PHL100 BDG Course Slides

8/13/2019 PHL100 BDG Course Slides

1/81

PHL100: Electromagnetic Waves and

Quantum Mechanics

Examinations:1. Minor I: 25 Marks

2. Minor II: 25 Marks

3. Major: 50 Marks

Books:

1. Introduction to Electrodynamics by David Griffiths

2. Introduction to Quantum Mechanics by David Griffiths

3. Quantum Mechanics by Robert Eisberg and Robert Resnick

Teachers:

1. Prof. B.D. Gupta 2. Dr. Kedar Khare 3. Prof. Ajit Kumar


2/81

Course up to Minor I

Electric and magnetic field vectors in a medium,

Susceptibility and conductivity, Maxwellsequations,

Boundary conditions, EM wave equation, Plane

wave solutions, Polarization of the EM waves,Poynting vector and intensity of the EM wave,

Reflection and refraction of EM waves at a dielectric

interface, Wave packet, Phase and Group velocities


3/81

Differential CalculusGradient:

T(x,y,z): Function of 3 variables. Can be temperature at a point

(x,y,z) in a room.

How does T changes when we change 3 variables by infinitesimal

amounts dx, dy, dz?

A theorem on partial derivatives

Can be written as

is the gradient of T. It is a vector quantity.

T T TdT dx dy dz x y z

.T dl .T T T

dT x y z dxx dyy dzz x y z

T


4/81

Divergence

. x y zv x y z v x v y v z x y z

yx z

vv v

x y z

A scalar quantity

Tells how much the function v spreads out from the point in question.

a)large (+ve) divergence b)large (-ve) divergence c) Zero divergence d)large (+ve) divergence

A magnetic field has the property

An electrostatic field has the property

0B

E


5/81

Curl

x y z

x y z

vx y z

v v v

( ) ( ) ( )y y

x xz z

v vv vv vx y zy z z x x y

A vector quantity

Measure of how much the vector curls around the point in

question or how much is the rotational effect.v

A magnetic field has the property

An electrostatic field has the property

oB J

0E


6/81

Integral calculus

Line Integral v

ld

: Vector function

: Infinitesimal displacement vector

Integral is to be carried out along a path P from point a to b.

At each point on the path we take the dot product of with

the displacement to the next point on the path.

For closed loop path:

F

Example: Work done by a force

b

aP

v dl

v dl

W F dl

vdl


7/81

Integral calculus

Surface Integral v

: Vector function

: Infinitesimal path of area with

direction perpendicular to the

surface

For closed surface:

Outward is positive for closed surface.

Arbitrary for open surface.

v dada

v da

Let = flow of liquid (mass per unit area per unit time). Then

= total mass per unit time passing through the

surface

= flux

v

v da


8/81

Integral calculus

Volume Integral T: Scalar function

: Infinitesimal volume element

Tdd

In Cartesian coordinates :

If T=T(x,y,z) = density of substance

Then

dxdydzd

dzyxT ),,( = Total mass


9/81

Divergence or Gausss Theorem

If = Flow of incompressible fluid

= Total amount of fluid passing out through

the surface per unit time

The divergence measures the spreading out of the vectors from

a point.

If there are many faucets within a region filled with incompressiblefluid then an equal amount of liquid will be forced out through

the boundaries of the region. In essence, divergence theorem

states

( ) sv d v da v

s

v da

(faucets within the volume) = (flow out through the surface)


10/81

Stokes theorem

( )s p

v da v dl Flux of curl through the

surface = Total amount of

rotation

Line integral of vector

around the boundary


11/81

Electric Field in Matter

Large classes of matter i) conductor

ii) Insulator

Conductor Dielectric

-

Dielectrics: All charges are attached to specific atom

or molecules. Atom as a whole is electrically neutral.

Conductor: Contains anunlimited supply of charges

that are free to move about

through the material

Eext

Eext

When an atom is

placed in an electric

field the positive andnegative charges are

pulled in opposite

directions.

The negative and

positive chargesattract each other.

Two opposing forces, a balance is reached. Atom gets polarized.

The atom has a tiny dipole moment, , in the direction ofp E

p E is the atomic polarizability.E=0


12/81

Molecules in electric field

Molecules are more complicated because they are typically asymmetric.

Non-polar molecules (CO2)

+OO C

O- O C

Eext00 Ewhenp

= Polarizability when the field is along the axis.

= Polarizability when the field is to the axis.

For general case :

11

1111EEp


13/81

Polar molecules (like water)

What happens when polar molecules are placed inan electric field?

Uniform field:

Force on positive end,

Force on negative end,

Two forces exactly cancels. However, there will be atorque:

( ) ( ) ( ) ( ( ))2 2d dN r F r F qE qE

EpEdqEqd

)2

(2

0 0p when E

EqF

EqF

EThe dipole in a uniform field experiences a torque N

F

F

E

d

+q

-q

r

r


14/81

Non-uniform field

does not exactly balance

There is a net force on the dipole in addition to the torque.

