Somnath Bhardwaj Lect_notes

Physics I: Oscillations and Wvaes

Somnath Bharadwaj

Department of Physics and MeteorologyIIT Kharagpur, 721 302, India

February 16, 2007

Contents

1 Oscillations 5

1.1 Simple Harmonic Oscillators SHO . . . . . . . . . . . . . . . . 51.2 Complex Representation. . . . . . . . . . . . . . . . . . . . . . 71.3 Energy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.4 Why study the SHO? . . . . . . . . . . . . . . . . . . . . . . . 10

2 The Damped Oscilator. 15

2.0.1 Underdamped Oscillations . . . . . . . . . . . . . . . . 162.0.2 Over-damped Oscillations. . . . . . . . . . . . . . . . . 172.0.3 Critical Damping. . . . . . . . . . . . . . . . . . . . . . 182.0.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . 19

3 Oscillator with external forcing. 21

3.0.5 Effect of damping . . . . . . . . . . . . . . . . . . . . . 23

4 Resonance. 29

4.0.6 Electrical Circuits . . . . . . . . . . . . . . . . . . . . . 294.0.7 The Raman Effect . . . . . . . . . . . . . . . . . . . . 31

5 Coupled Oscillators 35

5.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

6 Sinusoidal Waves. 41

6.1 What is a(x, t)? . . . . . . . . . . . . . . . . . . . . . . . . . . 416.2 Angular frequency and wave number . . . . . . . . . . . . . . 426.3 Phase velocity. . . . . . . . . . . . . . . . . . . . . . . . . . . 436.4 Waves in three dimensions. . . . . . . . . . . . . . . . . . . . . 456.5 Waves in an arbitrary direction. . . . . . . . . . . . . . . . . . 46

3

4 CONTENTS

7 Electromagnetic Waves. 47

7.1 Electromagnetic Radiation. . . . . . . . . . . . . . . . . . . . 477.2 Electric dipole radiation. . . . . . . . . . . . . . . . . . . . . . 507.3 Sinusoidal Oscillations. . . . . . . . . . . . . . . . . . . . . . . 537.4 Energy density, flux and power. . . . . . . . . . . . . . . . . . 55

8 The vector nature of electromagnetic radiation. 61

9 The Electromagnetic Spectrum 65

9.1 Radiowaves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 659.2 Microwave . . . . . . . . . . . . . . . . . . . . . . . . . . . . 659.3 21cm radiation. . . . . . . . . . . . . . . . . . . . . . . . . . . 669.4 Cosmic Microwave Background Radiation. . . . . . . . . . . . 669.5 Molecular lines. . . . . . . . . . . . . . . . . . . . . . . . . . . 679.6 Infrared . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 679.7 Visible light . . . . . . . . . . . . . . . . . . . . . . . . . . . . 689.8 Ultra Violet . . . . . . . . . . . . . . . . . . . . . . . . . . . 689.9 X-rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 699.10 Gamma Rays . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

10 Interference. 71

10.1 Young’s Double Slit Experiment. . . . . . . . . . . . . . . . . 7110.1.1 A different method of analysis. . . . . . . . . . . . . . 7410.1.2 Effect of Finite Source Size. . . . . . . . . . . . . . . . 76

10.2 The Michelson Interferometer . . . . . . . . . . . . . . . . . . 77

11 Coherence. 85

Chapter 1

Oscillations

Oscillations are ubiqutous. In fact it would be difficult to find somethingwhich never exhibits oscillations. Atoms in solids, electromagnetic fields,multi-storeyed buildings and share prices all exhibit oscillations. In thiscourse we shall restrict our attention to only the simplest possible situations.The elementary analysis considered here provides insights into a diverse va-riety of phenomena and examples shall be provided as we proceed.

1.1 Simple Harmonic Oscillators SHO

We consider the spring-mass system shown in Figure 1.1. A massless spring,one of whose ends is fixed has its other attached to a particle of mass m whichis free to move. We choose the origin x = 0 for the particle’s motion at theposition where the spring is unstretched. The particle is in stable equilibriumat the origin, and it will continue to remain there if left at rest. We areinterested in a situation where the particle is disturbed from equilibrium. Theparticle experiences a restoring force from the spring if it is either stretched orcompressed. The spring is assumed to be elastic which means that it followsHooke’s law where the force is proportional to the displacement F = −kxwith spring constant k.

The particle’s equation of motion is

md2x

dt2= −kx (1.1)

which can be written asx+ ω2

0x = 0 (1.2)

5

6 CHAPTER 1. OSCILLATIONS

m

k

k

m

x

Figure 1.1:

where the dots¨denote time derivatives and

ω0 =

√k

m(1.3)

It is straightforward to check that

x(t) = A cos(ω0t + φ) (1.4)

is a solution to eq. (1.2).We see that the particle performs sinusoidal oscillations around the equi-

librium position when it is disturbed from equilibrium. The angular fre-quency ω0 of the oscillation depends on the intrinsic properties of the oscil-lator. It determines the time period

T =2π

ω0(1.5)

and the frequency ν = 1/T of the oscillation. Figure 1.2 shows oscillationsfor two different values of ω0.

Problem 1.: What are the values of ω0 for the oscillations shown in Fig-ure 1.2. What is k if m = 1 kg?

The amplitude A and phase φ are determined by the initial conditions.Two initial conditions are needed to completely specify a solution. Thisfollows from the fact that the governing equation (1.2) is a second orderdifferential equation. The initial conditions can be specified in a variety of

1.2. COMPLEX REPRESENTATION. 7

x

t

B

A−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

0 0.5 1 1.5 2 2.5 3−1

−0.5

0

0.5

1

0 0.5 1 1.5 2 2.5 3

Figure 1.2:

t

x

B

A

−1.5

−1

−0.5

0

0.5

1

1.5

0 0.5 1 1.5 2 2.5 3

Figure 1.3:

ways, fixing the values of x(t and x(t) at t = 0 is a possibility. Figure 1.3shows oscillations with different amplitudes and phases.

Problem 2.: What are the amplitude and phase of the oscillations shownin Figure 1.3?

1.2 Complex Representation.

Complex number are very useful in representing oscillations. The amplitudeand phase of the oscillation can be combined into a single complex number,namely the complexamplitude

A = Aeiφ . (1.6)


Note that we have introduced the symbol ˜ (tilde) to denote complex num-bers. The property that

eiφ = cos φ+ i sin φ (1.7)

allows us to represent any oscillating equantity x(t) = A cos(ω0t + φ) as thereal part of the complex number

x(t) = Aei(ω0t+φ) = A[cos(ω0t + φ) + i sin(ω0t+ φ)] (1.8)

We calculate the velocity v in the complex representation v = ˙x. whichgives us

v(t) = iω0x = −ω0A[sin(ω0t+ φ) − i cos(ω0t+ φ)] . (1.9)

Taking only the real part we calculate the particle’s velocity

v(t) = −ω0A sin(ω0t + φ) . (1.10)

The complex representation is a very powerful tool which, as we shall seelater, allows us to deal with oscillating quantities in a very elegant fashion.

Problem 3.: A SHO has initial position x0 and velocity v0 at t = 0. Showthat the oscillator’s motion is represented by x(t) = (x0 − iv0/ω0)e

iω0t. Whatare the amplitude and phase of the oscillator?

1.3 Energy.

A spring-mass system has a potential energy V (x) = kx2/2 (Figure 1.4).This energy is stored in the spring when it is either compressed or stretched.The potential energy of the system

U =1

4kA2 cos2(ω0t+ φ) =

1

4mω2

0A21 + cos[2(ω0t+ φ)] (1.11)

oscillates with angular frequency 2ω0 as the spring is alternately compressedand stretched. The kinetc energy mv2/2

T =1

2mω2

0A2 sin2(ω0t + φ) =

1

4mω2

0A21 − cos[2(ω0t+ φ)] (1.12)

salso has similar oscillations which are exactly π out of phase.

1.3. ENERGY. 9

V(x)

x

Figure 1.4:

In a spring-mass system the total energy oscillates between the potentialenergy of the spring (U) and the kinetic energy of the mass (T ). The totalenergy E = T + U has a value E = mω2

0A2/2 which remains constant.

The average of value of an oscillating quantity is often of interest. Theaverage of any time varying quantity Q(t), denoted using 〈Q〉 is defined as

〈Q〉 = limT→∞

1

T

∫ T/2

−T/2Q(t)dt . (1.13)

The basic idea here is to average over a time interval T which is significantlylarger than the oscillation time period.

It is very useful to remember that 〈cos(ω0t+ φ)〉 = 0. This can be easilyverified by noteing that the values sin(ω0t+ φ) are bound between −1 and

+1. We use this to calculate the average kinetic and potential energieswhich have the same values

〈U〉 = 〈T 〉 =1

4mω2

0A2 (1.14)

The average kinetic and potential energies and the total energy are veryconveniently expressed in the complex representation as

〈U〉 = 〈T 〉 =1

4mvv∗ =

1

4kxx∗ (1.15)

where ∗ denotes the conjugate of a complex number.Problem 3.: The mean displacement of a SHO 〈x〉 is zero. The root mean

square (rms.) displacement√〈x2〉 is useful in quantifying the amplitude of

oscillation. Show that the rms. displacement is√xx∗/2.


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Atomic Vibrations in a Crystal

333333333333333444444444444444555555

Figure 1.5:

1.4 Why study the SHO?

What happens to a system when it is disturbed from stable equilibrium?This is a question that arises in a large variety of situations. For exam-ple, the atoms in many solids (eg. NACl, diamond, steel) are arranged ina periodic crystal as shown in Figure 1.5. The periodic crystal is known tobe an equilibrium configuration of the atoms. The atoms are continuouslydisturbed from their equilibrium positions (shown in Figure 1.5) as a conse-quence of random thermal motions and external forces which may happento act on the solid. The study of oscillations of the atoms disturbed fromtheir equilibrium position is very interesting. In fact the oscillations of thedifferent atoms are coupled, and this gives rise to collective vaibrations ofthe whole crystal which can explain properties like the specific heat capacityof the solid. We shall come back to this later, right now the crucial pointis that each atom behaves like a SHO if we assume that all the other atomsremain fixed. This is generic to all systems which are slightly disturbed fromstatble equilibrium.

