Physics Formula and Concepts

8/16/2019 Physics Formula and Concepts

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1Formulas and Concepts

Physics pocket diary ofconcepts and formulas

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KinematicsNote: - Bold letter are used to denote vector quantityi, j ,z are the unit vector along x,y and z axis

Quick review of Kinematics formulasS.No. Type of

MotionFormula

1 Motion in onedimension

r=x i v=(dx/dt) i a =d v /dt=(d 2x/dt 2)i and a = vdv /d r v= u + a ts = u t+1/2 a t2 v2= u 2+2 as In integral form r=∫ vdtv=∫ a dt

2 Motion in twodimension

r=x i +y j v=d r /dt=(dx/dt) i + (dy/dt) j a =d v /dt=(d 2x/dt 2)i+(d 2x/dt 2) j and a = vdv /d r Constant accelerated equation same as above

3 Motion in threedimension

r=x i +y j +z k v=d r /dt=(dx/dt) i + (dy/dt) j +(dz/dt) k a =d v /dt=(d 2x/dt 2)i+(d 2x/dt 2) j +(d 2x/dt 2)ka = vdv /d r

Constant accelerated equation same as above4 Projectile Motion x=(v 0cosθ 0)t

y=(v 0sinθ 0)t-gt 2 /2vx= v0cosθ 0 and v y= v0sinθ 0t-gt , where θ 0 is the angle initialvelocity makes with the positive x axis.

5 Uniform circularmotion

a=v 2 /R , where a is centripetal acceleration whose directionof is always along radius of the circle towards the centre anda=4π 2R/T2 acceleration in uniform circular motion in termsof time period T

Concept of relative velocityFor two objects A and B moving with the uniform velocities VA and VB. Relative velocity is defined asVBA= VB-VA where VBA is relative velocity of B relative to ASimilarly relative velocity of A relative to BVAB= VA-VBSpecial cases: -S.No. Case Description1 For straight line

motion If the objects are moving in the same direction, relative velocity can beget by subtracting other. If they are moving in oppositedirection ,relative velocity will be get by adding the velocities examplelike train problems

2 For two dimensionsmotion

ifv a =v xa i + v ya j vb=v xbi + v yb j Relative velocity of B relative to A

=vxb

i + vyb j -(v

xai + v

ya j )

= i(v xb-v xa) + j (v yb-v ya)3 For three

dimensions motion v a =v xa i + v ya j +v zaz vb=v xbi + v yb j + v zbz Relative velocity of B relative to A

=v xbi + v yb j + v zbz -(v xai + v ya j +v zaz )= i(v xb-v xa) + j (v yb-v ya)+ z(v yb-v ya)

Free fall accelerationS.No. Point1 Freely falling motion of any body under the effect of gravity is an example of uniformly

accelerated motion.2 Kinematics equation of motion under gravity can be obtained by replacing acceleration 'a' in

equations of motion by acceleration due to gravity 'g' .3 Thus kinematics equations of motion under gravity are

v = v 0 + gt , x = v 0t + ½ ( gt 2 ) and v 2 = (v 0)2 + 2gx4 Value of g is equal to 9.8 m.s -2 .The value of g is taken positive when the body falls vertically

downwards and negative when the body is projected up against gravity.


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Laws of motion S.No. Term Description1 Newton's first law of

motion'A body continues to be in state of rest or uniform motion unless itis acted upon by some external force to act otherwise'

2 Newton's second lawof motion

'Rate of change of momentum of a body is proportional to theapplied force and takes place in the direction of action of force

applied Mathematically, F= d p /dt =m a where, p=m v , momentum of the bodya =acceleration

3 Impulse Impulse is the product of force and time which is equal to thechange in momentumImpulse = F∆t =∆ p

4 Newton's third lawof motion

'To every action there is always an equal and opposite reaction'FAB=- FBA

5 Law of conservationof linear momentum

Initial momentum = final momentumm 1v1+m 2v2=m 1v1 '+m 2v2 ' For equilibrium of a bodyF1+ F2+ F3=0

Some points to note

S.No. Point1 An accelerated frame is called non inertial frame while an non accelerated frame is called

inertial frame2 Newton first law are valid in inertial frame only3 Apparent weight of a body in the lift

Going Upward with acceleration aW=m(g + a)Going Down with acceleration aW=m(g-a)

4 Always draw free body diagram to solve the force related problems

Friction and Frame of referenceS.No. Term Description1 Friction Frictional force acts between the bodies whenever there is a relative

motion between them. When bodies slip, frictional force is called staticfrictional force and when the bodies do not slip, it is called kineticfrictional force.

2 KineticFrictional force

When bodies slip over each otherf=µ KNWhere N is the normal contact force between the surface and µ k is thecoefficient of kinetic Friction. Direction of frictional force is such thatrelative slipping is opposed by the friction

3 Static Frictionalforce

Frictional force can also act even if there is no relative motion. Suchforce is called static Frictional force. Maximum Static friction that a bodycan exert on other body in contact with it is called limiting Friction.f max =µ sN WhereN is the normal contact force between the surfaceAnd µ s is the coefficient of static Frictionf max is the maximum possible force of static Friction. Note that µ s > µ k and Angle of friction tanλ=µ s

4 Inertial FrameOf reference

Inertial frame of references is those attached to objects which are at restor moving at constant Velocity. Newton’s law are valid in inertial frame ofreference. Example person standing in a train moving at constantvelocity.

5 Non InertialFrame Ofreference

Inertial frame of references is attached to accelerated objects forexample: A person standing in a train moving with increasing speed.Newton’s law are not valid.To apply Newton’s law ,pseudo force has to be introduced in the equationwhose value will be F=-ma


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Work, Energy and PowerS.No. Term Description

1 Work 1. Work done by the force is defined as dot product of force anddisplacement vector. For constant Force

W= F.swhere F is the force vector and s is displacement Vector2. For variable ForcedW= F.ds or W=∫ F.dsIt is a scalar quantity

2 Conservative And NonConservative Forces

1. If the work done by the force in a closed path is zero, then itis called conservative Force2. If the work done by the force in a closed path is not zero,then it is called non conservative Force3. Gravitational ,electrical force are Conservative Forces andNon Conservative Forces are frictional forces

3 Kinetic Energy 1. It is the energy possessed by the body in motion. It is definedasK.E=(1/2)mv 2

2. Net work done by the external force is equal to the change inthe kinetic energy of the system W=K f -Ki

4 Potential Energy 1. It is the kind of energy possessed due to configuration of thesystem. It is due to conservative force. It is defined asdU=- F.drUf -Ui=-∫ F.drWhere F is the conservative forceF=-(∂U/∂x) i-(∂U/∂y) j -(∂U/∂z) kFor gravtitional Force2. Change in Potential Energy =mghwhere h is the height between the two points3. Mechanical Energy is defined as

E=K.E+P.E

5 Law Of conservation ofEnergy

In absence of external forces, internal forces being conservative,total energy of the system remains constant.K.E1+P.E 1=K.E 2+P.E 2

6 Power Power is rate of doing work i.e., P=work/Time. Unit of power isWatt. 1W=1Js -1 . In terms of force P= F.v and it is a scalarquantity.