F

F

E

= difference between the fields at the plus end and minus end.

( )x xE d E dxxE

E xx

( ) ( )F F F q E E q E

d

E

If the dipole is very small (i.e. vectoris small) then small change in x

component of is

Gradient: .dT T dl dl T dl is very small.


15/81

Ep

EdqF

EdE

)(

)(

)(

More compactly

( )x xE d E ( )y yE d E ( )z zE d E

( ) ( ) ( )

( )

x y z

x y z

E x E y E z E

d E x d E y d E z

d E


16/81

PolarizationWhat happens to a piece of dielectric material when it is

placed in an electric field?

Neutral atoms:Electric field Induces tiny dipole moment,pointing in the same direction as the field (stretching).

Polar molecules:Molecule experiences a torque, tending to

line it up along the field direction (rotating).

Results:In the case of a material lot of little dipoles pointalong the direction of the fieldthe material

becomes polarized.

A convenient measure of this effect is

= dipole moment per unit volume, called polarization.

Now there are two fields:

1. Original field responsible for

2. New field due to

P

P

P


17/81

Bound ChargesConsider a long string of dipoles

Along the line head of one effectively cancels the tail of its neighbour .

Only two charges at the ends are left.

The net charges at the ends are called bound charge.

Results due to polarization and hence

Consider a tube of dielectric parallel to uniform polarization .

E

P

Dipole moment of tiny chunkIn terms of charge, dipole moment =

The bound charge that piles at the right end of the tube,

The surface charge density,

For an oblique cut, charge is still the same, but

Therefore,

( )P Ad qdPAq

PA

qb

cosendAA

b P n


18/81

Nonuniform Polarization

Bound Volume Charges

If the polarization is non-uniform, we get accumulations of boundcharge within the material as well as on the surface.

Diverging and hence results in a pile up of negative charge.

The net bound charge in a given volume

is equal and opposite to the amount thathas been pushed out through the surfacedue to non-uniform polarization.

(Using and divergence or Gauss Theorem)

= Volume charge density

bV S V

d P da P d

E

P

b

b P Therefore,

b P n


19/81

Uniformly Polarized Sphere

Two spheres of charge: a positive sphere and a negative sphere.

Without polarization the two are superimposed and cancelcompletely.

But when the material is uniformly polarized, all the plus

charges move slightly upward and all the minus charges moveslightly downward.

The two spheres no longer overlapPerfectly.

At the top there is a cap of leftoverpositive charge and at the bottom acap of negative charge.

This leftover charge is precisely the bound surface charge .b


20/81

Conversion of Gausss law from integral

equation to differential one

For closed surface:

0

1enc

S

E da Q

= total charge enclosed

within the surface.

= permitivity of free space

encQ

0

Applying divergence theorem:

We can also write: enc

V

Q d = Volume chargedensity

0

( ) ( )V V

E d d

Hence

This gives

0

1

E Gausss law in differential form

( )S V

E da E d

Th El t i Di l t


21/81

The Electric DisplacementGausss Law in the Presence of Dielectric

The effect of polarization is to produce accumulation of bound charge,

within the dielectric and on the surface.Now we take the field caused by both bound charge and free charge.

Gausss law in terms of total (volume) charge is:

where is now the total field inside thedielectric, not just that portion generatedby polarization.

Let (electric displacement)

Gausss law reads

b

P b P n

0 b f fE P

0D E P

fD

E

0 fE P

Bound charges are those

that accumulate through the

displacements that occur on

a molecular scale in

polarization process. The

conduction electrons in a

conductor and the electrons

injected into a dielectric with

a high energy electron beam

are examples of free

charges.

The electric field produced by bound charges dilute the effect of external field

for points within the dielectric. Thus the field experienced by the molecules of

the dielectric is less than the external applied field.


22/81

Linear Dielectric

Susceptibility and Permittivity

For many substances, the polarizationis proportional to the field,

provided is not too strong.

: electric susceptibility of the medium (dimensionless)

Materials that obey above equation are called linear dielectrics.

The total field may be due (i) in part to free charges and (ii) in

part to the polarization itself.

e0 eP E

E

E

0 0 0 0(1 )e eD E P E E E E

0 (1 )e

er

1

0

Permittivity of thematerial

Relative permittivity ordielectric constant

In vacuum, there isno matter to polarize

o ,0

e


23/81

Magnetic Fields in Matter

All magnetic phenomena are due to electric charge in motion.

In a magnetic material, at atomic scale, tiny currentsexist.Electrons orbiting around nuclei and electrons spinningabout their axes.

For macroscopic level: these current loops are very small and hence

these can be treated as magnetic dipoles.Random orientation of atoms cancels these dipoles.

