vacuum (or air) carrying currents I1 and I2 respectively. It has been observed experimentally that when
the currents in the wire are in the same direction, they experience an attractive force and when they
carry currents in opposite directions, they experience a repulsive force.
The magnetic field produced by current I1 at any point on wire RS is
0I1
B
2
This field acts perpendicular to the wire RS and points into the plane of paper. It exerts a force on
current carrying wire RS. The force acting on length l of the wire RS will be
F
lB1
I 2 I2
0 I1
l.
2
0 I1 I 2 l
2
Force per unit length,
F2
0 I1
I2
f
l
2
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
Length of wire remains same, so NB = NA/2 Thus
mA/mB= 2/1
10.
11.
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.
12.
A circular coil of N turns and radius R carries a current I. It is unwound and rewound to make another
coil of radius R/2, current I remaining same. Calculate the ratio of the
magnetic moments of the now coil and the original coil. Ans:
M1= NIA= NIπR2
M2=2 πNI (R/2)2
M2/M1 = ½
13. 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. 1. State Biot-Savart law for magnetic field produced at a point due to a small current element. How will you
find the direction of magnetic field?
Ans. Consider a current element dl
position vector r
of a conductor carrying current I. let P be a point having
with respect to current element dl
. Let
be the angle between dl
and r
According to Biot-Savart law, the magnetic field dB at point P due to current element is
1
(i)
dB
I
(ii)
dB
dl
(iii)
dB
sin
(iv) dB
r2
Combining all the four factors, we get
I dl sin
dB
r2
I dl sin
dB = K
r2
The proportionality constant K depends on the medium between point P and the current element and
the system of units chosen. For free space and in SI units,
0
K
10
TmA
4
0
I dl sin
dB
r2
4
As the direction of dB is perpendicular to the plane of
equation, we get the vector form of the Biot-Savart law as
0
dB
I dl
dl and r , so from the above
r
4
r3
The direction of dB is the direction of the vector dl
r . It is given by right hand screw
rule. If we place a right handed screw at point P perpendicular to the plane of paper and
turn its handle from
dl to r , then the direction in which the screw advances gives the
direction of dB . Thus the direction of dB is perpendicular to and into the plane of paper, as has been
Q. 2. 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. Suppose a charge q moves with velocity v in a magnetic field
B at angle .
It is found that the charge q experiences a force F such that
1. The force is proportional to the magnitude of the magnetic field, i.e., F B
2. The force is proportional to the charge q, i.e.,F
3. The force is proportional to the component of the velocity v in the perpendicular direction
of the field B, i.e., F sin
Combining the above factors, we get
F
sin
or
F = kqvB sin
The unit of magnetic field is so defined that the proportionality constant k becomes unity in the above
equation. Then
F = qvBsin
As the direction of
F = q( v
F is perpendicular to both
v andB, so we can express
F as
B)
Q. 3. 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.consider a conductor of length l, area of cross-section A, carrying current I lies
perpendicular to a magnetic field B .
Each electron moving with drift velocity vd experiences a magnetic Lorentz force
e
B
f (vd
)
If n is the number of free electrons per unit volume, then total number of electrons in the conductor is
N = n x volume = nAl
Total force on the conductor is
F
Nf
nAl
e(vd B
enA
l vd
B]
If I l represents a current element vector in the direction of current, then vectors
l and
vdwill have opposite directions and we can take
l vd=
F
vd l
enAvd(l
B)
I
F (l
[ enAvd I ]
B)
The magnitude of the force is
(i)
F = I l B sin
If =0 or 180 , then F = I l B(0) = 0
Thus a current carrying conductor placed parallel to the direction of the magnetic field
does not experience any force.
(ii)
If =90 , then F = I l B sin90o or Fmax = I l B
Thus a current carrying conductor placed perpendicular to the direction of a magnetic field
experiences a maximum force.
Q. 4. 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 unitcurrent flows
through it.
NBA
I
Current sensitivity,
s
I
k
Voltage sensitivity: It is the deflection produced in the galvanometer when a unitpotential
difference is applied across its ends.
Voltage sensitivity, Vs=α/V
=NBA/k RG
Factors affecting the sensitivity:
1. Number of turns N in its coil, Is
2. Magnetic field B
Is
3. Area A of the coil,
Is
1
4. Torsion constant k of the spring, Is
k
Q. 5. How will you convert a galvanometer into an ammeter of range 0 - I amperes? What is the effective
Ans.
resistance of an ammeter?
An ammeter is connected in series to a circuit. So it must have very small resistance so that it does not
affect the current. Therefore to convert a galvanometer into an ammeter, a low resistance, called shunt,
is connected in parallel with the galvanometer coil.
Let RG be the resistance of galvanometer and Ig be the current with which galvanometer gives full scale
deflection. A shunt of resistance S is connected in parallel with it. To measure maximum current I, the
maximum current through galvanometer must be Ig and hence rest current I - Ig should pass through
the shunt.
As galvanometer and shunt are in parallel, the potential difference across them is equal. So,
I gG
I
Ig S
S= IgG/I-Ig
The effective resistance of the ammeter becomes
RA GS
G S
Q. 6. 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. To insure it, a large resistance 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
V
Ig
V
R
I g(R
RG
V
RG)
or
R
RG
Ig
Q. 7. 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
Q. 8. What are magnetic lines of force? Give their important properties.
Ans. A magnetic line of force may be definred as the curve, the tangent to which at any point gives the
direction of the magnetic field at that point. It may also be defined as the path along which a unit
north pole would tend to move if free to do so.
Properties of lines of force: These are as follows :
(i) Magnetic lines of force are closed curves which start in air from the N-pole and end at the S-pole
and then return to the N-pole through the interior of the magnet.
(ii) The lines of force never cross each other.
(iii) They start from and end on the surface of the magnet normally.
(iv) The lines of force have a tendency to contract lengthwise and expand sidewise. This explains
attraction
between
unlike poles
and
repulsion
between
like poles.
(v) The
closeness
of the
lines
of force
is a measure
of the strength of the magnetic
field which
is relative
Maximum at the poles.
FIVE MARKS QUESTIONS
Q. 1. 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.
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.
B
If s be the position vector of point P relative to the elementdl, then from Biot-Savart
law, the field at point P due to the current element is
I dl
sin
0
dB
4
r2
Since dl
= 90 , therefore
s, i.e.,
Idl
0
dB
4
s2
The field
dB lies in the plane of paper and is perpendicular to
s , as shown by PQ. Let
be the angle between OP and CP. Then dB can be resolved into two rectangular
components.
1. dB sin along the axis,
2. dBcos
perpendicular to the axis.
For any two diametrically opposite elements of the loop, the components perpendicular to the axis of the
loop will be equal and opposite and will cancel out. Their axial components will be in the same direction,
i.e., along CP and get added up.
Therefore, total magnetic field at the point P in the direction CP is
= dB sin
a
0
sin
0
and
Idl a
. .
4
Since
0
As,
4
s2
s
and I are constant, and s and aare same for all points on the circular loop, we have
0Ia
=
4
s
2
Idl
dB
3
0Ia
dI
3 .2
2s
4
r2
s
I a2
0
B
2 23
2 (r a )
0
2
21
a)2
I a2
3
[
dI Circumference
]
NIa2
0
If the coil consists on N turns, then
B
2 (r2
3 .
a 2) 2
Magnetic field at the centre: For the field at the centre of the loop,r= 0. Therefore
0N I
B
2a
Q. 2.
Ans.
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.
Ampere’s circuital law: This law states that the line integral of the magnetic fieldBaround any closed
circuit is equal to 0 (permeability constant) times the total current I threading or passing through this closed
circuit.
0
B.dI I
Mathematically
Application of Ampere’s law to an infinitely long straight conductor:
Figure shows a circular loop of radius r around an infinitely long straight wire carrying a current I.
As the field lines are circular, the field B at any point of the circular loop is directed
along the tangent to the circle at that point. By symmetry, the magnitude of field B is same at every point
of the circular loop. Therefore,
B.dl
B dl
B dl
B.2
Form Ampere’s circuital law,
B.2
B
0I
I
0
2
Q. 3. 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 inside a closely wound long solenoid is uniform everywhere and zero outside it.
Figure shows the sectional view of a long solenoid. At various turns of the solenoid, current comes out
s the plane of paper at points marked .
To determine the magnetic field B at any inside point, consider a rectangular closed path abcdas the
Amperean loop. According to Ampere’s circuital law,
B.dl
Total current through the loop abcd ----------------------------(1)
0
b
B.dl
c
B.dl
d
B.dl
a
B.dl
B.dl
Now
a
c
b
c
d
c
B.dl
B dl
But
b
b
a
a
B.dl
B dl
d
d
d
B.dl
As B = 0 for points outside the solenoid.
c
b
B.dl
b
B.dl
b
B dl
B
dI
B l -------(2)
a
a
a
Where, l = length of the side ab of the rectangular loop abcd.
Now calculation for total current threading the loop abcd
Let number of turns per unit length of the solenoid = n
Then number of turns in length l of the solenoid = nl
Total current threading the loop abcd = nlI----------(3)
Putting the value of (2) and(3) in (1)
We get
Bl =
or
B
0nlI
nI
Q 4. 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
experiences a force
F q(v
v enters a magnetic field
B , it
B)
The direction of this force is perpendicular to both
v and B . The magnitude of this force
is
F = qv B sinθ
Following three cases are possible:
1. When the initial velocity is parallel to the magnetic field:
Here =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:
Figure shows a magnetic field B directed normally into the plane of paper, as shown by small crosses.
A charge +qis projected with a speed v in the plane of the paper. The velocity is perpendicular to the
magnetic field. A force F = qvB acts on the particle perpendicular to
both v and B . This force continuously deflects the particle sideways without changing its
speed and the particle will move along a circle perpendicular to the field. Thus the
magnetic force provides the centripetal force. Let r be the radius of the circular path.
mv2
Centripetal force,
= Magnetic force, qvB
r
mv
r=
qB
2
2
mv
Period of revolution, T=
.
v
The frequency of revolution is
2
v qB
qB
1
qB
fc
T
2
This frequency is independent of v and r and is called cyclotron frequency.
3. When the initial velocity makes an arbitrary angle with the field direction:
Consider a charged particle q entering a uniform magnetic field B with velocity
inclined at an angle with the direction of
B ,as shown in figure.
v =vsinθ of the initial velocity makes the charge move
The perpendicular component,
along a circular path of radius,
The period of revolution is
2
2
v
v sin
mvsin
2
T
qB
qB
The parallel component, vll = vcosθ of the initial velocity makes it move along the direction of the
magnetic field. Hence the resultant path of the charged particle will be ahelix, with its axis along the
direction of B , as shown in figure.
The distance moved along the magnetic field in one rotation is called pitch of the helical path.
Therefore
V|| =vcosθ x 2πm/qB = 2πm vcosθ/qB
Q. 5. 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, - particles, etc., to very
high energies.
Principle: A charged particle can be accelerated to very high energies by making it passthrough a moderate
electric field a number of times. This can be done with the help of a perpendicular magnetic field which throws
the charged particle into a circular motion, the frequency of which does not depend on the speed of the particle
and the radius of the circular orbit.
Construction: As shown in figure, a cyclotron consists of two small, hollow, metallichalf- cylinders D1 and
D2, called dees. An alternating voltage is applied across the gap between the two dees. The dees are placed
between the poles of a strong electromagnet. A source of charged particles is placed near the centre of the dees.
These ions move on a circular path in the dees, D1 and D2, on account of the uniform perpendicular magnetic
field B. The whole arrangement is evacuated to minimize collisions between the ions and the air molecules.
Theory: As a particle of chargeqand massmfollows a circular path under the effect ofperpendicular magnetic
field B, so
Magnetic force on charge = Centripetal force
mv2
qv Bsin
r
Or
mv
or
r
qB
Period of revolution of the charged particle is given by
2
2
T=
mv
2
.
v
v qB
qB
1
Hence frequency of revolution of the particle will be
qB
fc
T
2
Clearly, this frequency is independent of both the velocity of the particle and the radius of the orbit and
is called cyclotron frequency of magnetic resonance frequency.
Working: Suppose a positive ion enters the gap between the two dees and finds dee D1tobe negative. It
gets accelerated toward dee D1. As it enters the dee D1, it does not experience any electric field due to
shielding effect of the metallic dee. The perpendicular magnetic field throws it into a circular path. At
the instant the ion comes out of dee D1, it finds dee D2 negative. It now gets accelerated towards deeD2 .
It moves faster through D2 describing a larger semicircle than before. Thus if the frequency of the
applied voltage is kept exactly the same as the frequency of revolution of the ion, then every time the ion
reaches the gap between the two dees, the electric field is reversed and ion receives a push and finally it
acquires very high energy.
