tạp chí physics for you tháng 1 năm 2016

Energy Density and intensity The electric field and magnetic field in a plane electromagnetic wave propagating along x-axis are given by E E= 0sinwt xc− , B B= 0sinwt xc−  Here,

Physics Musing Problem Set 30 8

Core Concept 12 Thought Provoking Problems 22

PMT Practice Paper 25 JEE Accelerated Learning Series 31

Brain Map 46 Ace Your Way CBSE XI 57

JEE Workouts 64 Ace Your Way CBSE XII 68 Exam Prep 2016 75 Physics Musing Solution Set 29 81

Live Physics 83 You Ask We Answer 84

Crossword 85

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Volume 24 No 1 January 2016

single oPtion correct tyPe

1 Two batteries of emf e1 and e2 having internal

resistance r1 and r2 respectively are connected in

series to an external resistance R Both the batteries

are getting discharged The above described

combination of these two batteries has to produce a

weaker current than when any one of the batteries is

connected to the same resistor For this requirement

e21 must not lie between r r R1+2 and r R2r1

2 A particle of charge per unit mass a is released from

origin with velocity v v i= 0 in a magnetic field

ac supply of 100 V and 40 Hz, then the current in series combination of above resistor and inductor is

5 In a regular polygon of n sides, each corner is at

a distance r from the centre Identical charges are placed at (n – 1) corners At the centre, the magnitude of intensity is E and the potential is V The ratio V/E is

2 Manmohan Krishna (Bihar)

3 Meena Chaturvedi (New Delhi)

4 Naresh Chockalingam (Tamil Nadu)

subjective tyPe

6 In figure, a long thin wire carrying a varying current

I = I0 sin wt lies at a distance y above one edge of a

rectangular wire loop of length L and width W lying

in the X-Z plane What emf is induced in the loop?

W L X

Y

y

I

Z

7 A wire is wrapped N times over a solid sphere of

mass m near its centre, which is placed on a smooth

horizontal surface A horizontal magnetic field of

induction Bis present Find the

angular acceleration experienced

by the sphere Assume that the

mass of the wire is negligible

compared to the mass of the

sphere

R m



8 A stone is dropped from a balloon going up with

a uniform velocity of 5.0 m s–1 If the balloon was 50 m high when the stone was dropped, find its height when the stone hits the ground Take

g = 10 m s–2

9 A body weighs 98 N on a spring balance at the north pole What will be its weight recorded on the same scale if it is shifted to the equator? Use

g = GM/R2 = 9.8 m s–2 and the radius of the earth

R = 6400 km

10 A wheel of radius r and moment of inertia I about

its axis is fixed at the top of an inclined plane of inclination q as shown in figure A string is wrapped round the wheel and its free end supports a block of

mass M which can slide on the plane Initially, the

wheel is rotating at a speed w in a direction such that the block slides up the plane After some time, the wheel stops rotating How far will the block move before stopping?

M





nn

When two or more waves of similar kind, simultaneously

arrive at a point then the resultant disturbance (or

displacement in case of particles) at the point of meeting

is given by a vector sum of the disturbances produced

by each of the arriving waves

After superposition, the waves pass through as if they

did not encounter each other, hence there is no change

in the properties of either of the arriving waves after

superposition

If the disturbances are produced along same line, vector

sum becomes algebraic sum

Let us see what we understand from this

Consider two disturbances travelling in opposite

directions meet each other as shown

We are going to draw the shape of the resultant waveform at

(i) t = 2 s (ii) t = 2.5 s (iii) t = 4 s.

To do this, imagine individual waves travelling as if

they are not meeting each other and then we apply

Applications of Superposition of waves

of both the waves are identical else (w1 – w2)t will

come out to be time dependent expression

Coming back to interference; when waves of similar kind from two or more coherent sources simultaneously arrive at a point then the resultant intensity at the point of superposition is different from the sum of intensity of the arriving waves and is dependent on the phase difference between the arriving waves which

is directly dependent on path difference between the arriving waves

To put this in simple words, let me put a simple statement, imagine two coherent sources of light made

to interfere There might be a situation wherelight + light = darkness

Isn't it opposite to our common sense? Yes, it is since

we expect more bright light when two light sources are made to superimpose But this logic is true only for non-coherent sources

Let us see this through an

The displacement equations for the two sources are say

of SHM, we now understand

that the resultant amplitude of

oscillation can be found out

\ A R= A12+A22+2A A1 2cos( )Df

where A R = resultant amplitude of oscillation.

Now we understand that A1 and A2 are fixed for wave

but if we change the location of sources from the point P,

x1 and x2 changes, hence Dx changes, hence Df changes

Therefore A R becomes dependent on Dx.

Now, considering intensity of waves, as we know,

I ∝ A2

\ from, A R2 = A12 + A22 + 2A1A2cos(Df)

we have ,

I R= + +I I1 2 2 I I1 2cos( )Df

where I R = intensity of the resultant wave due to arriving

waves of intensities I1 and I2

Now extreme cases of interference may arise

Constructive interference:

such that crest of one wave coincides with the crest

of the other, the waves are said to be in phase and

in such case the amplitude and hence intensity of

resultant wave is maximum

A1

A2

+  A + A = A1 2 R

This clearly is possible only if one wave is shifted

with respect to the other wave by an integral

multiple of complete wavelength (l)

\ Dx n= l ⇒ Df= 2 ( ) = n(2p)lp nl

\ A Rmax = A1 + A2

I Rmax=( I1+ I2)2

Destructive interference:

that crest of one wave coincides with the trough

of the other, the waves are said to be out of phase and in such case the resultant amplitude and hence intensity of wave is minimum

Now, this is what I was talking about when I said light + light = darkness!

In general at any point in such case,

A R=2A  

2

0cos Df , I R= I 

 4

2

0cos2 DfLet us now see an application of interference

Q1 Two coherent sources

(i) the number of maximas and minimas detected

in one complete rotation

(ii) the angular position of the 1st maxima after

starting

Soln.: (i) In such

question analyse one

quadrant, since all

quadrants are identical

Let us see how S1 3S2 B( = 3x )

\ Dx = 0, since the same path length for waves

arriving from S1 and S2

At point B, S1B – S2B = 3l

\ Dx = 3l

Hence A and B are locations of maximas but in

between A and B, there must be other locations

where Dx would gradually increase from Dx = 0 to

Dx = 3l.

\ In between 0 and 3l, there would be locations

of Dx = l and Dx = 2l too.

