Classical Optics
Prof. D. Rich
Study of light: Wave / Particle duality of photons
E = h Einstein: Photoelectric Effect
h = 6.63 x 10-34 J·s
 
Wave aspect of light stems from the unification of E & B
Through Maxwell’s equations in vacuum:
  
E  ,
Gauss’s Law:
o
Magnetic Flux Law: (i.e.
no magnetic monopoles)
Faraday’s Law of
Induction:
Generalized
Circuital
(Ampere’s) Law:
 
  B  0,

 
B
 E   ,
t

 

E
  B   o j   o o
t
  1
 E  ds   dv
o
 
 B  ds  0

 
B 
 E  dr   t  dS

 
 
E 
 B  dr  o  j dS  o o  t  dS
Development of the idea of E-M wave propagation:
     
  E  (  B)  B  (  E )


 E  B
 
  o j  E   o o E 
 B
t
t
 
  o E 2 B2 
 o j  E  o 


t  2
2o 
 
  
  
Let’s use the differential vector identity:   (Q  R )  R  (  Q)  Q  (  R )
   
Let Q  E , R  B
Let
then


  
     o E 2 B2 

    E  B / o  j  E  

o
t  2
2o 
1
 
  3
Take an integral  d and use the Divergence theorem:
   Gd r   G  ds
3r
uE
2
1



uB
3

 
 

   o E 2 B2 
dV
E  B / o  ds   j  EdV   

t V  2
2o 
V
1
Rate at which
in V increases; Note
  total energy

that   E  B /  o  ds = rate at which energy
flows out of V across the boundary .


V

2
Rate at which the Kinetic Energy of the particles change.
3
Rate at which Energy stored in the Fields increase; units in
Joules/sec or Watts

G
V



   
j  nqv   n(r ) F (r )  v dV
Define:
  
S  E  B / o
(Units W/m2)
Power = Force
Velocity
Poynting vector; points in the direction in
which the fields E and B transport Energy.
We want to search for plane E-M wave solutions in a vacuum. So, from Maxwell’s
equations, using  = 0 and j = 0, we have
 
  E  0,
Let
 
  B  0,

 
B
 E   ,
t

 
E
  B   o o
t

E  Ex ( z, t )iˆ  E y ( z, t ) ˆj  Ez ( z, t )kˆ

B  Bx ( z, t )iˆ  By ( z, t ) ˆj  Bz ( z, t )kˆ
This assumption leads to conditions
on the E and B components:
 
E
E  0  z  0
z
 
B
B  0  z  0
z
Fields in a sinusoidal electromagnetic wave. The E-Field is parallel to
the x-axis and to the x-z plane and the B-Field is parallel to the y -axis
and to the y-z plane. The propagation direction is along z.

 
B
 E  
t


ˆj

y
Ey
 E y E x  ˆ Bz
ˆ
  k  
k 

y 
 t
 x
0
Similarly,
iˆ

x
Ex
kˆ


B

z
t
Ez

0


Bz
0
t
0

 
E
  B   o o
t

Ez
0
t
So, we can take Ez = Const. = 0 and Bz = Const. = 0 without a loss of generality.
Also, from above
(i )
(ii )
E y
E y
Bx
Bx

;
  o o
z
t
z
t
B y
B y
E x
E

;
   o o x
z
t
z
t
For (i) take /z and /t
2 Ey
z
2
2Ey
 2 Bx

  o o
tz
t 2
2Ey
 2 Bx
1  2 Bx


2
tz
t
o o z 2
;
We arrive immediately to expressions of the 1D Wave Equation.
We can now identify the constant representing the speed of light:
c
1
 o o

1
2
2


s

c
12
7
2
 8.85 10

4


10
m

kg
/
c
m3  kg 


(as predicted by Maxwell in the year 1861)
The 3D expressions are as follows:

2

 E
 2 E   o o 2
t
;

2

 B
 2 B   o o 2
t

 3.0 108 m / s
In general, the 3D wave equation has the form:


1   (r , t )
2
  (r , t )  2
v
t 2

 
 
