1

EM Waves

This Lecture

!More on EM waves

!EM spectrum

!Polarization

From previous Lecture!Displacement currents

!Maxwell’s equations

!EM Waves

2

Reminders on waves

! Traveling waves on a string along x obey the wave equation:

y=wave function

General solution : y(x,t) = f1(x-vt) + f2(x+vt) y = displacement

Traveling wave: superposition of sinusoidal waves (produced by a source that oscillates with simple harmonic motion): y(x,t) = A sin(kx-"t)

y(x,t) = sin(kx+ "t)

A = amplitudek = 2#/! = wave number ! = wavelength

f = frequency T = 1/f = period" = 2#f=2#/T angular frequency

!

!

" 2y(x, t)

"x 2=1

v2

" 2y(x, t)

"t 2

pulse traveling along +x pulse traveling along -x

v

Maxwell equations in vacuum

! in the absence of charges (q=0) and conduction currents (I=0)

!

E " dA = 0S# (Gauss' Law) B " dA = 0

S#

E " ds = $d%

B

dt (Faraday - Henry) B " ds

L# =

L# µ

0&0

d%E

dt (Ampere - Maxwell law)

From these equations we get EM wave equations traveling in vacuum!

4

!

EM waves from Maxwell equations

E x B direction of c

Solutions of these equations:

sinusoidal traveling transverse waves propagating along x

! E = Emax cos (kx – "t)

! B = Bmax cos (kx – "t)

•E and B are perpendicular oscillating vectors

E ! B = 0

5

E and B are orthogonal

! An easy way to understand this:!B = B A cos$

1. Max B flux $=0 => also

circular E is largest!2. Less flux

3. Null flux $=90° =>

circular E smallest!B parallel to area normal and E perpendicular to circuit so E % B

!

E " ds = -d#

B

dt$

E orthogonal to B!

1

2

3

$

I

Increasing B-field

E

E ! B = 0

Shown below is the E-field of an EM wave broadcast at 30 MHz and traveling to the right.

What is the direction of the magnetic field during the first !/2?

What is the wave length?

Quick Quiz on EM Waves

E

x

1) Into the page 2) out of the page

1) 10 m 2) 5 m

!/2

xz

y

E

B

loop in xy plane

loop in xz plane

loop in

yz

plane

A B C

Which orientation will have the largest induced emf?

Quick Quiz pn EM waves Important Relation between E and B

! E = Emax cos (kx – "t)

! B = Bmax cos (kx – "t)

! First derivatives:

!

"E

"x= #kEmax sin(kx #$t)

"B

"t=$Bmax sin(kx #$t)

"E

"x= #

"B

"t

This relation comes from Maxwell’s equations!

From:

EM Waves generators: Antennas

! 2 rods connected to alternate current generator; charges oscillate between the rods (a)

! As oscillations continue, the rods become less charged, the field near the charges decreases and the field produced at t = 0 moves away from the rod (b)

! The charges and field reverse (c)

! The oscillations continue (d)

Sources of EM waves: oscillating charges, accelerated/decellerated charges, electron transitions between energy levels in atoms, nuclei and molecules

The EM Spectrum

X-rays: ~10-12 -10-8 m

source: deceleration of high-energy

electrons striking a metal target

Diagnostic tool in medicine

Source: atoms and molecules Human eyeVisible range from red (700 nm) to violet (400 nm)

Radio: ! ~ 10 - 0.1 m

Sources: charges accelerating through conducting wires Radio and TV

Microwaves: ! ~10-4 -0.3 m

sources: electronic devices

radar systems, MW ovens

Infrared: ! ~ 7 x 10-7-10-3 m

Sources: hot objects and molecules

UV !~ 6 x 10-10 - 4 x 10-7 m

Most UV light from the sun is absorbed in the stratosphere by ozone

Gamma rays: !~ 10-14- 10-10 m

Source: radioactive nuclei, cause serious damage to living tissues

Poynting vector

! Rate at which energy flows through a unit area perpendicular to direction of wave propagation

! This is the power per unit area (J/s.m2 = W/m2)

! Its direction is the direction of propagation of the EM wave

! Magnitude:

! Magnitude is time dependent

! reaches a max at the same instant as E and B do

!

S =EB

µ0

=E2

cµ0

E/B=c

Energy density of E and B field

! In a parallel plate capacitor:

! Similarly for a solenoid with current:

!

C ="0A

d

!