Hence total maximas

= 2 × 4 (each quadrant) + 1 × 4(each point on

x and y axis)

= 12

For minimas, Dx=(2 1n+ )

2l

Hence between each successive maxima, there

would be minima too, so,

Dx =l2, 32l, 52l in each quadrant

\ Total minimas = 3 × 4 = 12

(ii) Given : R >> 3l hence, from all points on

circumference S1 and S2 will appear to be close by

Standing Waves/Stationary Waves

When two waves of similar kind, same frequency

and amplitude travelling from opposite directions

superimpose we get standing waves Now to understand

better what this actually means, let us start by considering

two such waves

y1(x, t) = Asin(wt – kx)

y2(x, t) = Asin(wt + kx)

Hence, considering superposition,

y R (x, t) = y1(x, t) + y2(x, t) = A[sin(wt – kx) + sin(wt + kx)]

= A[2sin(wt) cos(kx)]

= 2Acos(kx) sin(wt) = A x sin(wt)

where A x = 2Acos(kx)

This resultant wave clearly shows that it is not a travelling wave atleast

The quantity, A x = 2Acos(kx) is constant for a location

and the value is position dependent whereas

y(x, t) = A x sin(wt) indicates that the displacement along y-axis of a particle located at x varies simple harmonically with time period, T = 2wp

Hence, sin(wt) being same for all particles, each

But amplitude indicates magnitude of maximum

particle displacement Hence –2A is same as 2A

when we talk of amplitude

particles are not oscillating at all, i.e they are stationary

at their respective points It is due to these particles that the name stationary wave is coined

The locations where amplitude of oscillation is

(i) maximum (A x = ±2A) are said to be antinodes.

(ii) minimum (A x = 0) are said to be nodes

A x = 0 = 2Acos(kx)

⇒ cos(kx) = 0

⇒ kx=(2 1n+ )

2p

54

Let us try to draw the envelope of the particle's

displacement (for standing waves in a stretched

string) with these locations of nodes and antinodes

The separation between two consecutive nodes or

antinodes is said to be loop length which is l

2.The bold lines drawn above indicate the boundary

within which the particles are confined to move

How would standing waves in stretched string appear?

As below,

All the particles located between two consecutive

nodes oscillate in same phase, i.e., they reach their

extreme ends and mean position together Particles

located on opposite sides of a node always move

opposite to each other, hence are 180° out of phase

with respect to each other

Hence, unlike travelling wave, the energy does not

propagate from one end to the other, it gets confined

between a loop

Application of standing waves

Standing waves in stretched string fixed at both ends:

The fixed ends will always

be a displacement node

The smallest frequency

with which standing

waves can be set up is

said to be fundamental frequency (f0)

l

Velocity of wave, v= T

m

where, T = tension in string

m = mass per unit length

Since velocity is only medium dependent, v = f l

suggests that for minimum frequency, wavelength l should

be maximum Hence keeping the two extreme ends fixed (nodes) the largest wavelength that can be drawn

by inserting an antinode in between is shown here

\ v = f0 l0

l = 0 2

l

0

0 2l

l

0 2The frequencies higher than the fundamental frequency with which standing waves can be set up in the system are said to be overtones, whereas all integral multiples of fundamental frequency are said to be harmonics

f n = (n + 1)f0 = (n + 1)th harmonic = +(n 1)2v l

Standing waves in organ pipe:

The open ends of the organ pipe behave as displacement antinode whereas the closed end behaves as displacement node

It has both ends open It has one end closed

Fundamental mode

l = 0 2

0l

l

0 2

l = 0 4

0l

l

0 4

When two sound waves of frequency close to each

other (but not equal) superpose at a point, then at the

point of superposition the phase difference keeps on

changing with respect to time

Df = (w1 – w2)t

Hence the amplitude of resultant wave will also keep

on changing at a location as time passes by

But everytime Df = n(2p) we get to hear a maxima since

waves become in phase and everytime Df = (n + 1)p we

get to hear a minima

So, the appearance is a sound of alternating intensity,

alternating between a certain maximum and minimum

intensity

This phenomenon is beats and the number of maximas

heard per second is beat frequency

To find beat frequency (f b)

12

and Y = 7.8 × 1010 N m–2 for aluminium Find the frequency of sound emitted

Soln.: Velocity of wave

30 10

Q3 A tuning fork of unknown frequency produced

4 beats per second when sounded with another

of frequency 254 Hz It gives the same number

of beats per second whenever unknown tuning fork is loaded with wax Find its frequency before waxing

Soln.: Remember, two terms with tuning forks(i) Waxing : Here wax is added to the prongs due

to which frequency of wave decreases

(ii) Waning : Here the prongs are filed due to which frequency of wave increases

Coming back to the question, having 4 beats/second gives two possible situations

254 +4 258

f < 258

waxed

–4 250 waxed

f < 250

Here the difference with

254 will initially decrease

as long as > 254 but iff

< 254 then it might

f

happen that it reaches

= 250 Hz and again get

= 4 Hz, hence 258 Hz

f b

is the required frequency.

Here difference with

254 will clearly be greater than 4 Hence

> 4 not possible.

f b which is

Q4 For a certain organ pipe three successive frequencies

are observed at 425, 595, 765 Hz respectively

Taking speed of sound in air to be 340 m s–1, find

the fundamental frequency

Soln.: Note the ratio,

425 : 595 : 765 = 5 : 7 : 9

\ Odd multiples ⇒ closed organ pipe

\ 5f0 = 425 ⇒ f0 = 85 Hz

Q5 A tuning fork vibrating with a sonometer having a

wire of length 20 cm produces 5 beats per second The

beat frequency does not change if the length of wire is

changed to 21 cm Find frequency of tuning fork

Soln.: For a sonometer wire, f s∝ 1l

Hence if length increases, f s decreases

Hence we conclude that initially f s was greater than

f t (frequency of tuning fork) and later it became

smaller by same amount

2120

f

f t t

⇒ f t = 205 Hz

Q6 A tuning fork and an air column whose temperature

is 51°C produce 4 beats in one second when sounded

together When the temperature of air column is 16°C

only one beat is detected per second Find frequency

of tuning fork

Soln.: v∝ T

\ Decrease in temperature

⇒ decrease in velocity of wave

⇒ decrease in frequency of air column (f a)

Two situations were possible

f a – f t = 4 or f t – f a = 4

The second equation suggests that if f a decreases,

f b > 4 (increases) which is a contradiction

1817

A vehicle is moving towards a large building with

a speed v0 and blows a horn of frequency f0 which

travels with a speed v w Find the number of beats

detected by A who is moving with the vehicle and B

who stands in between vehicle and building

Soln.: Using method of images, we create a virtual source behind the building

f0

v w A

Speaking of A, he hears two frequencies (i) f1 = f0 (since there is no relative motion between

him and S) (ii) f v v w v v f

v f

nn

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1 A large tank A of water is kept at a constant

temperature q0 It is connected to another body B

of specific heat c and mass m by a conductor of

length l, area of cross-section A and thermal

conductivity K Initial temperature of body B is q1

Find the variation of temperature q of the body B

at time t.

t = 0 K

Tank of water

2 One end of a rod of length L and cross-section

area A is kept in a furnace at temperature T1 The

other end of rod is kept at a temperature T2 The

thermal conductivity of the material of the rod

K and emissivity of the rod is e It is given that

T2 = T S + DT, where DT < < T S , T S is the temperature

of surroundings If DT ∝ (T1 – T S), find the

3 N molecules each of mass m of gas A and 2N

molecules each of mass 2m of gas B are contained

in the vessel which is maintained at a temperature T The mean square of velocity of the molecules of B type is denoted by v2 and the mean square of the

x-component of the velocity of A type is denoted

by w2 What is the ratio of w

v

2

2 ?