But, when magnetic field is applied, a net alignment of these magneticdipoles occursmedium becomes magnetically polarized, or magnetized.

Three types of materials:1) Paramagnets: magnetization is parallel to

2) Diamagnetic: magnetization is opposite to

3) Ferromagnetic: retain their magnetization even after the external field

has been removed

B

B

T d F M ti Di l


24/81

Torques and Forces on Magnetic Dipoles

Electric dipole experiences a torque in an electric field.Magnetic dipole experiences a torque in a magnetic field.

Torque on a rectangular current loop in uniform magnetic field

Any current loop could be built up from

infinitesimal rectangles, with all the

internal side cancelling.

Centre of the loop at the origin.Loop is tilted by an angle from

the z-axis towards the y-axis.

is in the z-direction.

B

a = slanted sideb = side normal to BNet force on the loop is zero.


25/81

Torque on the loop (tending torotate it about x-axis is

sinN aF x

The force on each segment is ofthe magnitude

b= length of the horizontal sides of the loop (along x-axis)a = length of the slanted sides of the loop

Or

where = magnetic dipole moment of the loop.

F IbB

sin sinN IabB x mB x BmN

m Iab

magF I dl B

M ti ti


26/81

Each tiny loop has area a and thickness t

Surface current

Let is the outward-drawn unit vector then

bound surface current

m Mat Iam

b

IK M

t

nMKb

n

M

Magnetization

In the presence of magnetic field, matter becomes magnetized.

= magnetic dipole moment per unit volume called magnetization.

Plays a role analogues to the polarization in electrostatics.P

Tiny current loops are dipoles.

Equivalent to a simple ribbon of currentI flowing around the boundary

Slab of uniformly magnetized material

MatIa


27/81

It is a peculiar kind of current because no singlecharge makes the whole trip, each charge moves

only in a tiny little loop within a single atom.

The net effect is a macroscopic current flowing

over the surface of the magnetized object. This is

called bound current.

Every charge is attached to a particular atom, but

it is a perfectly genuine current and it produces a

magnetic field in the same way any other current

does.

Non Uniform magnetization


28/81

Non Uniform magnetization

The internal current no longer cancel.

Two adjacent chunks of magnetized materials.

Thick arrows - greater magnetization at that point.

When they join there is a net current in the x-direction.Non-uniform magnetization in z-direction:

Volume current density is

[ ( ) ( )] zx z zM

I M y dy M y dz dydzy

yMJ zxb

)(

A non-uniform magnetization in y-direction would contribute an amount

(-ve because when they join excess current flows in opposite direction)

so that

The right side is just the x-component of a curl.

Extending to 3-dimensions

z

My

( ) yz

b x

MMJ

y z

MJb

I Mt


29/81

What we have learnt:

The effect of magnetization is to establish bound currents

within the material

andon the surface

The field due to magnetization of the medium is just thefield produced by these bound currents.

Field = field due to bound currents+ field due to everythingelse( or free current)

Total current:

= due to magnetization results from the conspiracyof many aligned atomic dipoles.

= due to supply of current or transport of charge.

MJb

nMkb

bJ

fJ

b fJ J J


30/81

Conversion of Amperes law from integral

equation to differential one

0 encB dl I encI = Total current enclosed

by the integration path

encI J da Flow of charge is represented

by a volume current density (J).

Integral is taken over the surface bounded by the loop.

Applying Stokes theorem:

0B da J da

0B J Amperes law in differential form

( )s p

v da v dl

The Auxiliary Field H


31/81

1( ) f b f

o

B J J J J M

1

( ) foB M J

Amperes Law:

0

1H B M

Let

fH J

is auxiliary field or magnetic intensity.

is analogous to in electrostatics.D

H

Or,

( ) (1 )o o mB H M H B H (1 )o m

mM H

0,m o

For linear media :

Thus

where,

In vacuum, there is no matter to magnetize:

= magnetic susceptibility

(dimensionless)m

= permeability ofthe material

The Auxiliary Field H

H

C i f F d l f i t l ti


32/81

Conversion of Faradays law from integral equationto differential form

Changing magnetic field induces an electric field.

dE dl

dt

Emf

Flux B da

BE dl da

t

Faradays law

in integral form

Using Stokes theorem :

t

BE

( )s p

v da v dl


33/81

Boundary ConditionsElectrostatics

Electric field always undergoes a discontinuity when we

cross a surface charge .What is the amount of that changes at such boundary?

E

Thin Gaussian pill box A = area of the pill box lid

Gausss law states:

enc

o o

Q AE da

If thickness of the pill box tends to zero, the sides of pill box contributenothing to the flux. Thus we are left with

above below

o

AE A E A

The normal component of is discontinuous by an amount at anyboundary. It is continuous if .