Momentum and Collision S.No. Term Description1 Linear Momentum The linear momentum p of an object of mass m moving with

velocity v is defined as p=m vImpulse of a constant force delivered to an object is equal to thechange in momentumof the objectF∆t = ∆p = mv f - mv iMomentum of system of particles is the vector sum of individualmomentum of the particlep total =∑ v iMi

2 Conservation ofmomentum

When no net external force acts on an isolated system, the totalmomentum of the system is constant. This principle is calledconservation of momentum.if ∑Fext =0 then ∑ v iMi=constant

3 Collision Inelastic collision - the momentum of the system is conserved,but kinetic energy is not.Perfectly inelastic collision - the colliding objects stick together.Elastic collision - both the momentum and the kinetic energy ofthe system are conserved.

4 Inelastic collision While colliding if two bodies stick together then speed of the


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composite body is

21

2211

mmumum

v ++=

Kinetic energy of the system after collision is less then that beforecollison

5 Elastic collision inone dimension

Final velocities of bodies after collision are

2

21

21

21

211 umm

m2umm

mmv

++

+−

=

2

21

121

21

12 umm

mmu

mmm2

v

+−+

+

=

also1221 vvuu −=−

Special cases of Elastic Collision

S.No. Case Description1 m 1=m 2 v1=u 2 and v 2=u 12 When one of the bodies is at rest say u 2=0

1

21

211 umm

mmv

+−= and

1

21

12 umm

m2v

+

=

3 When m 1=m 2 and u 2=0 i.e., m 2 is at rest 0v 1 = and 12 uv =4 When body in motion has negligible mass i.e.

m 1m 2 11

uv = and 22 u2v =

Mechanics of system of particlesS.No. Term Description1 Centre of mass It is that point where entire mass of the system is imagined to be

concentrated, for consideration of its translational motion.2 position vector of

centre of mass R cm=∑ r iMi /∑Mi where r i is the coordinate of element i and Mi is mass ofelement i

3 In coordinatesystem

xcm=∑x iMi /∑Mi ycm=∑y iMi /∑Mi zcm=∑z iMi /∑Mi

4 Velocity of CM vCM=∑ v iMi /∑Mi The total momentum of a system of particles is equal the total masstimes the velocity of the centre of mass

5 Force When Newton’s second law of motion is applied to the system ofparticles we find F tot =Ma CM with a CM=d 2RCM /dt 2 Thus centre of mass of the system moves as if all the mass of thesystem were concentrated at the centre of mass and external forcewere applied to that point.

6 Momentumconservation inCOM motion

P= MvCM which means that total linear momentum of system ofparticles is equal to the product of the total mass of the system andthe velocity of its centre of mass.

Rigid body dynamicsS.No. Term Description1 Angular Displacement -When a rigid body rotates about a fixed axis, the angular

displacement is the angle ∆θ swept out by a line passingthrough any point on the body and intersecting the axis of rotation perpendicularly-It can be positive (counter clockwise) or negative (clockwise).-Analogous to a component of the displacement vector.-SI unit: radian (rad). Otherunits: degree, revolution.

2 Angular velocity -Average angular velocity, is equal to ∆θ/∆t .Instantaneous Angular Velocity ω =dθ/dt-Angular velocity can be positive or negative.


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-It is a vector quantity and direction is perpendicular to theplane of rotation-Angular velocity of a particle is different about different points-Angular velocity of all the particles of a rigid body is sameabout a point.

3 Angular Acceleration Average angular acceleration= ∆ ω /∆tInstantaneous Angular Acceleration

α=d ω /dt4 Vector Nature ofAngular Variables

-The direction of an angular variable vector is along the axis.- positive direction defined by the right hand rule.- Usually we will stay with a fixed axis and thus can work in thescalar form.-angular displacement cannot be added like vectors. Angularvelocity and acceleration are vectors

5 Kinematics of rotationalMotion

ω = ω 0 + α tθ = ω 0t+1/2 α t2 ω .ω = ω 0 .ω 0 + 2 α .θ ;Alsoα=dω/dt=ω(dω/dθ)

6 Relation BetweenLinear and angularvariables

v= ω Xr Where r is vector joining the location of the particle and pointabout which angular velocity is being computed

a = αXr 7 Moment of Inertia Rotational Inertia (Moment of Inertia) about a Fixed AxisFor a group of particles,I = ∑mr 2 For a continuous body,I = ∫ r2dmFor a body of uniform densityI = ρ∫r 2dV

8 Parallel Axis Therom Ixx=I cc+ Md 2 Where I cc is the moment of inertia about the centreof mass

9 Perpendicular AxisTherom

Ixx+I yy=I zz It is valid for plane laminas only.

10 Torque τ= rXF a lso τ=Iα where α is angular acceleration of the body.

11 Rotational KineticEnergy KE=(1/2)Iω2

where ω is angular acceleration of the body12 Rotational Work Done -If a force is acting on a rotating object for a tangential

displacement of s = rθ (with θ being the angular displacementand r being the radius) and during which the force keeps atangential direction and a constant magnitude of F, and with aconstant perpendicular distance r (the lever arm) to the axis of rotation, then the work done by the force is:W=τθ-W is positive if the torque τ and θ are of the same direction,otherwise, it can be negative.

13 Power P =dW/dt=τω

14 Angular Momentum L = rXp = rX(mv)=m( rXv)

For a rigid body rotating about a fixed axisL=Iω and d L /dt= τ if τ=0 and L is constantFor rigid body having both translational motion and rotationalmotionL= L1+ L2 L1 is the angular momentum of Centre mass about an stationaryaxisL2 is the angular momentum of the rigid body about Centre ofmass.

15 Law of Conservation OnAngular Momentum

If the external torque is zero on the system then Angularmomentum remains contants

dL /dt= τext if τext =0then d L /dt=0


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16 Equilibrium of a rigidbody

Fnet =0 and τext =0

17 Angular Impulse ∫τdt term is called angular impulse. It is basically the change inangular momentum

18 Pure rolling motion ofsphere/cylinder/disc

-Relative velocity of the point of contact between the body andplatform is zero-Friction is responsible for pure rolling motion

-If friction is non dissipative in natureE = (1/2)mv cm2+(1/2)Iω 2+mgh

GravitationS.No. Term Description1 Newton’s Law of

gravitation 221

r mGm

F = where G is the universal gravitational constant

G=6.67 ×10 -11 Nm2Kg-22 Acceleration due

to gravityg=GM/R 2 where M is the mass of the earth and R is the radius of theearth

3 Gravitationalpotential energy

PE of mass m at point h above surface of earth is

)( h RGmM

PE +−=

4 Gravitationalpotential )( h R

GM V +

−=

Law oforbits

Each planet revolves round the sun in an elliptical orbitwith sun at one of the foci of elliptical orbit.