4 An ideal monoatomic gas is confined in a cylinder

by a spring-loaded piston of area of cross-section

8 × 10–3 m2 Initially, the gas is at 300 K and occupies a volume of 2.4 × 10–3 m3 and the spring

is in its relaxed state as shown in figure The gas is heated by a small electric heater until the piston moves out slowly by 0.1 m Calculate the final temperature of the gas and the heat supplied (in J)

by the heater The force constant of the spring is

8000 N m–1, atmospheric pressure is 1 × 105 N m–2 The piston is massless and no friction exists between the piston and the cylinder

Heater

Piston

Rigid support

5 Figure shows three isotherms at temperatures

T1 = 4000 K, T2 = 2000 K, and T3 = 1000 K When

1 mole of an ideal monoatomic gas is taken through

paths AB, BC, CD and DA, (a) find the change in

*Randhawa Institute of Physics, S.C.O 208, First Fl., Sector-36D & S.C.O 38, Second Fl., Sector-20C, Chandigarh, Ph 09814527699

internal energy DU, (b) find the work done by the

gas W.

B

C D

1 Let the temperature of the body B be q, in time

dt, let a heat dQ flow through the conductor

The rate of increase in thermal energy of the

−

0 ( 0 1)e

KA mlc t

2 Rate of heat conduction through rod = rate of heat

lost from right end of the rod

Using binomial expansion, we have

T S

1 3

w2 = v

kT

m x

2

33

kT m kT m

2

2 32

23

4 Since F = kx = 8000 × 0.1 = 800 N

Pressure exerted on the piston by the spring is

DP F= =A

−800

5 N m 2The total pressure of the gas inside the cylinder

is

P2 = P1 + DP = Patm + DP

P2 = 1 × 105 + 1 × 105 = 2 × 105 N m–2

Since the piston has moved outwards, then increase

in volume of the gas is

Let the heat supplied by the heater be Q, then

22

32 = × ×

constant pressure is

P(V B – V A ) = R(T B – T A) = 8.314 × 2000 = 16.628 × 103 J

No work is done during BC and DA.

Work done on the gas during CD is

R(T C – T D) = 8.314 × 1000 = 8.314 × 103 J Net work done by the gas during the cycle is(16.628 × 103 – 8.314 × 103) J = 8.314 × 103 J

nn

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1 A body of mass 7m initially at rest explodes into two

fragments of masses 4m and 3m If the momentum

of the lighter fragment is p then the kinetic energy

released in the explosion will be

(a) 7p2

24m (b)

916

p m

2

p m

2 A particle of mass m describes a circle of radius r

The centripetal acceleration of the particle is 4

r2

The momentum of the particle is

3 Figure shows a block of mass m1 on a smooth

horizontal surface pulled by a

string which is attached to a

block of mass m2 hanging over a

frictionless pulley which has no

mass The blocks will move with

an acceleration

m g

1 2 2

the second case this force

is applied to block m2 The

forces of contact between

m2 m1

the blocks in the first and the second cases, respectively are

5 A uniform ladder is in equilibrium against a rough

wall as shown Points A and

B respectively are the point

of contact of ladder with wall

and with the ground Point C

is the CM of the ladder

1 Torque due to friction f2 about point A is not

zero

2 Torque due to friction f2 about point B is zero.

3 Torque due to weight is zero about A, B and C.

4 Torque due to f1 and N1 is not zero about A.

(a) Only 1 and 3 are correct(b) Only 2 and 3 are correct(c) Only 3 and 4 are not correct(d) Only 1 and 4 are correct

6 The moment of inertia of an annular disc (a disc

with concentric cavity) of mass M radius R and cavity radius r about an axis passing through its CM

and normal to its plane will be(a) 12M R r( 2+ 2) (b) 1

N2

This paper contains 45 multiple choice questions Each question has four choices (a), (b), (c) and (d), out of which only

PRACTICE PAPER CLASS

XI

Contributed by : K P Singh, KP Institute of Physics, Chandigarh, 09872662552

8 A body of mass 3 kg is under a force which causes

displacement in it, given by s t= 2

3 in metre, with

time t in seconds What is the work done by the

force between time t = 0 and t = 2 s ?

(a) 8 J (b) 5.2 J (c) 3.9 J (d) 2.6 J

9 The linear momentum p of a body varies with time

as p = 5a + 7bt2 where a and b are constants The

net force acting on the body for one dimensional

motion varies as

(a) t2 (b) t–1 (c) t–2 (d) t

10 A body is acted upon by a force proportional to

square of distance covered If the distance covered

is denoted by x, then work done by the force will

constants If it is known that the system has only

one stable equilibrium configuration, the possible

values of a and b are

of the pulley, which one of the

following is the force F required to

lift a 100 N load in the system of

pulleys as shown in the figure?

13 A body is falling freely under the action of gravity

alone in vacuum Which of the following quantities

remain constant during the fall?

(a) Kinetic energy

(b) Potential energy

(c) Total mechanical energy

(d) Total linear energy

14 A motor drives a body along a straight line with a

constant force The power P developed by the motor

must vary with time t as shown in figure

(c)

t

t P

15 Velocity-time graph of a particle V(m s )–1

20

2 t(s)

of mass 4 kg moving in a straight line is as shown in figure Work done by all forces on the particle is(a) 400 J (b) – 800 J (c) – 400 J (d) 200 J

16 An ideal spring with spring constant k is hung from the ceiling and a block of mass M is attached to

its lower end The mass is released with the spring initially unstretched Then the maximum extension

in the spring is(a) 4 Mg

2

17 A block of mass 5 kg is resting on a smooth surface

At what angle a force of 20 N be acted on the body

so that it will acquire a kinetic energy of 40 J after moving 4 m ?

