40 YEARS OF-

EMINENCE
IN EDUCATION

A HANDBOOK OF

PHYSICS
Chapterwise Formulae and Concepts
Rememberable Facts in Tabular Form

Class XI
 IMPORTANT FORMULAE AND CONCEPTS
UNIT-1: Electrostatics
1. Coulomb's force F between two point charges kept in a medium of
electric constant, F= 1 942
4TEgK
For air between the charges, dielectric constant K = 1.

Faiv4TtE0
In vector form Fo, = 2
4TE221
Where F1 is the force on charge q, due to q2 and f21 is the unit vector
in the direction from qto
q2
2. Electric field strength E at any point in the field where F is the force

experienced by a test charge qo kept at that point = lim F

9o-090

(a) Electric field strength due to a point charge at a distance r

F-4TEo2= 4TE0 rT where

r is unit vector along
-
(6) Due to sphere charge

)Inside point (r s R) E4ErEar o P3 ar E

E*1/2

(i) Outside point (r2 R) E= ;E« K

4TE2
(ii) On the surface (T =R) B= 1
4TE R2
(c) Due to hollow sphere of charge
() Inside point (r s R), E = 0 E

1 Ex1/
(ii) Outside, E=
47TE
E=0
(ii) On the surface, E 1_9-9
4TE.
47TE p26
R2 E
(c) Electric field strength due to infinite line charge having linear
charge density () coulomb/metre.
2A
E .= 1
47TE T

A Handbook of Physics-XII 1
 thin sheet of
d) Electric field
strength n e a r an infinite
harge.
E:28 (ii)at an equitorial point E= 4
4TET
()Electric field
strength n e a r a
conductor E
= n, where i is a
( Electric potential
to the surface in the outer direction. El.
unit vector normal ) at axial point V
inside a conductor E 0.
TE
=
field strength
3. Electric P o t e n t i a l :
V kq/R (i) at an equatorial point V= 0
conducting sphere or charged Va1/r
Due to a charged g)Total electric flux, o= |E.dS =x net charge enclosed by the
radius R.
spherical shell of
(r>R closed surface.
side.,
) Inside,msude 4TE R
SR) )Outside, Vont 4
electric field and potential, CAPACITORS
4. Relation between
E-- dr (numerically) 8. Capacitance for isolated conductor, C=
C medium
5. Work done in taking a charge q from one point to another in electrie 9. Dielectric constant K= Cair
field. 10. Capacitance of parallel plate capacitor
W q(V2- V) joule C a A in air
d
where V potential at initial point,
V,= potential at final point. (i C=ACo when medium of dielectric constant K fills the space
d
6. Work done in carrying a charge on equipotential surface is always between plates.
zero.
(ii) When the space between the plates is partly flled with a dielectric
7. Electric Dipole:
of thickness t, then capacitance C =- EA
(a) Dipole moment |p= q2 (21 being the separation from-q to +g)
-
6) Torque on a dipole in uniform electric field t = pxE 11. (a) Combination of Capacitors:
(c) Potential energy of dipole, U -p. E= -pE ) Capacitors in series: Net capacitance C is given by
cos
=
where 0 is the angle between
pand E ,1,L
