ak lecture5

8/11/2019 AK Lecture5

1/16

PHL 100

Instructor: Arun KumarLecture - 5

Electromagnetic Waves and Quantum Mechanics

1


2/16

2

1/ 22

, 1 1

2r ik k

= +

K is complex

0

0

exp( z) exp [ ( )]exp( z) exp [ ( )]

i r

i r

E E k i k z tB B k i k z t

= =

Real part of k determines the wavelength, phase velocityand refractive index:

2r p r

r p

c

k v n k k v

= = =

EM Waves in conductors:Recap:


3/16

3

Poor conductor >

,

2

r ik k

1 1

2i r

d

k k

= = =and

1

id k

=

Skin Depth


4/16

4

Skin depth for silver at optical frequencies

0 0

7 1 15

,

10 ( ) , ~ 10m Hz

= >>

21 2 10 o

i

d Ak

=


5/16

5

Skin depth for salt water at radio frequencies

21d cm

0

10 2 2

1

6 10 /

5 ( ) , 10

C N m

m GHz

= =

Acoustic communication or direct cable link to sea

surface is required


6/16

6

0

0

exp( z) exp [ ( )]

exp( z) exp [ ( )]

i r

i r

E x E k i k z t

kB y E k i k z t

=

=

Relation between E and B fields:

Sincek

is complex

1tan i

r

k

k

=

i

k k e

=

This means E and B are not in phase B E =

0

0

exp( z) cos [ ( )]

exp( z) cos [ ( )]

i r E

i r E

E x E k i k z t

kB y E k i k z t

= +

= + +


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7

E and B are not in phase


8/16

8

Reflection/ Transmission at conducting surface

iE

1 1,

2 2, ,

tE

rE

0 1

0 11

exp[ ( ) ]

exp[ ( ) ]

i i

ii

E x E i k z t

EB y i k z tv

=

=

x y

zIncident beam

Reflected beam

Transmitted beam

0 1

01

1

exp[ ( ) ]

exp[ ( ) ]

r r

rr

E x E i k z t

EB y i k z t

v

=

=

0 2

20 2

exp[ ( ) ]

exp[ ( ) ]

t t

t t

E x E i k z t

kB y E i k z t

=

=

8

k2 is complex valued

Dielectric

Conductor


9/16

9

Boundary conditions

E-parallel condition at boundary z = 0

0 0 0i r tE E E+ =

H-parallel condition at boundary z = 0

20 0 0

1 1 2

1( )i r t

kE E E

v =

Solve simultaneous equations to get reflected and

transmitted field amplitude in terms of incident field.


10/16

10

Solve simultaneous equations

1 12

2

vk

=Define

0 0 0 0

1 2,

1 1r i t iE E E E

= =

+ +

2 2,

2

r ik k

is large for

good conductor

0 0 0, 0r i tE E E Almost all the energy

is reflected


11/16


12/16

12

Energy Density:

( )

( )

20

2

2202

2 20

exp( 2 z)4

11exp( 2 z)

4

1 exp( 2 z)4

E i

B i

i

U E k

U E k

E k

=

+=

= +

For good conductors, magnetic contribution dominates:

20

1exp( 2 z)

4B iU E k


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13

0

2

1( )

exp( 2 z) cos ( ) cos ( )i r r

S E B

kz E k k z t k z t

=

= +

Poynting Vector in Metals:

2 cos ( ) cos ( )cos{2 ( ) } cos ( )

r r

r

k z t k z t k z t

+= + +

Using

0

0

2

2

1 ( ) exp( 2 z) cos2

exp( 2 z) cos2

i

r

i r

kS E B z E k

kS z E k where k k

= =

= =


14/16

14

Problem: As the wave propagates in a metal, the energy

density decreases because of the attenuation factorexp( 2 z)ik

Where does this lost energy go? This energy is equal to theJoule Heating in the medium due to the current density

generated by the electric field of the EM Wave.

The above was also discussed in the class.


15/16

15

Free electrons in metals and plasma

2

02 exp ( )q Ed x d x i td t md t

+ =

2

2

total driving dampingF F F

d x d x

m q E m d td t

= +

=

Steady state solution 0( ) exp ( )x t x i t=

2 0

0 0

q E

x i x m =

0

0 2

/q E mx

i =

+


16/16

16

0( ) exp ( )x t x i t=

00

2

/q E mx

i =

+

Current Density:

( )d x

J N f q

d t

=

2 /N f q mJ E

i

=

J and E are

not in phase

0 00

/ /

exp ( )

q E m q E md xi x i t id t i i = = =+

2 /N f q m

i

=

Conductivity

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