(a) 30° (b) 45° (c) 60° (d) 120°

18 The moment of inertia of a body about a given axis

is 1.2 kg m2 Initially, the body is at rest In order

to produce a rotational kinetic energy of 1500 J, an

angular acceleration of 25 rad s–2 must be applied about that axis for a duration of

(a) 4 s (b) 2 s (c) 8 s (d) 10 s

19 A particle performs uniform circular motion with an

angular momentum L If the frequency of particle’s

motion is halved and its kinetic energy doubled, the angular momentum becomes

(a) 2 L (b) 4 L (c) L/2 (d) L/4

20 The moment of inertia of two spheres of equal masses about their diameters are equal If one of them is solid and other is hollow, the ratio of their radii is

22 A uniform rod of mass m and length L

is suspended by means of two light

inextensible strings as shown in figure

Tension in one string immediately

after the other string is cut is

(a) mg

2 (b) mg (c) 2 mg (d)

mg

4

23 A circular platform is free to rotate in a horizontal plane

about a vertical axis passing through its centre A

tortoise is sitting at the edge of the platform Now

the platform is given an angular velocity w0 When the

tortoise moves along a chord of the platform with a

constant velocity with respect to the platform, the angular

velocity of the platform will vary with the time t as

24 A wheel has angular acceleration of 3.0 rad s–2 and

an initial angular speed of 2.00 rad s–1 In a time of

2 s it has rotated through an angle (in radian) of

25 The moment of inertia of a rod about an axis through

its centre and perpendicular to it is 1

12ML where, 2

M is the mass and L the length of the rod The rod

is bent in the middle so that the two halves make an

angle of 60° The moment of inertia of the bent rod

about the same axis would be

26 A body of mass 15 kg is suspended by the strings

making angles 60° and 30° with the horizontal as

shown in figure Then (take g = 10 m s–2)

27 A car of mass 1000 kg negotiates a banked curve of

radius 40 m on a frictionless road If the banking

angle is 45°, the speed of the car is

28 A 5000 kg rocket is set for vertical firing The exhaust speed is 800 m s–1 To give an initial upward acceleration of 20 m s–2, the amount of gas ejected per second to supply the needed thrust will be

(g = 10 m s–2)(a) 127.5 kg s–1 (b) 187.5 kg s–1(c) 185.5 kg s–1 (d) 137.5 kg s–1

29 For shown atwood machine m1 = 8 kg,

(a) push forward twice as hard (b) push forward four times as hard(c) push backward four times as hard(d) push backward twice as hard

31 A block of mass m is placed on a smooth wedge

of inclination q The whole system is accelerated horizontally so that the block does not slip on the wedge The force exerted by the wedge on the block has a magnitude

32 A system consists of 3 particles each of same mass and located at points (1, 2), (2, 4) and (3, 6) The co-ordinates of the center of mass are

34 Consider a two particle system with particles having

masses m1 and m2 If the first particle is pushed

towards the center of mass through a distance d,

by what distance should the second particle be moved so as to keep the center of mass at the same position ?

35 Two particles of mass m1 and m2 (m1 > m2) attract

each other with a force inversely proportional to the

square of the distance between them The particles are

initially held at rest and then released Then the CM

(a) moves towards m1 (b) moves towards m2

(c) remains at rest

(d) moves at right to the line joining m1 and m2

36 Two skaters A and B of masses 50 kg and 70 kg

respectively stand facing each other 6 m apart Then

they pull on a rope stretched between them How

far has each moved when they meet?

(a) both have moved 3 m

(b) A moves 2.5 m and B 2.5 m

(c) A moves 3.5 m and B 2.5 m

(d) A moves 2 m and B 4 m

37 A thin uniform circular disc of mass M and radius

R is rotating in a horizontal plane about an axis

passing through its center and perpendicular to its

plane with an angular velocity w Another disc of

same dimensions but of mass 1

38 A wheel is rolling uniformly along a level road (see

figure) The speed of translational motion of the

wheel axis is v What are the speeds of the points

A and B on the wheel rim relative to the road at the

instant shown in the figure?

39 If a force 10 15 25^i+ ^j− k^ acts on a system and gives

an acceleration 2 3 5^i+ −^j k^ to the centre of mass of

the system, the mass of the system is

40 Two particles A and B

are situated at a distance

d = 3 m apart Particle A

has a velocity of 5 m s–1 at

an angle of 60° and particle

B has a velocity v at an angle of 30° as shown in A

41 The ratio of the dimensions of Planck’s constant and that of the moment of inertia is the dimension of

(c) Angular momentum (d) Time

42 A ball is dropped to the ground from a height of

8 m The coefficient of restitution is 0.5 To what height will the ball rebound?

v g

02

v g

02

v g

0212

44 A ball is projected in vacuum as shown Average power delivered by gravitational force

(a) for A to C is positive

4 In a one dimensional elastic collision the fraction

of kinetic energy transferred by a projectile to a stationarytarget is 4 1 2

= 12 +1

222

p m

p m

22

14

1

3 = 724

2

p m

5 (c) : Torque of a force about a point is zero only

when the line of action of the force passes through

that point

6 (a) : Check: When r 0, it becomes disc and

when r R, it becomes ring.

16 (b) : Let the maximum extension in the spring be x

By work energy theorem, W = DK

L L

R H S = 53

21 (b): L = pr ⇒ log L = log (pr)

log L = log p + log r; y = mx + C

So, straight line with positive intercept

22 (a): Before cutting, T A + T B = mg

For rotational equilibrium about center

2= 2 ⇒ T A = T B

So, 2T A = mg or T A=mg

2

If string B is cut, just after cutting tension in

A remains same i.e., mg

2 .

23 (c): Moment of inertia first decreases and then increases, thus by law of conservation of angular momentum,

26 (b) : Horizontally: T2 cos 60° = T1 cos 30°

N a



35 (c) : Internal forces cannot change velocity of CM

of a system As CM was initially at rest, it will remain at rest

36 (c) : Two bodies under mutual internal forces always meet at their CM From theorem of moment

of masses, we can write

123

[ ][ ]

h I

L I

v g

e

nn

ElEctromagnEtic wavEs

Maxwell predicted the existence of electromagnetic

waves on the basis of his equations in 1865 According

to him, an accelerated charge produces a sinusoidal

time varying magnetic field, which in turn produces a

sinusoidal time varying electric field The two fields so

produced are mutually perpendicular and are sources

to each other Such a combination of electric and

magnetic fields constitute electromagnetic waves which

can propagate through empty space in the direction

perpendicular to both fields

Displacement current

It is that current which comes into play in the region,

whenever the electric field and hence, the electric flux is

changing with it It is given by I D= e f0d dt E

where e0 = absolute permittivity of free space,

d

dtf = rate of change of electric flux.E

In case of a steady electric flux linked with a region, the

displacement current is zero

maxwell’s Equations

 

 ∫s E dS q⋅ =e0,

This law gives the total electric flux in terms of charge

enclosed by the closed surface This law states that the

electric lines of force start from positive charge and end

at negative charge i.e., the electric lines of force do not

form a continuous closed path

 

 ∫s B dS⋅ =0,

This law shows that the number of magnetic lines of force entering a closed surface is equal to number

of magnetic lines of force leaving that closed surface.This law tells that the magnetic lines of force form a continuous closed path This law also predicts that the isolated magnetic monopole does not exist

 

 ∫E dl⋅ = −d dtfB,

induction This law gives a relation between electric field and a changing magnetic flux This law tells that the changing magnetic field is the source of electric field

 

 ∫B dl⋅ =m0I+m e f0 0d dt E,

This law states that the magnetic field can be produced

by a conduction current as well as by displacement current This law also states that the conduction current and displacement current together have a property of continuity At an instant, in a circuit, the conduction current is equal to displacement current

Production of Electromagnetic waves

The electromagnetic wave is emitted when an electron orbiting in higher stationary orbit of atom jumps to one of the lower stationary orbit of that atom

The electromagnetic waves are also produced when fast moving electrons are suddenly stopped by the metal of high atomic number

1 For plane electromagnetic waves propagating in the

z direction, which one of the following

combination gives the correct possible direction

for EandB field respectively?