We now show that any potential V (x) is well represented by a SHOpotential in the neighbourhood of points of stable equilibrium. The originof x is chosen so that the point of stable equilibrium is located at x = 0.For small values of x it is possible to approximate the function V (x) using aTaylor series

V (x) ≈ V (x)x=0 +

(dV (x)

dx

)

x=0

x +1

2

(d2V (x)

dx2

)

x=0

x2 + ... (1.16)

1.4. WHY STUDY THE SHO? 11

V(x) = x

V(x) = exp(x /2) −1

V(x) = x − x2 3

2

2

x

V(x)

Figure 1.6:

where the higher powers of x are assumed to be negligibly small. We knowthat at the points of stable equilibrium the force vanishes F = −dV (x)/dx =0 and V (x) has a minima

k =

(d2V (x)

dx2

)

x=0

> 0 . (1.17)

This tells us that the potential is approximately

V (x) ≈ V (x)x=0 +1

2kx2 (1.18)

which is a SHO potential. Figure 1.6 shows two different potentials whichare well approximated by the same SHO potentail in the neighbourhoodof the point of stable equilibrium. The oscillation fequency is exactly thesame for particles slightly disturbed from equilibrium in these three differentpotentials.

The study of SHO is important because it occurs in a large variety of sit-uations where the system is slightly disturbed from equilibrium. We discussa few simple situations.

Simple pendulum

The simple possible shown in Figure 1.7 is possibly familiar to all of us. Amass m is suspended by a rigid rod of length l, the rod is assumed to bemassless. The gravitations potential energy of the mass is

V (θ) = mgl[1 − cos θ] . (1.19)


676767676676767676676767676676767676878787878878787878878787878878787878

m

θ

l

g

Figure 1.7:

For small θ we may approximate cos θ ≈ 1 − θ2/2 whereby the potentialis

V (θ) =1

2mglθ2 (1.20)

which is the SHO potential. Here dV (θ)/dθ gives the torque not the force.The pendulum’s equation of motion is

Iθ = −mglθ (1.21)

where I = mgl2 is the moment of inertia. This can be written as

θ +g

lθ = 0 (1.22)

which allows us to determine the angular frequency

ω0 =

√g

l(1.23)

LC Oscillator

The LC circiut shown in Figure 1.8 is an example of an electrical circuitwhich is a SHO. It is governed by the equation

LI +Q

C= 0 (1.24)

where L refers to the inductance, C capacitance, I current and Q charge.This can be written as

Q+1

LCQ = 0 (1.25)

1.4. WHY STUDY THE SHO? 13

i

C L

q

Figure 1.8:

which allows us to identify

ω0 =

√1

LC(1.26)

as the angular frequency.

Chapter 2

The Damped Oscilator.

Damping usually comes into play whenever we consider any motion. Westudy the effect of damping on the spring-mass system. The damping forceis assumed to be proportonal to the velocity, acting to oppose the motion.The total force acting on the mass is

F = −kx− cx (2.1)

and the equation of motion is

mx = −kx− cx (2.2)

where c is a proportionality constant. Recasting this in terms of more con-venient coefficients we have

x + 2βx+ ω20x = 0 (2.3)

This is a second order homogeneous equation with constant coefficients. Weproceed to solve this by taking a trial solution

x(t) = Aeαt . (2.4)

Putiing the trial solution into equation (2.4) gives us the quadratic equation

α2 + 2βα+ ω20 = 0 (2.5)

This has two solutionsα1 = −β +

√β2 − ω2

0 (2.6)

15

16 CHAPTER 2. THE DAMPED OSCILATOR.

andα2 = −β −

√β2 − ω2

0 (2.7)

The nature of the solution depends critically on the value of the dampingcoefficient β, and the behavour is quite different depending on whether β <ω0, β = ω0 or β > ω0.

2.0.1 Underdamped Oscillations

We first consider the situation where β < ω0 which which is referred to asunderdamped. Defining

ω =√ω2

0 − β2 (2.8)

the two roots which are both complex have values

α1 = −β + iω and α2 = −β − iω (2.9)

The resulting solution is a superposition of the two roots

x(t) = e−βt[A1eiωt + A2e

−iωt] (2.10)

where A1 and A2 are constants which have to be determined from the initialconditions. The term [A1e

iωt+A1eiωt] is a superposition of sin and cos which

can be written asx(t) = Ae−βt cos(ωt+ φ) (2.11)

This can also be expressed in the complex notation as

x(t) = Ae(iω−β)t (2.12)

where A = Aeiφ is the complex amplitude which has both the amplitudeand phase information. Figure 2.1 shows the underdamped solution x(t) =e−t cos(20 t).

In all cases damping reduces the frequency of the oscillator ie. ω < ω0.The main effect of damping is that it causes the amplitude of the oscillationto fall exponentially with time. It is often useful to quantify how muchthe amplitude decays during the time period T = 2π/ω corresponding toa single oscillation. This is quantified through the logarithmic decrement λwhich defined as

λ = ln

[x(t)

x(t + T )

]=

2πβ

ω(2.13)

17

t

x

x(t)= e cos(20 t)−t

−1

−0.5

0

0.5

1

0 1 2 3 4 5

Figure 2.1:

Problem 4.: An under-damped oscillator with x(t) = Ae(iω−β)t has initialdisplacement and velocity x0 and v0 respectively. Calculate A and obtainx(t) in terms of the initial conditions.

2.0.2 Over-damped Oscillations.

This refers to the situation where

β > ω0 (2.14)

The two roots areα1 = −β +

√β2 − ω2

0 = −γ1 (2.15)

and

α2 = −β −√β2 − ω2

0 = −γ2 (2.16)

where both γ1, γ2 > 0 and γ2 > γ1. The two roots give rise to exponentiallydecaying solutions, one which decays faster than the orther

x(t) = A1e−γ1t + A2e

−γ2t . (2.17)

The constants A1 and A2 are determined by the initial conditions. For initialposition x0 and velocity v0 we have

x(t) =v0 + γ2x0

γ2 − γ1e−γ1t −

v0 + γ1x0

γ2 − γ1e−γ2t (2.18)


Overdamped

Critically damped

0

0.2

0.4

0.6

0.8

1

0 0.5 1 1.5 2 2.5 3

Figure 2.2:

The overdamped oscillator does not oscillate. Figure 2.2 shows a typicalsituation.

In the situation where β ω0

√β2 − ω2

0 = β

√

1 −ω2

β2≈ β

[1 −

1

2

ω2

β2

](2.19)

and we have γ1 = ω20/2β and γ2 = −2β.

2.0.3 Critical Damping.

This corresponds to a situation where β = ω0 and the two roots are equal.The governing equation is second order and there still are two independentsolutions. The general solution is

x(t) = e−βt[A1 + A2t] (2.20)

The solutionx(t) = x0e

−βt[1 + βt] (2.21)

is for an oscillator starting from rest at x0 while

x(t) = v0e−βtt (2.22)

is for a particle starting from x = 0 with spped v0. Figure 2.2 shows thelatter situation.

19

Overdamped

γ

γ

γUnderdamped

1

2

Critical

β

0

1

2

3

4

5

6

0 0.5 1 1.5 2 2.5 3

Figure 2.3:

2.0.4 Summary

There are two physical effects at play in a damped oscillator. The first isthe damping which tries to bring any motion to a stop. This operates on atime-scale Td ≈ 1/β. The restoring force exerted by the spring tries to makethe system oscillate and this operates on a time-scale T0 = 1/ω0. We haveoverdamped oscillations if the damping operates on a shorter time-scale thanoscillations Td < T0 which completely destroys the oscillatory behaviour.

Figure 2.3 shows the behaviour of a damped oscillator uder different com-binations of damping and restoring force. The plot is for ω0 = 1, it can beused for any other value of the natural frequency by suitably scaling thevalues of β. It shows how the decay rate for the two exponentially decayingoverdamped solutions varies with β. Note that for one of the modes thedecay rate tends to zero as β is increased. This indicates that for very largedamping a particle may get stuck at a position away from equilibrium.

Chapter 3

Oscillator with external forcing.

We consider the motion of an oscillator under the influence of an externalsinusoidal force F = cos(ωt + ψ). Why consider this particular form of theforce? This is because an arbitrary time varying force F (t) can always bedecomposed into the sum of sinusoidal forces of different frequencies

F (t) =∑

ω

Fω cos(ωt+ ψω) (3.1)

Here Fω and ψω are respectively the amplitude and phase of the differentfrequency components. Such an expansion is called a Fourier Series. Thebehaviour of the oscillator under the influence of the force F (t) can be de-termined by separately solving

mxω + kxω = Fω cos(ωt+ ψω) (3.2)

for a force with a single frequency and then superposing the solutions

x(t) =∑

ω

xω(t) . (3.3)

Let us focus our attention on equation (3.2) which has a sinusoidal force ofa single frequency and drop the subscript ω from xω and Fω. It is convenientto switch over to the complex notation

¨x + ω20x = f eiωt (3.4)

where f = Feiψ/m.

21

22 CHAPTER 3. OSCILLATOR WITH EXTERNAL FORCING.

ω

Resonance

π PHASE

AMPLITUDE

−φ

|x|

f= ω 0

=1

ω 02

2ωf/

f/

0

2

4

6

8

10

0 0.5 1 1.5 2

0

Figure 3.1:

The solution is a sum of two parts, the complementary function and theparticular integral. The complementary function is a solution to the homo-geneous equation without the external force. This, as we already know, isx(t) = Aeiω0t an oscillating function of frequency ω0. The particular integralis the extra ingredient in the new solution. This arises solely due to theexternal force. It is obvious that the particular integral is of the form

x(t) = Beiωt (3.5)

whereby equation (3.4) becomes

[−ω2 + ω20]B = f (3.6)

which gives a solution

x(t) =f

ω20 − ω2

eiωt (3.7)

The solution has an amplitude

| x |=f

| ω20 − ω2 |

. (3.8)

The phase of the oscillations relative to the applied force si is φ = 0 forω < ω0 and φ = −π for ω > ω0. The amplitude and phase are shown inFigure 3.1.

The first point to note is that the amplitude increases dramatically asω → ω0 and the amplitude blows up at ω = ω0. This is the phenomenon

23

of resonance. Under the external of an external force, the response of theoscillaotr is maximum when the frequency of the external force matches thenatural frequency of the oscillator. In a real situation the amplitude is regu-lated by the presence of damping which ensures that it does not blow up toinfinity at ω = ω0.

We next consider the low frequency ω ω0 behaviour

x(t) =f

ω20

eiωt =F

kei(ωt+ψ) , (3.9)

The oscillations have an amplitude F/k and are in phase with the externalforce.

This behaviour is easy to understan if we consider ω = 0 which is aconstant force. We know that the spring gets extended (or contracted) byan amount x = F/k in the direction of the force. The same behaviour goesthrough if F varies very slowly with time. The behaviour is solely determinedby the spring constant k and this is referred to as the “Stiffness Controlled”regime.