OR

o

E

above below

o

E E

0


34/81

For tangential component of we apply:E

0E dl to the thin rectangular loop.

The ends give nothing as

thickness tends to zero.

The sides give

0above belowE l E l

E is the component of parallel to the surface.EThe boundary conditions on can be combined into the

following single formula :

E

above below

o

E E n

is a unit vector perpendicular to the surface pointing from below to above.n

above belowE E

Using Stokes theorem and curl of E equal

to zero, the integral around a closed path

for E is zero.

Magnetostatics


35/81

Magnetostatics

Magnetic field is discontinuous at a surface current.

0 0B B da

Applying to a wafer-thin pill box,we get 0above belowB A B A

ORFor tangential components, consider an amperian loop running

perpendicular to the current

above belowB dl B B l

o enc oI Kl

above belowB B

above below oB B K

The component of parallel to the

surface but perpendicular to the current

is discontinuous in the amount

B

oK

above below oB B K n

( )s

v d v da


36/81

Problems

1. A sphere of radius R carries a polarization

where k is a constant and is the vector from the center.

(a) Calculate the bound charges and .

(b) Find the field inside and outside the sphere.

( )P r krr

b b


37/81

2. A thick spherical shell (inner radius a, outer radius b) is

made of dielectric material with a frozen-in

polarization

where kis a constant and ris the distance from the center.

(There is no free charge in the problem.) Find the

electric field in all three regions by two differentmethods:

(a) Locate all the bound charge,

and use Gausss law to calculatethe field it produces.

(b) Find Dand then get E.

( ) kP r rr

3 The space between the plates of a parallel plate


38/81

3. The space between the plates of a parallel-plate

capacitor is filled with two slabs of linear dielectric material.

Each slab has thickness a, so the total distance between

the plates is 2a. Slab 1 has a dielectric constant of 2, and

slab 2 has a dielectric constant of 1.5. The free chargedensity on the top plate is + and on the bottom plate - .

(a) Find the electric displacement D in each slab.

(b) Find the polarization P in each slab.

(c) Find the location and amount of all bound charge.


39/81

4. An infinitely long circular cylinder carries a uniform

magnetization Mparallel to its axis. Find the magnetic field

(due to M) inside and outside the cylinder.

5. A long circular cylinder of radius R

carries a magnetization

where kis a constant, sis the

distance from the axis, and is theusual azimuthal unit vector. Find the

magnetic field due to M, for points

inside and out side the cylinder.

2 M ks


40/81

6. An infinitely long cylinder, of radius R, carries a frozen-

in magnetization, parallel to the axis,

M kszwhere kis a constant, sis the distance from the axis; there

is no free current anywhere. Find the magnetic field inside

and outside the cylinder by two different methods:

(a) Locate bound currents, and calculate the field they

produce.

(b) Use Amperes law to find H, and then get B.


41/81

7. A coaxial cable consists of two very long cylindrical

tubes, separated by linear insulating material of magnetic

susceptibility . A current I flows down the linear

conductor and returns along the outer one; in each case,the current distributes itself uniformly over the surface.

(a) Find the magnetic field in the region between the tubes.

(b) Calculate the bound currents, and confirm that

(together, of course, with free currents) they generate thecorrect field.

m


42/81

8. An iron rod of length L and square cross section (side

a) is given a uniform longitudinal magnetization M, and

then bent around a circle with a narrow gap (w). Find the

magnetic field at the centre of the gap assuming

w


43/81

The flow of charge distributed throughout a three dimensional

region is described by volume current density .

Consider a tube of infinitesimal cross

section , running parallel to the flow.

= current in the tube

Thus

J

da

dI

dIJ

da

Jis the current per unit area perpendicular to flow.

We can also writeS S

I Jda J da Total charge per unit time leaving a volume V is

S VJ da J d

V V Vd

J d d ddt t

Jt

Since charge is conserved, whatever flows

out through the surface must come at the

expense of that remaining inside. Therefore

Continuity equation

Thus

Continuity Equation


44/81

Electrodynamics before Maxwell

i) (Gausss law)

ii) (no name)

iii) (Faradays law)

iv) (Amperes law)

01 E

0 B

BEt

0B J

There is a inconsistency in these formulas.


45/81

The divergence of curl is always zero.

( ) ( ) ( )y yx xz z

v vv vv vx y z

y z z x x y

v

( ) 0v

Let us apply this to Faradays law.

( ) B

E Bt t

Its both sides are zero and hence is satisfied.

Let us apply this to Amperes law.

0( )B J Its left side is zero but the right side is, in general, not zero

hence is not satisfied. For steady currents, the divergence of

J is zero. The problem is on the right side.

Amperes law is not right.

How Maxwell Fixed Amperes Law


46/81

How Maxwell Fixed Ampere s Law

0 0E

J Et t t

If we combine with J in Amperes law, it would be just right

to kill off the extra divergence. Thus

0

E

t

0 o oEB Jt

Such modification changes nothing, as far as magnetostatics is concerned:

when Eis constant, we still have .