a Work done in rotating the dipole in uniform electric field from C C C
orientation 6, to 6, is i) In series charge is the same on all capacitors
W= U-
U= pElcos 0,-cos 0,)
Work done in
to
rotating the dipole from equilibrium position
0
(ii)Net potentialdifference V= V,+ Va+ Vs
orientation 0 is (b) Capacitors in Parallel:
W=pE (1- cos 0) ) Net capacitance C=G +C, +C
e)Electric field due to short (i) Potential difference is same across all capacitors
a
dipole.
() at axial point V= V2 =
V V(same for all)
=
Ea
4Te (iii) Charge q= 4 +9 +9s
2A A Handbook of Physics-XIl 3
Handbook of
Physics-Xll
 stored in a capacitor, 7. Electrical conductivity d=
12. Energy P
8. J= o E (alternative forms of Ohm's law)
energy
density, 9. (i) Resistances in series
Electrostatic
13.
(in medium) Net resistance Rs= R+ R2+ R3
U,=eEinair) andeE Current is the same in each resistance V= Vi+ V2 + V3
a dielectric between plates of a (ii) Resistances in parallel: Net resistance R, is given by
Effect of Introducing
charged parallel plate capacitor
Physical When battery When battery is removed R R, R, R,
S.
Quantity
remains before introduction of Voltage is the same across each resistance
No. connected dielectric
IFh+h+
(1) Capacitance (C) increases K-times increases K-times 10. Temperature dependence of resistance
(2) Charge(Q) increases K-times remains constant R = Ro (1+ aAt)
(3) Electric Field remains constant decreases times where a is the temperature coefficient of resistance
or Ra R, [1+ a (tz-t,)
(4) Electric Potential () remains constant decreases K times
11. Internal resistance of a cell r =
(5) Electrostatic Energy Stored | increases K-times decreasestimes
K
where E is emf of cell, V= terminal p.d. across external resistance R.
Combination of Cells
UNIT-2: Current Electricity () When n-identical cells are connected in series
1. Drift nE
velocity, v, =
- m Current,iR+R R+n
where E is electric field strength, t is relation time, e is the charge on For useful series combination, the condition is Rext >> Rint
electron and m is the mass of electron. i) Whenm-identical cells are connected in parallel
2. Relation between Current and Drift
Velocity: E
i
I= -neAua R+Rnt R+rm
where n= number of free electrons per m", A Condition of useful parallel combination is R < rlm.
=
cross-sect1onal are
3. Ohm's law (ii) When N= mn, cells are connected in mixed grouping (m-rows in
V= RI
parallel, each row containing n cells in series)
4. Resistance
R- S nCurrent, i=
nE mnE
mR +nr
A
5. Specific resistance
p-
m
Condition for useful mixed grouping is Rext = Rint
ne t
6. Current density R=
i.e.
4A A Handbook of Physics-XIl 5
Handbook of
Physics-XII
 emfs B, and Ba and different int
internal
cells of different
two shown in Magnetic field due to a straight current
iv) When
resistances
of
r and r2
combination