At high frequencies ω ω0

x(t) = −f

ω2eiωt = −

F

mω2ei(ωt+ψ) , (3.10)

the amplitude is F/m and the oscillations are −π out of phase with respectto the force. This is the “Mass Controlled” regime where the spring does notcome into the picture at all. It is straight forward to verify that equation(3.10) is a solution to

mx = F cos(ωt+ ψ) (3.11)

when the spring is removed from the oscillator. Interestingly such a particlemoves exactly out of phase relative to the applied force. The particle movesto the left when the force acts to the right and vice versa.

3.0.5 Effect of damping

Introducing damping, the equation of motion

mx + cx + kx = F cos(ωt+ ψ) (3.12)

written using the notation introduced earlier is

¨x+ 2βx+ ω20x = feiωt . (3.13)


−φ

π

f= =1ω0β=0.2

1.0

0.6

0.4

2.0

β=0.20.6

ω

|x|

0

0.5

1

1.5

2

2.5

3

0 0.5 1 1.5 2 2.5 3 0

0.5

1

1.5

2

2.5

3

0 0.5 1 1.5 2 2.5 3 0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2 2.5 3 0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2 2.5 3 0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2 2.5 3 0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2 2.5 3

0

2

Figure 3.2:

Here again we separately discuss the complementary functions and the par-ticular integral. The complementary functions are the decaying solutionswhich, as discussed earlier, arise in the absence of the external force. Theseare shortlived transients which are not of interest when studying the longtime behaviour of the oscillations. The particular integral is important whenstudying the long time or steady state response of the oscillator. This solu-tion is

x(t) =f

(ω20 − ω2) + 2iβω

eiωt (3.14)

which may be written as x(t) = Ceiφf eiωt where φ is the phase of the oscil-lation relative to the force f .

This has an amplitude

| x |=f√

(ω20 − ω2)2 + 4β2ω2

(3.15)

and the phase φ is

φ = tan−1

(−2βω

ω20 − ω2

)(3.16)

Figure 3.2 shows the amplitude and phase for different values of the damp-ing coefficient β. The damping ensures that the amplitude is finite for allvalues of ω. Also the change in the phase is more gradual.

25

The low frequency and high frequency behaviour are exactly the sameas the undamped situation. The changes due to damping are mainly in thevicinity of ω = ω0. The amplitude is maximum at

ω =√ω2

0 − 2β2 (3.17)

For mild damping (β ω0) this is approximately ω = ω0.We next shift our attention to the energy of oscillator. The average

energy E is the quantity of interest. As we have shown earlier this can becalculated as E = kxx∗/2. Calculating this as a function of the frequency ofthe externalforce we have

E(ω) =k

2

f 2

(ω20 − ω2)2 + 4β2ω2

(3.18)

In the mild damping limit β ω0, for frequencies close to ω0 we have

(ω20 − ω2)2 = (ω0 + ω)2(ω0 − ω)2 ≈ 4ω2

0(ω0 − ω)2 (3.19)

which gives

E(ω) ≈k

8

f 2

ω20[(ω0 − ω)2 + β2]

(3.20)

This has a maxima at ω ≈ ω0 and the maximum value is

Emax =kf 2

8ω20β

2. (3.21)

We want to find ∆ω so that E(ω0 + ∆ω) = Emax/2 ie. half the maximumvalue. Using equation (3.20) we see that this requires ∆ω = β. The FWHM(Full Width at Half Maxima) is defined as 2∆ω and it has a value FWHM= 2β. as shown in Figure 3.3. The FWHM quantifies the width of the curveand it records the fact that the width increases with the damping coefficientβ.

We finally consider the power drawn by the oscillator from the externalforce P (t) = F (t)x(t)

P (t) = [F cos(ωt)][− | x | ω sin(ωt+ φ)] (3.22)

The average power is the quantity of interest, we study this as a functionof the frequency. Calculating this wehave

〈P 〉(ω) = −ωF | x | sin(φ)

2. (3.23)


FWHM

ω

ωE( )

0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2

Figure 3.3:

Using equation (3.14) we have

| x | sin φ =−2βω

(ω20 − ω2)2 + 4β2ω2

(F

m

)(3.24)

which gives the average power

〈P 〉(ω) =βω2

(ω20 − ω2)2 + 4β2ω2

(F 2

m

). (3.25)

Figure 3.4 shows the average power as a function of the frequency. Thedissipated power peaks at ω = ω + 0. The mild damping situation withβ ω0 is of considerable interest. In this limit the power falls quite fastaway from ω0 and we can restrict our analysis to the neighbourhood of thisvalue ie. ω ≈ ω0. We can then us eq. (3.19 ) which gives is

〈P 〉(ω) =β

(ω0 − ω)2 + β2

(F 2

4m

). (3.26)

Such a profile occurs in a large variety of situations where we have reso-nance and it is referred to as a Lorentzian profile. It has a peak at ω = ω0

and its width is determined by the value of β, the FWHM being 2β.

27

FWHM

ω

P( )ω

0

0.5

1

1.5

2

0 0.5 1 1.5 2

Figure 3.4:

Chapter 4

Resonance.

4.0.6 Electrical Circuits

Electrical Circuits are the most common technological application where wesee resonances. The LCR circuit shown in Figure 4.1 characterizes the typicalsituation. The circuits includes a signal generator which produces an ACsignal of voltage amplitude V and angular frequency ω. Applying Kirchoff’sLaw to this circuit we have

LdI

dt+q

C+RI = V cos(ωt+ φ) (4.1)

which may be written solely in terms of the charge q as

Lq +Rq +q

C= V cos(ωt+ φ) (4.2)

We see that this is a damped oscillator with an external sinusoidal force. Theequation governing this is

¨q + 2β ˙q + ω20 q = veiωt (4.3)

where ω20 = 1/LC, β = R/2L and v = (V/L) eiφ.

We consider the power that the circuit draws from the signal generator.The resistance is the only circuit element which draws power. We proceedto calculate the dissipated power by calculating the impedence

Z(ω) = iωL−i

ωC+R (4.4)

29

30 CHAPTER 4. RESONANCE.

C

RL

V

Figure 4.1:

which varies with frequency. The voltage and current are related as V = IZ,which gives the current

I =V

i(ωL− 1/ωC) +R. (4.5)

The average power dissipated may be calculated as 〈P (ω)〉 = RII∗/2. Wefind that the power drawn from the signal generator, as a function of fre-quency, is described by a a Lorentzian profile.

〈P (ω)〉 =ω2

(ω20 − ω2)2 + 4β2ω2

(RV 2

2L2

)(4.6)

A large variety of electrical devices are quite adequately modeled as LCRcircuits. For exaple, consider a speaker which is attached to a music system.The speaker can be modeled as a combbination of a resistance, inductanceand capacitance. The speaker dissipates power in two forms namely soundand heat, both of these are included in the resistance. The music systemproduces signals across nearly the entire audible range 20 Hz to 20, 000 KHz,and this is fed into the speaker. The speaker should be designed so as tohave nearly equal response at all frequencies across the entire audible band.Our study of the LCR circuit show that the response will be sharply peakedin a narrow frequency range if the damping is low. It is necessary to have alarge damping if the speaker is to respond to a large frequency range.

Problem For an Electrical Oscillator with L = 10mH and C = 1µF ,

31

a. what is the natural (angular) frequency ω0? (10KHz)

b. choose R so that the oscillator is critically damped. (200 Ω)

c. for R = 2 Ω, what is the maximum power that can be drawn from a10V source? (25W )

d. what is the FWHM of the peak? (200Hz)

e. at what frequency is half the maximum power drawn? (10.1KHz and9.9KHz)

f. what is the value of the quality factor Q? (Q = ω0/2β = 50 )

g. what is the time period of the oscillator? (T = 2π/ω where ω =

ω√

1 −2β /ω

20 ≈ 10KHz and T = 2π10−4sec.)

h. what is the value of the log decrement λ? (x(t) = [Ae−βt]eiωt , λ =ln(xn/xn+1) = βT = 2π10−2)

4.0.7 The Raman Effect

We consider the Raman effect as an example of a situation where we seea resonance. The Raman effect, discovered by C.V. Raman and K.S. Kr-ishnan in 1928, is one of the most important scientific discoveries made inIndia in recent times. Raman and Krishnan found that if a beam of veryintense light which is nearly monochromatic at a frequency ν is incident ona liquid or a gas, the scattered light, emitted in directions different fromthe incident direction, has components at two frequencies different from theincident frequency. The components at the two new frequencies ν − ∆ν andν + ∆ν known as the Stokes and anti-Stokes lines respectively (Figure 4.2).Sir. C.V. Raman received the Nobel prize for this discovery which is knownas the Raman effect.

As an example, we consider light of frequency ν = 6.0× 1014Hz incidenton benzene C6H6 which is aliquid. Figure 4.3 shows the the Stokes lines thatare produced in this situation. The three different Stokes lines can be asso-ciated with three different modes of vibration of the benzene molecule whosestructure is shown in Figure 4.4. As we have discussed earlier, the hydro-gen and carbon atoms in the benzene molecule experience simple harmonicforces F = −kx if they are displaced from their equilibrium positions. The


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Incident

ν

Scattere

dν Rayleigh

Anti−Stokesν+∆ν

ν−∆ν Stokes

Figure 4.2:

vibrations of a benzene molecule can be analysed in terms of coupled simpleharmonic oscillators. As we shall discuss later, the oscillations of coupledsimple harmonic oscillators can be decomposed into normal modes, each ofwhich oscillates like an uncoupled simple harmonic oscillator. For the ben-zene molecule, in the Breating Mode the whole benzene molecule expandsand contracts maintaining its original structure. In the Bending Mode pairsof hydrogen atoms alternately approach one another and move away, whilein the C-C Stretching or Kekule Mode pairs of carbon atoms alternatelyapproach one another and move away. Each of these modes of vibrationhave their own characteristic frequency. When a benzene molecule interactswith the incident light some of the incident energy can be used to excitethe vibrational modes of the molecule, this gives rise to the Stokes lines.The anti-Stokes lines arise due to the transfer of energy from the vibrationalmodes to the radiation. The exchange of energy between the vibrationalmodes and the radiation is a quantum mechanical phenomena requiring usto interpret the radiation in terms of wave-packets called photons.

A closer look at the spectral lines which arise in Raman scattering showsthem to have a finite width with a Lorentzian frequency profile. Figure4.5 shows the Raman line corresponding to the Kekule mode of benzene.It is quite natural to interpret the Lorentzian line profile in terms of theresonance of a damped simple harmonic oscillator, and surprisingly such aninterpretation provides a very simple and useful model for the interaction ofthe vibrational modes of the benzene molecule with radiation. Indeed, theresonance of a damped simple harmonic oscillator is an adequate working

33

Figure 4.3:

Figure 4.4:


ht

Figure 4.5:

model for a large variety of spectral lines including those arising in atomictransitions.

