EM Waves: Lecture Notes on Electromagnetic Waves, Study notes of Geology

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Lecture 32

Chapter 34

Electromagnetic Waves

Review

ï^

EM Waves ññ Wavelengths of 10

8

to

10

-

16

meters (10-

24

Hz)

ñ Traveling wave of both

E

and

B

fields

ñ

E

field is

⊥⊥⊥⊥

B

field

ñ Wave moves in direction

⊥⊥⊥⊥

to both

E

and

B

fields

ñ

E

and

B

vary sinusoidally

with same frequency ñ At large distances fields

are in phase

B

E

r

r

×

)

sin(

t

kx

E

E

m

ω −

=

)

sin(

t

kx

B

B

m

ω −

=

Review

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Poynting vector,

S

ñ rate of

energy transported per unitarea ï^

Instantaneous energy flow rate ï^

Defined intensity

I

to be time

averaged value of

S

2

0

rms

avg

E

c

S

I

μ

ave

ave

avg

powerarea

area

time

energy

S I^

  

  

  

  

=

/

B

E

S

r

r

r^

×

EB

S

0

=

EM Waves (12)

ï^

Problem ñ Isotropic point light source as powerof 250 W. You are 1.8 meters away. Calculatethe rms values of the

E

and

B

fields.

ï^

To find

E

rms

need

ï^

Find intensity

I

from

2

0 1

rms E

c

I

μ

=

2

r

P

I^

s π

(^02)

0

4

c r P

Ic

E

s

rms

π

μ

μ

=

=

m V

E

rms

2

8

8

×

×

−

π

EM Waves (14)

ï^

Look at sizes of

E

rms

and

B

rms

ï^

This is why most instruments measure

E

ï^

Does not mean that

E

component is stronger

than

B

component in EM wave

ñ Canít compare different units ï^

Average energies are equal for

E

and

B

T

B

rms

7

10

(^6). 1

−

×

=

m

V

E

rms

/

(^1).

48

EM Waves (15)

ï^

The energy density of electric field,

u

E^

is equal

to energy density of magnetic field,

u

B

B

c

E

=

2 0 1 2

E

u

E

ε

=

2 2 0 1 2

2

0 1 2

)

(^

B c

cB

u

E

ε

ε

=

=

0

1 ε^0

μ

c

2 0

2 0 0 0

ε^

B

B

u

E^

2 0 2

B^ μ

u

B^

B

E

u

u

=

EM Waves (17)

ï^

Just defined intensity,

I

as power

per unit area

A

so power is

ï^

Change in energy is amount ofpower

P

in time

t

ï^

Want force of radiation on object ï^

For total absorption ï^

Find force is

I

A

P

t

IA

t

P

U

p^ t

F

∆ ∆

U c

p

IA^ c

t

c

t

IA

t

c

U

p t

F

EM Waves (18)

ï^

For total reflection back along original path ï^

Express in terms of radiation

pressure

p

r^

which is force/area

ï^

SI unit is N/m

2

called pascal

Pa

U c

p

IA c

t c

t

IA

t c

U

p t

F

2

2

2

=

∆

∆

=

∆ ∆

=

∆ ∆

F^ A

p

r^

=

I c

p

r^

=

I c

p

r

2

ï^

Total absorption

ï^

Total reflection

EM Waves (20)

ï^

Source emits EM waveswith random planes ofoscillation (

E

field changes

direction) is unpolarizedñ Example, light bulb or Sun ï^

Resolve

E

field into

components ï^

Draw unpolarized light assuperposition of 2 polarizedwaves with

E

fields

to

each other

EM Waves (21)

ï^

Transform unpolarizedlight into polarized byusing a polarizing sheet ï^

Sheet contains longmolecules embeddedin plastic which wasstretched to align themolecules in rows ï^

E

field component || to polarizing direction of sheet is passed (transmitted), but

component is absorbed

EM Waves (23)

ï^

For polarized light, resolve Einto components ï^

Transmitted || component is ï^

Use definition of intensity ï^

Cosine-squared rule: Intensity ofpolarized wave changes as cos

2 θθθθ

θ 2

0

cos I

I^

=

θ

cos E

E

y^

=

θ

θ

μ

μ

2 0 2 2 0 2 0

cos

cos

I

E

c

E

c

I^

EM Waves (24)

ï^

Have 2 polarizing sheetsñ First one called polarizerñ Second one called analyzer ï^

Intensity of unpolarizedlight going throughpolarizer is ï^

Light is now polarized andintensity of light afteranalyzer is given by

2

0

cos I

I

=

0

1 2

I

I^

=

EM Waves (26)

ï^

Look at relative orientation of polarizationdirection between the 2 sheets. ï^

What is the intensity if the sheets areÖñ Polarized || ñ all light passesñ Polarized

⊥⊥⊥⊥

to each other ñ no light passes

ñ For angles in between ñ get more light if closer to ||

a,d,b,c