Q. 6. 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
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 theplane 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
= Force x perpendicular distance =
IbB x asinθ = IBAsinθ
If the rectangular loop has N turns, the torque increases N times i.e.,
NIBAsinθ
But NIA = m, the magnetic moment of the loop, so
mBsinθ
In vector notation, the torque
=
=
is given by
=mxB
The direction of the torque t is such that it rotates the loop clockwise about the axis of suspension.
Q. 7. 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
= Force x perpendicular distance = NIbB x a sin 90 = NIB (ab) = NIBA
Here =90 , because the normal to the plane of coil remains perpendicular to the field
B
in all positions.
The torque deflects the coil through an angle . A restoring torque is set up in the coil due to the elasticity of
the springs such that
restoring
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
NBAI /
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.
Q. 8. 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.
Consider two long parallel wires AB and CD carrying currents
I1 and I2 in the same direction. Let r be the separation between them.
The magnetic field produced by current I1 at any point on wire CD
is
B
0I1
2
This field acts perpendicular to the wire CD and points into the plane of paper.
It exerts a force on current carrying wire CD. The force acting on length l of
the wire CD will be
F
IB1
I 2 I2
I
l. 0 1
2
0 I1 I 2 l
2
Force per unit length,
F2
0 I1
I2
f
l
2
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:
If I1=I2 =1A and r =1m, then
0
f
Nm
2
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 metre of their length.
Q. 9. Define the terms (i) magnetizing field, (ii) magnetic intensity, (iii) magnetisation, (iv) relative
permeability and (v) magnetic susceptibility. Give their S I unit, if any
Ans. (i) Magnetising field - The magnetic field that exists in vacuum and induces magnetism
is called magnetising field.
The magnetising field of a solenoid is
(ii) Magnetising field intensity or magnetic intensity - It is the number of ampere-turns (nI)
flowing round the unit length of the solenoid required to produce a magnetising field.
-1
Thus H = nI Also, B0
0nI
0H Its SI unit is Am .
(iii) Magnetisation or intensity of magnetization - It is the magnetic momentdeveloped per unit
volume of a material when placed in a magnetising field. It is a vector quantity.
𝑀=
𝑀
𝑀
Its SI unit is Am-1.
(iv) Relative permeability - It is the ratio of permeability of the medium to thepermeability
of free space.
r
0
It is unit less quantity.
(v) 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.
M
H
As magnetic susceptibility is the ratio of two quantities having the same units, so it has no unit.
Q. 10. Explain how does an atom behave as a magnetic dipole. Derive an expression for the magnetic
dipole moment of the atom. Also define Bohr magneton.
Ans.Magnetic dipole moment of a revolving electron
In hydrogen-like atoms, an electron revolves around the nucleus. Its motion is equivalent to a current
loop, which possesses a magnetic dipole moment = IA. As shown in fig., consider an electron revolving
anticlockwise around a nucleus in an orbit of radius r with speed v and time period T.
Equivalent current,
charg e
e
e
ev
I
time
T 2 /v 2
Area of the current loop
2
A
The orbital magnetic moment of the electron is
ev
evr
m IA
2
2
2
The angular momentum of the electron due to its orbital motion is
l mevr
m
e
So,
2me
l
Or, m el
2me
According to Bohr's quantisation condition, the angular momentum of an electron in any permissible
orbit is,
nh
where n = 1, 2, 3, ……………
l
2
100
eh
So, m n
4 e
This equation gives orbital magnetic moment of an electron revolving in nth orbit.
Bohr magneton - It is the magnetic moment associated with an electron due to itsorbital motion in the first
orbit of hydrogen atom. It is the minimum value of ‘m’, which can be obtained by putting n =l in the above
equation. Thus Bohr magneton is given by
eh
m
mi
n
4
e
Q11.What Does a toroid consist of? Show that for and ideal toroid of closely wound turns the magnetic field.
a. Inside the toroid is constant.
b. In the open space inside an exterior to toroid is zero.
Ans. A solenoid bent into the form of close loop is called toroid. The magnetic field B has a constant magnitude
everywhere inside the toroid.
1.for points in the open space interior to the toroid let B1 be the magnitude of the magnetic field along the
Amperian loop 1 of radius r1
Length of the loop 1 L1=2πr1
As the loop enclose no current so I=0
Appling ampere circuital law
B1L1=µ0I OR B1 x 2πr1= µ0 x 0 or B1=0
Thus the magnetic field at any point P in the open space interior to the toroid is zero.
2.For inside the toroid Let B be the magnitude of the magnetic field along the Amperean loop of radius r.
Length of loop 2
L2 = 2πr
If N is the total number of turns in the toroid and I the current in the toroid then the total current is enclosed by
loop 2 = NI
Appling ampere circuital law Bx2πr = µ0 x NI or B=µ0 NI/2πr
If be the average radius of the toroid and n the number of turns per unit length
Then, N = 2πrn
Therefore, B=µ0nI
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
The SI unit of magnetic flux is Tm2 or Wb. It is a Scalar quantity.
The dimension of magnetic flux is [ML2T-2A-1].
2. 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.”
3. 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
4. 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
5. 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
6. 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
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.”
These currents are also known as Foucault’s currents after the name of the discoverer of eddy 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.
ɛ=
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 
I
or L 

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
17. The average value of alternating emf / alternating current for the ‘half’ cycle is 0 or 0


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 = ɷL
Reactance of a Capacitor (or Capacitive Reactance) XC = 1/ɷC
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 
1  V  VC
When the source frequency f >fr   tan 1  L
 or   tan  L
R


 VR



 X  XL 
1  V  VL
When the source frequency f<fr   tan 1  C
 or   tan  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.”
L
1 L r
=
= 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 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.
Q=
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
Efficiency of Transformer:- η =
vs I s
P
 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.
Ans.
Q3.
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.
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.
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?
Ans.
L = φ/I
A
B
Φ
I
For given I, A has larger value of φ, so A has larger self inductance.
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.
S
N
S
N
G
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. 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?
Ans: The capacitance increases, decreasing the reactance Xc. Therefore the rms current increases.
Q11. 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?
Ans: V0=100V I0=100A Ф= π/3 P=VrmsIrmsCos Ф=2500W
Q12 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?
Ans: 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) State he condition for resonance to occur in a series LCR a.c. circuit and derive an expression for the
resonance frequency.
(b)
Draw a plot showing the variation of the peak current (i0) with frequency of the a.c. source used. Define
the quality factor, Q of the circuit.
Ans.(a) Electrical resonance in Series LCR Circuit.
Electrical resonance takes place in a series LCR circuit when the circuit allows maximum alternating
current for a frequency at which capacitive reactance becomes equal to the inductive reactance.
Impedance of a series LCR circuit,
Z = √[R2+(XL-XC)2]
And current in circuit,
I = E/Z = E/√[R2+(XL-XC)2]
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),
(b)
Refer to NCERT text book part 1 fig. 7.16 page no 248.
Quality Factor (Q-factor of Resonance Circuit is defined as 2π times the ratio of the energy stored in the
circuit to the energy dissipated in resistance per cycle of a.c. supply.
Q = 2π x energy stored in the circuit per cycle/energy dissipated per cycle.
Q.2.
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?
Ans.
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
Secondary e.m.f.
Es = -Nsdφ/dt
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.
The electric mains in a house are marked 220 V, 50 Hz. Write down the equation for instantaneous
voltage.
Ans.
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.
Average power over a full cycle, PL = average of PL
Average of sin 2ωt for full cycle =
𝑇
= ∫ sin 2𝜔𝑡 𝑑𝑡 =
0
=
1
[cos 2𝜔𝑇
2𝜔𝑇
= 1/2𝜔𝑇[cos
=
1
𝑇
[cos 2𝜔𝑡/2𝜔]
0
𝑇
− cos 0 O
2𝜋
𝑇
𝑋 𝑇 − cos 0O]
𝐼
[1 − 1] = 0
2𝜔𝑡
Thus, average power = 0
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.
(Delhi 2013)
2. Predict the direction of induced current in metal rings 1 and 2 when current I in the wire is steadily
decreasing?
(Delhi 2012)
1
I
2
3. A bar magnet is moved in the direction indicated by the arrow between two coils PQ and CD. Predict the
direction of induced current in each coil.
(All India 2012)
N
S
P
Q
C
D
4. 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?
(All India 2013)
5. How can the self- inductance of a given coil having N number of turns area of cross-section A and length
l be increased?
(Delhi 2012)
6. Define self-inductance of a coil. Write its SI unit.
(All India 2010)
7. 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 self-inductance?
(Delhi 2010)
ø
A
B
I
8. Mention any two useful applications of eddy current.
(Delhi 2009C)
9. 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.
(All India 2010C)
S
N
2 marks questions
10. 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.
(Delhi 2012)
11. A current is induced in coil C1 due to the motion of current carrying coil C2. (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.
(Delhi 2011)
C1
C2
12. Predict the polarity of the capacitor in the situation described by adjoining figure. Explain the reason too.
(Delhi 2011, All India 2011)
S
N
S
N
13. 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.
(All India 2010)
14. (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.
(Delhi 2010)
S
P
Opposition
to current
15. A coil Q is connected to low voltage bulb B and placed near another coil P 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.
(Delhi 2010)
B
A. C. Source
Opposition
to current
Q
P
16. The Circuit arrangement given below shows that when an A C passes through the coil A the current
starts flowing in the coil B. (i) State the underlying principle involved. (ii) Mention two factors on which
the current produced in the coil B depends.
(All India 2008)
A
B
17. The figure given below shows an arrangement by which current flows through the bulb X connected
with coil B when AC is passed through coil A (i) Name the phenomenon involved. (ii) If a copper sheet is
inserted in the gap between the coils explain how the brightness of the bulb would change?
(All India 2008)
A
B
X
18. 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.
(Delhi 2006C)
19. 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.
(Delhi 2006C)
20. 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.
(All India 2012)
21. (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?
(All India 2011C)
O
Frequency
O
Frequency
(I)
22.
23.
24.
25.
26.
27.
(II)
3 marks questions
(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?
(All India 2009)
(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.
(All India 2009)
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.
(Delhi 2008)
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.
(All India 2010C)
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.
(All India 2010C)
A metallic square loop ABCD of size 15 cm and resistance 1 Ω is moved at a uniform velocity of v m/s in a
uniform magnetic field of 2 T the field lines being normal to the plane of paper. The loop is connected to
an electrical network of resistors each of resistance 2 Ω. Calculate the speed of the loop for which 2 mA
current flows in the loop.
(All India 2006C)
x
x
x
x
x
P
D
C
2Ω
Q
2Ω 2Ω
x
S
x
x
x
2Ω
2Ω x
x
x
x
x
B
X
x
x
R
A
x
15 cm
x
x
5 marks questions
28. 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
29. (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.
(All India 2009)
30. (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.
(All India 2009)
x
x
x
x
x R x
x
P
x
x
x
x
x
x
x
l x
x
x
x
x
x
x v
x
x
x
x
x
x
x
x
x
x
x S
x
x
Q
x
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.
1.
Name
Gamma rays
Frequency
Wavelength
Production
Uses
Range (Hz)
Range
1019 – 1023
10-11 to 1014
m
Emitted by radioactive nuclei,
In medicine, to
destroy cancer
cells.
Produced in nuclear reaction
2.
X – rays
1016 - 1020
10-8 to 1012
m
Generated by bombarding a
metal target by high energy
electron
Used as
diagnostic tool
in medicine, to
study crystal
structures
3.
Ultraviolet
rays
1015 – 1017
(4 × 10-7 to 6
× 10-10) m
Produced by special lamps &
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.
microwaves
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
In radio &
motion of charges in conducting television
communication
wires.
system, in
cellular phones
to transmit
voice
communication.
Question Answer related to this topic
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.
(1 mark)
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.
mark)
(1
X – rays
Q.10
Write the expression for the velocity of e.m. waves in terms of
permittivity and permeability. (1 mark)
Q.11
Q.12
Sketch a schematic diagram depicting electric and magnetic fields
for an electromagnetic wave propagating along z-direction.
(2
marks)
The oscillating magnetic field in a plane electromagnetic wave is given by BY=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) Radiowaves (b)Gammarays (c)Visiblelight (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
Q.20
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)
Arrange the following electromagnetic radiation in ascending order of their
frequencies.
(a)Microwaves (b)Radiowaves (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 x 107
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)
(i)10-3 nm
(ii)10-3 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
castings
(ii) Sterilization
Q.27
A plane electromagnetic wave of frequency 25MHz travels in free space along the xdirection at a particular point in space and time
this time. (3 marks)
. Determine the
at
Q.28
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.
Mirror
A smooth and highly polished reflecting surface is called a mirror.
Plane Mirror:
A highly polished plane surface is called a plane mirror
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.
Spherical Mirror:
A highly polished curved surface whose reflecting surface is a cut part of a hollow glass sphere is called
a spherical mirror.
Spherical mirrors are of two types
(b) Convex Mirror
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 vacuum (c) to the speed of light in any medium (u) is
c
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
n12 =
1
.