are
connected
is
in parallel as
fig
2.
carrying wire, Nr*****.. . . **.
then net emf Ho M 90 2 P
wwwww-
1
B
4TR SIn
tsin 6) 6 P
E E,+E2 where and 62 are the angles substended by ******
E = ww 2 ends of the conductor at the reference point with
the normal. For infinitely long wire
2TR
Net internal
resistance rint 3. Magnetic field due to a current carrying circular coil
)At centre B =Ho
L-+ 2R
(ii) At a point on the axis B.axas = o aw h e r e a = radius of coil)
12. Joule's Law of heating
effect of current: 2(a + 2
and x is the distance of the point)
?
W=l'RI==V
R
joule. 4. Ampere's circuital law o8.dd 4,/
13. Electric Power
5. Magnetic field strength within solenoid
P=Vl =l'R =watt. B=Honl where n = number of turns per metre length.
R
6. Magnetic field due to toroid
Value of External Current from the Terminal Power Consumed
Potential in External ) Within the coils B=HoN
Resistance Cell
2Ttr
Difference Resistance
(i) Outside the toroid B= 0.
R V E-Ir P = I'R 7. Magnetic force on a moving charge in a magnetic field
R =0
(Short carcuit)
V=E- P 0 8. Magnetic force on a current carrying conductor
(Maximum) V =0) F I lxB
R r 9. Force per unit length between parallel currents:
V=E-
V= Maximum
EHobde
21tr
N/m
10. Torque experienced by a current carrying loop in a uniform
Open circuit, I=0 V= E-00 P 0
magnetic field
R= c
V E T= NI Ax B = MxB
11. Magnetic moment of a current loop
UNIT-3: Magnetic Effects of Current and M =NIA ampere X metre
Magnetism 12. Deflection in moving coil galvanometer
1. Biot-Savart
Law: Magnetic field due to a current element
h- NAB,
dB= ldlx
4TT
C NAB
Current sensitivity of a galvanometer S=
06 A Handbook of 07
A Handbook of Physics-XII
Physics-XII
 conversion of
galvanometer into ammeter, UNIT-4: Electromagnetic Induction and Afternating Current
13. For
Shunt resistance required S=7 1. Magnetic flux sB.A =BA cos
into voltmeter,
where 0 is the angle between A and B.
conversion of galvanomter
14. For
2. Induced emf in a coil -N
Series resistance required R=-6
3. EMF induced in a moving conductor, e = Bul
moment of a n orbital electron
15. Magnetic where B, v, l are mutually perpendicular
2m 4. Magnetic flux
16. Magnetic field due to a short magnetic dipole where Lis the coefficient of self-induction.
B 2M
() At axis 5. If L is self inductance, emf induced e = -L
47t
=
HoM 6. Self inductance of a solenoid
(ii) At equator. Bequator 4T L 4 4 2 Al = H4 N"A
17. Elements of earth's magnetic field 7. Mutual Inductance E= -M
H= B,cos 6 A
Horixontal component
=HoN4
where angle of dip. 8. Mutual inductance of solenoid coil system M
Vertica component V=B,cos6
where N, =number of turns/metre in solenoid, N, = number of turns
tan 6=and B =vH*+V
H n coil.
9. Energy stored in inductance
18. Magnetic susceptibility x. =
19. Curie's law of magnetic susceptibility Z.* for paramagnetie
10. For an alternating current circuit
materials.
Distinction between Dia-, Para-and Ferromagnetics V V sin ot; I=Io sin (ot +)
Remark 11. RMS value of an alternating current
Property Diamagnetic Paramagnetic Ferromagnetic
Bo is magnetic Vo
)Magnetic B <Bo B Bo B>>Bo in
induction B induction
free space 12. Maxwell's Equation
(i) Intensity of small and small and very high and m is magnetic
There are four maxwel' equation given below:
magnetisation negative positive positive moment
M Gauss law in electrostatics: E.ds o
(im) Magnetic small and small and Very large and (i) Gauss law magnetism: Bds =0
suscepibility negative positive positive iii) Faraday's law of electromagnetic induction:
HM
(iv) Relative
d
emf=fE.d=- dt
4,>1(of (iu) Maxwell-Ampere's circuital law:
permeability the order the
thousands)
B.a-vo+
A Handbook of Physics-XII 09
08 A Handbook of
Physics-X|
Unit-5: Electromagnetic Waves
Parts of Electromagnetic Spectrum
E E
2
10 A Handbook of A Handbook of Physics-Xil 11
Physics-XII
13. Peak emf in a rotating coil of generator
E = NBAo Supply Voltage V V sin ot V Vo sin ot
V=Vo sin ot
I=I, sin ot
14. For a
Transformer =r (transformation ratio) Current