Energy Density and intensity

The electric field and magnetic field in a plane

electromagnetic wave propagating along x-axis are

given by

E E= 0sinwt xc− , B B= 0sinwt xc− 

Here, E0 and B0 are the amplitudes of the fields and c is

the speed of light

The energy of electric field is given by U E= 12e0E dV2

and the energy of the magnetic field is given by

0

2m

The average energy density of electric field is

u E=12 <E > =12 E  = E

2

14

0

0 0

2

020

u av=u E+u B=2u E=2u B=12e0 0E2 = 1

2 0

B

m

The unit of u E and u B are J m–3

Intensity of electromagnetic wave is defined as energy

crossing per unit area per unit time perpendicular to

the directions of propagation of electromagnetic wave

The intensity I is given by the relation

I u c= av =12 0 0E c2 =12B c02

0

SELFCHECK

2 A red LED emits light at 0.1 watt uniformly around

it The amplitude of the electric field of the light at a distance of 1 m from the diode is

(JEE Main 2015)

momentum of Electromagnetic wave

The electromagnetic wave also carries linear momentum with it The linear momentum carried by the portion of

wave having energy U is given by p = U/c.

If the electromagnetic wave incident on a material

surface is completely absorbed, it delivers energy U and momentum p = U/c to the surface.

If the incident wave is totally reflected from the surface, the momentum delivered to the surface is

U/c – (–U/c) = 2U/c It follows that the electromagnetic

wave incident on a surface exert a force on the surface

radiant flux of electromagnetic wave

According to Maxwell, when a charged particle is accelerated it produces electromagnetic wave The total radiant flux at any instant is given by

c

0 26

field strength, i.e to the surface integral of the poynting

vector formed by the component of the field in the plane

of the surface

The average value of poynting vector S( ) over a convenient time interval in the propagation of electromagnetic wave is known as radiant flux density When energy of electromagnetic wave is incident on

a surface, the flux density is called intensity of wave

(denoted by I) Thus I = S.

Electromagnetic spectrum

The orderly distributions of electromagnetic radiations according to their wavelength or frequency is called the electromagnetic spectrum

0.1 m They can be produced by electrons oscillating

in wires of electric circuits and antennas can be used

to transmit or receive radiowaves that carry AM or

FM radio and TV signals

Microwaves

with typical wavelength in the range 1 mm to 0.1 m

They are commonly produced by oscillating electric

circuits, as in the case of microwave oven

Infrared radiation,

than the visible (from 7 × 10–7 m to about 1 mm),

is commonly emitted by atoms or molecules when

they change their vibrational motion or vibration

with rotation Infrared radiation is sometimes

called heat radiation

Visible light,

electromagnetic wave, is that part of the spectrum

the human eye can detect The limits of wavelength

of the visible region are from 430 nm (violet) to

740 nm (red)

Ultraviolet light

from 4.3 × 10–7 m (430 nm) down to 6 × 10–10 m (0.6 nm) Ultraviolet rays can be produced by electrons in atoms as well as by thermal sources such as the sun

X-rays

• typical wavelength in the range of about

10–9 m (1 nm) to 10–13 m (10–4 nm) can be produced with discrete wavelength in atoms by transitions involving the most tightly bound electrons and they can also be produced in continuous range

of wavelengths when charged particles such as electrons are decelerated

Gamma rays

the electromagnetic spectrum (less than 10 pm), are emitted in the decays of many radioactive nuclei and certain elementary particles

Electromagnetic Spectrum

Radiowaves 104 to 109 > 0.1 m Rapid acceleration and

decelerations of electrons in aerials Different specialised uses in radio communication.Microwaves 109 to 1011 0.1 m to

1 mm Klystron valve or magnetron valve (a) Radar communication.(b) Analysis of fine details of

molecular and atomic structure.Infrared

11 to

5 × 1014 1 mm to 700 nm Vibration of atoms and molecules (a) Useful for molecular structure.(b) Useful for haze photography.Visible light 4 × 1014 to

7 × 1014 700 nm to400 nm Electrons in atoms emit light when they move from one

energy level to a lower energy level

Detected by stimulating nerve endings of human retina

(a) Absorbed by glass

(b) Can cause the tanning of the human skin

(c) Ionize atoms in atmosphere, resulting in the ionosphere

10–3 nm X-ray tubes or inner shell electrons (a) Penetrate matter(b) Ionize gases

(c) Cause fluorescence(d) Cause photoelectric emission from metals

Gamma

18 to 1022 < 10–3 nm Radioactive decay of the

Light is a form of energy which makes objects visible

to our eyes The branch of physics, which deals with

nature of light, its sources, properties, measurement,

effects and version is called optics It is divided in to

two parts, ray optics and wave optics

reflection

A ray of light is composed of packets of energy

known as photons The photons have the ability of

colliding with any surface Thus, the photons transfer

momentum and energy in the same way as the transfer

of momentum and energy take place between any two

particles during collision When a photon is incident

on a plane surface, it gets rebounded This is known as

reflection

Laws of Reflection : A light ray is reflected by a

smooth surface in accordance with the two laws of

reflection

The angle of incidence is equal to the angle of

•

reflection i.e., ∠i = ∠r.

The incident ray, the reflected ray and the normal

•

to the reflecting surface are coplanar

reflection by Plane mirror

If the object is real, the image formed by a plane mirror

is virtual, erect, of same size and at the same distance

from the mirror

characteristics

If keeping the incident ray fixed, the mirror is rotated

by an angle q, about an axis in the plane of the mirror,

the reflected ray is rotated through an angle 2q

As every part of a mirror forms a complete image of

an extended object and due to superposition of images

brightness will depend on its light reflecting area, a

large mirror gives more bright image than a small one

This in turn also implies that if a portion of a mirror

is obstructed, complete image will be formed but of

reduced brightness

If two plane mirrors are kept facing each other at an

angle q and an object is placed between them, multiple

images of the object are formed as a result of successive

reflections The number of images, n is given by

Case-I :

q is even integer, number of

images formed n = (m – 1) for all positions of

object

Case-II :

• If m =360q is odd integer, number of °images formed n = m – 1, if the object is on the bisector of mirrors and n = m, if the object is not

on the bisector of mirrors

Case-III :

• If m =360q is a fraction, number of °

images formed will be equal to its integral part.When two plane mirrors are placed parallel to each

other, then q = 0° ⇒ n = ∞ Therefore, infinite number

of images are formed

If the mirror moves away or towards an object by a

distance d, then the image moves away or towards the object by a distance 2d.