However, it plays a crucial role in the propagation of electromagnetic waves.

Apart from curing the defect in Amperes law, Maxwells term implies:

A changing electric field induces a magnetic field.

0B J


47/81

Maxwells Equations

i) (Gausss law)

ii) (no name)

iii) (Faradays law)

iv) (Amperes law)

0

1 E

0 B

BE

t

Maxwells

correction to

Amperes law

0 o o

EB J

t

A changing electric field induces a magnetic field.

Electromagnetic Waves


48/81

Electromagnetic WavesWave

Disturbance of a continuous medium that propagate with a fixed shape

at constant velocity.In the presence of absorption, wave diminishes in size as it moves.If the medium is dispersive different frequencies travel at the differentspeeds.

Standing waves do not propagate at all.

Sinusoidal Waves: The most familiar waveform.

])(cos[),( vtzkAtzf

A= Maximum amplitude or displacement of the wave

= Phase constant (02)

Central maximum when

phase is zero.

0)( vtzk

2k

= wave number


49/81

As time passes, the entire wave train proceeds to the right, at

speed v. At any fixed point z, the string vibrates up and down,

undergoing one full cycle in a period2

T kv

Frequency (no. of oscillations/time) is

More convenient unit is angular frequency .

Sinusoidal wave in terms of is

)cos(),( tkzAtzfIf it is travelling to left then

)cos(),( tkzAtzf

1

2

kv v

T

2 kv

Complex Notations


50/81

Complex Notations

Exponentials are much easier to manipulate than sines

and cosines.Eulers formula:

Sinusoidal wave can be written as

represents the real part of the complex number

cos sinie i

, Re

i kz t

f z t Ae

Re

Let us introduce a complex wave function:

=

complex amplitude absorbing phase constant

Actual wave function is:

itkzi

eeAtzf )(

~

),(

)(~ tkzieA

iAeA~

( , ) Re[ ( , )]f z t f z t

Interface : Boundary conditions


51/81

Interface : Boundary conditions

(Reflection and Transmission)

Incident wave: (z < 0)

Reflected wave:(z < 0)

Transmitted wave: (z > 0)2~

( )( , ) i k z t TT

z t A ef

)(

~

1~),( tzkiII

eAtzf

)(

~

1~

),( tzkiRR

eAtzf

All parts of the system are oscillating at the same frequency .

The wave velocities are different in two regions. The

wavelength and wave numbers are also different.

)(

)()(

2

11

~

~~),(

~

tzki

T

tzki

R

tzki

I

eA

eAeAtzf

The net disturbance is(z < 0)

(z > 0)

At the boundary (z = 0) f(z t) should be continuous


52/81

At the boundary (z 0), f(z,t) should be continuous.

Therefore~ ~

0 0

(0 , ) (0 , )

z z

f t f t

f f

z z

Applying above conditions:1

1 2

1 2

1 2

2T I

R I

kA Ak k

k kA A

k k

2

2 1

2 1

2 1

2T I

R I

vA A

v v

v vA A

v v

In terms of velocities:

kv const

The real amplitudes and phases are related by


53/81

p p y

If the second medium is heavier than the first the

reflected wave is out of phase by 180 deg .

In other words, since

1 2v v

12 vv

2 1

2 1

R Ii i

R I

v vA e A e

v v

2

2 1

2T Ii i

T I

vA e A e

v v

If the second medium is lighter than the first all threewaves have the same phase angle and the

corresponding amplitudes are( )R T I

2 1

2 1R I

v v

A Av v

2

2 1

2T I

vA A

v v

( )R T I

1 1cos cosI Ik z t k z t the reflected wave is upside down. The amplitudes are

1 2

2 1

R I

v vA A

v v

2

2 1

2

T I

v

A Av v

Polarization


54/81

Polarization

Vertical polarization

xeAtzf tzki

Iv

~

),(

~ )( 1

yeAtzf tzkih ~

),(~ )( 1 Horizontal polarization

neAtzf tzki

I

~),(

~ )( 1 Plane of vibrationalong n

Since the waves are transverse, is perpendicular to the direction of

propagation and hence

n0. zn

yxn sincos

yeAxeAtzf

tzkitzki

sin

~

cos

~

),(

~ )()(

Superposition of two waves- one horizontally polarized and the other

vertically:

Electromagnetic waves in vacuum (Q = 0, J = 0)


55/81

Electromagnetic waves in vacuum (Q 0, J 0)

Maxwells equations areEquations 3 and 4 are coupled, first

order partial differential equations for

. They can be decoupled byapplying curl to equations 3 and 4.