Problem For the Raman line shown in Figure 4.2

a. What is the natural frequency and the corresponding ω0?

b. what is the FWHM?

c. what is the value of the quality factor Q?

Chapter 5

Coupled Oscillators

Consider two idential simple harmonic oscillators of mass m and spring con-stant k as shown in Figure 5.1 (a.). The two oscillators are independentwith

x0(t) = a0 cos(ωt+ φ0) (5.1)

andx1(t) = a1 cos(ωt+ φ1) (5.2)

where they both oscillate with the same frequency ω =√

km

. The amplitudesa0, a1 and the phases φ0, φ1 of the two oscillators are in no way interdepen-dent. The question which we take up for discussion here is what happens ifthe two masses are coupled by a third spring as shown in Figure 1 (b.).

The motion of the two oscillators is now coupled through the third springof spring constant k

′

. It is clear that the oscillation of one oscillator affectsthe second. The phases and amplitudes of the two oscillators are no longerindependent and the frequency of oscillation is also modified. We proceed tocalculate these effects below.

The equations governing the coupled oscillators are

md2 x0

dt2= −kx0 − k

′

(x0 − x1) (5.3)

and

md2 x1

dt2= −kx1 − k

′

(x1 − x0) (5.4)

The technique to solve such coupled differential equations is to identifylinear combinations of x0 and x1 for which the equations become decoupled.

35

36 CHAPTER 5. COUPLED OSCILLATORS

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k m km

x0 1x

(a.)

(b.)

k’

k m km

x0 1x

Figure 5.1: This shows two identical spring-mass systems. In (a.) the twooscillators are independent whereas in (b.) they are coupled through an extraspring.

In this case it is very easy to identify such variables

q0 =x0 + x1

2and q1 =

x0 − x1

2. (5.5)

These are referred to as as the normal modes of the system and the equationsgoverning them are

md2 q0dt2

= −kq0 (5.6)

and

md2 q1dt2

= −(k + 2k′

)q1 . (5.7)

In this case the normal modes lend themselves to a simple physical interpre-tation where.

The nornal mode q0 represents the center of mass. The center of massbehaves as if it were a particle ofmass 2m attached to two springs (Figure 5.2)and its oscillation frequency is the same as that of the inidividual uncoupled

oscillators ω0 =√

2k2m

.

37

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[W[W[[W[W[[W[W[[W[W[[W[W[[W[W[[W[W[[W[W[

\W\\W\\W\\W\\W\\W\\W\\W\k k2m

Figure 5.2: This shows the spring mass equivalent of the normal mode q0

which corresponds to the center of mass.

The normalmode q1 represents the relativemotion of he two masses whichleaves the center of mass unchanged. This can be thought of as the motionof two two particles of mass m connected to a spring of spring constant k =(k + 2k

′

)/2 as shown in Figure 5.3. The oscillation frequency of this normal

mode ω1 =√

k+2k′

mis always higher than that of the individual uncoupled

oscillators (or the center of mass).

~k

]^]^]]^]^]]^]^]]^]^]]^]^]_^_^__^_^__^_^__^_^__^_^_

`^`^``^`^``^`^``^`^``^`^àââaââaââaââaââ

m m

Figure 5.3: This shows the spring mass equivalent of the normal mode q1

which corresponds to two particles connected through a spring.

Equivalently, we may interpret q0 as a mode of oscillation where the twomasses oscillate with exactly the same phase, and q0 as a mode where theyhave a phase difference of π (Figure 5.4). Recollect that the phases of the twomasses are independent when the two masses are not coupled. Introducinga coupling causes the phases to beinterdependent.


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x

t1

q0

q

Figure 5.4: This shows the motion coresponding to the two normal modes q0

and q1 respectively.

5.1 Solution

For the normal modes we have

q0(t) = A0 ei ωot (5.8)

q1(t) = A1 ei ω1t (5.9)

where it should be borne in mind that A0 amd A1 are complex numbers withboth amplitude and phase ie. A0 = A0e

iψ0 etc. We then have the solutions

x0(t) = A0 ei ω0t + A1 e

i ω1t (5.10)

x1(t) = A0 ei ω0t − A1e

i ω1t (5.11)

where the A0 and A1 are determined by the initial position and velocities ofthe two oscillators.

We consider an example where the two particles are initialy at rest. Theparticle x0 is given a small displacement a0 and then left to oscillate. Thefact that x1 is initially at the equilibrium position with zero velocity, and the

5.1. SOLUTION 39

fact that x0 is initially at rest together tell us that A0 = A1 and both arereal with value a0/2. Using this we have

x0(t) =a0

2[cos ω0t+ cos ω1t] (5.12)

andx1(t) =

a0

2[cos ω0t− cos ω1t] (5.13)

where the motion of both the particles is the superposition of two simple har-monic motions of different frequencies ω0 and ω1. This can also be expressedas

x0(t) = a0 cos[(ω1 − ω0

2

)t]cos

[(ω0 + ω1

2

)t]

(5.14)

x1(t) = a0 sin[(ω1 − ω0

2

)t]sin

[(ω0 + ω1

2

)t]

(5.15)

which is the product of two harmonic motions with angular frequencies (ω1 +ω1)/2 and (ω1 − ω1)/2.

The situation where the two oscillators are weakly coupled (k′ k) isinteresting. We now have

ω1 =

√√√√ k

m

(1 +

2k′

k

)≈ ω0 +

k′

kω0 (5.16)

which allows us to express the particle motions as

x0(t) =

[a cos

(k′

2kω0t

) ]cos ω0t (5.17)

x0(t) =

[a sin

(k′

2kω0t

) ]sin ω0t (5.18)

We see that both the particles execute a rapid oscillation with angularfrequency ω0, the amplitude undergoes a slow modulation at a smaller angu-lar frequency k′

2kω0. The oscillations are slowly transferred from the particle

which receives the initial displacement to the particle originally at rest, andthen back again. This is shown in Figure 5.5.


x0x 1

tt

−1

−0.5

0

0.5

1

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 −1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Figure 5.5: This shows the motion of x0 and x1.

Chapter 6

Sinusoidal Waves.

We shift our attention to oscillations that propagate in space as time evolves.This is referred to as a wave. The sinusoidal wave

a(x, t) = A cos(ωt− kx + ψ) (6.1)

is the simplest example of a wave, we shall consider other possibilities laterin the course. It is often convenient to represent the wave in the complexnotation introduced earlier. We have

a(x, t) = A ei(ωt−kx) (6.2)

6.1 What is a(x, t)?

The wave phenomena is foud in may different situations, and a(x, t) repre-sents a different physical quantity in each situation. For example, it is wellknown that disturbances in air propagate from one point to another as wavesand are percieved by us as sound. Any source of sound (eg. a loud speaker)produce compressions and rarefactions in the air, and the patterns of com-pressions and rarefactions propagate from one point to another. Using ρ(x, t)to denote the air density , we can express this as ρ(x, t) = ρ+∆ρ(x, t) whereρ is the density in the absence of the disturbance and ∆ρ(x, t) is the changedue to the disturbance. We can use equation (6.1) to represent a sinusoidalsoud wave if we identify a(x, t) with ∆ρ(x, t).

The transverse vibrations of a stretched string is another examples. Inthis situation a(x, t) corresponds to y(x, t) which is the displacement of thestring shown in Figure 6.1’

41

42 CHAPTER 6. SINUSOIDAL WAVES.

x

y

Figure 6.1:

6.2 Angular frequency and wave number

The sinusoidal wave in equation (6.2) has a complex amplitude A = Aeiψ.Here A, the magnitude of A determines the magnitude of the wave. We referto φ(x, t) = ωt− kx+ ψ as the phase of the wave, and the wave can be alsoepressed as

a(x, t) = Aeiφ(x,t) (6.3)

If we study the behaviour of the wave at a fixed position x1, we have

a(t) = [Ae−ikx1]eiωt = A′eiωt . (6.4)

We see that this is the amiliar oscillation (SHO) discussed in detail in Chap-ter 1. The oscillation has amplitude A′ = [Ae−ikx1] which includes an extraconstant phase factor. The value of a(t) has sinusoidal variations. Startingat t = 0, the behaviour repeats after a time period T when ωT = 2π. Weidentify ω as the angular frequency of the wave related to the frequency ν as

ω =2π

T= 2πν . (6.5)

We next study the wave as a function of position x at a fixed instant of timet1. We have

a(x) = Aeiωt1e−ikx = A′′

e−ikx (6.6)

where we have absorbed the extra phase eiωt in the complex amplitude A′′

.This tells us that the spatial variation is also sinusoidal as shown in Figure

6.3. PHASE VELOCITY. 43

a(x)

x

λ

Figure 6.2:

6.2. The wavelength λ is the distance after which a(x) repeats itself. Startingfrom x = 0, we see that a(x) repeats when kx = 2π which tells us thatkλ = 2π or

k =2π

λ(6.7)

where we referr to k as the wave number. We note that the wave numberand the angular frequency tell us the rate of change of the phase φ(x, t) withposition and time respectively

k = −∂φ

∂xand ω =

∂φ

∂t(6.8)

6.3 Phase velocity.

We now consider the evolution of the wave in both position and time together.We consider the wave

a(x, t) = Aei(ωt−kx) (6.9)

which has phase φ(x, t) = ωt− kx. Let us follow the motion of the positionwhere the phase has value φ(x, t) = 0. We see that at the initial time t = 0we have φ = 0 at x = 0, and after a time ∆t this moves to a position

∆x =(ω

k

)∆t (6.10)


x

a(x,0)

φ=0

a(x, t)∆

∆x

−1.5

−1

−0.5

0

0.5

1

1.5

−2 −1 0 1 2 3

Figure 6.3:

a

x

t=0 t=1

t=3

−1.5

−1

−0.5

0

0.5

1

1.5

−1 −0.5 0 0.5 1

Figure 6.4:

shown in Figure 6.3. The point with phase φ = 0 moves at speed

vp =(ω

k

). (6.11)

It is not difficult to convince oneself that this is true for any constant valueof the phase, and the whole sinusoidal pattern propagates along the +xdirection (Figure 6.4) at the speed vp which is called the phase velocity ofthe wave.

6.4. WAVES IN THREE DIMENSIONS. 45

x

x

y

y

z

z

t=0

t=1

φ=0

φ=−1φ=0φ=−2

φ=−1 φ=−2

Figure 6.5:

6.4 Waves in three dimensions.

We have till now considered waves which depend on only one position coordi-nate x and time t. This is quite adequate when considering waves on a stringas the position along a string can be described by a single coordinate. It isnecessary to bring three spatial coordinates (x, y, z) into the picture whenconsidering a wave propagating in three dimensional space. A sound wavepropagating in air is an example.