Law ofareas

The straight line joining the sun and the planet sweepsequal area in equal interval of time.

5 Kepler’s Law ofplanetary motion

Law ofperiods

The squares of the periods of the planet are proportionalto the cubes of their mean distance from sun i.e.,T2 R 3

6 Escape velocity Escape velocity is the minimum velocity with which a body must beprojected in order that it may escape earth’s gravitational pull. Its

magnitude is v e=√(2MG/R) and in terms of g v e=√(2gR)

OrbitalVelocity

The velocity which is imparted to an artificial satellite fewhundred Km above the earth’s surface so that it maystart orbiting the earth v 0=√(gR)

7 Satellites

PeriodicTime

T=2π√[(R+h) 3 /gR 2]

Withaltitude

−=

Rh

g g h2

1

With

depth

−=

Rd

g g d 1

8 Variation of g

Withlatitude

φ φ 2cos037.0−= g g

ElasticityS.No. Term Description1 Elasticity The ability of a body to regain its original shape and size when deforming

force is withdrawn2 Stress Stress=F/A where F is applied force and A is area over which it acts.3 Strain It is the ratio of the change in size or shape to the original size or shape.

Longitudinal strain = ∆l/l volume strain = ∆V/V and shear strain is due to

change in shape of the body. 4 Hook’s Law Hook's law is the fundamental law of elasticity and is stated as “for smalldeformations stress is proportional to strain".


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Thus, stress proportional to strainor, stress/strain = constantThis constant is known as modulus of elasticity of a given materialYoung's Modulus of Elasticity Y=Fl/A∆lBulk Modulus of Elasticity K=-V∆P/∆V

5 ElasticModulus

Modulus of Rigidity η=F/Aθ6 Poisson's

Ratio

The ratio of lateral strain to the longitudinal strain is called Poisson’s ratio

which is constant for material of that body. σ=l∆D/D∆l7 Strainenergy

Energy stored per unit volume in a strained wire is E=½(stress)x(strain)

HydrostaticsS.No. Term Description1 Fluid pressure It is force exerted normally on a unit area of surface of fluid P=F/A. Its

unit is Pascal 1Pa=1Nm -2 .2 Pascal’s Law Pressure in a fluid in equilibrium is same everywhere.3 Density Density of a substance is defined as the mass per unit volume.4 Atmospheric

pressureWeight of all the air above the earth causes atmospheric pressure whichexerts pressure on the surface of earth. Atmospheric pressure at sea

level is P 0=1.01x10 5Pa5 Hydrostatic

pressureAt depth h below the surface of the fluid is P=ρgh where ρ is the densityof the fluid and g is acceleration due to gravity.

6 Gauge pressure P=P 0+ ρgh , pressure at any point in fluid is sum of atmosphericpressure and pressure due to all the fluid above that point.

7 Archimedesprinciple

When a solid body is fully or partly immersed in a fluid it experience abuoyant force equal to the weight of fluid displaced by it.

8 Upthrust It is the weight of the displaced liquid.9 Boyle’s law PV=constant10 Charle’s law V/T=constant

Hydrodynamics

S.No. Term Description1 Streamline

flowIn such a flow of liquid in a tube each particle follows the path of itspreceding particle.

2 Turbulentflow

It is irregular flow which does not obey above condition.

3 Bernoulli’sprinciple t cons ghu p tan

21 2 =++ ρ ρ

4 Continuity offlow 2211

v Av A = where A 1 and A 2 are the area of cross section of tube of variable cross section and v 1 and v 2 are the velocity of flow of liquidscrossing these areas.

5 Viscosity Viscous force between two layers of fluid of area A and velocity gradient

dv/dx isdxdv

A F η −= where η is the coefficient of viscosity.

6 Stokes’ law Viscous force on a spherical body of radius r falling through a liquid ofviscosity η is rv F πη 6= where v is the velocity of the sphere.

7 Poiseuilli’sequation

Volume of a liquid flowing per second through a capillary tube of radius rwhen its end are maintained at a pressure difference P is given by

l Q

η π

8Pr 4= where l is the length of the tube.

Simple Harmonic MotionS.No. Term Description1 SHM In SHM the restoring force is proportional to the displacement from

the mean position and opposes its increase. Restoring force is F=-Kx Where K=Force constant , x=displacement of the system from itsmean or equilibrium position


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Differential Equation of SHM isd2x/dt 2 + ω 2x=0Solutions of this equation can both be sine or cosine functions .Weconveniently choosex=Acos(ωt+φ) where A ,ω and φ all are constants

2 Amplitude Quantity A is known as amplitude of SHM which is the magnitude ofmaximum value of displacement on either sides from the equilibrium

position3 Time period Time period (T) of SHM the time during which oscillation repeats itselfi.e, repeats its one cycle of motion and it is given byT=2π/ω where ω is the angular frequency

4 Frequency Frequency of the SHM is the number of the complete oscillation perunit time i.e., frequency is reciprocal of the time periodf=1/T. Thus angular frequency ω=2πf

5 Velocity Velocity of a system executing SHM as a function of time is v=-ωAsin(ωt+φ)

6 Acceleration Acceleration of particle executing SHM isa=-ω 2Acos(ωt+φ). So a=-ω 2x This shows that acceleration is proportional to the displacement but inopposite direction

7 Kinetic energy At any time t KE of system in SHM isKE=(1/2)mv 2 =(1/2)mω 2A2sin 2(ωt+φ)

which is a function varying periodically in time8 Potential energy PE of system in SHM at any time t isPE=(1/2)Kx 2 =(1/2)mω 2A2cos 2(ωt+φ)

9 Total energy Total Energy in SHME=KE+PE =(1/2)mω 2A2 and it remain constant in absence of dissipative forces like frictionalforces

10 Oscillations of aSpring masssystem

In this case particle of mass m oscillates under the influence of hook’slaw restoring force given by F=-Kx where K is the spring constantAngular Frequency ω=√(K/m)Time period T=2π√(m/K)frequency is =(1/2π)√(K/m)Time period of both horizontal and vertical oscillation are same butspring constant have different value for horizontal and vertical motion

11 Simple pendulum -Motion of simple pendulum oscillating through small angles is a case

of SHM with angular frequency given byω=√(g/L)and Time period T=2π√(L/g)Where L is the length of the string.-Here we notice that period of oscillation is independent of the mass mof the pendulum

12 CompoundPendulum

- Compound pendulum is a rigid body of any shape,capable of oscillating about the horizontal axis passing through it.-Such a pendulum swinging with small angle executes SHM with thetimeperiodT=2π√(I/mgL)Where I =Moment of inertia of pendulum about the axis of suspensionL is the lenght of the pendulum

13 DampedOscillation

-SHM which continues indefinitely without the loss of the amplitudeare called free oscillation or undamped and it is not a real case- In real physical systems energy of the oscillator gradually decreaseswith time and oscillator will eventually come to rest. This happensbecause in actual physical systems, friction(or damping ) is alwayspresent-The reduction in amplitude or energy of the oscillator is calleddamping and oscillation are call damped

14 ForcedOscillations andResonance

- Oscillations of a system under the influence of an external periodicforce are called forced oscillations- If frequency of externally applied driving force is equal to the naturalfrequency of the oscillator resonance is said to occur

WavesS.No. Term Description1 Wave -It is a disturbance which travels through the medium due to repeated

periodic motion of particles of the medium about their equilibriumposition.