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 a symmetrical convex lens of focal length f is cut into two parts along its optic axis, then focal length
of each part (a plano convex lens) is 2f.
If a symmetrical convex lens of focal length f is cut into two parts along the principal axis, then focal
length of each part remains unchanged as f
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
The refractive index of material of prism
⌈n21 =
A+δ
sin ( 2 m )
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 bluish colour predominates in a clear sky, since blue has a shorter wavelength than red and is scattered
much more strongly.
Clouds are generally white
 Clouds have droplets of water with a >> λ

 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.
Magnifying power
When final image is at least distance vision
𝑚=
𝑣𝑜 1 + 𝐷
(
)
𝑢𝑜
𝑓𝑒
When final image is at infinity
𝑚=
𝐿𝐷
𝑓𝑂 𝑓 𝑒
L is the tube length
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)
Spberical 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
(ii)
spherical aberration.
Chromatic Aberration Image of a white object formed by lens is usually coloured and
blurred. 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 near by objects,
is called its resolving power.
Limit of Resolution
The minimum distance between two near by 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 which is
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)
IMPORTANT DERIVATIONS
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 refractive index of the medium.
Ray of light is travelling from denser(water) to
rarer (air)medium.
According to Snells law
𝑛12 =
𝑠𝑖𝑛 𝑖
𝑠𝑖𝑛 𝑟
If angle of incidence i =C
Then r= 900
𝑛12 =
𝑠𝑖𝑛 𝑐
𝑠𝑖𝑛 90
⌊𝑛21 =
1
⌋
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
𝑛21 =
𝑛1
𝑛2 𝑠𝑖𝑛 𝑖 𝑖
𝑛21 =
=
=
𝑛1 𝑠𝑖𝑛 𝑟 𝑟
𝑛1 𝑖 = 𝑛2 𝑟
In triangle MOC 𝑖 = 𝛼 + 𝛾
In triangle MCI 𝛾 = 𝑟 + 𝛽
𝑛1 (𝛼 + 𝛾) = 𝑛2 (𝛾 − 𝛽) _______________________(1)
Where 𝛼 =
𝑀𝑃
𝑃𝑂
, 𝛽=
𝑀𝑃
𝑃𝐼
and 𝛾 =
𝑀𝑃
𝑃𝐶
Substituting values of α, β and γ in eq. (1)
𝑀𝑃
𝑛1 ( 𝑃𝑂 +
𝑀𝑃
𝑃𝐶
𝑀𝑃
) = 𝑛2 ( 𝑃𝐶 −
𝑀𝑃
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
XP1Y, 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
𝑛2
𝑓
𝑛1
Hence[ = (
− 1) (
1
𝑅1
−
1
𝑅2
)]
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
𝑖=(
𝐴+𝛿𝑚
2
)_________________________________(6)
Substituting values of i and r in eq. (1)
⌈𝑛21
𝐴+𝛿
𝑠𝑖𝑛 ( 2 𝑚 )
=
⌉
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
Let the waves from two coherent source of light be represented by
OF LIGHT
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 (t + ∅/2)
The amplitude of the resultant displacement is 2a cos (∅/2) and therefore the intensity at that
point will be
I=4a2cos2 /2) = 4I0cos2 /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.
Hence condition for destructive interference ∅ =
(2𝑛+1)𝜋
2
_________(5)
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
2
(𝑆2 𝑃) − (𝑆1 𝑃) = 𝐷 + (𝑦 + ) − 𝐷 − (𝑦 − ) = 2𝑦𝑑
2
2
2
2
𝑆2 𝑃 − 𝑆1 𝑃 =
2𝑦𝑑
2𝑦𝑑
=
𝑆2 𝑃 + 𝑆1 𝑃
2𝐷
path difference ∆𝑥 =
𝒚=
𝒏𝝀𝑫
𝒅
𝑦𝑑
𝐷
= 𝑛𝜆 for constructive interference
𝒘𝒉𝒆𝒓𝒆 𝒏 = 𝟎, 𝟏, 𝟐, 𝟑 … …position of bright band
path difference ∆𝑥 =
𝒚=
2
(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
IMPORTANT FORMULAE
1. Mirror Formula
1 1 1
= +
𝑓 𝑣 𝑢
2. Magnification
(m) = I/O = -v/u
𝑛21 =
3. Rrefractive index
𝑠𝑖𝑛 𝑖 𝑛2
=
𝑠𝑖𝑛 𝑟 𝑛1
𝑅𝑒𝑎𝑙𝑑𝑒𝑝𝑡ℎ
𝑡
=
𝐴𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝑑𝑒𝑝𝑡ℎ 𝑡𝑎𝑝𝑝
𝑡
𝑥=𝑡−
𝑛
𝑛21 =
4. Apparent Shift
1
5. Relation between critical angle and
refractive index
𝑛12 =
6. Refraction at spherical surface
𝑛2 𝑛1 𝑛2 − 𝑛1
−
=
𝑣
𝑢
𝑅
7. Lens formula
1 1 1
= −
𝑓 𝑣 𝑢
8. Lens makers formula
1
n2
1
1
= ( − 1) ( − )
f
n1
R1 R 2
9. Magnification
(m) = I/O = v/u
10. Power of lens
𝑃=
𝑠𝑖𝑛 𝐶
.
1
𝑓
1 1 1
= +
𝑓 𝑓1 𝑓2
11. Combination of lens
𝑃 = 𝑃1 + 𝑃2
𝑚 = 𝑚1 × 𝑚2
12. Relation between i, e and δm
δm = 2i − A
A+
sin ( 2 )
=
A
sin 2
13. Prism formula
n21
14. Magnifying power of simple
microscope
(iii)
(
final image at d) 𝑚 =
1+𝑑
𝑓
(iv) m = d/f
𝑣
1+𝐷
)
𝑓𝑒
𝑜
(final image at d) 𝑚 = 𝑢𝑜 (
15. Magnifying power of simple
microscope
𝐿 𝐷
𝑂 𝑓𝑒
(final image at infinity) 𝑚 = 𝑓
16. Magnifying power of Telescope
Length of tube
17. Condition for constructive interference
18. Condition for destructive interference
𝑚=
𝑓𝑜
𝑓𝑒
𝐿 = 𝑓0 + 𝑓𝑒
∅ = 2𝑛𝜋
∅=
(2𝑛+1)𝜋
2
𝜃=
Angular width interference
(2𝑛+1)𝜆
2
𝑑
𝜆
𝑑
20. Resultant intensity in interference
I=4a2cos2
21. Position of nth maxima
Position of nth minima
𝑦𝑚𝑎𝑥 =
22. Condition for minima in diffraction
𝑎𝑠𝑖𝑛𝜃 = 𝑛𝜆
23. Condition for maxima in diffraction
𝑎𝑠𝑖𝑛𝜃 = (
24. Width of central Bright fringe
diffraction
𝛽=
2𝜆𝐷
𝑑
𝜃=
2𝜆
𝑑
25. Angular width diffraction
∆𝑥 =
𝜆𝐷
𝛽=
19. Fringe width interference
∆𝑥 = 𝑛𝜆
= 4I0cos2
(2𝑛 + 1)𝜆𝐷
𝑛𝜆𝐷
𝑦𝑚𝑖𝑛 =
𝑑 .
2𝑑
2𝑛 + 1
)𝜆
2
26. Malus law
𝐼 = 𝐼0 𝑐𝑜𝑠 2 𝜃
27. Brewsters law
𝑛 = tan 𝑖𝑝
28. Resolving power of microscope
𝑅𝑃 =
2𝑛𝑆𝑖𝑛𝛽
1.22𝜆
29. Resolving power of telescope
𝑅𝑃 =
30. Fresnels distance
𝑧=
𝐷
1.22𝜆
𝑎2
𝜆
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. It is known that the refractive index, , of the material of a prism, depends on the wavelength , ,
of the incident radiation as per the relation
𝜇 =𝐴+
𝐵
𝜆2
where A and B are constants. Plot a graph showing the dependence of µon λ.
Ans:
4. How does power of lens vary when incident red light is replaced by blue light?
Ans: Wavelength decreases hence power increases
5. 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.
6. 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.
7. 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
8. 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.
9. 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.
10. 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
11. 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 thiscombination 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. The following table gives data about the single slit diffraction experiment:
Find the ratio of the widths of the slits used in the two cases. Would the ratio of the half angular
widths of the first secondary maxima, in the two cases, be also equal to q?
Ans: Let d and d’ be the width of the slit in the two cases
𝜆
𝑝𝜆
𝑑
𝑑′
Hence𝜃 = 𝑎𝑛𝑑𝑞𝜃 =
Or
3.
𝑑
𝑑′
=
𝑞
𝑝
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.
4. Define resolving power of a compound microscope. How does the resolving power of a compound
microscope change when
(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
5. 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.
6. 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; therefore, angular separation remains unchanged if screen is moved away
from the slits. But the actual separation between fringes β = D λ/d increases, so visibility of fringes
increases.
(ii) When width of source slit is increased, then the angular fringe width 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.
7. 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
8. 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?
Ans
𝐴
R=1.22𝜆
𝑅1 𝐴1
=
𝑅2 𝐴2
A1 because it has greater resolving power
9. 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)
10. (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).
11. Write the condition necessary for diffraction to occur. Draw the intensity distribution curve in
single slit diffraction pattern.
Ans: Size of aperture or obstacle should be comparable to wavelength of light used.
12. 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.
13. 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.
14. 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.
15. Draw a ray diagram of compound microscope. Write the expression for its magnifying power.
Ans:
𝑣
1+𝐷
)
𝑓𝑒
𝑚 = 𝑢𝑜 (
𝑜
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) Draw a ray diagram for the formation of image by a compound microscope. Define its
magnifying power. Deduce the expression for the magnifying power of the microscope.
(ii) Explain.
(a)Why must both the object and the eyepiece of a compound microscope have short focal
lengths?
(b)While viewing through a compound microscope, why should our eyes be positioned not on the
eyepiece but a short distance away from it for best viewing?
Q5.
(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?
Q6.
(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.
In a Young’s double slit experiment,
Q7.
(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)
2. 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:
U 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.
V 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:
10. 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.
11. F=distance of screen from the lens = 40cm
12. 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:
11. 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?
12. The object is moved to a distance of 40cm in front of the mirror. How is focal length of
mirror affected?
13. What values of life do you learn from the mirror formula?
Answer:
12. 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)
Focal length (f) of mirror remains unaffected. On changing u; v changes and not f.
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.
a) 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 3D effect.
Read the above passage and answer the following questions:
What qualities do these boys possess?
What do you mean by Polarization?
Mention the other applications of polarization.
Answer:
Curiosity to learn, approaching the teacher to learn new things, inquisitiveness.
The phenomenon of restricting the oscillations of a light wave (electric vector) in a
particular direction is called polarization of light.
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.
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.
It’s experiment was performed by Heinrich Hertz.
It was explained by Einstein and he also got Nobel Prize for this.
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 nature of
o depends
on(i) The frequency of incident light and (ii)
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
Law of Photoelectric effect-
De-Broglie Hypothesis-
According 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 K-
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
vary if frequency is
0 in a photocell .How will the stopping potential
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 W 2 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 λ 0 .Does photo emission
occur if wavelength (λ) of incident radiation is more than λ 0 , 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
Q.21
Calculate the number of photons emitted per second by transmitter of 10 KW
power; radio waves of frequency 6×105 Hz. (2 marks)
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. (3 marks)
For ‘A’, Energy incident is less than work function.
For ‘B’, Energy incident is equal to work function of metal.
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. (3 marks)
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 electro magnetic 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
As E= h = Φo + K
.
(5 marks)
Q.29
IF the frequency of incident radiation an 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. (5 marks)
E = h = Φo + eVo

Topic to be covered for supportive learner :ATOMS






Schematic arrangement of the Geiger-marasden experiment.
Alpha partical scattering and Rutherford model.
Bhor’s Postulate (statements only)
Energy levels in hydrogen atoms
Line spectra of hydrogen atoms.
Summary given in N.C.E.R.T text book.
NUCLEI






Isotopes, isobars, isotones.
Mass defect.
Mass energy relation.
Binding per nucleon /mass number graph.
Potential energy of nucleons / seprations of nucleon graph.
Radio active decay law.
𝑑𝑁
𝑑𝑡
= 𝜆𝑡
N= N0e-λt
 Half life ,mean life and decay constant and their relation
T1/2 = log2 = 𝜏 log2
𝜏
 Radio active decay
Alpha decay, beta decay, gamma decay.
 Nuclear fission and fusion Reaction
Important Formula for Numerical :1 Nucleus consist of protons and neutrons.