If the mirror moves with speed v towards or away from

fixed object then image appears to move towards or

away from the object with speed 2v.

If the object moves with speed v towards a fixed mirror,

the image also moves towards the mirror with speed

v The speed of the image relative to the object in this

case is 2v.

KEYPOINT

The image formed by a plane mirror is laterally

•

inverted and the linear magnification is unity

reflection from a spherical surface

There are two types of spherical mirrors, concave and convex mirror as shown in figure

For a concave mirror, the centre C of the sphere of

which the mirror is a part is in front of reflecting

surface, for a convex mirror it is behind C is centre of curvature of the mirror, and P, the centre of the mirror surface is called pole The line CP produced is the principal axis AB is the aperture of the mirror.

A narrow beam of rays, parallel and near to the principal

axis, is reflected from a concave mirror so that all rays

converge to a point F on the principal axis In figure, F

is called the principal focus of the mirror and it is a real

focus since light actually passes through it Concave

mirrors are also known as converging mirrors because

of their action on a parallel beam of light

A narrow beam of rays, parallel and near to the principal

axis, falling on a convex mirror is reflected to form a

diverging beam which appears to come from a point

F behind the mirror A convex mirror has a virtual

principal focus and it is also a diverging mirror

sign conventions

All distances have to be measured from the pole of the mirror Distances measured in the direction of incident light are positive, and those measured in opposite direction are taken as negative

Heights measured upwards and normal to the principal axis of the mirror are taken as positive, while those measured downwards are taken as negative

Image formation by concave mirror

S

No

Position of

6 Between focus F

mirror formula

If an object is placed at a distance u from the pole of a

mirror and its image is formed at a distance v (from the

pole) then, 1 1 1f v u= +

In this formula, to calculate unknown, known quantities

are substituted with proper sign

KEYPOINT

Mirror formula is independent of the angle the

•

incident ray makes with the axis Therefore all

paraxial rays from point object must, after reflection,

pass through image to give point image

Power of mirror

Power of mirror, P f= (in m)1 = f(in cm)100 where f should

be taken with proper sign i.e., –ve for concave and

+ve for convex mirror The unit of power is dioptre

(D) or m–1

lateral magnification

A real image is always inverted one and a virtual one

is always erect Keeping these points in mind and that

the real object and its real image would lie on the same

sides in case of mirrors, we define magnification (m) in

case of reflection by spherical mirror as m= − v u

SELFCHECK

3 You are asked to design a shaving mirror assuming

that a person keeps it 10 cm from his face and views

the magnified image of the face at the closest

comfortable distance of 25 cm The radius of curvature

of the mirror would then be

2 Between pole P

4 A thin convex lens of focal length f is put on a plane

mirror as shown in the figure When an object is kept

at a distance a from the lens - mirror combination, its image is formed at a distance a

Light travels through vacuum at a speed of c = 3 × 108 m s–1

It can also travel through many materials, such as air, water and glass However, atoms in the material absorb, re-emit, and scatter the light Therefore, light travels

through the material at a speed that is less than c, the

actual speed, depending on the nature of the material The change in speed as a ray of light goes from one material to another causes the ray to deviate from its incident direction The index of refraction or refractive

index m of a material is the ratio of the speed of light (c)

in vacuum to the speed of light in the material (v).

Mathematically, refractive index is given by the relation

m =Speed of light in the materialSpeed of light in vacuum =c v The absolute refractive index of a substance is the relative refractive index for light travelling from vacuum into the substance It is commonly referred to as the refractive index

Relative refractive index is a measure of how much

light bends when it travels from any one substance into

any other substance

Laws of refraction (snell’s Law)

When light strikes the interface between two transpanent

materials, such as air and water, the light generally

divides into two parts Part of the light is reflected,

with the angle of reflection being equal to the angle

of incidence The remainder is transmitted across the

interface If the incident ray does not strike the interface

at normal incidence, the transmitted ray has a different

direction than the incident ray The ray that enters the

second material is said to be refracted and behaves in

one of the following two ways:

When light travels from a medium with smaller

•

refractive index into a medium with larger refractive

index, the refracted ray is bent toward the normal,

as shown in figure (i)

When light travels from a medium with larger

•

refractive index into a medium with smaller

refractive index, the refracted ray is bent away from

the normal, as shown in figure (ii)

These two possibilities illustrate that both the incident

and refracted rays obey the principle of reversibility

Thus, the directions of the rays in part (i) of the drawing

can be reversed to give the situation depicted in part (ii)

In part (ii) the reflected ray lies in the water rather than

in the air In both parts of figure, the angles of incidence,

refraction, and reflection are measured relative to the

normal

The angle of refraction q2 depends on the angle of

incidence q1 and on the indices of refraction, m2 and m1,

of the two media The relation between these quantities

is known as Snell’s law of refraction

According to Snell’s law, when light travels from a

material with refractive index m1 into a material with

refractive index m2, the refracted ray, the incident ray,

and the normal to the interface between the materials all

lie in the same plane The angle of refraction q2 is related

to the angle of incidence q1 by m1sinq1 = m2sinq2

Normal

rayIncident

ray

11

2 Refracted

ray(i)

Air( )1

Water( )2

Normal

2Air( )1

Refractedray

1ray

Incidentray

(ii)

Refractive Index of a Material Depends Upon Four Factors as Follows

Material : Refractive index is maximum for diamond

Its value is 2.42 It is rated to be one for vacuum or almost unity for air This value of refracting index is minimum

Colour of light : Refractive index of violet colour is

maximum while that for red light is minimum mV > mR.VIBGYOR denotes seven colours in descending order

of refractive index

Thus mV > mI > mB > mG > mY > mO > mR.m

Temperature and density : mr− =1 constant

where r denotes density of material Since density

decreases with rise of temperature, refractive index also decreases with rise of temperature

i.e m is inversely proportional to temperature,

m ∝



 T 1

Surrounding media : Refractive index of medium 2

with respect to medium 1 is represented as 1m2 or m21

and here the ray of light travels from 1 to 2.