BandE

0

0

o o

E

B

BE

t

EBt

2

2

22

2

22

2

( ) ( )

( ) 0

( )o o

o o

E E E

BE

t

B EE

t t

EE

t

22

2o o

BB

t

Similarly

In vacuum, each Cartesian component of E and B satisfies the three

dimensional wave equation: 22

2 2

1 ff

v t

Maxwells equations imply that empty space supports

the propagation of EM waves travelling at a speed :

8

0 0

13 10 /v x m s

M h ti Pl W


56/81

Monochromatic Plane Wave

Fixed frequency

Wave is travelling in z-direction and has no x and ydependence

Fields are uniform over every plane perpendicular to

the direction of propagation.

Consider following forms of the fields:( )

0

( )

0

( , )

( , )

i kz t

i kz t

z t e

B z t B e

where and are thecomplex amplitudes with no x

and y dependence.

and should, apart from

wave equation, follow

Maxwells equations

0E 0B

BE

( ) [( ) ( ) ( ) ] i kz t E E x E y E z e


57/81

0 0 0

( )

0

0

0 z

[( ) ( ) ( ) ]

0

E= 0 + 0+ ik ( ) 0

( ) 0

Applying B=0

gives (B ) 0

x y z

x y z

yx z

i kz t

z

z

E E x E y E z e

E E x E y E z

EE EE

x x x

E e

E

Electromagnetic waves are transverse.

The electric and magnetic fields are perpendicular

to the direction of propagation.

( )( ) i kz t B z t B e


58/81

y xZ

yx Z

y x z

BE

t

E BE

y z t

BE E

z x t

E E B

x y t

0( , )B z t B e

0 0

0 0 0 0

0

0 0

0 0 0 0

0 0

0

0 0

0 0 ( ) ( ) 0

( ) 0 ( ) ( ) ( )( )

0 0 ( ) 0

( ) 0 ( ) 0

0 ( ) ( ) ( ) ( )( )

0 0 ( ) ( ) 0

0 0 ( ) 0

( ) ( )( )

(

x x

x y y x

z

y x

y x x y

y y

z

y x

i B B

kik E i B B E

i B

LetE const and E

then

kik E i B B E

i B B

i B

kB E

0 0) ( )( )x yk

B E

Let us assume . 0ox

E cons 0oy

E

Magnetic vector is

perpendicular to

electric vector.

Maxwells equation:( )

0 0 0 [( ) ( ) ( ) ]

i kz t

x y zB B x B y B z e

( ) ( ) ( )B B B B


59/81

0 0 0 0

0 0 0 0

0 0 0

0 0 0

0 0

0 0 0

( ) ( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( ) ( )

=( ) ( )

1

( ) ( )

x y z

x y z

x y

y x

x y

B B x B y B z

E E x E y E z

z E E y E x

k k k

z E E x E y

B x B y

k

B z E k E

and are in phase and mutually perpendicular. Their amplitudes (real)

are related by

E B1

o o o

kB E E

c

2

2

k

Generalization:


60/81

Generalization:

There is nothing special about z-direction. It can travel in an arbitrarydirection. We consider a propagation (or wave) vector, pointing inthe direction of propagation, whose amplitude is the wave number k.

is the appropriate generalization of kz.

k

,

is the polarization vector.

Since is transverseE

zzyyxxr

k r

( )( , ) i k r t oE r t E e n

( ) 1 ( , ) ( ) ( )i k r t oEB r t e k n k Ec c

n

0n k

0

0

( , ) cos( . )

1

( , ) cos( . )( )

E r t E k r t n

B r t E k r t k n

c

(Real)

Energy in Electromagnetic Waves


61/81

0 0

2 2 2

0 0 0

2 2

0 0

1

1( )

2cos ( )

B E Ec

u E E E

E kz t

)1

(

2

1 2

0

2

0 BEu

0

1( )S E B

22 2 2

0 0 02

0 0 0

2 2

0 0

1

cos ( )

E c cS E E c E

c c

S c E kz t z

gy gEnergy per unit volume stored in e.m. field is

In the case of monochromatic plane wave

Electric and magnetic contributions areequal.

As the wave travels, it carries thisenergy along with it.

Energy per unit area per unit time (energy flux density) transportedby the fields is given by the Poynting vector:

For monochromatic plane wave propagatingin z-direction:

2

0 0

1

2I s c E

Average power per unit

area transported:

2

0

1 1cos ( 2 / ) 2

T

kz t T dt T

Electromagnetic Waves in Matter


62/81

g

(Q=0 , J=0 region only)

0

0

E

B

B

E t

EB

t

Maxwells equations are

BH

ED

1

1 cv

n

Difference with vacuum is only inthe replacement of by .0 0

0 0

n

n is the refractive index of the material.1

( )

1

S E B

B E

v

2

0

1

2

I vEIntensity:

Poynting

vector:

Boundary Conditions


63/81

1 1 2 2

1 2

1 2

1 2

1 2

(1)

(2)

(3)

1 1(4)

E E

B B

E E

B B

y

Medium 2Medium 1

1

2 21

above below

o

E E

above below oB B K

Reflection and transmission at normal incidence


64/81

Reflection and transmission at normal incidence

A plane wave of frequency

, traveling in z-direction

and polarized in x-directionis incident on the interface.