Nucleus of protons in a nucleus zXA is Z and number of neutrons ,N =A-Z
2 Radius of Nucleus :- R= R0A1/3 where R0 = 1.2 x 10-15m
3 Density of Nuclear Matter :-
Dn = 1017kg/m3
(2)
4 Einstein ‘s Mass Energy Equivalence Relation is E =mc2
1amu =1u =931 MeV
5.Mass Defect =mass of nucleons in given nucleus –mass of nucleus
=Zm p + (A - Z )mn– M nucleus
6.Rutherford –Soddy formula :(i) Number of atoms undecayed after time t
N=N0e-λt
N/N0 =[1/2]n
(i)
Where n = t\T is number of half lives.
7. Relation between half –life (T) mean life (𝝉 ) and disintegration constant (λ ) is
𝜏 =1/λ and T = 0.693 𝜏
= 0.693/λ
8. Displacement Laws:
(i)
For α -particle
zXA
(ii)
For β - particle
zXA
z-2YA-4 + 2He4
z+1Y
A
+
0
-1β
+ ν
(iii)
For gamma– ray
zXA
zXA + γ
9. In nuclear fission a heavy nucleus break into lighter nuclei .Nearly 0.1 % mass is converted
energy .In each fission of 92 U 235 with slow neutron 200 MeV energy is released .
into
10 In nuclear fusion two lighter nuclei combine to form a heavy and 0.7 % mass is converted into
energy .
ATOMS AND NUCLEI
 Question bank for the supporting learner
 Very Short answer Questions :Q:-1
Ans
Q:-2
Ans
(4)
(1 Mark)
Among alpha ,beta and gamma radiations ,which get affected by electric field ?
Alpha and beta radiations are charged ,so they are affected by electric field.
What will be ratio of two nuclei of mass numbers A1 and A2?
Radius of nucleus R=R0A1/3
1/3
𝑅1
𝑅2
=[
𝐴1
𝐴2
]
Q:-3
Compare the radii of two nuclei numbers 1 and 27 respectively?
Ans
1/3
𝑅1
𝑅2
=[
𝐴1
𝐴2
]
=
(1/ 27 )1/3 = 1/3
Q:-4
What is the nuclear radius of 125 Fe if that of 27 Al is 3.6 fermi?
Ans
Nuclear radius ,R= R0A1/3 ⇒
R
(R Fe/RAl) = (A Fe/AAL)1/3
A1/3
= (125/ 27)1/3
R Fe = ( 5/3 ) RAl = (5/3) x 3.6 = 6.0 fermi
Q:-5
Assuming the nuclei to be spherical in shape ,how does the surface area of a nucleus of
mass number A1 compare with that of a nucleus of mass number A2?
Ans
Radius of nucleus of mass number A is
R= R0A1/3
Surface area of the nucleus , S= 4 πR2 ∝ R2
(S1/S2) = (R1/R2) 2 = (A1/A2)2/3
Q:-6
What is the ratio of nuclear densities of the two nuclei having mass number in the ratio 1:4?
Ans
Nuclear Density is independent of mass number so ratio of nuclear density is 1:1.
Q:-7
What happens to the neutron to proton ratio after the emission of α-particle?
Ans
Neutron to proton ratio increases after the emission of α-particle .
Q:-8 Two nuclei have mass number in the ratio of 2: 5 . What is the ratio of their nuclear
densities ?
Ans
Nuclear density in independent of mass number ,so ratio of nuclear densities is 1:1
Q:-9
Which has greater ionizing particle : α-particle or β-particle .
Ans
α-particle has greater ionizing power than β-particle.
Q:-10 What is the difference between an electron and a β-particle?
Ans β-particle are simply very fast moving electrons. The specific charge of electron is higher
than that of βparticle.
Q:-11 Why do α-particle have high ionizing power?
Ans α-particle are heavier ,they move slowly ; so posses large momentum .Due to this property
they come in contact with large number of particles ; so they possess high ionizing power .
Q:-12 A nucleon of mass number A , has a defect Δm .Give the formula ,for the binding energy
per nucleon , of this nucleus .
Ans
B.E per nucleon ,
Bn =
Total binding energy
=
Number of nucleon
∆m.c2
A
Where ¢ is the speed of light in vacuum.
Q:-13 The binding energy per nucleon of the two nuclei A and B are 4 MeV and 8 -2 MeV .Which of
accompanied
by release of energy ,Which of the two nucleus is more stable?
Ans
The nucleus (B) having larger binding energy is more stable.
Q:-14
Four nuclei of an element fuse together to form a heavier nucleus .If the process is
accompanied by release of energy ,which of the two the parent or the daughter nucleus would have a
higher binding energy /nucleon ?
Ans
The daughter nucleus will have higher binding energy per nucleon .
Q:-15 Name the reaction responsible for energy production in the sun
Ans
Fusion reaction.
Q:-16 Write a typical nuclear reaction in which a large amount of energy is released in the process
of number Fission?
Ans
Nuclear fission reaction is
92U
235
+ 0n1
141
56Ba
+ 36Kr92 + 3( 0n1) + 200MeV
Q:-17 Write any one equation representing nuclear fusion reaction?
Ans
Equation of fusion reaction
2
1H
+
2
1H
3
1H
+
1
1H
+4.0MeV
Q:-18 Give the mass number and atomic number of elements on the right hand side of the decay
process?
220
86Ru
Ans
Po + He
The complete equation representing mass number and atomic number is give below
220
216
86Ru
84Po
+ 2 He4
For polonium Z=84,A =216
For helium ( α-particle )Z = 2 ,A =4 .
Q:- 19 State condition for controlled chain reaction in a nuclear reactor ?
Ans
In nuclear fission ,two or three neutron are released per fission .If on the average one
neutron causes further , the chain reaction is said to be controlled .
Q:-20 Name the
reactor?
Ans
absorbing material used to control the reaction rate of neutron in a nuclear
Cadmium is the absorbing material for neutrons produced in a nuclear reactor.
Q:-21 What a percentage of a given mass of a radioactive substance will be undecayed after four
half periods?
Ans
Percentage of mass of radioactive substance undecayed after n=4 half –lives
= ( ½ )4x 100% = 100/16 = 6.25%
Q:-22 Write the nuclear decay process for nuclide β-decay of
Ans
15P
32
16S
32
32
15P
?
+ - 1e0 + v (Antineutrino )
Q:-23 Define the term activity of a radio nuclide .Write its S.I Unit .
Ans
. It is
Activity to the radionuclide is defined as the rate disintegration of given radioactive sample
proportional to the number of undecayed radio nuclides in that sample.
It is denoted R =
𝑑𝑁
𝑑𝑡
. Its S.I unit is Becquarel
.
Q:-24 If the nucleon a nucleus are separated far apart from each other,the sum of masses of all
these nucleons is larger than the mass of the nucleus .where does this mass difference
come from?
Ans
According to mass –energy equivalence relation E = mc2, this mass difference in nucleus
remain in the form binding energy . When nucleon are separated this binding energy of
nucleus is converted into mass.
Q:25
The radioactive isotope D –decays according to the sequence .
D
α
D1
β
D2
If the mass number and atomic number of D2 are 176 and 71 respectively ,what is the
I.
II.
Mass number
Atomic number ,of D?
Ans The sequence is represented as
z
α
DA
I.
Given A-4 = 176 ⇒ mass number of D, A =180
II.
Z -1 =71 ⇒ atomic number of D, Z =72
A-4
z-2D1
β
z-1D2
A-4
Q:-26 Write two characteristic features of nuclear force which distinguish it from Coulomb’s force?
Ans
Characteristic features of Nuclear Force:
1. Nuclear Force are short range attractive force (range 2 to 3 fm ) while Coulomb’s
force has range upto infinity and may be attractive or repulsive.
2. Nuclear forces are charge independent forces ; while Coulomb force acts only between
charged particle .
 Short Answer Questions:Q:-1
( 2, 3 Marks)
The trajectories ,traced by different ά –particles, in Geiger –Marsden experiment were
observed as shown in the
figure .
(a) What names are given to the symbols ‘ b’ and ‘θ’ shown here .
(b) What can we say about the value of b for (1) ‘θ’=00 (ii) ‘θ’=π radians.
θ
b
0 Target nucleus
Ans (a) The symbol ‘b’ represents impact parameter and ‘θ’ represents the scattering angle .
(b) When ‘θ’=00 ,the impact parameter will be maximum and represent the atomic size .
(c) When ‘θ’ = π radians ,the impact parameter ‘b’ will be minimum and represent the
nuclear size .
Q:-2
Define ionization energy .What is its value for hydrogen atom?
Ans
The minimum energy required to removed an electron from atom to infinitely for away is
called the ionization energy.
The ionization energy for hydrogen atom is 13.6 eV.
Q:-3 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)
γ-rays are similar to X-rays
α –rays are easily absorbed by matter .
(9)
(iii)
(iv)
γ-rays travel with greatest speed
β –rays are similar to cathode rays .
Q:-4 Define the term ‘ Activity ‘ of a radioactive substance .State its SI unit .Give plot of activity
of a radioactive
species versus time. Two different radioactive elements with half lives T1 and T2 have N1
and N2
(undecayed ) atoms respectively present at a given instant .Determine the ratio of their
activities at this
instant.
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 .
Activity R = dN
= λN
A0
dt
A
A/2
Decay constant λ = loge2/T
A/4
0
T
2T
t
Activity R = ( loge2) N/T
For two elements R1/R 2 = ( N1/N2 ) X ( T2 / T1 )
Q:-5
The half –life of 6C14 is 5700 years. What does it mean? Two radioactive nuclei X and Y
initially contain an equal number of atoms .Their half lives are 1 hour and 2 hours
respectively .Calculate the ratio of their rates of disintegration after two hours?
Ans
The half –life of 6C14 is 5700 years. It means that one half of the present number of
radioactive nuclei of 6C14 will remain undecayed after 5700 years.
Number of nuclei Y after 2 hour . Nx = N0
1
2
Number of nuclei Y after 2 hour . Ny = N0
1
2
2/1
= N0
4
2/2
= N0
2
Radio of rates of disintegration
Rx
= No/4
Ry
=
N0/2
1
2
Q:-6 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:-
Binding
Fe56
Energy 9
Per
016
U238
8
N
Nucleon7
(MeV) 6
5
Li7
4
3
2
1
H2
0
20
40
60
80
120
140
160
180
200
220 240
Mass number
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:-7
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:-8 A radio nuclide sample has N0 nuclei at t=0 .Its number of undecayed nuclei get reduced to
N0 /e at t= 𝜏.
What does the term 𝜏 stand for ? write the term of 𝜏 , the time interval ‘T’ in which half of
the original number of nuclei of this radio nuclei would have got decayed ?
Ans
𝜏 is the mean life time of radio nuclei T is the half –life period of ratio nuclide ,the relation is
𝜏 =1.44 T
i.e Mean life period =1.44 x half life period .
Q:-9
Explain with example ,whether the neutron –proton ratio in a nucleus in increases or
decreases due to β-decay?
Ans
In β-decay a neutron is converted into a proton ,so the neutron – proton ratio decreases .
Equation of β-decay is
zX
A
90Th
234
z+AY
+A
91Pa
234
+ -1β0 + ν
+ -1β0 +
Neutron to proton ratio before β-decay
= (234 -90 ) / 90 =1.60
ν
Neutron to proton ratio after β-decay
= (234-91 ) / 91 = 1.57
So Neutron to proton ratio in β decay decreases
Q:-10 Whit the help of an example how the neutron to proton ratio changes during α –decay of a
nucleus ?
38
92U .
Ans Let us take the example of α –decay of
38
92U
90Th
234
The decay scheme is
+ 2α 4
Neutron to proton ratio before α –decay = (238 -92 ) / 92 = 146 / 92 = 1.59
Neutron to proton ratio after α –decay =
(234 – 90 ) / 90 = 144 /90 = 1.60
This shows that the neutron to proton ratio increases during α –decay of a nucleus .
Q:-11 Distinguish between isotops and isobars. Give one example for each of the species?
Ans
Isotopes
Isobars
The nuclides having the same The nuclides having the same atomic
atomic number Z but different mass (A) but different atomic numbers (Z)
atomic mass (A) are called isotopes are called isobars .
Examples :
Example :1
2
3
3
1H ,1H
1H , 2He
Q:-12 Group the following six nuclides into three pairs (i) isotones (ii) isotopes(iii) isobars ?
12
3
6C , 2He
Ans Isotones:-
, 80Hg198 , 1H3 ,79Au197 , 6C14
198
80Hg
Isotopes :
12
6C
Isobars :-
2He
3
and
and
and
197
79Au
14
6C
3
1H
(same number of neutrons A-Z )
(same atomic number )
( same mass number )
Q:-13 In a radioactive decay as follows
A
0
+1e
A1
α
A2
The mass number and atomic number of A2 are 176 and 71 respectively ,what are the mass number
and atomic numbers of A1 and A .Which of these elements are isobars?