We use the vector ~r = xi+yj+zk to denote a point in three dimensionalspace. The solution which we have been discussing

a(~r, t) = Aei(ωt−kx) (6.12)

can be interpreted in the context of a three dimensional space. Note thata(~r, t) varies only along the x direction and not along y and z. Consideringthe phase φ(~r, t) = ωt − kx we see that at any particular instant of timet, there are surfaces on which the phase is constant. The constant phasesurfaces of a wave are called wave fronts. In this case the wave fronts areparallel to the y − z plane as shown in Figure 6.5. The wave fronts movealong the +x direction with speed vp as time evolves. You can check this byfollowing the motion of the φ = 0 surface shown in Figure 6.5.


n

∆ r

Figure 6.6:

6.5 Waves in an arbitrary direction.

Let us now discuss how to describe a sinusoidal plane wave in an arbitrarydirection denoted by the unit vector ~n. A wave propagating along the idirection can be written as

a(~r, t) = Aei(ωt−~k·~r) (6.13)

where ~k = ki is called the wave vector. Note that ~k is different from k whichis the unit vector along the z direction. It is now obvious that a wave alongan arbitrary direction n can also be represented by eq. (6.13) if we change

the wave vector to ~k = kn. The wave vector ~k carries information aboutboth the wavelegth λ and the direction of propagation n.

For such a wave, at a fixed instant of time, the phase φ(~r, t) = ωt− ~k · ~rchanges only along n. The wave fronts are surfaces perpendicular to n asshown in Figure 6.6. The phase difference between two point (shown in

Figure 6.6) separated by ∆~r is ∆φ = −~k · ∆~r.

Problem For a wave with ~k = (4i + 5j) m−1 and ω = 100 Mhz, what arethe values of the following?

a. wavelength

b. frequency

c. phase velocity

d. phase difference between the two points (x, y, z) = (3, 4, 7) m and (4, 2, 8) m.

Chapter 7

Electromagnetic Waves.

What is light, particle or wave? Much of our daily experience with light,particularly the fact that light rays move in straight lines tells us that wecan think of light as a stream of particles. This is further borne out whenplace an opaque object in the path of the light rays. The shadow, as shownin Figure 7.1 is a projected image of the object, which is what we expect iflight were a stream of particles. But a closer look at the edges of the shadowreveals a very fine pattern of dark and bright bands or fringes. Such a patterncan also be seen if we stretch out our hand and look at the sky through athin gap produced by bringing two of our This cannot be explained unlesswe accept that light is some kind of a wave.

It is now well known that light is an electromagnetic wave. We shall nextdiscuss what we mean by an electromagnitic wave or radiation.

7.1 Electromagnetic Radiation.

What is the electric field produced at a point P by a charge q located ata distance r as shown in Figure 7.2? Anybody with a little knowledge ofphysics will tell us that this is given by Coulumb’s law

~E =−q

4πε0

err2

(7.1)

where er is an unit vector from P to the position of thecharge.In the 1880s J.C. Maxwell proposed a modification in the laws of elec-

tricity and magnetism which were known at that time. The change proposed

47

7.1. ELECTROMAGNETIC RADIATION. 49

by Maxwell unified our ideas of electricity and magnetism and showed bothof them to be manifestations of a single underlying quantity. Further it im-plied that Coulumb’s law did not tells us the complete picture. The correctformula for the electric field is

~E =−q

4πε0

[er′

r′2+r′

c

d

dt

(er′

r′2

)+

1

c2d2

dt2er′

](7.2)

This formula incorporates several new effects. The first is the fact that noinformation can propagate instantaneously. This is a drawback of Coulumb’slaw where the electric field at a distant point P changes the moment theposition of the charge is changed. This should actually happen after sometime. The new formula incorporates the fact that the influence of the chargepropagates at a speed c. The electric field at the time t is determined by theposition of the charge at an earlier time. This is referred to as the retardedposition of the charge r′, and er′ also refers to the retarded position.

The first term in eq. (7.2) is Coulumb’s law with the retarded position.In addition there are two new terms which arise due to the modificationproposed by Maxwell. These two terms contribute only when the chargemoves. The magnetic field produced by the charge is

~B = −er′ × ~E/c (7.3)

A close look at eq. (7.2) shows that the contributioin from the firsttwo terms falls off as 1/r′2 and these two terms are of not of interest atlarge distances from the charge. It is only the third term which has a 1/rbehaviour that makes a significant contribution at large distances. This termpermits a charged particle to influence another charged particle at a greatdistance through the 1/r electric field. This is referred to as electromagneticradiation and light is a familiar example of this phenomena. It is otbviousfrom the formula that only accelerating charges produce radiaddotion.

The interpretation of the formula is substantially simplified if we assumethat the motion of the charge is relatively slow, and is restricted to a regionwhich is smtion. all in comparison to the distance r to the point where wewish to calculate the electric field. We then have

d2

dt2er′ =

d2

dt2

(~r

′

r′

)≈~r

′

⊥

r(7.4)

where ~r′⊥

is the acceleration of the charge in the direction perbendicular toer′ . The parallel component of the acceleration does not effect the unit vector


θ

r a’E

Figure 7.3:

er′ and hence it does not make a contribution here. Further, the motion of thecharge makes a very small contribution to r′ in the denominator, neglectingthis we replace r′ with the constant distance r.

The electric field at a time t is related to a(t− r/c) which is the retardedacceleration as

E(t) =−q

4πε0c2ra(t− r/c) sin θ (7.5)

where θ is the angle between the line of sight er to the charge and thedirection of the retarded acceleration vector. The electric field vector isin the direction obtained by projecting the retarded acceleration vector onthe plane perpendicular to er as shown in Figure 7.3.

7.2 Electric dipole radiation.

We next consider a situation where a charge accelerates up and down along astraight line. The analysis of this situation using eq. (7.5) has wide applica-tions including many in technology. We consider the device shown in Figure7.4 which has two metallic wires A and B connected to an oscillating voltagegenerator. Positive charge accumulates at the free tip of wire A and electronsat the free tip of wire B when a positve and negative voltage are applied toA and B respectively. The positive and negative charges get interchangedwhen the voltage is reversed. The charges oscillate up and down the twowires A and B as if they were a single wire. In the situation where the timetaken by the electrons to move up and down the wires is much larger than

7.3. SINUSOIDAL OSCILLATIONS. 51

A

B

Vl

Figure 7.4:

the time taken for light signal to cross the wire, this can be thought of as anoscillating electric dipole. Note that here we have many electrons oscillatingup and down the wire. Since all the electrons have the same acceleration,the electric fields that they produce adds up. The electric field produced isinversely proportional to the distance r from the oscillator. At any time t,the electric field is proportional to the acceleration of the charges at a timet− r/c in the past.

It is possible to measure the radiation using a another electric dipoleoscillator where the voltage generator is replaced by a detector, say an os-cilloscope. An applied oscillating electric field will give rise to an oscillatingcurrent in the wires which can be converted to a voltage and measured. Adipole can measure oscillating electric fields only if the field is parallel tothe dipole and not if they are perpendicular. A dipole is quite commonlyused as an antenna to receive radio waves which is a form of electromagneticradiation.

Figure 7.5 shows an experiment where we use a dipole with a detector(D) to measure the electromagnetic radiation produced by another oscillatingelectric dipole (G). The detected voltage is maximum at θ = 90 and fallsas sin θ in other directions. At any point on the circle, the direction of theelectric field vector of the emitted radiation is along the tangent.


θ

G

D

Figure 7.5:

yE

yE

xz

B

Ek

Dipole

θ

Figure 7.6:

7.3. SINUSOIDAL OSCILLATIONS. 53

7.3 Sinusoidal Oscillations.

We next consider a situation where a sinusoidal voltage is applied to thedipole oscillator. The dipole is aligned with the y axis (Figure 7.6). Thevoltage causes the charges to move up and down as

y(t) = y0 cos(ωt) (7.6)

with acceleration

a(t) = −ω2y0 cos(ωt) . (7.7)

producing an electric field

E(t) =qy0ω

2

4πε0c2rcos[ω(t− r/c)] sin θ . (7.8)

It is oftern useful and interesting to represent the oscillating charge interms of other equivalent quantities namely the dipole moment and the cur-rent in the circuit. Let us replace the charge q which moves up and down asy(t) by two charges, one charge q/2 which moves as y(t) and another charge−q/2 which moves in exactly the opposite direction as −y(t). The electricfield produced by the new configuration is exactly the same as that producedby the single charge considered earlier. This allows us to interpret eq. (7.8)in terms of an oscillating dipole

dy(t) = q y(t) = d0 cos(ωt) (7.9)

which allows us to write eq. (7.8) as

E(t) =−1

4πε0c2rdy(t− r/c) sin θ . (7.10)

Returning once more to the dipole oscillator shown in Figure 7.4, we notethat the excess charge which resides at the tip of A is balanced by exactlythe opposite charge at the tip of the wire B. This has a dipole momentdy(t) = l q(t) where l is the length of the dipole oscillator and q(t) is theexcess charge accumulated at one of the tips. This allows us to write dy(t)in terms of the current in the wires I(t) = q(t) as

dy(t) = lI(t) . (7.11)


We can then express the electric field produced by the dipole oscillator interms of the current. This is particularly useful when considering technolog-ical applications of the electric dipole oscillator. For a current

I(t) = −I sin(ωt) (7.12)

the electric field is given by

E(t) =ωlI

4πε0c2rcos[ω(t− r/c)] sin θ . (7.13)

where I refers to the peak current in the wire.

We now end the small detour where we discussed how the electric field isrelated to the dipole moment and the current, and return to our discussionof the electric field predicted by eq. (7.8). We shall restric our attention topoints along the x axis. The electric field of the radiation is in the y directionand has a value

Ey(x, t) =qy0ω

2

4πε0c2xcos

[ωt−

(ω

c

)x]

(7.14)

Let us consider a situation where we are interested in the x dependence ofthe the electric field at a great distance from the emittter. Say we are 1 kmaway from the oscillator and we would like to know how the electric fieldvaries at two points which are 1m apart. This situation is shown schemati-cally in Figure 7.6. The point to note is that a small variation in x will makea very small difference to the 1/x dependence of the electric field which wecan neglect, but the change in the cos term cannot be neglected. This isbecause x is multiplied by a factor ω/c which could be large and a change inωx/c would mean a different phase of the oscillation. Thus at large distancesthe electric field of the radiation can be well described by

Ey(x, t) = E cos [ωt− kx] (7.15)

where the wave number is k = ω/c. This is the familiar sinusoidal plane wavewhich we have studied in the previous chapter and which can be representedin the complex notation as

Ey(x, t) = Eei[ωt−kx] (7.16)

7.4. ENERGY DENSITY, FLUX AND POWER. 55

<U>

cx

Figure 7.7:

We next calculate the magnetic field ~B. Referring to Figure 7.6 we see thatwe have er = −i. Using this in eq. (7.3) with ~E = Ey(x, t) j we have

~B(x, t) = i× ~Ey(x, t)/c =E

ccos(ωt− kx) k (7.17)

The magnetic field is perpendicular to ~E(x, t) and its amplitude is a factor1/c smaller than the electric field. The magnetic field oscillates with thesame frequency and phase as the electric field.