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-Examples are sound waves travelling through an intervening medium,water waves, light waves etc.

2 Mechanicalwaves

Waves requiring material medium for their propagation are MECHANICALWAVES. These are governed by Newton's law of motion.-Sound waves are mechanical waves in atmosphere between source andthe listener and require medium for their propagation.

3 Non mechanical

waves

-Those waves which does not require material medium for their

propagation are called NON MECHANICAL WAVES.-Examples are waves associated with light or light waves , radio waves,X-rays, micro waves, UV light, visible light and many more.

4 Transversewaves

These are such waves where the displacements or oscillations areperpendicular to the direction of propagation of wave.

5 Longitudinalwaves

Longitudinal waves are those waves in which displacement or oscillationsin medium are parallel to the direction of propagation of wave forexample sound waves

6 Equation ofharmonic wave

-At any time t , displacement y of the particle from it's equilibriumposition as a function of the coordinate x of the particle is y(x,t)=Asin(ωt-kx) where,A is the amplitude of the wavek is the wave numberω is angular frequency of the waveand (ωt-kx) is the phase.

7 Wave number Wavelength λ and wave number k are related by the relationk=2π/λ 8 Frequency Time period T and frequency f of the wave are related to ω by

ω/2π = f = 1/T 9 Speed of wave speed of the wave is given by

v = ω/k = λ/T = λf 10 Speed of a

transversewave

Speed of a transverse wave on a stretched string depends on tension andthe linear mass density of the string not on frequency of the wavei.e, v=√T/µT=Tension in the stringµ=Linear mass density of the string.

11 Speed oflongitudinalwaves

Speed of longitudinal waves in a medium is given byv=√B/ρ B=bulk modulusρ=Density of the medium

Speed of longitudinal waves in ideal gas isv=√γP/ρ P=Pressure of the gas ,ρ=Density of the gas and γ=C p /C V

12 Principle ofsuperposition

When two or more waves traverse thrugh the same medium,thedisplacement of any particle of the medium is the sum of thedisplacement that the individual waves would give it.y=Σy i(x,t)

13 Interference ofwaves

If two sinusoidal waves of the same amplitude and wavelength travel inthe same direction they interfere to produce a resultant sinusoidal wavetravelling in that direction with resultant wave given by the relationy′(x,t)=[2A mcos(υ/2)]sin(ωt-kx+υ/2) where υ is the phase

difference between two waves.-If υ=0 then interference would be fully constructive.-If υ=π then waves would be out of phase and there interference wouldbe destructive.

14 Reflection ofwaves

When a pulse or travelling wave encounters any boundary it getsreflected. If an incident wave is represented byy i(x,t)=A sin(ωt-kx) then reflected wave at rigid boundary is y r(x,t)=Asin(ωt+kx+π)=-Asin(ωt+kx)and for reflections at open boundary reflected wave is given byy r(x,t)=Asin(ωt+kx)

15 Standing waves The interference of two identical waves moving in opposite directionsproduces standing waves. The particle displacement in standing wave isgiven by y(x,t)=[2Acos(kx)]sin(ωt) In standing waves amplitude ofwaves is different at different points i.e., at nodes amplitude is zero and

at antinodes amplitude is maximum which is equal to sum of amplitudesof constituting waves.


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16

Normal modesof stretchedstring

Frequency of transverse motion of stretched string of length L fixed atboth the ends is given byf=nv/2L where n=1,2,3,4,.......-The set of frequencies given by above relation are called normal modesof oscillation of the system. Mode n=1 is called the fundamental modewith frequency f 1=v/2L. Second harmonic is the oscillation mode withn=2 and so on.

-Thus the string has infinite number of possible frequency of vibrationwhich are harmonics of fundamental frequency f 1 such that f n=nf 117 Beats Thus beats arises when two waves having slightly differing frequencies ν 1

and ν 2 and comparable amplitude are superposed.-Here interfering waves have slightly differing frequencies ν 1 and ν 2 suchthat|ν 1-ν 2|


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11 Volume coefficient of gast V V V t

v

0

0−=γ

12 Ideal gas equation PV=nRT where n is number of moles of gas and R isuniversal constant.

Measurement of heat and temperatureS.No. Term Description1 Relation between Celsius,

Fahrenheit and Kelvin scale oftemperature 100

273180

32100

−=−= K F C

2 Principle of electricalresistance thermometer

Rt=R 0(1+αt) where R t is resistance at t 0C, R 0 isresistance at ) 0C and α is temperature coefficient ofresistance.

3 Joule’s mechanical equivalentto heat(J)

W=JH where value of J is J=4.2J/cal.

4 Heat capacity C=∆Q/∆t where ∆Q is the amount of heat absorbed and∆t is rise in temperature.

5 Specific heat It is the amount of heat required to raise thetemperature of 1Kg of substance by 1 0C

6 Molar specific heat Molar specific heat of the substance is the amount ofheat required to raise the temperature of 1 mole of thesubstance through 1 0C.

7 Relation between C p and C v CP-Cv=R where R is universal gas constant.8 Latent heat of fusion Heat energy required to convert a unit mass of

substance from the solid to the liquid state withoutchange in temperature.

9 Latent heat of sublimation Heat energy required to convert a unit mass ofsubstance from the liquid to the gaseous state withoutchange in temperature.

10 Principle of calorimetery Heat lost=Heat gain

ThermodynamicsS.No. Term Description1 Thermodynamic state It refers to the state of the system that is completely defined by

pressure , volume and temperature of the system.2 Zeroth law of

thermodynamicsIf two systems A and B are separately in equilibrium with thethird system C then system A and B are in thermal equilibriumwith each other

3 First law ofthermodynamics

Heat energy given to the system is equal to the increase ininternal energy of the system and the work done. Q=∆U+W

4 Second law ofthermodynamics

Heat can not flow from a colder body to a hotter body withoutsome work being done by the external agency.