1

1

1

1

2

2

( )

0

( )0

1

( )

0

( )0

1

( )

0

( )0

2

( , )

( , )

( , )

( , )

( , )

( , )

i k z t

I I

i k z t II

i k z t

R R

i k z t R

R

i k z t

T T

i k z t TT

E z t E e x

EB z t e y

v

E z t E e x

E

B z t e yv

E z t E e x

EB z t e y

v

Since the plane wave is incident

perpendicularly, the perpendicular

components of E and B will be zero

and only parallel components will be

there and hence the boundary

conditions will be1 2

1 2

1 2

1 1

E E

B B

EEE~~~

At z = 0:


65/81

TRI EEE 000 At z 0:

and 0 0 01 1 1 2 2

1 1 1 1 1( ) ( )I R TE E Ev v v

1 10 0 0 0

2 2

I R T Tv

E E E Ev

1 1 1 2

2 2 2 1

v n

v n

0 0

2( )

1T IE E

0 0

1( )

1R IE E

OR where

Solving

If permitivities are close to their values in vacuum,( ) 2

1

n

n

0I

21

10T

0I21

21

0R

E~

)2

(E~

E~

)(E~

nn

n

nn

nn

2 21 2

1 2

22 2

1 1

2

2 1 1 1 2

2 2

1 2 1 2 1 2

( ) ( )

( )

4 4

( ) ( )

0R

0I

0T

0I

=

R

I

T

I

EI n nR

I E n n

EI vT

I v E

n n n n

n n n n n

and

20

1

2I vE

22 1 2 1n n

Reflection and transmission at oblique incidence


66/81

( )

0

1

( )

0

1

( )

0

2

( , )

1( , ) ( )

( , )1

( , ) ( )

( , )

1( , ) ( )

I

R

T

i k r t

I I

I I I

i k r t

R R

R R R

i k r t

T T

T T T

E r t E e

B r t k Ev

E r t E e

B r t k Ev

E r t E e

B r t k E

v

q

211 vkvkvk TRI

TTRI kn

nk

v

vkk

2

1

1

2

All three waves have same

frequency

Z = 0

kI

kRkT

T

I

R

E E E Using boundary conditions: At z = 0


67/81

I R T

I R T

E E E

B B B

We get

I R Ti(k .r t) i(k r t) i(k .r t) e e e In the above equation x, y and t dependence is confined tothe exponents.

Since the boundary conditions must hold at all points on theplane (i.e., independent of values of x and y on z=0) and forall times, the exponents must be equal. Otherwise a slightchange in x, say, would destroy the equality.

I R Tk r k r k r

( ) ( ) ( ) ( ) ( ) ( )I x I y R x R y T x T yk x k y k x k y k x k y

Therefore, when z = 0

Th b ti l h ld if th t


68/81

The above equation can onlyhold if the componentsare separately equal.

If x=0, then

( ) ( ) ( )I y R y T yk y k y k y (y-z plane, x = 0)

and if y = 0, then

( ) ( ) ( )I x R x T xk x k x k x (x-z plane, y = 0)

First law: The incident, reflected and transmittedwave vectors lie in a plane that includes normal tothe surface (interface) and is called plane of

incidence.

Second Law :


69/81

( ) ( ) ( )I x R x T xk x k x k x

TTRRII kkk sinsinsin

I R

I R

k k

Snells Law of Reflection

If y = 0 then

Since

Therefore


70/81

Third Law :

sin sinI I T Tk k

1

2

sin sinT I T T n

k k

n

OR

1 2sin sinI Tn n

Snells Law of Refraction

These are the three fundamental laws of geometrical optics.

After taking care of exponential factors which cancel, the


71/81

boundary conditions are:

1 0 0 2 0

0 0 0

0 0 , 0 ,

0 0 , 0 ,

1 2

( )

1 1( ) ( )

I R z T z

I R z Tz

I R x y T x y

I R x y T x y

(E E ) (E )

B B B

(E E ) (E )

B B B

Perpendicular

Parallel to interface

where in each case.

The last two equations represent pairs of equations, one

for the x-component and one for the y-component.

0 0

1B v k E

1.

2.

3.

4.

Case : Polarization of the incident wave is parallel to the


72/81

plane of incidence ( lies in x-z plane)

The reflected and transmitted waves are also polarized in

this plane.

E

1 0 0 2 0

2 2 10 0 0 0

1 1 2

2

1 1 2 1 10 0 02

2 2 1 2 2

sin sin sin

sinsin

.