Ans The reaction may be expressed as
zX
A
(A)
Z-1Y
A
+
(A1)
Given Z-3 =71 ⇒ Z =74 and Z -1 =73
+1e
0+
ν
Z-3Y1
A-4
( A2 )
+ 2He4
(13)
Also A -4 =176 ⇒ A =180
Thus ,mass numbers of A1 and A are 180 each
The atomic numbers of A1 and A are 73 and 74 respectively
The elements A and A1 are isobars.
Q:-14 A radioactive nucleus A undergoes a series of decay according to following scheme:
A
α
A1
β-1
A2
α
A3
γ
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
72A
180
α
76
70A
β-1
71A
176
α
69A3
172
γ
69A4
172
Clearly ,mass number of A4 is 172 and atomic number is 69.
Q:-15 You are given two nuclides 3X7 and 3Y4
(ii)
(iii)
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:- 16 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 .Draw a plot of N as
a function of 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 .
(14)
𝑑𝑁
𝑑𝑡
𝑑𝑁
⇒
𝑁
= −λ N
where λ disintegration constant .
= −λ dt
Integration loge N =
λ t +C
…(1)
Where 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
N0
Substituting this equation in (1) ,we get
⇒
Loge N – loge N0 =
λt
Loge N / loge N0 =
λt
N
N=N0e—λt
t
The graph is shown in fig.
Q:-17 Derive expression for average life of a radio nuclei .Give its relationship with half life ?
Ans All the nuclei of a radio active do not decay simultaneously ; but nature of decay process is
statistical , i.e it can not be stated with certainly which nucleus will decay when .The time of decay
of a nucleus may be between 0 and infinity . The mean of lifetimes of all nuclei of a radioactive
elements is called its mean life .It is denoted by 𝜏.
Expression for Mean Life :According to Rutherford –Soddy Law .rate of decay of a radioactive elements
R( t ) =
𝑑𝑁
𝑑𝑡
=λN
Therefore ,the number of nuclei decaying in between time t and t + dt is
dN = λ Ndt
If N0 is the total number of nuclei at t= 0 , then mean lifetime
𝜏
=
Total life time of all the nuclei
Total number of nuclei
Also we have
N=N0e—λt
= ∑ t. dN
N0
∑t λ (N0e—λt ) dt
𝜏 =
= λ ∑t e -λt dt
N0
As nuclei decay indefinitely , we may replace the summation into integration with limits from t=0
to t= i.e.
∞
∫0
𝜏 = λ
te - λt dt.
Integrating by parts ,we get = 1
λ
Thus ,
𝜏 = 1
λ
i.e the mean lifetime of a radioactive elements is reciprocal of its decay constant .
Relation Between Mean Life and Half Life .
Half life
T=
0.6931
λ
Mean life
𝜏
= 1
λ
Substituting value of λ form (2) in (1) ,we get
T = 0.6931
𝜏
Q:18 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=
When
N0e—λt
t = T,N =
2
N0
…(1)
N0e—λt
N0 =
2
e-λt
Or
= 1/2
Taking log of both sides
….(2)
-
Or
λT loge e = loge 1- loge 2
λT = log e 2
T = loge 2
λ
….(3)
= 2.3026 log10 2 = 2.3026 X 0.3010
λ
or
λ
T = 0.6931
….(4)
λ
Q:-19 What are α –particles ? In the reaction
numbers of Y?
ZX
A
α+γ give the atomic number and mass
Ans
α –particles are doubly ionized helium atoms (or nuclei of helium ) .When a radioactive
nuclide emits an are α –particles ,its mass number is reduced by 4 and atomic number by 2 ; so
the atomic number of Y is Z-2 and mass number (A- 4).
Q:-20 Explain why is the energy distribution of β –rays continuous/
Ans During β-decay ,a neutron is converted into a proton with the emission of a β- particle with an
antineutrino
1
0n
1P
1
+ 1β0 + ν
The energy produced in this decay is shared by β-particle and antineutrino ; therefore β –particle
may have varying amount of energy starting from zero to a certain maximum value.
Q:-21 In a nuclear reactor give the function of the (i) moderator (ii) control rods (iii) coolant (iv)
heavy water .
In the reaction
92P
236
141
aQ
+
b
36R
+ 3( 0n1)
What are the values of a and b .
Ans (i) A moderator slows down the fast moving neutrons to convert them to thermal neutrons.
(2) Control rods absorb excess neutrons to control the chain reaction .
(3) Coolant transfers heat from the core of reactor to surroundings.
(4) Heavy water in a reactor is used as a moderator and coolant.
By conservation of mass number
236 = 141 +b +3x1 ⇒ b = 92
By conservation of charge
92 = a + 36 ⇒ a = 56
Q:-22 Why is the heavy water used as a moderator in a nuclear reactor ?
Ans The basic principal of mechanics is that momentum transfer is maximum when the mass of
colliding particle and target are equal . Heavy water has negligible absorption cross –section for
neutron and its mass is small ; so heavy molecules do not absorb fast neutrons ;but simply slow
them.
Q:-23 Draw a graph showing the variation of potential energy between a pair of nucleon as a
function of their separation .Indicate the regions in which the nucleus force is (i) attractive (ii)
repulsive?
Ans Part AB represent repulsive force and part BCD represents attractive force .
+ 100 MeV
Repulsive
0
-100MeV
AttrA
Attractive
r (fm)
Q:-24 Why is the mass of a nucleus always less than the sum of the masses of constituents ,
neutrons and protons?
If the total number of neutrons and protons in a nuclear reaction is conserved how then is the
energy absorbed or evolved in the reaction ? Explain ?
Ans When nucleon combine to form nucleus , some mass is converted into binding energy in
accordance with mass energy equivalence relation ∆E=∆m.c2 .Hence ,the mass of a nucleus is less
than that of nucleons.
2nd Part:-
In fact the number of protons and number of neutron are the same before and after
a nuclear reaction ,but the binding energies of nuclei present before and after a nuclear reaction
are different .This different is called the mass defect (∆M).
∆M =[Zmp+(A-Z)mn] –M
Einstein’s mass energy relation ,E =mc2
Can express this mass different in term energy as ∆Eb = ∆Mc2
This show that if a certain number of neutron and protons are brought together to from a nucleus
of a certain charge and mass , an energy ∆Eb will be released in the process .The energy ∆Eb is called
the binding energy of the nucleus .If we separate a nucleus into its nucleons we would have to
transfer a total energy equal to ∆Eb ,to the particles .
ELECTRONIC DEVICES
GIST OF THE UNIT
CLASSIFICATION OF MATERIALON THE BASIS OF CONDUCTIVITY AND RESISTIVITY
(i)
(ii)
(iii)
Metals: They possess very low resistivity (or high conductivity). ρ ~ 10–2 – 10–8 Ω m σ ~
102 – 108 S m–1
Semiconductors: They have resistivity or conductivity intermediate to metals and
insulators. ρ ~ 10–5 – 106 Ω m σ ~ 105 – 10–6 S m–1
Insulators: They have high resistivity (or low conductivity). ρ ~ 1011 – 1019 Ω m σ ~ 10–
11 – 10–19 S m–1 The values of ρ and σ given above are indicative of magnitude and could
well go outside the ranges as well. Relative values of the resistivity
TYPES OF SEMICONDUCTORS
(i)
Elemental semiconductors: Si and Ge
(ii)
Compound semiconductors: Examples are: •
(iii)
Inorganic: CdS, GaAs, CdSe, InP, etc. •
(iv)
Organic: anthracene, doped pthalocyanines, etc. •
(v)
Organic polymers: polypyrrole, polyaniline, polythiophene, etc. Most of the
currently available semiconductor devices are based on elemental semiconductors Si
or Ge and compound inorganic semiconductors.
1) Semiconductors are the basic materials used in present solid state devices
like diode, transistor, ICs,etc. they are elemental(Si, Ge)as well as
compounds(GaAs,CdS etc).
2) Lattice structure and atomic structure of consistent elements decide whether
a particular material will be a insulator, metal or semi conductors.
3) Pure semiconductors are called intrinsic. Semiconductors, ne=nh ie no. of
electrons is equal to no. of holes. Holes are electron vacancies with an
effective positive charge.
4) The separation between conduction band and valance band on the energy
level diagram is called Forbidden energy gap.
5) For Germanium the forbidden energy gap is 0.7ev while it in 1.1ev silicon.
6) The number of charge carriers can be changed by doping. Such
semiconductors are called extrinsic semiconductors these are of 2 types (ntype
and p-type).
7) Ne >>nh while in p-type semiconductors nh>>ne.
8) N-type is obtained by doping Si or Ge with pentavalent atomic(donors)like
As,Sb,P etc,while p-type is obtained by doping Si or Ge with trivalent
atoms(acceptors) like B,Al,In, etc.
9) nenh= ni2 , For insulators Eg > 3 eV. For semiconductors Eg is 0.2eV to 3 eV.
10) Formation of pn junction produces a depletion layer consisting of
immobile ion cores devoid of charge carriers with a width of 10-3 mm.
11) Potential barrier is produced about 0.7 V for a silicon pn junction and 0.3V
for Germanium pn junction.
12) In forward bias ,the barrier is decreased while ,it increases in reverse
bias.Hence forward current is more (mA) while it is very small (ìA) in
reverse bias.
13) Diodes can be used for rectifying ac voltage. With the help of a capacitor or
suitable filter ,a dc voltage can be obtained.
14) There are some special purpose diodes.
15) Zener diode is used as a voltage regulator.
16) P-n junctions have been used to obtain many photonic or optoelectronics
devices. Eg- photodiodes , Solar cells, LED and diode LASER
Concept map
Identification of important topics /concepts for slow learners
1) Difference between insulator ,conductor and semiconductor on the basis of Energy band diagram.
2) Difference between n-type and p-type semiconductor on the basis of
doping and energy band diagram.
3) Definition of important terms like depletion layer,forward bias,and reverse
bias,barrier potential, doping.
4) Graph forward Bias and reverse bias of a p-n junction diode.
5) Diode as rectifier-working and cicuit diagram with graph.
6) Use of Zener diode as a voltage regulator
7) CE Amplifier circuit with working and graph.
8) Transistor as an oscillator circuit and working
9) Logic gates-AND,OR,NAND,NOR,NOT, with symbols and truth tables.
10) Some simple digital circuits with combination of gates.
IMPORTANT DERIVATIONS COVERING WHOLE UNIT (3 & 5 Marks)
1. Distinguish between conductor , insulator and semiconductor on the basis of
energy band diagram ?
Ans.
* Conductor-If there is no energy gap in the energy band diagram, the solid
behaves as conductor.
* Insulator -If there is a large energy gap in the energy band diagram, the solid
behaves as insulator or bad conductor
* Semiconductor - If there is a small energy gap in the energy band diagram,
the solid behaves as a semiconductor. For Germanium Eg = 0.72 Ev, For
Silicon Eg = 1.1eV
Conductor insulator semiconductor
2. 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 contacts
provided at the ends for external voltage.
Forward bias: In forward bias, the p-type is connected with the positive
terminal and the n-type is connected with the negative terminal.
Reverse bias : In reverse bias , the p-type is connected with the negative
terminal and the n-type is connected with the positive terminal.
3. What is zener diode. Draw V-I characteristic curve of zener diode. Explain its
use as an voltage regulator with circuit diagram.
Ans. It is designed to operate in the reverse breakdown voltage region
continuously without being damaged.
A zener diode has unique feature that voltage drop across it , is independentof current through it.
When the input d.c.voltage across zener diode increase beyond a certain limit
i.e. zener voltage , the current through the circuit rises sharply , causing a
sufficient increase in voltage drop across the dropping resistor R. As a result
of it the voltage across zener diode remain constant and hence the output
voltage lower back to normal value.
4. What is junction transistor. Write its types with symbol. Giving circuit
diagram of p-n-p transistor in CE draw input & output characteristic curve.
Ans. A junction transistor is a three terminal solid state device obtained by
growing a thin layer of one type semiconductor in between two thick layers of
other 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
sandwiched between two thicker section of p-type semiconductors.
5.
What is amplifier. Discuss use of n-p-n transistor as an amplifier with circuit diagram. What is phase relation
between input & output waveform.
Ans. A device which increases the amplitude of the input signal is called amplifier. In common emitter
amplifier, input signal to be amplified is applied between base-emitter circuit and the output amplified signal
is taken across the load resistance in emitter- collector circuit.
There is a phase difference of π between input and output signal.
6. From the diagram shown below identify whether the diode is forward or reverse biased.
There is a phase difference of π between input and output signal.
6. From the diagram shown below identify whether the diode is forward or reverse biased.
Ans.
(a) Forward bias (b) Reverse bias.
7. 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 ‘full wave 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.
8. (a) Draw transfer characteristic curve of Base-biased C-E transistor.
(b) Mention the region where the transistor used as switch & where as
Amplifier.
.
vi ((b) Switch- ON Switch- saturation region
OFF-Switch-cut off region
Amplifier – active region
9. You are given two circuits as shown in Fig. Giving truth table identify the
logic operation carried out by the two circuitsElectronic Devices
Ans.