U =1

2CV

2=1

2

"0A

dE2d2# u

E=U

Ad=1

2"0E2 True for any geometry

True for any geometry

!

uB

=B2

2µ0

Energy carried by EM waves

!

uE

=1

2"0E2 = u

B=B2

2µ0

=E2

2c2µ

0

! Total instantaneous energy density of EM waves

u =uE + uB = 1/2 'oE2 + B2 /(2µo)

! Since B = E/c and

EM waves carry energy!In a given volume, the energy is shared equally by the two fields

!

uE

= uB

The average energy density over one or more cycles of oscillations is:

!

uav

= 2uE ,av

= "0E2

=1

2"0Emax

2 since <sin2(kx - "t)> = 1/2E x B

Intensity and Poynting vector

! Let’s consider a cylinder with axis along x of area A and length L and the time for the wave to travel L is !t=L/c

! The average power of the EM wave in the cylinder is:

! The intensity is

!!

Pav

=U

av

"t=uavAL

"t= u

avAc

E/B=c

!

I =Pav

A=1

2"0cE

max

2 =EmaxBmax

2µ0

average power per unit area (units W/m2)

!

I= Sav

I & E2

Radiation pressure and momentum

! Complete absorption on a surface: total transferred momentum p = U / c and prad=Sav/c

! Radiation momentum:

! Radiation Pressure

! Perfectly reflecting surface: p = 2U/c and prad = 2Sav/c

!

F = ma = mdv

dt=dp

dt

!

p = force " time = force " distance "time

distance=energy

velocity=U

c

the radiation momentum is the radiation energy/velocity

!

prad =F

A=

p

A"t=

U

cA"t=Power

cA=I

c

!

Sav

= I =Power

A

Radiation pressure from the Sun

! Solar intensity at the Earth (~1.5 x 108 km far from the Sun):

~1350 W/m2

! Direct sunlight pressure I/c ~4.5 x 10-6 N/m2

! If the sail is a reflecting mirror prad = 2 x 4.5 x 10-6 N/m2

! Can be used by spacecrafts for propulsion as wind for sailing boats!

! What is the sail area needed to accelerate a 104 kg spacecraft at a = 0.01 m/s2 assuming perpendicular incidence of the radiation on the sail?

! A = F/prad = ma/prad = 107 m2

!

prad =F

A=Power /A

c=I

c

!

=Sav

c

Polarization of Light Waves (34.8)

! Linearly polarized waves: E-field oscillates at all times in the plane of polarization

Linearly polarized light: E-field has one spatial orientation

Unpolarized light: E-field in random directions. Superpositionof waves with E vibrating in many different directions

Circular and elliptical polarization

! Circularly polarized light: superposition of 2 waves of equal amplitude with orthogonal linear polarizations, and 90˚ out of phase. The tip of E describes a circle (counterclockwise = RH and clockwise=LH depending on y component ahead or behind)

! The electric field rotates in time with constant magnitude.

! If amplitudes differ ( elliptical polarization

Producing polarized light

! Polarization by selective absorption: material that transmits waves whose E-field vibrates in a plain parallel to a certain direction and absorbs all others

This polarizationabsorbed

This polarizationtransmitted transmission axis

Polaroid sheet (E. Land 1928)Long-chain hydrocarbon

molecules

DEMO with MW generator and metal grid

MW generator

Metal grid

pick up antenna connected to Ammeter

If the wires of the grid are parallel to the plane of polarization the grid absorbs

the E-component (electrons oscillate in the wire).

The same thing happens to a polaroid: the component parallel

to the direction of the chains of hydrocarbons is absorbed.

If the grid is horizontal the Ammeter will measure a

not null current since the wave reaches the antenna

pick-up

This

polarization

absorbed

This polarization

transmitted transmission axis

Relative orientation of polarizers

! Transmitted amplitude is Eocos$

(component of polarization along polarizer axis)

! Transmitted intensity is Iocos2$

( square of amplitude)

! Perpendicular polarizers give zero intensity.

Polarizers and MALUS’ LAW

transmission axis

Polaroid sheet

Long-chain hydrocarbon molecules

If linearly polarized light (plane of polrization indicated by red arrow) of intensity I0 passes through a polarizing filter with transmission axis at

an angle $ along y

Einc = E0sin$ i + E0 cos$ j

After the polarizer Etransm = E0cos$ j

So the intensity transmitted isItransm = E0

2 cos2$ = )0cos2$

y

x

$

E0cos$

this is calledMalus’ law

! Transmitted intensity: I = I0cos2$ I0 = intensity of polarized beam on analyzer

(Malus’ law)

Unpolarized light on polarizers

Allowed componentparallel to analyzer axis

Polaroid sheets

Only I0/2 is transmitted of unpolarized light by a polarizer and it is polarized along the transmission axis.An analyzer rotated at an angle $ respect to the polarizer transmits

100% of the incident intensity when $ = 0 and zero when $ = 90°