Although our previous discussion was restricted to points along the x axis,the facts which we have learnt about the electric and magnetic fields hold atany position (Figure 7.6). At any point the direction of the electromagneticwave is radially outwards with wave vector k = kr. The electric and magneticfields are mutually perpendicular, they are also perpendicular to the wavevector k.

7.4 Energy density, flux and power.

We now turn our attention to the energy carried by the electromagnetic wave.For simplicity we shall initialy restric ourselves to points located along the xaxis for the situation shown in Figure 7.6.

The energy density in the electric and magnetic fields is given by

U =1

2ε0E

2 +1

2µ0B2 (7.18)


For an electromagnetic wave the electric and magnetic fields are related. Theenergy density can then be written in terms of only the electric field as

U =1

2ε0E

2 +1

2c2µ0E2 (7.19)

The speed of light c is related to ε0 and µ0 as c2 = 1/ε0µ0. Using this we findthat the energy density has the form

U = ε0E2 (7.20)

The instantaneous energy density of the electromagnetic wave oscillates withtime. The time average of the energy density is often a more useful quantity.We have already discussed how to calculate the time average of an oscillatingquantity. This is particularly simple in the complex notation where theelectric field

Ey(x, t) = Eei(ωt−kx) (7.21)

has a mean squared value 〈E2〉 = EE∗/2. Using this we find that the averageenergy density is

〈U〉 =1

2ε0EE

∗ =1

2ε0E

2 (7.22)

where E is the amplitude of the electric field.We next consider the energy flux of the electromagnetic radiation. The

radiation propagates along the x axis at the point where we want to calculatethe flux. Consider a surface which is perpendicular to direction in which thewave is propagating as shown in Figure 7.7. The energy flux S refers to theenergy which crosses an unit area of this surface in unit time. It has unitsWattm−2. The flux S is the power that would be received by collecting theradiation in an area 1m2 placed perpendicular to the direction in which thewave is propagating as shown in Figure 7.7.

The average flux can be calculated by noting that the wave propagatesalong the x axis with speed c. The average energy 〈U〉 contained in an unitvolume would take a time 1/c to cross the surface. The flux is the energywhich would cross in one second which is

〈S〉 = 〈U〉 c =1

2cε0E

2 (7.23)

The energy flux is actually a vector quantity representing both the directionand the rate at which the wave carries energy. Referring back to equation


y

x

z

θ

Ωd

φ

Figure 7.8:

(7.8) and (7.15) we see that at any point the average flux 〈~S〉 is pointedradially outwards and has a value

〈~S〉 =

(q2y2

0ω4

32π2c3ε0

)sin2 θ

r2r . (7.24)

Note that the flux falls as 1/r2 as we move away from the source. This isa property which may already be familiar to some of us from considerationsof the conservation of energy. Note that the total energy crossing a surfaceenclosing the source will be constant irrespective of the shape and size of thesurface.

Let us know shift our point of reference to the location of the dipole andask how much power is radiated in any given direction. This is quantifiedusing the power emitted per solid angle. Consider a solid angle dΩ along adirection r at an angle θ to the dipole as shown in Figure 7.8. The power dPradiated into this solid angle can be calculated by multiplying the flux withthe area corresponding to this solid angle

dP = 〈~S〉 · r r2 dΩ (7.25)

which gives us the power radiated per unit solid angle to be

d

dΩ〈~P 〉(θ) =

(q2y2

0ω4

32π2c3ε0

)sin2 θ . (7.26)

This tells us the radiation pattern of the dipole radiation, ie.. the directionaldependence of the radiation is proportionalto sin2 θ. The radiation is maxi-mum in the direction perpendicular to the dipole while there is no radiation


θ P

Figure 7.9:

emitted along the direction of the dipole. The radiation pattern is shown inFigure 7.9. Another important point to note is that the radiation depends onω4 which tells us that the same dipole will radiate significantly more powerif it is made to oscillate at a higher frequency, doubling the frequency willincrease the power sixteen times.

The total power radiated can be calculated by integrating over all solidangles. Using dΩ = sinθ dθ dφ and

∫ 2π

0dφ

∫ π

0sin3θ dθ =

8π

3(7.27)

gives the total power P to be

〈~P 〉 =q2y2

0ω4

12πc3ε. (7.28)

It is often convenient to express this in terms of the amplitude of the currentin the wires of the oscillator as

〈~P 〉 =I2l2ω2

12πc3ε. (7.29)

The power radiated by the electric dipole is proportional to the square of thecurrent. This behaviour is exactly the same as that of a resistance exceptthat the oscillator emits the power as radiation while the resistance convertsit to heat. We can express the radiated power in terms of an equivalentresistance with

〈~P 〉 =1

2RI2 (7.30)


where

R =

(l

λ

)2

790 Ω (7.31)

l being the length of the dipole (Figure 7.4) and λ the wavelength of theradiation.

Chapter 8

The vector nature of

electromagnetic radiation.

Consider a situation where the same electrical signal is fed to two mutuallyperpendicular dipoles, one along the y axis and another along the z axis asshown in Figure 8.1. We are interested in the electric field at a distant pointalong the x axis. The electric field is a superposition of two components

~E(x, t) = Ey(x, t)j + Ez(x, t)k (8.1)

One along the y axis produced by the dipole which is aligned along the yaxis, and another along the z axis produced by the dipole ariented along thez axis.

In this situation where both dipoles receive the same signal, the twocomponents are equal Ey = Ez and

~E(x, t) = E(j + k) cos(ωt− kx) (8.2)

If we plot the time evolution of the electric field at a fixed position (Figure8.2) we see that it oscillates up and down along a direction which is at 45

to the y and z axis.The point to note is that it is possible to change the relative amplitudes of

Ey and Ez by changing the currents in the oscillators. The resultant electricfield is

~E(x, t) = (Eyk + Ez j) cos(ωt− kx) (8.3)

The resultant electric field vector has magnitude E =√E2y + E2

z and it

61

62CHAPTER 8. THE VECTOR NATURE OF ELECTROMAGNETIC RADIATION.

x

z

y E

Figure 8.1:

45

z

yθο

Figure 8.2:

63

y

E

zzE

y

Left Circular Right Circular

Figure 8.3:

oscillates along a direction at an angle θ = tan−1(Ez

Ey

)with respect to the y

axis (Figure 8.2).Under no circumstance does the electric field have a component along the

direction of the wave i.e along the x axis. The electric field can be orientedalong any direction in the y-z plane. In the cases which we have considereduntil now, the electric field oscillates up and down a fixed direction in they-z plane . Such an electromagnetic wave is said to be linearly polarized.

The polarization of the wave refers to the time evolution of the electricfield vector.

An interesting situation occurs if the same signal is fed to the two dipoles,but the signal to the z axis is given an extra π/2 phase . The electric fieldnow is

~E(x, t) = E[cos(ωt− kx)j + cos(ωt− kx+ π/2)k

](8.4)

= E[cos(ωt− kx)j − sin(ωt− kx)k

](8.5)

If we now follow the evolution of ~E(t) at a fixed point, we see that it moveson a circle of radius E as shown in Figure 8.3

We call such a wave left circularly polarized. The electric field wouldhave rotated in the opposite direction had we applied a phase lag of π/2. Wewould then have obtained a right circularly polarized wave .

Oscillations of different amplitude combined with a phase difference ofπ/2 produces elliptically polarized wave where the ellipse is aligned with they-z axis as shown in Figure 8.4. The ellipse is not aligned with the y − z

64CHAPTER 8. THE VECTOR NATURE OF ELECTROMAGNETIC RADIATION.

y

z

E

y

E

z

Figure 8.4:

z

y

E

Figure 8.5:

axis for an arbitrary phase differences. This is the most general state ofpolarization (Figure 8.5). Linear and circularly polarized waves are specificcases of elliptically polarized waves.

Chapter 9

The Electromagnetic Spectrum

Electromagnetic waves come in a wide range of frequency or equivalentlywavelength. We refer to different bands of the frequency specrtum usingdifferent names. These bands are often overlapping. The nomenclature ofthe bands and the frequency range are typically based on the properties ofthe waves and the techniques to generate or detect them.

9.1 Radiowaves

This spans wavelengths less than 1GH3 . There is no theoretical lower limit,and the familiar 50H3 power supply lines emit radio waves at this frequencywhich corresponds to a wavelength.

λ =c

v=

3 × 105km/s

50 51= 6000km (9.1)

The upper frequency part of this band is used for radio and televisiontransmission.

9.2 Microwave

This refers to the frequency range from 1GH3 to 3× 1011H3. or wavelengths30cm to 1mm.

The communication band extends into the microwave.The earths atmosphere is largely transparent at frequencies ranging from

less than 1cm to 30cm . Radiowaves and microwaves are both very useful

65

66 CHAPTER 9. THE ELECTROMAGNETIC SPECTRUM

for Space communication and astronomy. This branch of astronomy is calledradio astronomy.

There is a large variety of electical circuits and antennas used to producewaves in these two bonds.

We briefly discuss the origin of some of the astronomical souces of radioand microwave radiation.

9.3 21cm radiation.

Hydrogen is the most abundant element in the universe. Atomic Neutral hy-drogen, has two states, one where the proton and electron spins are alignedand another where they are opposite as shown in Figure - A transition be-tween these two states causes radiation at 1.420GH3 or 21cm to be emitted .This is a very important source of information about our galaxy and externalgalaxies. Interestingly, the 1 Giant Meterwave Radio Telescope (GMRT ) lo-cated in Narayan Gaon near Pune is the worlds largest low frequency radiotelescope.

9.4 Cosmic Microwave Background Radia-

tion.

(( MBR))

An object which is equally efficient in absorbing and emitting radiationat all frequencies is referred to as a black body. A metallic cavity with asmall hole to radiation in and out is an example of a black body. It is foundthat the radiation spectum inside a black body in thermal equilibrium at aatmosphere T cavity is completely specified by the temperature.

The energy density of the elecromagnetic waves in a frequency in dν isfound to be

duν = uν dν

where

uν =8πhν3

c31[

exp(hνkt

)− 1

] (9.2)

where h is the Plack constant and k is the Boltzmann Constant.