5 Work in volumechanges

If pressure remains constant while the volume changes, thenwork is

W = P(V 2-V1)6 Quasi static Processes In Quasi static process deviation of system from it's

thermodynamic equilibrium is infinitesimally small.7 Isothermal Process temperature of the system remains constant throughout the

process and thus ∆Q =∆W8 Adiabatic Process no heat enters or leaves a system thus ∆U=U 2 - U 1= - ∆W9 Isochoric process volume of the system remain uncharged throughout and U 2 - U 1

= ∆U =∆Q10 Isobaric Process This process takes place at constant pressure.11 Work done in

Isothermal processW=nRT ln(V 2 /V 1)Where n is number of moles in sample of gas taken

12 Ideal gas equation foradiabatic process

PVγ = K (Constant)Where γ is the ratio of specific heat (ordinary or molar) atconstant pressure and at constant volume γ = C p /C v

13 Work done in anAdiabatic process

W = (P 1V1-P2V2)/(γ-1) In and adiabatic process if W>0 i.e., workis done by the gas then T 2< T 1 . If work is done on the gas

(W T 1 i.e., temperature of gas rises14 Thermal efficiency ofheat engine

η = 1-(Q 2 /Q 1)


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15 Coefficient ofperformance ofrefrigerator

β= Q 2 /W =Q 2 /(Q 1-Q2)

16 Efficiency of carnotengine

β= Q 2 /W =Q 2 /(Q 1-Q2)

17 Carnot Theorem Carnot’s theorem consists of two parts

(i) No engine working between two given temperatures can bemore efficent than a reversible Carnot engine working betweensame source and sink.(ii) All reversible engines working between same source and sink(same limits or temperature) have the same efficiencyirrespective the working substance.

Heat transferS.No. Term description1 Thermal

Conductivity LT T kA

H )( 12 −= Where H is the quantity of heat flowing through the slab

and k is the constant called thermal conductivity of material of slab. 2 Convection Convection is transfer of heat by actual motion of matter3 Radiation Radiation process does not need any material medium for heat transfer4 Stefan

Boltzmann lawThe rate u rad at which an object emits energy via EM radiation depends onobjects surface area Aand temperature T in kelvin of that area and is given by u rad =σεAT4 Where σ= 5.6703×10 -8 W/m 2K4 is Stefan boltzmann constant and ε is emissivity of object's surface withvalue between 0 and 1

5 Wein'sdisplacementlaw

λmT = b Where b=0.2896×10 -2 mk for black body and is known asWien's constant

EmissivePower

It is the energy radiated per unit area per unit solidangle normal to the area.

E = ∆u/ [(∆A) (∆ω) (∆t)]where, ∆u is the energy radiated by area ∆A of surface insolid angle ∆ω in time ∆t.

AbsorptivePower

is defined as the fraction of the incident radiation that isabsorbedby the body

a(absorptive power) = energy absorbed / energyincident

6 Kirchoff's law

Kirchoff'sLaw

"It status that at any given temperature the ratio ofemissive power to the absorptive power is constant forall bodies and this constant is equal to the emissivepower of perfect B.B. at the same temperature.

E/a body =E B.B 7 Newton's Law

of Cooling For small temperature difference between the body and surrounding rateof cooling is directly proportional to the temperature difference andsurface area exposed i.e.,

dT/dt = - bA (T 1 - T 2). This is known a Newton's law of cooling

Kinetic theory of gasesS.No. Term Description

Boyle's law At constant temperature, the volume of a givenmass of gas is inversely proportional to pressure.Thus PV = constant

Charle's Law When pressure of a gas is constant the volume ofa given mass of gas is directly proportional to itsabsolute temperature.

V/T = Constant

1 Gas laws

Dalton's law ofpartialpressures

The total pressure of mixture of ideal gases issum of partial pressures of individual gases ofwhich mixture is made of

2 Ideal gasequation PV = nRT where n is number of moles of gas

3 Pressure of gas P = (1/3)ρv mq 2 or PV = (1/3) Nmv mq 2 where v mq 2 known as mean


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square speed4 rms speed vrms = √(3P/ρ) = √(3PV/M) =√(3RT/M) 5 Law of

Equipartition ofenergy

each velocity component has, on the average, an associated kineticenergy (1/2)KT

Monatomicgases

CV= (3/2)R , C P = 5/2 R and γ mono = C P /C V=5/3

6 Specific HeatCapacity

Diatomic gases CV =(5/2)R , CP=(9/2)R and γ=C P /C V=9/77 Specific heat

Capacity of Solids C=3R this is Dulang and Petit law

8 Mean free path If v is the distance traversed by molecule in one second then meanfree path is given by

λ = total distance traversed in one second /no. of collision sufferedby the molecules

=v/πσ 2vn=1/πσ 2n

Electric Charge, Force and FieldS.No. Term Description1 Charge Charges are of two types

(a) positive charge (b) negative chargelike charges repel each other and unlike charges attract each other.

2 Properties ofcharge

1. Quantisation : - q=ne where n=0,1, 2........ and e is charge of anelectron.2. Additivety : - q net =Σq3. conservation :- total charge of an isolated system is constant

3 Coulomb’s law The mutual electrostatic force between the charges q 1 and q 2 separatedby a distance r is given byForce on the charge q 1 F1=Kq 1q2r 12 /r 2 where r 12 is the unit vector in the direction from q 2 to q 1 For more then two charges in the system the force acting on anycharges is vector sum of the coulomb force from each of the othercharges. This is called principle of superpositionfor q 1,q 2,q 3......q n charges are present in the system. Force on charge q 1 F=Kq 1q2r 12 /r 12 2+Kq 1q3r 13 /r 13 2+Kq 1q4r 14 /r 14 2....Kq 1qnr 1n /r 1n 2 Similarly for the other charges...

4 Electric field -The region around a particular charge in which its electrical effects canbe observed is called the electric field of the charge-Electric field has its own existence and is present even if there is nocharge to experience the electric force.

5 Electric fieldIntensity

E= F /q 0 Where F is the electric force experience by the test charge q 0 at thispoint.It is the vector quantitySome point to note on this1. Electric field lines extend away from the positive charge and towardsthe negative charges2. Electric field produces the force so if a charge q is placed in theelectric field E the force experience by the charge isF=q E3 Principle of superposition also applies to electric field

so E= E1+ E2+ E3+ E4+......Electric field intensity due to point chargeE=KQ r /r 2 Where r is the distance from the point charge and r is the unit vectoralong the direction from source to point.Electric field for theUniformly charged ring

E=KQx/(r 2+x 2)3/2 Where x is the distance from thecentre of the ring. At x=0E=0

Electric Field due touniformly charged disc

E= (σ/2ε 0)(1- x/(√R 2+x 2))σ=Surface charge density of thedisc. At x=0 E=σ/2ε 0

6 Some usefulFormula

Electric Field Intensity due toInfinite sheet of the charge

E=σ/2ε 0

7 Charge density Linear charge density

λ=Q/L=dQ/dLSurface charge densityσ=Q/A=dQ/dA


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Volume charge densityρ=Q/V=dQ/dV

8 Electrostaticsin Conductor

1. E=0 inside the conductor2. All charge resides on the outer surface of the conductor3.Electric at the surface is Perpendicular to the surface