10. Giving circuit diagram discuss working of transistor as an oscillator.Ans.
An oscillator is a device which can produce undamped electromagnetic
oscillations of desired frequency and amplitude.
Tank circuit containing an inductance L and a capacitance C connected in
parallel can oscillate the energy given to it between electrostatic and magnetic
energies. However, the oscillations die away since the amplitude decreases
rapidly due to inherent electrical resistance in the circuit.
In order to obtain undamped oscillations of constant
amplitude, transistor can be used to give regenerative or positive feedback
from the output circuit to the input circuit so that the circuit losses can be
compensated
The frequency of oscillation is given by-
11. 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 between
the input and output voltage.
13.
What is half wave rectifier. Giving circuit diagram & input-output
waveform 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 in
second half cycle the diode is reverse bias & hence not conduct. Hence it gives
half dc
14. Distinguish between intrinsic & extrinsic semiconductor.
Ans.
15. 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 is
basically a p-n junction operating in a photovoltaic mode without external
bias.
Working: When light photons fall at the junction electron-hole pairs are
generated. those more in opposite direction due to junction field. These
charges accumulate at the two sides of the junction and photo voltage is
developed.
Use: It is used in calculators etc.
SOME IMPORTANT QUESTIONS FROM PREVIOUS YEAR PAPERS
(CHAPTER-ELECTRONIC DEVICES)
1.In a semiconductor the concentration of electorns is 8x1013cm-3 and that of holes
is 5x1012cm-3. Is a p-type or n-type semiconductor?
Ans : As concentration of electrons is more than holes, the given extrinsic semi
conductor is n-type.
2.The energy gaps in the energy band diagrams of a conductor, semiconductor and
insulator 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 an
insulator is ≥ 3eV. E1=0, E2=1eV, E3≥3eV . E1 < E2 < E3.
3. Find the truth table of following gates….
4. Find the truth table of following
gates….
gates.
Ans : OR, AND and NOT gates.
6.What is the width of dipletion layer in p-n junction diode?
5.Name the fundamental
Ans : The depletion layer in junction diode is of the order of micrometer (≈10-6m)
7.The current gain (α) of a transister in common base configuration is 0.98. What
does It physically mean?
Ans :The current gain α=0.98 means that 98% of charge carriers of an emitter
reach the collector and constitute the collector current
8.In the following diagrams, write which of the diodes are forward biased and
which are reverse biased
Ans : The junction diode in figs (b) and (d) are forward biased and (a) and (c) are
reverse biased.
9.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 fig
Truth table of NOR gate
10.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
11. Show the output waveforms(Y) for the following inputes A and B of(i) OR gate (ii) NAND gate
Ans:
12.A photodiode if
fabricated from a semiconductor with band gap of 2.8eV. Can
it detect a wavelength of 6000nm? JustifyAns : Energy corresponding to wave length 6000 nm is E==3.3X1020J=0.2 eV
The photon energy of given wavelength is much less than the band gap hence it
cant detect the given wavelength.
13. For CE transister 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
14.Two amplifiers are connected one after another in series (cascaded). The first
amplifier has a voltage gain of 10 and the second has a voltage gain of 20. If the
input 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
15.In a common emitter mode of a transister, the dc current gain is 20, the emitter
curent 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
16.A semiconductor has equal electron and hole concentration of 6X108/m3. On
doping 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
nh = ni2/ ne=4x104m-3
17.A semiconductor has equal electron and hole concentration of 2*108/m3. On
doping with certain impurity, hole concentration increases to 4*1010/m3.
(i) What type of semiconductor is obtained on doping.
(ii) Calculate the new electron and hole concentration of the semiconductor.
(iii) How does the energy gap vary with doping.
Ans : Given ni=2x108m-3, nh=4x1010m-3
(i) p-type semiconductor
(ii) ne nh = ni2
ne = ni2/ nh = 106 m-3
New electron concentration = 106 m-3
Hole concentration =4x1010m-3
(iii) Energy gap decreases on doping.
18. The V-I characterstic of a silicon diote is given in fig.
Calculate the diode resistance in forward bias atV=+2V.
Ans : At V=+2V, i=70mA=70x10-3A
Therefore R=V/i =28.6Ω
19. A change of 0.2 mA in the base current cause a change of 5mA in the collector
current for a common emitter amplifier.
(i) Find the a.c. current gain of the transistor
(ii) If the input resistance is 2kΩ, and its voltage gain is 75, calculate the load
resistor used in the circuit.
Ans : (i) AC current gain β= ΔIC/ΔIB = 5/0.2 = 25
(ii) Voltage gain, AV= β RL/Ri
Therefore, RL= AV Ri / β =6kΩ
20. Two signals A,B are given below, are applied as
input to (i) AND (ii)NOR and (iii) NAND gates.
Draw the output wave form in each case.
Ans:
PRECTICE QUESTIONS
Q 1.Draw the logic circuit of NOT gate and write its truth table.
Ans:
A
Y
A
Y
0
1
1
0
Q 2.
Name the type of biasing of p-n junction diode. So that the junction offers very high resistance.
Ans:
Reverse biasing
Q 3.
Name one imparity each, which when added to pure Si, produces (i) n- type and (ii) p- type
semiconductor.
Ans:
(i) for n- type – arsenic.
for p- type – indium
Q 4.
Draw energy band diagram of a p- type semiconductor.
Ans:
C.B.
Eg. o o o o
Acceptor energy level
o o V. B.
0.01 – 0.05 ev
(T > 0 K)
Refer to NCERT book fig no 14.9 page no . 477
Q 5.
Name two factors on which electrical conductivity of a pure semiconductor at a given temperature
depends.
Ans:
(i)
Band gap
(ii)
Biasing
Q 6.
Give the ratio of the number of holes and the number of electrons in an intrinsic semi-conductor.
Ans:
ηe = ηn So
Q 7.
How does the width of depletion layer change in reverse li of p-n. junction diode?
Ans:
Increases.
Q 8.
Draw the Circuits Symbol for npn transition.
1:1
Ans:
C
B
E
Q 9.
Draw the circuit symbol for LED.
Ans:
Q 10.
Which of the diodes is (i) forward biased and (ii) reverse biased in the following circuits? Justify your
answer.
+3 V
Ans:
-10 V
A
B
-10 V
Ans:
Fig (a) diode is reverse biased because p-type of the diode is at lower potential.
+4 V
C
Fig (b) diode is reverse biased,
-8 Vp-type of the diodes is at lower potential.
Fig (c) diode is forward biased, because p- type of the diode is at higher potential.
Fig (d) diode is forward biased because p- type of the diode is at higher potential.
Q 11.
Write two character feature to distinguish between n-type and p-type semiconductor.
D
Ans:
Two distinguish feature are –
(i)
(ii)
n-type
Majority charge carriers are electrons
Mobility is high
(i)
(iii)
p-type
Majority charge carriers are holes.
Mobility is low.
Q 12.
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)
(ii)
Q 13.
Low operational voltage and
Fast on-off switches capability
Draw a circuit diagram to slow how a photo diode is viewed. Draw its characteristics curres for two
different illumination intensities.
N𝜗
Ans:
NA
n side
Refer to NCERT text book fig no. 14.23 Page no. 487
mA
Reverse line
I1
I1
I1
I1
Volt
𝜇A
p side
Q 14.
Which special type of device can act as a voltage regulator? Give the symbol of this divide and draw
the general shape of its V-I characteristics.
Ans:
Zener divide
I mA
Reverse Line
V2
Forward Line
V
I 𝜇A
Q 15.
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 gate
Q→
AND gate
Q 16.
How is an n-type semiconductor formed? Name the major charge carriers in it. Draw the energy
band diagram of an n-type semiconductor.
Ans:
When and intrinsic semiconductor (like Si, Ge) is doped with pentavalent impurity atoms (e.g.
arsenic , phosphorus etc) n-type semiconductor is formed. Each impurity atom donates an electron
majority charge carriers in n-type semiconductor are electrons.
C. B. .
Eg
.
O
.
.
.
V. B.
.
.
Donor energy level
O
Q 17.
Define the depletion layer and the potential barrier in a p-n junction diode.
Ans:
Refer to NCERT text book.
Q 18.
Ans.
Q 19.
Ans:
Q 20.
Ans:
Q 21.
Ans:
Q 22.
Ans:
Q 23.
Ans:
If the output of a two input NOR gate is fed as the input to a NOT gate (i) Name the new logic gate
abstained and (ii) write down its truth table.
(i) OR gate
A
B
Y
0
0
0
1
0
1
0
1
1
1
1
1
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.
50 Hz for half wave
100 Hz for full wave.
What is a solar cell? How does it works? Give its one use.
Solar cell is device for converting solar energy into electricity. It is basically a p-n junction operating
in a photovoltaic mode without external bias.
Working: When light photons fall at the junction electron-hole pairs are generated. those more in
opposite direction due to junction field. These charges accumulate at the two sides of the junction
and photo voltage is developed.
Use: It is used in calculators etc.
Draw the circuit diagram showing use of a transistor as an oscillator.
Refer to NCERT book fig no 14.33 (a)and (b) on page no 500.
What is a solar cell. Draw its circuit diagram. Draw I- V characteristics of a solar cell.
Definition given in Q no 19.
For Circuit and I-V graph.
Refer NCERT text book fig no 14.25 (a) and (b) on page no 489.
What is LED with the help of a diagram, show the biasing of a Light Emitting Diode (LED). Give its two
uses.
It is a heavily doped p-n junction which under forward bias emits spontaneous radiation.
h𝜗
n
n
p
Metalized contact
Q 24.
With the help of energy band diagram, distinguish between conductors, semiconductors and
insulators.
Ans:
Refer NCERT text book fig. 14.2 on page no. 471.
Q 25.
What is a rectifier. With the help of a labeled circuit diagram, explain the working of a full wave
rectifier. Draw the input and output waveforms.
Ans:
Refer NCERT text book, fig 14.19 on page no 484.
Q 26.
Give the symbol of a n-p-n transistor, show the biasing of a n-p-n transistor and explain transistor
action.
Ans:
Refer NCERT text book fig 14.27 and 14.28 on page no 492.
Q 27
What is Zener diode with the help of a circuit diagram explain the use of Zener diode as a DC voltage
regulator.
Ans:
Refer to NCERT text book fig. no 14.21 and fig. 14.22
Q 28.
Draw a circuit diagram of a common emitter amplifier, Deduce an expression for its voltage gain.
Explain briefly how the output voltage is out of phase by 180o with the input voltage.
Ans:
Refer to NCERT text book fig. no 14.32 on page no 498.
COMMUNICATION SYSTEMS
GIST OF THE UNIT:
Communication refers to the sending, receiving and processing of information by electrical means. A modern
communication system involves
1. Sorting, processing and storing of information.
2. Actual transmission after filtering of noise.
3. Reception which may include processing steps, example- decoding, storage and
interpretation
ELEMENTS OF A COMMUNICATION SYSTEM
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.
There
are two basic modes of communication: point-to-point and broadcast.
Point-to-point communication mode: In this mode, communication takes place over a link between a single
transmitter and a receiver. For example Telephony
Broadcast mode: In this mode, a large number of receivers corresponding to a single transmitter.
Radio and television are examples of broadcast mode of communication
BASIC TERMINOLOGY USED IN ELECTRONIC COMMUNICATION SYSTEMS
(i) Transducer: Any device that converts one form of energy into another can be termed as atransducer. For
example, a microphone converts a sound signal into an electrical signal.
(ii) Signal: Information converted in electrical form and suitable for transmission is called asignal.Signals can
be either analog or digital.
Analog signal: Analog signals are continuous variations of voltage or current. They are essentially singlevalued functions of time. Sine wave is a fundamental analog signal. All other analog signals can be fully
understood in terms of their sine wave components. Sound and picture signals in TV are analog in nature.
Digital signal: Digital signals are those which can take only discrete stepwise values. Binary system that is
extensively used in digital electronics employs just two levels of a signal. ‘0’ corresponds to a low level and
‘1’ corresponds to a high level of voltage/current.
(iii) Noise: Noise refers to the unwanted signals that tend to disturb the transmission and processing of
message signals in a communication system. The source generating the noise may be located inside or
outside the system.
(iv) Transmitter: A transmitter processes the incoming message signal so as to make it suitable for
transmission through a channel and subsequent reception. (v) Receiver: A receiver extracts the desired
message signals from the received signals at the
channel output.
(vi) Attenuation: The loss of strength of a signal while propagating through a medium is
known as attenuation.
(vii) 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 the signal in communication systems.
(viii) Range: It is the largest distance between a source and a destination up to which the signal
is received with sufficient strength.
(ix) Bandwidth: Bandwidth refers to the frequency range over which equipment operates or the portion of
the spectrum occupied by the signal.