9.5. MOLECULAR LINES. 67

It has been found, by poenting a redio reciever in different directionsan the sky, that there is a radiation with a black body spectrum at T =2.73 ± 0.06K arriving from all directions in the sky. This radiation is notteeristrial in origin and it is believed that we are actually seeing a radiationwhich pervades the whole universe and is a relic of a hot past referred to asthe hot Big Bang . This is referred to as the Cosmic Microwave Backgroundradiation which peaks in the microwave region of the Spectrum.

9.5 Molecular lines.

Polar molecules like water

H H++µ = 6.2 10

− 30cm

are influenced by electromagnetic radiation incident. and can be set intovibrations of rotation by the external, oscillating field.

The water molecules try to align their dipole movement with the externalelectric field which set the molecule into rotation. The rotational enegy levelsare quantized and the molecules have distant frequencies at which there areresonances where it emits or absorbs maximum energy. The Microwave ovensutilize a 2−45GH3 rotational transition of water. The water molecules absorbthe incident EN radiation and start rotating. The rotational energy of themolecules is converted to random motions or heat.

Most of the rotational and vibrational transitions of molecules lie in themicrowave infra Red (IR) bands.

The frequency range 50GH3 to 10TH3 is often calld Tera hertz radiationor T-rays. The water vapour in the atmosphere is opaque at much of thesefrequencies. Dry Substances like paper, plastic which do not have watermolecules are transparent to T- rays, whereas it is absorbed by substanceswith waves and is reflected by metals which makes it suitable for variousimaging applications.

9.6 Infrared

The frequency range from 3 × 1011H3 to ∼ 4 × 1014H3 is referred to as theinfrared band (IR). The frequency is just below that of red light. This is


again subdivided into

near IR 780 -3000nmintermediate IR 3000 - 6000 nmfar IR IR 6000 - 15000 nmextreme IR 15,000 nm - 1mm

The division are quite loose and veryThere are many molecular vibrational and rotational transitions which

produce radiation in the IR. Further, the black body radiation from hotobjects omit copious amounts of IR. as the spectrum typically peaks in thisbond .

For example the black body rediation from human beings peaks at around10,000 nm making the IR sutable for “ night vision”

Stars like the sun and incandescent lamps emit copiously in the IR. Infact an incandescent lamp radiates away more than 50% of its energy in theIR.

IR spy satellites are used to monitor sudden heat generation indicating arocket launch or posibly nuclear explosion.

Much of fiber optical communication also works in the IR.

9.7 Visible light

This corresponds to E-M radiation in the wavelength range 3.84 × 1014H3

to 7.69 × 1014H3 The atmosphere is transparent in this bond, which is whypossibly we developed vision in this frequency range.

Transition of the electrons in the outer shell of atoms producer radiation.These are used in visible sources as well as detectors.

Thermal radiation from objects at few thousand kelvin is also anothersource of visible light.

Colour is our perception of the relative contribution from different fre-quencies. Different combinations of the frequencies can produce the samecolour.

9.8 Ultra Violet

These are waves with frequency just higher than blue light and this bandspans from 8 × 1014H3 to 3.4 × 1016H3.

9.9. X-RAYS 69

UV has sufficient energy to ionize the atoms in the upper atmosphere,producing the ionosphere. Fortunately ozon O3 in the atmosphere absorbsUV as it is harmful. It is used to kill bacteria in many water purificationseptems.

Human eyes cannot see in UV as it is absorbed by the lens. People whoselens have been removed due to cataract can see in UV to some extant. Manyinsects can see in part of the UV band.

UV is producd through more energetic electronic transitions in atoms.For example, The transition of the hydrogen atom from the first excited tothe ground slate produces UV.

UV is used in lithography for making ICS; erasing EPROMS.etc.

9.9 X-rays

2.4 × 1016 to 5 × 1019H3

These are produced by very fast moving electrons when they encounterthe positively changed nuclei of atoms and are accelerated as a consequence.This occurs when electrons are bombarded on a copper plate. Also in hotplasma found in many astrophysical situations like around our sun or aroundblack holes.

Also produced in inner shell electron transitions in atoms.

9.10 Gamma Rays

The highest frequency electromagnetic radiation is refered to as GammaRays. These are produced in nuclear transitions when the neucleus goes froman excited state to a lower energy state. Also produced when an electron andpositron annhilate.

This radiation will ionize most gases and is not very difficult to detect.

Chapter 10

Interference.

In this chapter we consider what happens when we superpose two waves.Naively, we would expect the intensity (energy density or flux) of the re-sultant to be the sum of the individual intensities. For example, a room isilluminated by two light bulbs A and B. If we switch on both A and B, wewould expect the room to be twice brighter than if only A or B were on.Actually, this does not always hold. A wave, unlike the intensity, can have anegative value. If we add two waves whose values have opposite signs at thesame point, the total intensity is less than the intensities of the individualwaves. This is an example of the phenomena of interference.

10.1 Young’s Double Slit Experiment.

We begin our discussion of interference with a situation shown in Figure 11.1.

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Point Source Slits

Screen

Figure 10.1:

71

72 CHAPTER 10. INTERFERENCE.

θd

Source Slits Screen

P

Figure 10.2:

Light from a distant point source is incident on a screen with two thin slits.The seperation between the two slits is d. We are interested in the image ofthe two slits on a screen which is at a large distance from the slits. Note thatthe point source is aligned with the center of the slits as shown in Figure10.2. Let us calculate the intensity at a point P located at an angle θ on thescreen.

The radiation from the point source is well described by a plane waveby the time the radiation reaches the slits. The two slits lie on the samewavefront of this plane wave, thus the electric field oscillates with the samephase at both the slits. If E1(t) and E2(t) be the contributions from slits 1and 2 to the radiation at the point P on the screen, the total electric fieldwill be

E(t) = E1(t) + E2(t) (10.1)

Both waves originate from the same source and they have the same frequency.We can thus express the electric fields as E1(t) = E1e

iωt, E2(t) = E2eiωt and

E(t) = Eeiωt. We then have a relations between the amplitudes

E = E1 + E2

It is often convenient to represent this addition of complex amplitudes graph-ically as shown in Figure 10.3. Each complex amplitude can be representedby a vector in the complex plane, such a vector is called a phasor. The sumis now a vector sum of the phasors E1 = E1e

iφ1 and E2 = E2eiφ2 shown in

the figure. The intensity of the wave is

I = 〈E(t)E(t)〉 =1

2EE∗ (10.2)

where we have dropped the constant of proportionatily in this relation. Thesquare of the length of the resultant phasor E in Figure 10.3 gives us the

10.1. YOUNG’S DOUBLE SLIT EXPERIMENT. 73

φ2 1

φ −Imaginary

E~

E~ 2

E1~

~

Imaginary

Real

1φ1

E~

E~ 2

E

φ2

Real

Figure 10.3:

intensity. It is also clear from the figure that this has a value

I = E21 + E2

2 + 2E1E2 cos(φ2 − φ1) (10.3)

We now repeat the same exercise, ie. calculating the intensity, doing italgebraically. The intensityis

I = I1 + I2 + E1E∗

2 + E∗

1E2 (10.4)

= I1 + I2 + E1E2

[ei(φ1−φ2) + ei(φ2−φ1)

](10.5)

I = I1 + I2 + 2√I1I2 cos(Φ2 − Φ1) (10.6)

The intensity is maximum when the two waves have the same phase

I = I1 + I2 + 2√I1I2 (10.7)

and it is minimum when Φ2 − Φ1 = π i.e the two waves are exactly out ofphase

I = I1 + I2 − 2√I1I2 . (10.8)

The intensity is the sum of the two intensities when the two waves are π/2out of phase.

In the Young’s double slit experiment the waves from the two slits arriveat P with a time delay because the two waves have to traverse different paths.The resulting phase difference is

Φ1 − Φ2 = 2πd sin θ

λ(10.9)

and

I(θ) = 2I

[1 + cos

(2πd sin θ

λ

)](10.10)

= 4I cos2

(πd sin θ

λ

)(10.11)


λ / d

θ

I( )θ

θ 0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Figure 10.4:

k

k 1

2

Ar∆

B

θ/2

Figure 10.5:

For small θ we have

I(θ) = 2I

[1 + cos

(2πdθ

λ

)](10.12)

There will be a pattern of bright and dark lines (fringes ) seen on thescreen. The fringes are straight lines parallel to the slits, and the spacingbetween two succissive bright fringes is λ/d radians as shown in Figure 10.4

10.1.1 A different method of analysis.

Consider two waves, originally from the same source incident on a plane asshown in Figure 10.5. A possible means to achieve this is to place two prismsin the path of a plane wave as shown in the figure. The two waves are in twodifferent directions ~n1 and ~n2 with ~k1 = k~n1 and ~k2 = k~n2 as shown in thefigure.

Let A be a point on the screen where both waves arrive at the samephase.ie.

E1 = E2 = EeiΦ(A)

10.1. YOUNG’S DOUBLE SLIT EXPERIMENT. 75

The intensity at this point will be maximum. Now at any neighbouringpoint B a distance ~∆r from A, the phase of the two waves will, in general,be differne. The first wave has a phase

Φ1(B) = Φ(A) − ~k1 · ∆~r (10.13)

and the second waveΦ2(B) = Φ(A) − ~k2 · ∆~r (10.14)

The phase difference between the two waves at the point B is

Φ2(B) − Φ1(B) = −(~k2 − ~k1

)· ∆~r . (10.15)

The intensity at the point B will be minimum ie. the point will be dark if(~k2 − ~k1

)· ∆~r = (2N + 1)π with N = 0,±1,±2, ...

and the intensity at the point B will be maximum if(~k2 − ~k1

)· ∆~r = 2Nπ with N = 0,±1,±2, ...

If both the wave vector ~k1 and ~k2 are coplanar with the normal to thesurface, the dark and bright patterns form straight line fringes. We shallanalyze a situation where the two waves are in the x−y plane and the screenis normal to the x axis. Further, To simplify matters we shall assume thatboth thw waves are nearly normal to the surface making a small angle θ/2as shown in Figure 10.5. We then have

~k1 =2π

λ[cos(θ/2)i+ sin(θ/2)j] ≈

2π

λ(i+ θ/2 j)

and~k1 ≈

2π

λ(i− θ/2 j) .

Further, the displacement ∆~r is in the y − z plane so we can write it as

∆~r = ∆yj + ∆zk .

We then have the condition for a maxima

−(~k2 − ~k1

)· ∆~r =

2π

λθ∆y = 2Nπ


θd

Source Screen

βα/2

−α/2Slits

Figure 10.6:

We thus have periodic pattern of bright lines at a spacing of

∆y =λ

θ

further it is also clear that there will also be dark lines half-way betweenthe bright lines. These are straight line fringes parallel to the z directionas shown in Figure 10.4. It should also be clear by now that the situationconsidered here is identical to the Young’s double slit experiment.