9 Electric Flux dφ= E.da da is the area vector to the surface and it is taken positive along the

outward normal to the surfacedφ=Edacosθφ=∫ E.da

10 Gauss’sTheorem

Flux in closed surface is equal net charge inside divided by ε∫ E.da =q in /ε

11 Some points a. E is the electric field present due to all charges in the system not justthe charge insideb. Flux crossing a closed surface does not depend on the shapes andsize of Gaussian surface

Electric potentialS.No. Term Description1 Electric

Potentialenergy

∆U=-W Where ∆U = Change in Potential energy and W= Work done by the electriclines of forcesFor a system of two particlesU(r)=q 1q2 /4πεrwhere r is the separation between the charges.We assume U to be zero at infinitySimilarly for a system of n chargesU=Sum of potential energy of all the distinct pairs in the system For example for three chargesU=(1/4πε)(q 1q2 /r 12+q 2q3 /r 23+q 1q3 /r 13 )

2 Electric PE ofa charge

=qV where V is the potential there

3 ElectricPotential

Liken Electric field intensity is used to define the electric field; we can alsouse Electric Potential to define the field. Potential at any point P is equal tothe work done per unit test charge by the external agent in moving the testcharge from the reference point(without Change in KE)Vp=W ext /qSo for a point chargeVp=Q/4πεrwhere r is the distance of the point from charge

4 Some pointsaboutElectricpotential

1 . It is scalar quantity2 .Potential at point due to system of charges will be obtained by thesummation of potential of each charge at that pointV=V1+V2+V3+V43 .Electric forces are conservative force so work done by the electric forcebetween two point is independent of the path taken4 . V2-V1=-∫ E.dr5 . In Cartesian coordinates systemdV=-E.drdV=-(E xdx+E ydy+E zdz)

So E x=∂V/∂x , E y=∂V/∂y and E z=∂V/∂zAlsoE=-[(∂V/∂x)i+(∂V/∂y)j+(∂V/∂z)k]6 . Surface where electric potential is same everywhere is call equipotentialsurfaceElectric field components parallel to equipotential surface is always zero

5 Electricdipole

A combination of two charge +q and -q separated by the distance dp=q d Where d is the vector joining negative to positive charge

6 Electricpotential dueto dipole

V=(1/4πε)(pcosθ/r 2)where r is the distance from the center and θ is angle made by the line fromthe axis of dipole

7 Electric fielddue to dipole

Eθ=(1/4πε)(psinθ/r 3)Er=(1/4πε)(2pcosθ/r 3)Total E=√E θ2+E r2


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=(p/4πεr 3)(√(3cos 2θ+1))Torque on dipole= pXE Potential EnergyU=- p .E

8 Few morepoints

1. ∫ E.dl over closed path is zero2.Electric potential in the spherical charge conductor is Q/4πεR where R isthe radius of the shell and the potential is same everywhere in the

conductor3. Conductor surface is a equipotential surface

CapacitanceS.No. Term Description1 Capacitance C of the

capacitorC=q/Vor q=CV-Unit of capacitance is Farads or CV -1 capacitance of a capacitoris constant and depends on shape, size and separation of thetwo conductors and also on insulating medium being used formaking capacitor.

2 Capacitance of parallelplate cap C=(ε 0A)/dwhere,C= capacitance of capacitorA= area of conducting plated= distance between plates of the capacitorε0=8.854× 10 -12 and is known as electric permittivity in vacuum.

3 parallel plate aircapacitor in presenceof dielectric medium

C=εA/d

4 Capacitance ofspherical capacitorhaving radii a, b (b>a)

(a) air as dielectric between themC=(4πε 0ab)/(b-a)(b) dielectric with relative permittivity εC=(4πεab)/(b-a)

5 Parallel combination ofcapacitors

C=Q/V= C 1+C 2+C 3 , resultant capacitance C is greater then thecapacitance of greatest individual one.

6 Series combination ofcapacitors 1/C=1/C 1+1/C 2+1/C 3, resultant capacitance C is less then thecapacitance of smallest individual capacitor.7 Energy stored in

capacitorEnergy stored in capacitor isE=QV/2or E=CV 2 /2or E=Q 2 /2Cfactor 1/2 is due to average potential difference across thecapacitor while it is charged.

8 Force between platesof capacitor A K

Q F

02 ε =

9 Force per unit area ofplates

0

2

2 ε σ K

f = Where σ is charge per unit area.

Electric current and D.C. circuitsS.No. Term Description1 Electromotive force EMF of a cell is the total energy per unit charge when the cell is

on an open circuit i.e., when the current through the cell iszero.

2 Electric current I=q/t it is the rate of flow of electric charge. Unit of current isampere.

3 Drift speed of electronin a conductor neA

I vd =

Where I is the current, n is the number of electrons

per unit volume and A is the area of cross section of conductor.4 Resistivity of conductor

τ ρ

2nem= where m is the mass of electron and τ is the

relaxation time5 Ohm’s law V=IR where R is the resistance of the given conductor and unitof resistance is ohm(Ω)


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6 Electrical resistivity Ρ=RA/l where l is the length of the wire and A is its area ofcross-section

7 Resistors in series R=R 1+R 2+R 3+….8 Resistors in parallel

.....1111

321

+++= R R R R

9 Terminal voltage It is equal to emf of battery minus potential drop acrossinternal resistance r across the battery. Terminal voltage = E-Ir

10 Kirchoff’s first law The algebraic sum of current at any junction in a circuit is zero.11 Kirchoff’s second law The algebraic sum of the products of the current and

resistances and the emf in a closed loop is zero.12 Heating effect of current Heat energy delivered by current when it flows through

resistance of R ohm for t sec. maintained at potential differenceV is H=V 2t/R

13 Electrical power P = VI = I 2R = V 2 /R14 Variation of resistance

with temperatureR=R 0(1+α(T-T 0))

15 Variation of resistivitywith temperature

ρ=ρ 0(1+α(T-T 0))

Magnetic effect of currentS.No. Term Description1 Biot-Savart law Magnetic fiels dB at any point whose position vector is r wrt

current elemant d l is given by3

(4 r

d I d

r)l B 0

×=π

µ

2 Magnetic field due tolong current carryingconductor r

I B

π µ 4

20=

3 Magnetic field atcentre of a circularloop

r I

B2

0µ =

4 Magnetic field atcentre of coil of nturns

r In

B20µ =

5 Magnetic field on theaxis of a circular loop 2/322

2

0

)(2 xr Ir B += µ

if there are n turns in the coil then

2/322

20

)(2 xr Inr

B += µ

6 Ampere’s circuital law I d 0 I B µ =⋅∫7 Field due to toroidal

solenoidnI B 0µ =

8 Field inside straightsolenoid

nI B 0µ = and direction of field is parallel to the axis of solenoid

9 Force on movingcharge in magneticfield

) BvF ×= q( Direction of force is perpendicular to both v and B

10 Force on currentcarrying conductor inthe magnetic field

) I Bl F ×= ( where l is the length of the conductor in the

direction of current in it11 Force between two

parallel wires carryingcurrent

R I I

F π

µ 2

210=

12 Torque on a currentcarrying loop

) Bmτ ×= ( Where m is the magnetic moment of the dipole andmagnitude of magnetic moment is m=NIB where A is the area ofthe loop and N is the number of turns in the loop.