(x) Modulation: The original low frequency message/information signal cannot be transmitted to long
distances Therefore, at the transmitter; information contained in 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.
(xi) Demodulation: The process of extracting the information from the carrier wave at the receiver is termed
demodulation. This is the reverse process of modulation.
(xii) Modulation Index:μ=Am/Ac is called modulation index, in practice; μ is kept ≤ 1 to avoid distortion.
(xiii) 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
(xiv) Antenna:
An antenna is basically a small length of a conductor that is used to radiate or receive electromagnetic
waves. It acts as a conversion device.
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. The type of communication system needed for a given
signal depends on the band of frequencies which is considered essential for the communication process.
For speech signals, frequency range 300 Hz to 3100 Hz is considered adequate. 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
Similar to message signals, different types of transmission media offer different bandwidths.
The commonly used transmission media are wire, free space and fiber optic cable.
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. This range of frequencies is further subdivided and allocated for
various services as indicated in Table
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.
SOME IMPORTANT WIRELESS COMMUNICATION FREQUENCY BANDS
IONOSPHERE
The upper portion of atmosphere about 65 km to 400 km above the earth’s surface is called ionosphere. The
ionosphere is so called because of the presence of a large number of ions or charged particles. It extends
from a height of ~ 65 Km to about 400 Km above the earth’s surface.
PROPAGATION OF ELECTROMAGNETIC WAVES
Ground wave Propagation In ground wave propagation, the radio waves (AM) travel along the surface of the
earth. These waves are called ground waves or surface waves.
Ground wave propagation can be sustained only at low frequencies (~500 kHz to 1500 kHz).The maximum
range of ground wave propagation depends on two factors:
(i)The frequency of the transmitted wave
(ii)The power of the transmitter
Sky waves Propagation IONOSPHERE
The upper portion of atmosphere about 65 km to 400 km above the earth’s surface is calledionosphere. The
ionosphere is so called because of the presence of a large number of ions orcharged particles. It extends
from a height of ~ 65 Km to about 400 Km above the earth’ssurface.
PROPAGATION OF ELECTROMAGNETIC WAVES
Ground wave Propagation
In ground wave propagation, the radio waves (AM) travel along the surface of the earth. Thesewaves are
called ground waves or surface waves.
Ground wave propagation can be sustained only at low frequencies (~500 kHz to 1500kHz).Themaximum
range of ground wave propagation depends on two factors:
(i)The frequency of the transmitted wave
(ii)The power of the transmitterSky waves Propagation
In the frequency range from a few MHz up to 30 MHz, long distance communication can beachieved by
ionospheric reflection of radio waves back towards the earth. This mode ofpropagationis called sky wave
propagation and is used by short wave broadcast services.
(i) Critical frequency. The highest frequency above which the ionosphere no longer returns theskywave back
to earth when transmitted in vertical direction is called critical frequency.Above this frequency, the radio
wave will penetrate the ionosphere and is not reflected by it. It is given by f c=√ 9(Nmax ) , where Nmax is
maximum electron density of the ionosphere.
(ii) Critical angle. For a given frequency, the vertical angle above which the sky wave no longer returns to
earth but travels outward into space is called critical angle.
(iii) Skip distance. The distance between the transmitting aerial and the point where the sky wave is first
received after returning to earth is called skip distance. In Fig 3 the ground distance AC is skip distance. The
ground-wave range here is AB.
(iv) Skip zone. The ground distance BC is called skip zone. No signal can be picked up in the skip zone.
(v) Fading. It is the fluctuation in signal strength at the receiver end and it may be rapid or slow. It may be for
all frequencies or for any particular frequency.
Space wave Propagation
Another mode of radio wave propagation is by space waves. 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. At frequencies above 40 MHz, communication is essentially limited to
line-of-sight paths.
RANGE OF TV TRANSMISSION
The TV signals are in the 100–200 MHz range. Therefore, transmission of such signals via ground waves or
sky waves is not possible. In such situations, we useline-of-sight transmission i.e., TV signals are transmitted
by direct waves.
d= √2Rh
Thus, the range of TV transmission depends upon the height of thetransmitting antenna. The greater the
height of the antenna, thelarger is the range of TVtransmission. For this reason, TV broadcasts are made
from talltransmitting antennas.Range of Space wave propagation between two antennas on earth’s surface:
Aspace wave travels in a straight line from transmittingantenna to receiving antenna i.e. it is a line ofsight
(LOS) communication.
The maximum line-of-sight distanced between the two antennas having heights Ht and Hr above the earth
(see Fig.6) is given by d =√2RHt +√2RHr
Television broadcast, microwave links and satellite communication are some examples of Communication
systems that use space wave mode of propagation.
MODULATION AND ITS NECESSITY
Most of the message, information or speech signals are of low frequency which cannot be transmitted to
long distance.(Audio frequency signal’s range 20Hz to 20kHz).This is because – (i)For efficient radiation and
reception, the transmitting antennas should have heights comparable to a quarter wavelength (λ/4) of
frequency used.
(ii)Audio frequency range being so small, there would be so much overlapping and confusion.
(iii)Energy carried by low frequency audio waves is too small.
Effective power radiated by an antenna
A theoretical study of
radiation from a linear antenna (length l) shows that the power radiated is proportional to (l/λ)2. For a good
transmission, we need high powers and hence this also points out to the need of using high frequency
transmission.
Modulation
For translating the original low frequency baseband message or information signal into high frequency wave
before transmission such that the translated signal continues to possess the information contained in the
original signal. In doing so, we take the help of a high frequency signal, known as the carrier wave, and a
process known as modulation which attaches information to it.
A sinusoidal carrier wave can be represented as c (t ) = Ac sin (ωct + φ), Where c(t) is the signal strength
(voltage or current), Ac is the amplitude ,ωc ( = 2πνc) is the angular frequency and φ is the initial phase of
the carrier wave.
During the process of modulation, any of the three parameters ,viz Ac ,ωc and φ, of the carrier wave can be
controlled by the message or information signal. This results in three types of modulation:
(i) Amplitude modulation (AM), (ii) Frequency modulation (FM) and (iii) Phase modulation (PM)
AMPLITUDE MODULATION
In amplitude modulation the amplitude of the carrier signal is varied in accordance with the information
signal/modulating signal.A carrier signal is modulated only in amplitude value.The modulating signal is the
envelope of the carrierThe required bandwidth is 2B, where B is the bandwidth of the modulatingsignal
Here ωc –ωm and ωc +ωm are respectively called the lower side band (LSB) and upper side band (USB)
frequencies. The modulated signal now consists of the carrier wave of frequency ωc plus two sinusoidal
waves each with a frequency slightly different from, known as side bands. The frequency spectrum of the
amplitude modulated signal is shown in .fig
The difference between the highest and the lowest frequencies present in the A.M. wave is called its
bandwidth.
PRODUCTION OF AMPLITUDE MODULATED WAVE
Amplitude modulation can be produced by a variety of methods. A conceptually simple method is shown in
the block diagram.
DETECTION OF AMPLITUDE MODULATED WAVE
The transmitted message gets attenuated in propagating through the channel. The receiving antenna is
therefore to be followed by an amplifier and a detector. In addition, to facilitate further processing, the
carrier frequency is usually changed to a lower frequency by what is called an intermediate frequency (IF)
stage preceding the detection. The detected signal may not be strong enough to be made use of and hence
is required to be amplified.
DETECTION
the modulating signal from the modulated carrier wave.
Detection is the process of recovering
We just saw that the modulated carrier wave contains the frequencies and ωc ±ωm.
All possible formulae
Q 1.
Ans:
BASIC QUESTIONS
Name the type of communication in which the signal is a discrete and binary coded version of the
message of information.
Digital communication.
Q 2.
Ans:
Name the types of communication systems according to the mode of the transmission.
1. Analog communication
2. Digital communication
Q 3.
Ans:
Name the device which can represent digital data by analog signal and vice-versa.
Modem.
Q 4.
Ans:
How does the effective power radiated by an antenna vary with wavelength?
Power radiated by an antenna is inversely proportional to the wavelength.
Q 5.
Ans:
What is transducer.
Any device that convert one form of energy into another can be termed as a transducer.
Q 6.
Ans:
Bolck diagram of a receiver is as shown.
Recive
Signal
Ans:
X = I F Stage
1
output
Amplifier
X
Detector
Identify X and Y
Y
Y = Amplifier
Q 7.
Ans:
What is the requirement of transmitting microwaves from one to another on the earth?
The transmitting and receiving antenna must be in line of sight.
Q 8.
Ans:
What type of modulation is required for television broadcast?
Frequency modulation.
Q 9.
Ans:
What type of modulation is required for television broadcast of voice signal.
Amplifier Modulation.
Q 10.
Ans:
What do you mean by modulation.
Superimposition of baseband signals over carrier wave is called modulation.
Q 11.
What is ground wave? Why is ground wave transmission of signals restricted to a frequency of 1500
KHz?
Low frequency radio waves propagating along ground is called ground wave. Energy losses due to
absorption of energy by the earth are frequency dependent. For frequencies above 1500 KHz, it is
quite high.
Ans:
Q 12.
Ans:
Q 13.
Ans:
Q 14.
Ans:
Q 15.
Ans:
Q 16.
Ans:
What is sky wave proportion of waves? Explain why sky wave transmission of em wave can not be
used for TV transmission.
In Sky wave propagation, transmitted wave goes up in the sky and is reflected back from the
ionosphere. Waves transmitted towards ionosphere suffer total internal reflection. The radio waves
up to frequency 40 MHz can be reflected back by ionosphere .
TV transmission takes place at frequencies higher than 40 MHz, so sky wave propagation can
not be used for this.
What is space wave propagation? Give two examples of communication system which use space
wave mode. A TV tower is 80 m tall, calculate the maximum distance upto which the signal
transmitting from the tower can be received.
It is the mode of propagation in which waves travel in space in a straight line from the transmitting
antenna to the receiving antenna.
Example: Television broadcast, Microwave links , satellite communication.
(c)
given
h = Som.
Formula
d = √2Rh
Draw a schematics diagram showing the (i) Ground wave (ii) Sky wave and (iii) Space wave
propagation modes for e m waves.
Write the frequency range for each of the following.
(i)
Standard AM broadcast
(ii)
Television
(iii)
Satellite Commubication
Refer NCERT text book fig 15.6 on page no 522.
Write any two factors which justify the need for modulating a signal. What do you mean by
amplitude modulation? Draw a diagram showing an amplitude modulation wave by superposing a
modulating signal over a sinusoidal carrier wave.
(i)
Antenna size become practically viable.
(ii)
It avoids mixing up of signals from different transmitters.
Refer NCERT text book fig 15.8.
Define modulation index. Give its physical significance.
What do you mean by amplitude modulation with the help of block diagram explain how AM signal
is achieved.
Refer to NCERT book fig no 15.10 on page 525.
Q 17.
Ans:
Q 18.
Ans:
What do you mean by demodulation with the help of block diagram explain detection of AM signal is
carried out.
Refer NCERT text book fig no 15.13 on page 527.
Draw a plot of the variation of amplitude versus W for an amplitude modulated wave. Define
modulation index, State its importance for effective amplitude modulation. What is the role of a
band pass filter in Amplitude Modulation.
Refer NCERT text book, fig no 15.9 on page 525.
ASSIGNMENT-ELECTRONIC DEVICES
ASSIGNMENT– 1
1. The forbidden energy band in germanium that lies between the highest filled band
and theempty band above it, has a width of 0.7eV. Compare the conductivity of Ge
with that ofSi at: (a) Very low temperature and (b) room temperature.
Ans :
2. What element other than As and Sb can be used as am impurity with germanium
to forman n-type semiconductor?
Ans:
3. What element other than gallium and indium can be used to form a p-type
semiconductor?
Ans:
4. Why is a semiconductor damaged by a strong current?
Ans :
5. Name two factors on which electric conductivity of a pure semiconductor at a
giventemperature depends.
Ans :
6. Zener diodes have higher dopant densities as compared to p-n junction diodes.
How doesit effect the:
(i) Width of the depletion layer
(ii) Junction field.
Ans :
7. State the factors which control:
(i) Wavelength of light and
(ii) Intensity of light emitting by an LED.
Ans :
8. Visible photons are known to have energies ranging (nearly) from 1.8eV to 2.8eV.
Usethe information to reason why silicon is not a suitable semiconductor for
designingvisible light LEDs?
Ans:
9. (i) With the help of circuit diagrams, distinguish between forward biasing and
reversebiasing of a p-n junction diode.
(ii) Draw V-I characteristics of a p-n junction diode in (a) forward bias and (b)
reversebias.
10. Explain the use of p-n junction diode as a rectifier. Draw the circuit diagram of a
fullwave rectifier and explain its working. Draw the input and output waveforms.
11. Explain with the help of a circuit diagram how a Zener diode works as a dc
voltageregulator. Draw its I-V characteristics.