I=I,sin a- Isin +
Peak Current -
>1 R DL o1/oc VgoC
For a step up transformer r
Impedance(2) = R Vo
For a step down transformer, r = < 1 o
Z= s
R Resistance X = Inductive
Nc Capacitive
reactance
Direction of Current Induced in Some Cases reactance
Phase zero (in same phase)
System Primary Current Induced Current difference
+( leads ) leads 1)
1. Straight wire-coil system (i) Current Clockwise current Phasor
increasing
Diagram
(i) Current Anticlockwise
decreasing current
Variation of
Zwith v
2. Self inductive circuit i) Key is pressed Opposite to
direction of main
currents
(i) Key is released In the direction of
main current c
R does not depend
on v
3. Magnetic-coil system (i) North pole Anticlockwise
approaching current
i) coil
Combination of Components (RL or RC or LC)
N (i) North pole Clockwise current
(Gi) receding coil TERM RL RC LC
Circuit i s same in R & C in L & C
OO00000 / is same in R &L |is same
Man observing
direction of current
- R
R
h L h
Individual Components (R or L or C)
TERM R L C Phasor
V
diagram No
R
Circuit
V
Ve
VY -****
V= V-Vc(V> V)
=V+V =V+VE V= Ve-V, (V¢> VD
12 A Handbook of A Handbook of Physics-XII 13
Physics-XI
 1
4. Thin Lens formula:
Supply V Vsin ot V V, sin ot V= V, sin ot
Voltage
Current I = I sin (cot - ¢)
I=I, sin (ot +¢) = lh sin (ot + (ii) Linear magnification,m =
Phase V lags I (ii) Lens maker's formula,
in between
and I
Vleads(40toags=0to -J.7 N¢ >X.)
V leads I
(iv) Power of a lens P- diopter (f is in metres)
() Lens immersed in a liquid of refractive index n
Impedance|z =/R^+ xi
Variation As v increases,
Z=/R+
As v increases,
Z=X-Xl
As v increases, Z
L-1Randf -1
of Z with v Z increases Z decreases first decreases then
where f is focal length of lens in air.
Z Zt increases
(vi) Lenses in contact
R o r or P-P+P
R
Important Information:
() Refractive index n is maximum for violet and minimum for red
UNIT-6: Optics colour.
1. Refraction (ii) Critical angle increases with increase of wavelength.
Sini (ii) Minimum distance between an object and its real image is 4f.
() Snell's law, F
sin r (iv) When a lens immersed in a liquid is invisible, then refractive
index of liquid = refractive index of lens material.
(ii)n,=
5. () Refraction through a Prism:
(ii) n == speed of light in vacuum
vspeed of light ina medium tr A A angle of prism
nedium 4+ A+ô 8=angle of deviation
(iv) If object is in medium of refractive index n.
Sin sin i
Real depth Ni N2
'n Sin sin
Apparent depth app
For minimum deviation
Apparent shift --=(1-) i i = i and r =r2 =r E N
2. Critical angle for total internal reflection Angle of minimum deviation R
sin C = m 2i -A
1+,
Sin
3. A fish diver in water at 2
or
depthh sees the outside world in a
n
Sin
=-
horizontal circle of radius r given by Sinr
sin)
(i) For a thin prism ö =(n - 1) A
15
A Handbook of Physics-XII
14 A Handbook of Physics-XIll
 10. (a) Condition of Maxima for Young's Double Slit Experimen
6. Simple Microscope: on Interference of light:
at D)
Magnifying power, M =
1+(For final image
Phasedifference =2n
1,2,3,..
(For final image at infinity) Path difference A=nà
(b) Condition of Minima:
7. Compound Microscope: Phase difference o= (2n -1)n|
() Magnification M= mo X m
Path difference A=(2n-1)A"=.2.3,.
(i) M = - + D (For final image at D)
11. If sources of amplitude
of
a, and a, are coherent, intensity I at
where
a
point in the region superposition difference phase between
waves isso
I = ai +a, + 24a, coso=, +I, +2/(7,1,) coso
Maximum Intensity, 1max (a, t a,)"
tii) M = - D|
4 Minimum Intensity, Imin (a, -a,)
(for final image at o)
LD max
I min (, -a,)
transferred from
8. Astronomical Telescope: In interference energy is conserved. It is simply
minima to maxima.
() M-- (for final image at infinity),
12. Young's double slit experiment:
L =o+f. Position of maxima, y, = nD
(i) M = (For final image at distinct vision)
Position of minima, y, -
L=fo+u DA
9. Resolving Power:
Fringe width, B= yn+ 1 - Ynd
Resolving limit of telescope
Angular fringe width, B,=