I I R R T T

TI R T T

I

T T T

( E E ) ( E )

nE E E En

v v vE E E

v v v

Boundary condition 1 gives:

where 1 1

2 2

v

v

1v

1 2

2 1

n v

n v

0 0 0I R TE E E Equation (1)

1 0 0 2 0I R z T z(E E ) (E )

Boundary condition 2 gives nothing ( 0 = 0 ) because magnetic


73/81

fields have no z-components.


0 0 0

0 0 0 0

cos cos cos

cos

cos

I I R R T T

TI R T T

I

E E E

E E E E

coscos

T

I

0 0 0I R TE E E Equation (2)


0 0 01 1 2 2

1 1

( )I R TE E Ev v along y-direction

0 0 0I R TE E E Or, Same as Equation (1)

Solving Equations (1) and (2) gives

0 0 0

0 0 , 0 ,

( )I R z Tz

I R x y T x y

B B B

(E E ) (E )

0 0 , 0 ,

1 2

1 1( ) ( )I R x y T x yB B B

2( )E E


74/81

0 0( )R IE E

0 0( )T IE E

Fresnels equations for the case of polarization in the

plane of incidence.

Note:

1. The transmitted wave is always in phase with the

incident one.

2. The reflected wave is in phase if , and 180

deg out of phase if .

3. The amplitudes of the transmitted and reflected

waves depend on the angle of incidence because

is a function of :

I2 21

2

2

1 ( ) sin1 sincos

cos cos cos

I

TT

I I I

n

n

Three cases:


75/81

1. Normal incidence ( ), , we get same equations

that we have got earlier.

2. Grazing angle of incidence ( ), diverges and thewave is totally reflected.

3. There is an intermediate angle, , called Brewsters

angle, at which the reflected wave is completely lost. This

occurs when .

0I 1

90o

I

B

2 21

2

1 ( ) sin

cos

B

B

n

n

22

221

2

2

1sin B

n

n

For the case and hence :1 2 2 1n n2

2

2sin

1B

This gives 2

1

tan Bn

n


76/81

For:

1

2

1

1.5

n

n

21 1 01 cos

2I I II v E

2

1 1 0

1cos

2R R RI v E

2

2 2 0

1cos

2T T TI v E

Intensities:

Reflection and transmission coefficients for waves polarized


77/81

parallel to the plane of incidence are:

2 2

0

0

RR

I I

EI

R I E

2 2

02 2

1 1 0

cos 2

cos

TT T

I I I

EI vT

I v E

Group and Phase Velocities


78/81

Consider the waves having same or comparable amplitudes

but different frequencies:

1 0 1 1

2 0 2 2

cos

cos

E E k z t

E E k z t

Superposition of these waves traveling together in a given

medium is

1 2 0 1 1 2 2cos cosRE E E E k z t k z t

1 1cos cos 2cos cos2 2 Using

1 2 1 2 1 2 1 202 cos cos

2 2 2 2R

k k k k E E z t z t

Let1 2 1 2 1 2k kk

1 2k kk


79/81

1 2

2p

1 2

2g

1 2

2p

k 1 22

gk

Then

02 cos cos

R p p g gE E k z t k z t

Product of two cosine waves having frequencies

and and wave propagation constants and .p

g gkpk

Its application is in the phenomenon

of dispersion. Due to dispersion, lightcomponents of different wavelengths

travel with different speeds through a

refractive medium. The resultant of

waves E1and E2is shown.

The velocity of higher frequencywave is called the phase velocity:

1 2

1 2

p

p

p

vk k k k

The velocity of the envelope wave is

called the group velocity:

1 2

1 2

g

g

g

dv

k k k dk

Group velocity determines the speed with which energy

i t itt d


80/81

is transmitted.

Relation between phase and group velocities:

( )p pg p

d kv dvdv v kdk dk dk

In non-dispersive medium, velocity of a wave does not

depend on wavelength, therefore

0pdv

dk and hence g pv v

In dispersive medium:

( )p cv

n k

n= refractive index

2

p pdv vc dn dn

dk n dk n dk

1 1p

g p p p

kv dn k dn dnv v v v

n dk n dk n d

Problems


81/81

1. An electromagnetic wave is given by

where E0is a constant. Prove that the Poynting vector is .

2. Show that the energy density corresponding to the field

is a constant everywhere.

3. Calculate and in a medium characterized by

4. In free space a plane wave with mA/mis incident normally on a lossless medium in

region . Determine the reflected wave and the transmitted

wave

1 2 2

0 E z

1 2 1 20 0 cos sinE xE z t yE z t

0 0 cos sinE xE z t yE z t

H

80 0 0, , 4 , 20sin 10E t z y

0z 8

10cos 10H x t z 0 08 , 2

0z ,r rH E

H E

Documents

PHL100 BDG Course Slides