Refractive index of medium 1 with respect to medium 2

is denoted by 2m1

2 1

m = velocity of light in watervelocity of light in glass

w g

a g a w

m

= = 3 24 3// =98

 Intensity of light This is due to the partial

reflection and partial absorption of light

apparent Depth

If an object is placed below the surface of water or under

a glass slab, it appears to be raised, i.e the apparent depth

is less than the real depth This is due to refraction

Refractive index (m) can also be given as

m =Apparent depthReal depth

Lateral shift due to slab

The perpendicular distance between the incident ray and

the emergent ray, when the light is incident obliquely

on a parallel sided refracting slab, is called lateral shift

In figure, distance d is lateral shift and is given as

d=cos sin(t r i r t− ), =thickness of slab

total internal reflection

The medium whose refractive index is greater is known

as optically denser medium whereas one which has

smaller value is known as optically rarer medium

In the figure shown, light travels from medium 1 to 2, of

corresponding refractive indices m1and m2 and m1 > m2

For any angle of incidence q1 in medium 1, corresponding angle of refraction in medium 2 is q2

As m1 > m2, sinq2 > sinq1 and hence q2 > q1

In this way for a certain value of q1, q2 will be equal to 90° This value of q1 is known as critical angle (qc) If

q1 > critical angle, q2 should be greater than 90° This implies that the ray will return to the previous medium For the angle of incidence greater than critical angle, light does not transmit into medium 2 rather it reflects back to the medium 1 This phenomenon is called total internal reflection

Applications of Total Internal Reflection

Totally reflecting prisms

refraction at curved surface

For a spherical surface, m2 m1 m2 m1

−The symbols should be carefully remembered as, m2

is refractive index of the medium in which light rays are entering, m1 is refractive index of the medium from which light rays are coming Care should also be taken

while applying the sign convention to R.

lens

Lens is a transparent medium bounded by two surfaces,

at least one of them must be curved

Thin Lens : A lens is said to be thin if the gap between

two surfaces is very small

lens maker’s formula

If the refractive index of the material of the lens

is m, R1 is the radius of curvature of the surface

facing the object and R2that of the other surface then focal length is

f =(m− )R −R 

lens formula

1 1 1

v u f− =

Here, u is the distance of the object from the lens, v is

the distance of the image from the lens and f is the focal

length Values to be introduced with proper sign

limitation of the lens maker’s formula

The lens should be thin so that the separation

•

between the two refracting surfaces is small

The medium on either side of the lens should be

•

same

If any of the limitation is violated, then we have to use the refraction at the curved surface formula for both the surfaces

laws of formation of images by lens

The rays coming parallel to principal axis of lens

•

pass through the focus after refraction

The rays coming from the focus of lens go parallel

•

to the principal axis of lens after refraction

The rays of light passing through optical centre

•

go straight after refraction without changing their path

Image formation by Convex Lens

focus (F2) Real, inverted and extremely diminished

2F2 Real, inverted and diminished

of same size as the object

Power of the lens

Power of lens is, P = 1 f , where f should be in meters with

proper sign i.e., +ve for convex and –ve for concave, P

is in dioptre

combination of the lenses

For a system of lenses, the net power of the system is

P = P1 + P2 + P3 +

provided all the thin lenses are in close contact The

focal length of the net system can be written as

f = f + f + f +

f1, f2, f3 should be taken with proper sign

When a convex and a concave lens of equal focal length

are placed in contact, the equivalent focal length is equal

to infinite Therefore, the power becomes zero

If the lenses are kept at a separation d, then the effective

If the lens is converging, put +ve for its focal length and

–ve for the diverging lens

silvering of lenses

If any surface of a lens is silvered, it will ultimately

behave as a mirror and the power of mirror thus formed

will be equal to the sum of powers of the optical lenses

and the mirrors in between

change of focal length of a lens

When a lens (convex lens) of refractive index m, is kept

in any medium other than air, the focal length is given

is less than that of the lens

mm < m ⇒ f > f0 and P < P0

Therefore, the focal length increases and consequently

the power of the lens decreases The nature of the lens

remains unchanged, since f is still positive, it remains

converging

Case-II :

is equal to that of the lens

i.e., m m = m ⇒ f = ∞ and P f= =1 0Therefore, the focal length of the lens becomes infinity and the power becomes zero

Consequently, the lens behaves like a glass slab (plate)

Case-III :

is greater that of the lens,

i.e., m m > m ⇒ f < f0.Therefore, the focal length of the lens decreases Consequently, the power of the lens increases Numerically,

When a bi-(equi) convex lens is cut transversely into two equal parts, the radius of curvature of the lens in the cutting side increases to infinity Now the focal length

of each part increases to twice the previous value.When a lens is cut into two equal halves parallel to principal axis the focal length of each part remains constant

Prism

A prism has two plane surfaces inclined to each other

Angle of prism, A = r1 + r2

Deviation of incident ray, d = i + e – A

Minimum Deviation : From (i – d) graph, we conclude

that deviation d is minimum, when ray of light passes symmetrically through the prism

2

22

If the angle of prism is very small, sin A ≈ A

White light is actually a mixture of light of different

colours Therefore, when white light gets refracted

under certain conditions, its components bend by

different amounts and separate out This phenomenon

of splitting of light into its component colours, due to

different refractive indices for different colours, is called

where d is deviation of mean ray (yellow)

As, dV = (mV – 1)A, d R = (mR – 1)A,

∴ w=mmV −−mR

Y 1

Here, mY =mV+2mR

Deviation without Dispersion

This means an achromatic combination of two prisms in

which net or resultant dispersion is zero and deviation

is produced For two prisms,

Dispersion without Deviation

A combination of two prisms in which deviation produced for the mean ray by the first prism is equal and opposite to that produced by the second prism is called a direct vision prism This combination produces dispersion without deviation

For deviation to be zero, (d + d′) = 0

⇒ (m – 1)A + (m′ – 1)A′ = 0

⇒ A′ = −((mm′ −−11))A

(–ve sign implies prism A′ has to be kept inverted).

Total angular dispersion

SELFCHECK

5 Monochromatic light is incident on a glass prism of

angle A If the refractive index of the material of the

prism is m, a ray, incident at an angle q, on the face

AB would get transmitted through the face AC of

the prism provided

distance of distinct vision (D) D is about

25 cm to 30 cm

an image of object situated between focus and

optical centre of lens A magnified, erect and virtual

v o and u o refer to objective lens, f e refers to eye-lens

Magnifying power for normal vision

m = Refractive index of medium

q = Angle subtended by the ray with the axis of the

f o and f e respectively refer to focal lengths of objective

and eyepiece D = 25 cm to 30 cm for normal eye.