13 Lorentz force Force on electron moving with velocity v in presence of bothuniform electric and magnetic field is )( BvF ×+−= e

14 Magnetic dipolemoment of barmagnet

m= q(2 a ) where q is the pole strength and (2 a ) is the length ofthe bar magnet. It is the vector pointing from south to north poleof the magnet.

15 torque on the barmagnet

Bmτ ×=

16 Potential energy of amagnetic dipole

U=-mBcosθ


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Electromagnetic inductionS.No. Term Description1 Magnetic flux θ φ cos BA=⋅= A B where θ is the angle between normal to the plane

of the coil and magnetic field s.I. unit of flux is Weber

2 Faraday’s lawdt d e φ −= and for a closely packed coil of N turns

dt d e φ −= where e is

the induced EMF3 Lenz’s law The induced current has such a direction such that magnetic field of the

current opposes the change in the magnetic flux that produces current.4 Inductance L of

inductori

L

φ = Where N is windings of inductor, I is the current associated

with each winding of inductor.5 Self induction Phenomenon by which an opposing EMF is introduced in the coil

because of varying current in coil itself. Self induction EMF isdt di

Le −=

6 Series RL circuitRise of current )1( / Lt e

R

ei τ −−= where R L L /=τ is inductive time

constant of the circuit. On removing emf current decays from a value i 0according to equation Lt eii τ /0

−=7 Magnetic energy Energy stored by inductor’s magnetic field is U=½Li 2.8 Density of

stored magneticenergy

u=B 2 /2µ 0

9 Mutual induction It refers to the phenomenon by which a current I is induced in a coilwhen current in a neighbouring coil circuit is changed. It is described by

dt di

M e 12 −= and

dt di

M e 21 −= where M is the mutual induction for the

coil arrangement.

Magnetism of matterS.No. Term Description1 Gauss’s law for

magnetic fields0=⋅= ∫ A B d φ i.e., net magnetic flux through any closed Gaussian

surface is zero.2 Spin magnetic

dipole moment S µ me

s −=

where S is spin angular momentum.

3 Orbital magneticdipole moment L µ m

e L 2

−= where L is orbital angular momentum

4 Diamagnetism Diamagnetic materials are those materials which on being placed inmagnetic field get feebly magnetised in the direction opposite to themagnetic field.

5 Paramagnetism In Paramagnetic materials each atom has permanent magneticmoment but dipole moments are randomly oriented and material asa whole lacks property of magnetism but dipoles can be aligned inthe presence of external magnetic field to give net dipole momentand material gets feebly magnetised in the direction of the field.

6 Ferromagnetism Ferromagnetic materials when placed in external magnetic field getsstrongly magnetised in the direction of the magnetic field.

7 Maxwell’sextension ofampere’s law

∫ =⋅dt d φ

ds B 0 0ε µ

Transient Currents

S.No. Term Description1 Growth of charge inCR circuit )1(0

CRt

eqq−

−= and current in CR circuit is CRt

eii−

=0


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2 Decay of charge in CRcircuit CR

t

eqq−

=0

and current in CR circuit is CRt

eii−

−=0

3 Capacitive timeconstant

CR has dimensions of time and is called capacitive time constantfor circuit

4 Energy stored ininductor

( )221 LiU =

5 Energy stored incapacitor ( ) ( ) E qCE U 0

2

2121 == Where E is the maximum value of potential

difference set up across the plates.6 LC oscillations Frequency of oscillations is

LC f

π 21=

Alternating CurrentS.No. Term Description1 Alternating

currentIt is current whose magnitude changes with time and direction reversesperiodically. I=I 0sinωt where I 0 is the peak value of a.c. and ω=2π/T isthe frequency

2 Mean value of

a.c.

Im=2I 0 /π = 0.636I 0

3 RMS value I rms =I 0 /√24 a.c. through

resistorAlternating emf is in phase with current

5 a.c. throughinductor

Emf leads the current by an phase angle π/2

6 a.c. throughcapacitor

Emf lags behind the current by an phase angle π/2

7 Inductivereactance

Opposition offered by inductor to the flow of current mathematically, fL L X L π ω 2==

8 Capacitivereactance

Opposition offered by capacitor to the flow of current mathematically,

fC C X C π ω 2

11 ==

9 a.c. throughseries LR circuit

Emf leads the current by an phase angle φ given by R

Lω φ =tan and

impedance of circuit is ( )222 L R Z ω +=10 a.c. through

series CR circuitEmf lags behind the current by an phase angle φ given by

RC ω φ

1tan =

and impedance of circuit is

+=

22

2 1C

R Z ω

11 a.c. throughseries LCR circuit

Emf leads/lags behind the current by an phase angle φ given by

RC L ω ω φ

1tan

−= emf leads the current when

C L

ω ω

1> and lags behind

whenC

Lω

ω 1< and impedance of circuit is 22 1

−+=

C L R Z

ω ω

12 Average powerof an a.c. circuit φ cos

2rmsrmsrmsavg I E R I P == Where φ is called power factor of the circuit.

13 Transformer It is a device used to change low alternating voltage at high current intohigh voltage at low current and vice-versa. Primary and secondaryvoltage for a transformer are related as

P

S P S

V V =

and current through

the coils is related as

S

P P S

I I =

Electromagnetic wavesS.No. Term Description1 Conduction current It is the current due to the flow of electrons through the

connecting wires in an electric circuit


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2 Displacement current It arises due to time rate of change of electric flux in some partof circuit

dt d

I D ε 0=

3 Modified ampere’scircuital law )()(. 0 dt

d I I I E C DC

ϕ ε µ µ +=+=∫ 00dl B

where I C is the conduction

current.Gauss’s law in electrostatics

0

qds E ε =⋅∫

Gauss’s law in magnetism 0=⋅∫ ds BFaraday’s law of EMinduction dt

d Bφ −=⋅∫ dl E

4 Maxwell’s Equations

Ampere-Maxwell’s circuitallaw )(. 0 dt

d I E C

ϕ ε µ +==∫ 0dl B

5 Velocity of EM wavesin free space

smv /1031 8

00

×==µ ε

Huygens’ Principle and Interference of LightS.No. Term Description1 Wave front It is the locus of points in the medium which at any instant

are vibrating in the same phase.1 Each point on the given primary Wavefront acts as a

source of secondary wavelets spreading out disturbancein all direction.

2 Huygens’ Principle

2 The tangential plane to these secondary waveletsconstitutes the new wave front

3 Interference It is the phenomenon of non uniform distribution of energy inthe medium due to superposition of two light waves.