12. For an n-p-n transistor in the common-emitter configuration, draw a labeled
circuitdiagram of an arrangement for measuring the collector current as a function of
acollector-emitter voltage for at least two different values of base current. Draw the
shapeof the curves obtained. Define the terms: (i) Output resistance and (ii)
currentamplification factor.
Answers/Hints of Worksheet 1
Ans:1 (a) Both Ge and Si are bad conductors at very low temperatures.
(b) On account of the smaller width of the forbidden band in Ge (for Si it is 1.1 eV), it
isbetter conductor than Si at room temperature.
Ans:2 Phosphorus which is a pentavalent atom, can also be used to form an n-type
semiconductor.
Ans:3 Boron which is a trivalent atom can also be used to form a p-type
semiconductor.
Ans:4 When we pass a strong current through a semiconductor, it gets heated. Due
toexcessive heat, the covalent bonds are broken and crystal breakdown occurs. This
makesthe semiconductor useless.
Ans:5 (i) Width of the forbidden gap (Eg)(ii) Intrinsic charge carrier concentration (ni)
Ans:6 (i) The width of depletion layer is small.
(ii) The junction field is high.
Ans:7 (i) Nature of semiconductor used in LED
(ii) Forward bias applied to LED
Ans:8 For Si, band gap (Eg) is 1.1 eV, which is less than 1.8 eV. Hence, it is not suitable
formaking LEDs emitting visible light.
ASSIGNMENT– 2
1. Carbon and silicon are known to have similar lattice structures. However, the
fourbonding electron of carbon are present in second orbit while those of silicon are
presentin third orbit. How does this difference result in a difference in their
electricalconductivities?
Ans
2. Why is the conductivity of n-type semiconductor greater than that of p-type
semiconductor even when both of these have same level of doping?
Ans :
3. At what temperature would an intrinsic semiconductor behave like a perfect
insulator?
Ans :
4. Why should a photodiode be operated at a reverse bias?
Ans:
5. How is the band gap 'Eg' of a photodiode related to the maximum wavelength m
that canbe detected by it?
Ans:
6. If the base region of a transistor is made large as compared to a usual transistor,
how doesit affect: (i) the collector current and (ii) current gain of this transistor?
Ans:
7. How will you test in a simple way whether a transistor is spoilt or is in working
order?
Ans :
8. Show diagrammatically a forward biased and reverse biased p-n junction.
Ans:
9. With the help of a suitable diagram, explain the formation of depletion region in a
p-njunction. How does its width change when the junction is (i) forward biased and
(ii)reverse biased?
Ans :
10. With the help of a diagram, show the biasing of a light emitting diode (LED). Give
itstwo advantages over conventional incandescent lamps.
Ans :
11. Explain through a labeled diagram the working of a transistor as an amplifier
(commonemitterconfiguration).
12. (a) Draw the circuit arrangement for studying the input and output characteristics
of an nptransistor in CE configuration. With the help of these characteristics define (i)
inputresistance (ii) current amplification factor
(b) Describe briefly with the help of a circuit diagram how an n-p-n transistor is used
toproduce self-sustained oscillations.
Answers/Hints of Worksheet 2
1. The energy required to take out an electron from Si atom is less than that required
to takeout an electron from carbon atom. Hence, number of free electrons for
conduction in Si is
significant but negligibly small for C.
2. Electron (majority carries in an n-type semiconductor) have greater mobility than
holes
(majority carriers in a p-type semiconductor) for a given applied electric field.
3. At 0(Zero) K when there is no thermal agitation of electrons
4.
5. .
6. When the base region becomes large, most of the charge carriers coming from
emitter
will be neutralized in the base by electron-hole combination. Due to this: (i) collector
current decreases and as a result of this (ii) the current gain also reduces.
7. Take one of the junctions of the transistor. Note its impedance by first biasing it in
the
forward direction and then in the reverse direction. If the transistor is not spoilt, the
impedance of the junction is less when it is reverse biased than when it is forward
biased.
Similarly test the other junction.
376
ASSIGNMENT– 3
1. How does the energy gap in an intrinsic semiconductor vary when doped with a
pentavalent impurity?
Ans:
2. An n-type semiconductor has a large number of electrons but still it is electrically neutral.
Explain the reason.
Ans:
3. How does the thickness of the depletion layer in a p-n junction diode vary with increase
in reverse bias?
Ans:
4. State the reason, why GaAs is most commonly used in making a solar cell.
Ans:
5. In a transistor base is made very thin and dopped with little impurity atoms. Why?
Ans:
6. Draw the transfer characteristics of a base biased transistor in its common emitter
configuration. Explain briefly the meaning of the term ‘active region’ in these
characteristics. For what particle used, do we use transistor in this active region.
7. Give a circuit diagram of a common emitter amplifier using n-p-n transistor. Draw input
and output wave forms of the signal. Write the expression for its voltage gain.
8. Draw a labeled circuit diagram of a common emitter transistor amplifier. Explain clearly
how the input and out signals are in opposite phase.
9. State briefly the underlaying principle of a transistor oscillator. Draw a circuit diagramshowing how the
feedback is accomplished by inductive coupling. Explain the oscillatoraction.
10. Compare semi-conductor devices with vacuum tube devices.
11. Give the logic, truth table and Boolean expression for AND gate. How is it obtained inpractice?
12. Explain how the heavy dopping of the p and n sides of a p-n junction diode helps ininternal field emission
even with a reverse bias voltage of a few volts only.Draw thegeneral shape of the V-I characteristics of a
zener diode. Discuss how the nature of thesecharacteristics led to the use of a zener diode as a voltage
regulator.
Answers/Hints of Worksheet 3
1. When doped with a pentavalent impurity, donor energy levels are created just
below thebottom of conduction band, thereby, decreasing the energy gap.
2. In a n-type semiconductor, the large number of free electrons belong to those
electrons ofthe impurity (pentavalents) atoms which have not been able to form
covalent bonds. (Outof five electrons of the pentavalent impurity atoms, four form the
covalent bonds and thefifth is left free). Since atoms on the whole are neutral, the ntype semiconductor is alsoneutral.
3. The thickness of depletion layer increases with increase in reverse bias of the pnjunction.
4. Since GaAs has a suitable band gap (Eg=1.53 ev) and has high optical
absorptioncoefficient, it is used for making solar cell.
5. This is done so that only a small fraction of
(i)The holes (in p-n-p transistor) combined with the electrons in the base to get
neutralized.
(ii)The electrons (in n-p-n transistor) combined with the holes in the base to get
neutralized.
iii. Thus, infact base is made thin and dopped with little impurity only to allow a
smallbase current.
ASSIGNMENT- COMMUNICATIONS
ASSIGNMENT-1
are the thrtee basic units of a communication system?
2. Identify the parts X and Y in the following block diagram of a generalised
communication
3.What type ofmodulation is requiredfor televiosion broad cast?
4. why is frequency modulation is prefered over amplitude modulation?
1. What
5. . Name te type of radio wave propagation involved when T V signal, broad costby a
tallantena are intercepted directly by the receiver antenna
6.
What is the purpose of modulating signal in tyransmission?7. A modulating signal is a
square wave as shown in figure.
a. Sketch the amplitude
modulated waveform.
b. What is the modulatiron index?
9. A Transmitting antenna at the top of the tower has height of 32m and the height of
thereceiving atenna is 50 m. What is the maximum distance between them for satis
factorycommunication in LOS mode?10. What does the tertm LOS communication
mean? Name the types of wave used forcommunication. Give typical examples with
the help of suitable figureof communicationsystem for all three wave.
11. A T V tower has a height of 75 km . What is the maximum distance and area up to
whichTV transmission can be received? Assume radius of earth is 6400km12. A
schematic arrangement for transmitting a message signal 20 Hz to 20 KHz is given
BelowGive two drawbacks from which this arrangement suffers. Describe briefly with
the helpof-A block diagram the alternative arrangement for the transmission and
reception of theMessage signal
ASSIGNMENT-2
1.. Why short waves band are used for long distance broadcast?
2. What should be the length of dipole antena for a cartrier wave of frequency
600Mhz ?
3.A message has a band width of 5 Mhz.Suggest a possible communicationcation
channelforits transmission.
4. . A Carrier wave of peak voltage 20 V is used totransmit amessage signal.
What should be the peak voltage of the modulatig signal in order to have a
modulation indexof 80%?
5. Name the type of communication systems according to mode of the
transmission.
6. a T V tower has a height of 500m at agiven place. If radius of earth is
6400km. what is its coverage range.
7.Distinguish between point to point and broadcast communication modes
with example.
8. By What percentage will the transmission rangof a T V tower be affected
when the height of the tower is increased by 21 %.
9. Explain Sky wave, space wave and ground wave propagation with suitable
example.
10. What does the process of dectection of amplitude modulated mean .
Explain the function of detecterwith suitable block diagram.
11. What is amplitude modulation.Show graphiocally. Write its two limitations
and two advantages.
12Define the term modulation. Explain Need of modulation. Name three
different type modulationused for message signalusinng carrier wave.
explain the meaning of any one of them.
ASSIGNMENT-3
One mark questions:
1. What do you mean by a transducer? Give one examples
2. Name the basic modes of communication systems.
3. What do you mean by the term demodulation?
4.For long distance Radio broadcast, we use short wave band only. Why?
5. Name the type of radio wave propagation involved with T.V. signals, broadcast by
a tallantenna are intercepted directly by the receiver antenna.
6. Name the types of communication systems according to the mode of the
transmission.
2 marks questions:
1. Give reasons for the following:
(I) Long distance radio broadcasts use shortwave bands.
(ii) Satellites are used for long distance TV transmission.
3. Suggest two methods by which range of T.V. transmission can be increased
4. . T.V. tower has a height of 100 m. How much population is covered by the T.V.
broadcast if The average population density around the tower is 1000 km2 (radius of
theearth = 6.37 x106 m).
5. Distinguish between ‘point to point’ and ‘broadcast’ communication modes. Give
oneexample of each.
6. Define the term ‘modulation index’ for an AM wave. What would be the
modulationindex for an AM wave for which the maximum amplitude is ‘a’ while the
minimumamplitude is ‘b’?
3 marks questions:
1.If the sum of the heights of transmitting and receiving antennae in line of sight of
communication is fixed at h,show that the range is maximum when the two antennae
haveheight h/2 each.
2. For an amplitude modulated wave, the maximum amplitude is found to be 10V
while theminimum amplitude Is found to be 2V. Determine the modulation index; μ.
What wouldbe the value of μ if the minimum amplitude is zero volt?
3. Explain the following terms:
(i) Ground waves (ii) Space waves (iii) Sky waves.
5 marks question:
1. Derive an expression for the range into which signals transmitted by a T.V. tower
canreceive. What does the term LOS communication mean? Name the types of
waves thatare used for this communication. Which of the two-height of transmitting
antenna andheight of receiving antenna - can affect the range over which this mode
ofcommunication remains effective?
ASSIGNMENT-4
1 mark questions
1. Sky wave transmission of electromagnetic wave cannot be used for TV
transmission why?
2. What is the function transmitter in communication system?
3. How does the effective power radiated by an antenna vary with wavelength?
4. What is a transponder?
5. Suggest a possible communication channel for the transmission of a message
signal which
has a bandwidth of 5MHz.
2 marks questions
1. Distinguish between analog and digital signals.
2. What is space wave communication? Write the range of frequency suitable for
space wavecommunication?
3. Why are the high frequency carrier wave used for transmission?
3 marks questions
1. Define modulation index for an AM? What would be the modulation index for
which themaximum amplitude is a while the minimum amplitude is b?
2. A transmitting antenna at the top of tower has a height of 36m and the height of
receivingantenna is 49m. What is the maximum distance between them for
satisfactorycommunication in the LOS mode? (radius of earth = 6400Km.)
3. A transmitting antenna has a height of 100m. If the radius of earth = 6400Km. Find:
(i) distance up to which communication is possible on the surface of the earth.
(ii) area covered by the signals.
ASSIGNMENT-5
1 mark questions
1. What is the function of repeater in communication system?
2. What is the function of receiver in communication system?
3. Name the type of communication system according to the mode of transmission.
4. What is purpose of modulating a signal in transmission?
5. What is length of dipole antenna to transmit signal of frequency 200MHz.
2marks questions
1. Define modulation index. Why the amplitude of modulating signal kept less than
amplitudeof carrier wave?
2. Write two factors justifying the need of modulation for transmission of signal.
3. Block diagram of a receiver is shown in the figure (i) Identify X and Y (ii) Write
theirfunctions.
Receiving antenna
3marks questions
1. A carrier wave of peak voltage 20V is used to transmit a message signal. What should be the
peak voltage of modulating signal in order to have modulation index of 80%?
2. Define amplitude modulation. Derive an equation for amplitude modulated carrier
Wave . What does this equation signify?
3. A schematic arrangement for transmitting a message signal (20 Hz to 20KHz) is given below