Magnifying power for normal vision

e

=Maximum length of tube = (

6 A telescope has an objective lens of focal length

150 cm and an eyepiece of focal length 5 cm If a

50 m tall tower at a distance of 1 km is observed through this telescope in normal setting, the angle formed by the image of the tower is q, then q is close to

of propagation of rays Light is fundamentally a wave and in some situations we have to consider its wave properties explicitly If two or more light waves of the same frequency overlap at a point, the total effect depends on the phases of the waves as well as their amplitudes The resulting patterns are as a result of the wave nature of light and can’t be understood on the basis of rays

Wave front : Particles of light wave which are equidistant

from the light source and vibrate in the same phase constitute a wavefront

Depending on the type of light source, we generally have three types of wavefronts spherical, cylindrical, and plane

Spherical wavefront :

of light As we move away from the point source

of light, the radius of the spherical wavefront

increases

Cylindrical wavefront :

is linear, such as a slit, the light propagates in

the medium by forming a cylindrical wavefront

The radius and height of a cylindrical wavefront

increases as we move away from the linear source

of light

Plane wavefront :

source of light, if the point of observation lies far

away from the source, a plane wavefront of light is

obtained

huygens Principle

According to Huygens principle :

Each and every point on the given wavefront,

•

called primary wavefront, acts as a source of new

disturbances, called secondary wavelets, that travel

in all directions with the velocity of light in the

medium

A surface touching these secondary wavelets

•

tangentially in the forward direction at any instant

gives a new wavefront at the instant, which is known

as the secondary wavefront

Let us consider two wavefronts AB, shown in figure, for

understanding Huygens principle

B2

B1B

a

b

c d e

Let us assume that AB is the primary wavefront at any

given instant of time One of them is a plane wavefront and the other is a curved wavefront Let us take points

a, b, c, d and e on the two primary wavefronts AB These

points act as secondary wavelets The distance travelled

by light in Dt seconds is cDt, where c is the speed of light Now with a, b, c, d, and e as centre and radius equal

touching these dotted circles tangentially as shown in figure Since light travels in forward direction, wavefront

A1B1 is the newly formed secondary wavefront Thus

by using Huygens principle, if we know the primary wavefront at any instant, we can calculate the secondary

wavefront after instant t.

change in wavelength and velocity

The frequency of a wave is the characteristic of its source Thus it does not change due to medium Thus the velocity of a wave changes from one medium to another medium due to its change in wavelength In a medium of absolute refractive index m, the velocity of

(a) bends downwards (b) bends upwards(c) becomes narrower(d) goes horizontally without any deflection

(JEE Main 2015)

coherent sources

Two sources are said to be coherent if they emit light waves of the same frequency, nearly the same amplitude and are always in phase with each other It means that the two sources must emit radiations of the same colour (wavelength) Presence of coherent sources is necessary

to observe the phenomenon like interference

Phase difference and path difference : If the path

difference between the two waves coming from two

coherent sources is l (i.e equal to the wavelength of

light coming from the sources) the phase difference is

2p Therefore for a path difference x, the phase difference

d is given as d = 2 x.lp

interference

The term interference refers to any situation in which

two or more waves overlap in space When this occur,

the total wave at any point at any instant of time is

governed by the principle of superposition, which states

that, when two or more waves overlap, the resultant

displacement at any point and at any instant may be

found by adding the instantaneous displacements that

would be produced at the point by the individual waves

if each were present alone

Interference effects are most easily seen when we

combine sinusoidal waves with a single frequency

u and wavelength l In optics, sinusoidal waves are

characteristic of monochromatic light (light of a single

colour) as we analyze characteristic of interference

constructive and Destructive interference

When waves from two or more sources arrive at a point

in phase, the amplitude of resultant wave is the sum of

the amplitudes of the individual wave, the individual

waves reinforce each other This is called constructive

interference

A crest of one wave arrives at the same time a trough of

the other wave The resultant amplitude is the difference

between the two individual amplitudes If the individual

amplitudes are equal, then the total amplitude is zero

This cancellation or partial cancellation of the individual

waves is called destructive interference

young’s Double slit Experiment

Here two coherent sources

Screen

are obtained by allowing

light from a source to pass

and S2 separated by a small

distance d, and equidistant

from a screen

For constructive interference (i.e for the formation of

bright fringes)

Path difference for

x n=nl=y dD n

where n = 1, 2, 3, , l = wavelength of light,

on the screen

d = separation between the two coherent sources.

D = perpendicular distance between screen and the two

coherent sources

Phase difference

For central maximum,

difference and phase difference are zero

Distance between two consecutive bright

= lD d

For destructive interference (i.e for the formation of

dark fringes)Path difference for

x n=(2 1 2n− )l=y dD n (where n = 1, 2, )

from centre of the interference pattern x n is an odd

number multiple of half wavelength

Phase difference

• dn=2lpx n=2pn−12 Distance of

pattern is given as

Alternately formed bright and dark fringes are parallel

Thus b is proportional to l and D and inversely proportional to d.

If the light waves coming from the two coherent

sources be of equal amplitude a then

If phase difference d = 0, 2p, 4p,

and x = 0, l, 2l Then, Intensity I = 4a2

Intensity is maximum for bright fringes

When phase difference is d = p, 3p,

and path difference is x = l/2, 3l/2,

I = 0 (i.e Intensity is minimum.)

Formation of interference fringes is in accordance with law of conservation of energy

intensity Distribution

If a, b are the amplitudes of interfering waves due

to two coherent sources and f is constant phase

difference between the two waves at any point P, then the resultant amplitude at P will be

R= a2+ +b2 2 cosfab

If a2 = I1, b2 = I2, then

Resultant intensity I = R2 = a2 + b2 + 2 ab cosf

I I= + +1 I2 2 I I1 2 cosfWhen cosf =1;Imax= + +I1 I2 2 I I1 2 =( I1+ I2)2

When cosf = −1,Imin=( I1 − I2)2

If the sources are incoherent, I = I1 + I2

conditions for sustained interference Pattern

Superimposing light waves should be from

The angle at which superimposing waves are

•

inclined should not be very large

Interfering waves should be in same state of

8 In a Young’s double slit experiment with light of

wavelength l the separation of slits is d and distance

of screen is D such that D >> d >> l If the fringe

width is b, the distance from point of maximum

intensity to the point where intensity falls to half of

maximum intensity on either side is

interference in thin films

The coloured pattern formed by a thin film on water

surface is due to the interference pattern between two

beams One is reflected from the top surface of the film

and the other from the bottom surface

For reflected light :

•

The correct path difference x=2mtcosr−l2

t = thickness of film; m = refractive index of the film,

r = angle at which light is refracted in film.

For bright fringes : 2mtcosr=(2 1 2n− )l

Actual path difference x = 2mtcosr

For bright fringes : 2mtcosr = nl

No opticals are required Opticals are in the form of collimating lens and

focusing lens are required.Fringes are not sharp

and well defined Fringes are sharp and well defined.Greater the wavelength of a wave, higher will be its degree

of diffraction All types of waves exhibit diffraction Appearance of a shining circle around the section of sun just before sun rise is due to diffraction of light When

an intense source of light is viewed with the partially opened eye, colours are observed in the light It is due

to diffraction

A single slit of width a gives a diffraction pattern with

a central maximum The intensity falls to zero at angles

of ± ±la, 2 etc., with successively weaker secondary almaxima in between