d0 - 22% D= distance between sources and screen,
a
a
where = wavelength of light d distance between slits
a = diameter of objective
Resovling power single slit of width 'a':
1.22 13. Diffraction at a
lens of telescope sin 0 =
t nà, n=1, 2, 3,..
For microscope,d6 = Directions of minima are, a
2nsin n=l,2,3,
of maxima are,a sin 0 t (2n - 1),
. .
=
Directions
2n sin6
Resolving power =
0 sin
of central maximum,
=
Angular half-width
where 0 is semiangle of cone of rays entering into objective, n sin 6 15
called the numerical aperture
Total angular width, 20 = 2 sin
17
Handbook of PhysiICS-XIl
16 A Handbook of A
Physics-XI|
 For small 0. linear half-width at a screen at distance D from slit, (i) Momentum of a photon, P =-
Rest mass of a photon =
zero
(iii)
(iv) Kinetic mass of a photon. m *5
Total linear width =
2D
2. Einstein's photoelectric equation is (E)= hv--W
14. Polarisation of light:
3. Work function W hv, =
(a) This phenomenon proves that light waves are transverse.
where vo= threshold frequency and i = threshold wavelength
(6) Brewster's law, = tan
ip i, polarisingangle.
=
(c)Malus Law: When polarised light passes through an analyser 4. Photoelectrons emitted have kinetic energy
ranging from zero to a
(Nicol prism, polaroid, tourmaline crystal), the intensity of certain maximum limit. The maximum kinetic energy is
emergent light I=I, cos0 E )ma mv=eV, where Vs= stopping potential.
where -angle between the pass axis
ofpolariser and the analyser. 5. For photoelectric emission to take place, energy of photon E2 Wor
Comparative study of three types of wavefront v2 Vo or
sh
S. Wavefront Shape of Shape of Variationof Variation of 6. () de Broglie wavelength associated with moving particle
No. wavefront
light source amplitude
with distance
intensity with h h
distance =
m mv 2mE, 2mqV
i (ii) For electrons = A
1. Spherical Point source A
UNIT-8: Atoms and Nuclei
1. Nucleus consists of protons and neutrons.
Number of protons in a nucleus z X* is Z and number of neutrons,
2. Cylinderical Linear
A 2.
N=A-Z
Radius of Nucleus: R R, A where R, =1.2 x 10
=
m
3. Density of Nuclear Matter: D,= 10*" kg/ m
Extended 4. Einstein's Mass Energy Equivalence Relation is E = mei
large source
situated at l amu =l u = 931 Me V
3. Plane A = constant I = constant
very large of nucleus
distance
. Mass Defect =
mass of nucleons in given nucleus mass
Am =Zm, + (A -2) m, - Mucleus
B.E. =Zm, +(4-Z)m,-M.C
UNIT-7: Dual Nature of Matter and Radiation 6. Rutherford-Soddy Decay Law:
1. Photon: () Number of atoms undecayed after time t,
) Energy of a photon, E = hv = N=N,e
Ifis in A, then energy of photon in eV is E=- 12373e
eV wheren=is number of
half lives.
A(in Ä) T
Physics-XII 19
Handbook of
18 A Handbook of Physics-XII A
 (T) mean life (7) and disintegratioon
7. Relation between half-life
constant (4) is
0.693
and T 0.693 t=-
T =

Radioactive Decay:
8. Displacement Laws for
() For a-particle X" 2_,Y+ , He'
(ii) For B-particle ,X' >
z1Y" +_, B° + V

(iit) For y-ray ,X" zX" + Y

on getting bombarded breaks
9. In nuclear fission, a heavy nucleus
converted into energy. In
into lighter nuclei. Nearly 0.1% mass is
each fission of 92U3 with slow neutron 200
MeV energy is released.

10. In nuclear fusion, lighter nuclei under extremely high temperature

combine to form a heavy nucleus and mass is converted into energy.

UNIT-9: Electronic Devices

1. Energy band gap: E, =hv=
2. In intrinsic semiconducotr: n, = nhn
number density of electron in conduction band
where, n, =

nnumber density of holes in valence band

n= intrinsic carrier concentration
3. Total current through the pure semiconductor:

I=l+lh
4. In n-type semiconductor:

n Na>> nh
where, Na= density of donor atoms.
5. In p-type semiconductor:

nN>> n,
where, N = density of acceptor atoms.

6. Mass action law:

n n
7. The conductivity of the semiconductor:
elne Hetnh Hh
where, H. = electron mobility and Hh = hole mobility

nelectron density and n, = hole density

e electronic charge

20 A Handbook of Physics-Xll