4 Condition of maximumintensity

π φ n2= or x=nλ where n=0,1,2,3,….6 Condition of minimum

intensityπ φ )12( += n or x=(2n+1)λ/2 where n=0,1,2,3,….

7 Ratio of maximum andminimum intensity 2

21

2

21

min

max

)()(

aaaa

I I

−+

=

8 Distance of nth brightfringe from centre of thescreen d

nD yn

λ = where d is the separation distance between two

coherent source of light, D is the distance between screen andslit, λ is the wavelength of light used.

9 Angular position of nthbright fringe d

n D yn

n

λ θ ==

10 Distance of nth darkfringe from centre of thescreen d

Dn yn 2

)12( λ +=′

11 Angular position of nthdark fringe d

n D yn

n 2)12( λ

θ +=′

=′

12 Fringe width

d

Dλ β =

Diffraction and polarisation of lightS.No. Term Description1 Diffraction It is the phenomenon of bending of light waves round the sharp corners

and spreading into the regions of geometrical shadow of the object.2 Single slit

diffraction Condition for dark fringes is anλ

θ =sin where n= ± 1, ± 2, ± 3, ± 4……. ,a is the width of slit and θ is angle of diffraction. Condition for brightfringes is

an2

)12(sin

λ θ

+=

3 Width of centralmaximum is a

Dλ θ

20 =

where D is the distance between slit and screen.

4 Diffraction The arrangement of large number of narrow rectangular slits of equal


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grating width placed side by side parallel to each other. the condition formaxima in the interference pattern at the angle θ is d sin θ=nλ wheren=0,1,2,3,4………

6 resolving power of the grating

For two nearly equal wavelengths λ 1 and λ 2 between which a diffraction

grating can just barely distinguish resolving power isλ λ

λ λ λ

∆=

−=

12

R

where λ=(λ 1+λ 2)/27 Diffraction of X-Rays by crystals

the condition for constructive interference is 2 d sin θ=nλ wheren=1,2,3,4………

8 Polarisation It is the phenomenon due to which vibrations of light are restricted in aparticular plane.

9 Brewster’s law ptan=µ where µ is refractive index of medium and p is angle of polarisation.

10 Law of Malus θ 20 cos I I = where I is the intensity of emergent light from analyser, I 0 is the intensity of incident plane polarised light and θ is the anglebetween planes of transmission of analyser and the polarizer.

Reflection and Refraction of lightS.No. Term Description

1. The incident ray, the normal at a point of incidenceand reflected ray all lie on the same plane.

1 Laws ofreflection

2. Angle of incidence is always equal to angle ofreflection.

2 Relation betweenf and R

f=R/2 both f and R are positive for concave mirror and negative forconvex mirror.

3 Mirror formula

f vu111 =+

4 Magnification f

v f O I

m −==

5 Refraction The phenomenon of change in path of light as it goes from onemedium to another

1 The incident ray, the normal at a point of incidenceand refracted ray all lie on the same plane.6 Laws ofrefraction2 The ratio of sine of angle of incidence to the sine of

angle of refraction is constant for any two givenmedia. This is known as Snell’s law. Mathematically

abr

iµ =

sinsin

where abµ is relative refractive index of

medium b w.r.t. medium a. 7 Lateral shift

r r it

d cos

)sin( −=

8 Sphericalrefractingsurface

Rvu1221 µ µ µ µ −=+− for object situated in rarer medium

Rvu

2112 µ µ µ µ −=+− for object situated in denser medium

9 Power ofsphericalrefractingsurface

R P 12

µ µ −=

10 Lens maker’sformula ( )

−−=

21

111

1 R R f

µ focal length of convex lens is positive and that of

concave lens is negative

11 Lens equation f vu111 =+−

12 Linearmagnification f

v f O I

m −==

13 Power of lens ( )

−−==

21

111

1

R R f

P µ

14 For thin lensesplaced in contact

Focal length21

111 f f f

+=


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Power ofequivalent lens

P=P 1+P 2

Magnification ofequivalent lens

m=m 1+m 2

15 Sphericalaberration

The inability of lenses of aperture to bring all the rays in wide beam oflight falling on it to focus a single point is called spherical aberration.

Dual nature of waves and matterS.No. Term Description

1 Energy of a photon ν ε h= where h is the plank’s constant.2 Photoelectric

effectφ ν += max KE h where 0ν φ h= is the work function of a metal and

0ν is the critical frequency for that metal.3 Compton effect )cos1( φ λ λ λ −=−′ C where λ c=h/m 0c is Compton wavelength4 De Brogli

wavelengthmvh=λ

5 De Brogli phasevelocity

vcv P

2

==νλ

6 Wave formula

−=

λ ν π

xt A y 2cos or )cos( kxt A y −= ω where πν ω 2= is

angular frequency and P v

k ω

λ π == 2 is the wave number.

7 Phase velocity

k v P

ω =

8 Group velocity

dk d

v g ω =

9 Uncertaintyprinciple

It is impossible to know both the exact position and exactmomentum of an object at same time. Mathematically ,

π 4h

p x ≥∆∆10 Uncertainty in

energy and time π 4h

t E ≥∆∆

Atomic structureS.No. Term Description

1 Velocity of electron in an orbit

mr

ev

04πε = where r is the orbit radius

2 Total energy of hydrogen atom

r

e E

0

2

8πε −=

Lyman

−=

22

1111

n R

λ where R is Rydberg

constantBalmer

−=

22

1211

n R

λ R=1.097x10 -7 m -1

Paschen

−=

22

1311

n R

λ Brackett

−=

22

1411

n R

λ

3 Atomic spectra

Pfund

−= 22 1511

n R

λ


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4 Orbital electron wavelengthaccording to Bohr atom model

mr

eh 04πε λ =

5 Condition for orbital stability nλ=2πr n where n=1,2,3…. And r n is radius of orbitthat contains n wavelengths.

6 Orbital radii in Bohr atom2

022

me

hnr n

π

ε =

Atomic nucleus and nuclear energyS.No. Term Description

1 Isotopes Nuclei having same atomic number but different mass number.2 Isobars Nuclei having different atomic number but same mass number.3 Atomic mass unit One atomic mass unit is defined as one twelfth part of the mass of 12C6

atom. Value of a mass unit is 1u=1.66054x10 -27 Kg=931 MeV4 Nuclear radius R=R 0A1/3 Where value of R 0 ≅ 1.2 x 10 -15 ≅ 1.2 fm and is known as nuclear

radius parameter5 Nuclear density The density of nuclear matter is approximately of the order of 10 17 Kg/m 3

and is very large compared to the density of ordinary matter.6 Nuclear forces It is the force which holds the nucleons together inside the nucleus.7 Mass defect mm Z A Zmm n p −−+=∆ )(

8 Binding energy [ ][ ]22 )( cmm Z A Zmmc BE n p −−+=∆=9 Binding energy

per nucleon=BE/A

10 Radioactivedecay law

t e λ −= 011 Half life T=ln2/λ12 Average life Tav=1/λ


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