10.1.2 Effect of Finite Source Size.

In our discussion of Young’s experiment we have assumed that the two slitsare illuminated by a point source (Figure 10.2) , or in other words we haveignored the angular extent of the source. For example, a distant star is a pointsource whereas the sun which has an angular extent of 0, 5 is an extendedsource. We now include this effect assuming that the source subtends a finiteangle α at the slits as shown in Figure 10.6.

The different points on the source are assumed to be incoherent sources.This means assuming that the waves originating from different points on thesource do not interfere with one another. To determine the intensity patternon the screen, we have to superpose the interference pattern produced byeach point on the source.

For a wave originating at an angle β on the source (Figure 10.6), thephase difference between the two slits at an angle θ on the screen is

φ2 − φ1 =2πd

λ[sin β + sin θ] . (10.16)

Using this in eq. (10.6) we calculate the interference pattern produced on

10.2. THE MICHELSON INTERFEROMETER 77

the screen by a point source at an angle β. For this source

I(θ, β) = 2I

[1 + cos

(2πd

λ(θ + β)

)](10.17)

gives the intensity at an angle θ on the screen. Note that we have assmuedthat the angles θ and β are small. The total intensity is calculated by addingup the contributions from all the points on the source

I(θ) =1

α

α/2∫

− α/2

I(θ, β) dβ , (10.18)

which gives

I(θ) = 2I

[1 +

λ

α2πd

sin

[2πd

λ

(θ +

α

2

)]− sin

[2πd

λ

(θ −

α

2

)]]

= 2I

[1 +

λ

πdαsin

(πdα

λ

)cos

(2πdθ

λ

)]

= 2I

[1 + sinc

(πdα

λ

)cos

(2πdθ

λ

)](10.19)

where the sinc(x) = sin(x)/x. The first point to note is that in the limitα→ 0 we recover eq. (10.12) which is the interference pattern producedby a point source. The second point is that the factor sinc(πdα/λ) fallsas α is increased. This causes the intensity difference between the maximaand the minima to fall (Figure 10.7), reducing the contrast of the fringepattern. The interference pattern is washed away as the sinc() term falls,and there is no interference when the source subtends a very large angle sothat sinc(πdα) → 0.

10.2 The Michelson Interferometer

Figure 10.8 shows a typical Michelson interferometer setup. A ground glassplate G is illuminated by a light source. The ground glass plate has theproperty that it scatters the incident light into all directions.


sinc()=0

0.2

0.02θ

θ

I( )

0

0.5

1

1.5

2

2.5

−10 −5 0 5 10

Figure 10.7:

Ó Ó Ó Ó Ó Ó Ó Ó Ó Ó Ó Ó Ó Ó Ó Ó Ó Ó Ó ÓÔ Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô Ô

G

B C

M 2

1M

T

Figure 10.8:

The light scattered forward by G is incident on a beam splitter B whichis at 45. The beam splitter is essentially a glass slab with one surface semi-silvered to inrease its reflectivity. It splits the incident wave into two parts E1

and E2, one which is transmitted (E1) and another (E2) which is reflected.The two beams have nearly the same intensity. The transmitted wave E1

is reflected back to B by a mirror M1. and a part of it is reflected into thetelescope T. The reflected wave E2 travels in a perpendicular direction. Themirror M2 reflects this back to B where a part of it is transmitted into T.An observer at T would see two images of G, namely G1 and G2 (shown inFigure 10.9) produced by the two mirrors M1 and M2 respectively. The twoimages are at a separation 2d where d is the difference in the optical pathsfrom B to G1 and from B to G2. Note that E2 traverses the thickness of thebeam splitter thrice whereas E1 traverses the beam splitter only once. Thisintroduces an extra optical path for E2 even when M1 and M2 are at the sameradial distance from B. It is possible to compensate for this by introducingan extra displacement in M1, but this would not serve to compensate for theextra path over a range of requencies as the refractive index of the glass inB is frequency dependent. A compensator C, which is a glass block identical


G 1T

θ

G 2

2d

S 1 S 2

Figure 10.9:

to B (without the silver coating), is introduced along the path to M1 tocompensate for this.

S1 and S2 are the two images of the same point S on the ground glass plate.Each point on the ground glass plate acts as a source emitting radiation inall directions. Thus S1 and S2 are coherent sources which emit radiation inall direction. Consider the wave emitted at an angle θ as shown in Figure10.9. The telescope focuses both waves to the same point. The resultantelectric field is

E = E1 + E2 (10.20)

and the intensity is

I = I1 + I2 + 2√I1I2 cos(φ2 − φ1) (10.21)

The phase difference arises because of the path difference in the two armsof the interferometer. There is an additional phase difference of π because E2

undergoes internal reflection at B whereas E1 undergoes external reflection.We then have

φ2 − φ1 = π +(

2π

λ

)2d cos θ (10.22)

So we have the condition

2d cos θm = mλ (m = 0, 1, 2, ...) (10.23)

for a minima or a dark fringe. Here m is the order of the fringe, and θm isthe angle of the mth fringe.

The fringes will be circular as shown in Figure 10.10. When the centralfringe is dark, the order of the fringe is

m =2d

λ. (10.24)


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Figure 10.10:

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Increasing d

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Figure 10.11:

Le us follow a dark fringe of a fixed order, say m (shaded in Figure 10.11),as we increase d the difference in the length of the two arms. The value ofcos θm has to decrease which implies that θm increases. As d is increased,new fringes appear at the center and the existing fringes move outwards andfinally move out of the field of view.

Let us now estimate the angular separtion between the fringe lines in theMichelson Interferometer. We consider the situation where the central spot isdark, and calculate θ the angle to the first circular dark ring (Figure 10.10).If the central dark spot is of order m then

2d = mλ , (10.25)

then the first dark ring will be of order m− 1 and it will satisfy

2d cos θ = (m− 1)λ . (10.26)

Using the fact that for small θ we can use cos θ ≈ 1 − θ2/2, subtracting the


two above equations gives us

θ =

√λ

d. (10.27)

We see that the fringe separation decreases as the difference in the arm lengthd is increased. The fringe spacing increases if d is reduced, and the centraldark spot occupies the entire field of view when the two arms have exactlythe same length. The whole field of view is dark in this situation.

The visible part of the electromagnetic spectrum has wavelengths in therange of 600 nm. These length-scale are much smaller than those which canbe directly percieved by us. The Michelson Interferometer can be used todetermine the wavelength of light. Consider a situation where the interfer-ometer is arranged so that there is a central dark spot. We now increase dby moving one of the mirrors so that the existing fringes move outward andnew fringes appear at the center. We increase d to d + ∆d so that 100 newfringes appear at the center. We initially have

2d = mλ , (10.28)

and we finally have2(d+ ∆d) = (m + 100)λ , (10.29)

which gives us

∆d = 50λ , (10.30)

The value of ∆d will be of the order of 30µm which can be measuredusing a micrometers screw gauge. If we can measure the distance one of themirrors has to be moved so as to introduce 100 new fringes, we can use thisto determine λ the wavelength of light.

There are many situations when we have two spectral lines whose wave-lengths are very close, say λ1 and λ1 + ∆λ with ∆λ λ1. For example,Sodium has two very close spectral lines one at 589.0 nm and another at589.6 nm. The Michelson Interferometer can also be used to determine ∆λthe differenc ein wavelengths. The fringe pattern corresponding to the twodifferent wavelengths are said to be in concordance if the dark fringes of boththe wavelengths coincide at the center. This can be achieved by suitably ad-justing d. The intensity of the fringe pattern is brightest when the two setof fringes are in concordance. This requires that the condition

2d = m1λ1 = m2(λ1 + ∆λ) (10.31)


be satisfied for two integers m1 and m2 with m2 < m1. Now if d is graduallyincrease to d + ∆d, then at the center m1 and m2 will increase by differentamounts ∆m1 = 2∆d/λ1 and ∆m2 = 2∆d/(λ1 + ∆λ) respectively with∆m2 < ∆m1. As ∆d is increased we will reach a situation where ∆m2 =∆m1−1/2. Under this condition the the dark fringes of λ1 coincide with thebright fringes of λ1 + ∆λ and vice-verse. The field of view has nearly uniformintensity and the fringe pattern is washed out. This situation is referred toas discordance. The two fringes are in concordance again if ∆d is increasedfurther so that ∆m2 = ∆m1 − 1. We then have

2(d+ ∆d) = (m1 + ∆m1)λ1 = (m2 + ∆m1 − 1)(λ1 + ∆λ) , (10.32)

subtracting this from the previous equation gives us

2∆d = ∆m1λ1 = (∆m1 − 1)(λ1 + ∆λ)

= (2 ∆d

λ1− 1)(λ1 + ∆λ) . (10.33)

On rearranging the terms we have

λ1

∆λ+ 1 =

2 ∆d

λ1

(10.34)

which, using the assumption that ∆λ/λ1 1 , gives us

∆λ =λ2

1

2 ∆d. (10.35)

It is possible to determine ∆λ by measuring ∆d the distance one of themirrors has to bemoved so that the fringe system goes from one concordanceto the next concordance.

The Michelson Interferometer finds a variety of other very interestingapplications. It was used by Michelson and Morley to show that the speedof light is the same in all directions. Since the Earth os moving, we wouldexpect the speed of light to be different along the direction of the Earth’smotion. Michelson and Morley established that the speed of light does notdepend on the motion of the observer, providing a direct experimental basisfor Einstein’s Special Theory of Relativity.

The fringe patter in the Michelson interferometer is very sensitive to dis-placements in the mirrors, and it can be used to measure very small dis-placements of the mirrors. A Michelson interferometer whose arms are 4km


long is being used in an experiment called LIGO which is an ongoing effortto detect Gravitational Waves, one of the predictions of Einstein’s GeneralTheory of Relativity.

Chapter 11

Coherence.

The Michelson interferometer works on the principle of division of amplitude.An incident wavefront is split into two parts which are then combined witha time delay. The time delay ∆t arises because the two wavefronts traversedifferent distances in the two arms of the interferometer. An interferencepattern will be seen as long as ∆t is less than the coherence time T of thelight, i.e. 2d < cT . The coherence time is determined by the spread infrequencies or “Bandwidth” of the light. If the light has a bandwidth ∆νi.e. if the light is a superposition of waves of different frequencies spread over∆ν, the coherence time is

T ≈1

∆ν. (11.1)

It is common to refer to cT as the coherence length.

∆ω/ ω ∼ 0.2

t

E

−8000

−6000

−4000

−2000

0

2000

4000

12 13 14 15 16 17 18

Figure 11.1:

85

Somnath Bhardwaj Lect_notes

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