 (r , t )  c1 f (r  k / k  vt)  c2 g (r  k / k  vt)
2
For case (i) above
E y ( z, t )  E yo cos( kz  t )
i.e., a simple harmonic
plane wave solution
Insertion into the 1D wave eq. yields
 k 2 E yo cos( kz  t )   2 o o E yo cos( kz  t )

k    o o ; k   / c ; c   / k
Note that k is the wavevector and is the
direction of propagation; k without the arrow is
the magnitude and is called the wavenumber.

k  kkˆ  kzˆ
Again use the result
Bx E y

  E yo k sin( kz  t )
t
z
E yo k

Bx  
cos( kz  t )


Thus, if E  E yo cos( kz  t ) ˆj
then
 

 1  
k
1
B   E yo cos( kz  t )iˆ  k  E ; B  k  E (in general)



Consider a 1D propagation of an arbitrary wave of the form: f = f (z-vt)
f(z)
If z-vt = const. then
f(const.) = const.1
z1
z2
z3
t1
t2
t3
z
z 3> z 2> z1
t 3> t 2> t1
The same analysis can be performed of course for harmonic waves:
    ˆ
ˆ
E  E yo cos( kz  t ) j  E yo cos k  z  t   j
k 
 
Ey

2
v ph   c 
 
k
2 / 
If z-vt = const. then
Ey(const.) = const.1
z
z1
z2
z3
t1
t2
t3
Same analysis can be applied in 3D for a plane wave: Surfaces on which
the amplitude has a constant phase form a set of planes which are
perpendicular to the propagation direction. For harmonic plane waves:

  

 (r )  Aei ( k r t )  Aei ( k ( r k / k )t )
z 3> z 2> z1
t 3> t 2> t1
e i k  1  e i 2 
k  2 / 
Planes are such that the phase
defines a set of planes:
 
k  r  const.

kˆ  r  const.'
Use the vector identity
  
  
  
A  ( B  C )  B( A  C )  C ( A  B)
  1   
  
 k2
1   
k  B  k  (k  E )  k (k  E )  E (k  k )   E


using



   c  
E   2 k  B  Bk
k
k

0

 1  
kE
1
B  k E  B 
 By  Ex  c  E / B


c
As shown before, it’s possible to express in 3D using a
complex field representation:
 i ( kr t )
 
E (r , t )  Eo e
 i ( kr t )
 
B (r , t )  Bo e


E  Re E


B  Re B
With the complex representations, it is possible to derive
explicit relations between E, B and k:


 
 
  E  0  Re   E  0
 
 
    i ( kr t )
  E  ik  Eo e
 k  Eo  0  k  Eo
 
 
 
also
  B  0  k  Bo  0  k  Bo


 


 B 
B
and
 E  
 Re   E  Re  

t
 t 
1  
 Bo  k  Eo
Let’s examine the flow of energy again using the



Poynting vector S:
1  
S
EB
o

E  E yo cos( kz  t ) ˆj
 

 1  
k
1
B   E yo cos( kz  t )iˆ  k  E ; B  k  E




1
2 k
ˆj  iˆ
S   cos 2 (kz  t )E yo 


o
1
o

cos (kz  t )E yo 
2
2
because
k

zˆ
1
2
cos (...) 
t
2
c 
2
1
 o o
  ck
2

1 E yo
1
2
 S 
zˆ  c o E yo zˆ
t
c 2o
2

 Energy J
S  A 
  Watt  W
Sec.
s
A

A
Therefore, we can define
irradiance as
I S
t
c o 2 Average Energy

Eo 
Area·time
2
In older texts (and in discussion) the term “intensity” is also used.
The energy per unit volume or energy density stored in the fields
can be written as before
u  uE  uB 
oE2
B2  o E 2
E2
1 2
2



 oE 
B
2
2
2o
2
2o c
o
Note, again a factor of ½ must be added for the time averages:
u t  uE  uB
The units
oE2
c

t

 o Eo 2
2

1
2o
Bo
2
J 1 momentum

3
m c
volume
Thus, similar to the Poynting vector, the E-M momentum P per
unit volume that exerts a radiation pressure is given by:

  o
2
P   o E  B  E yo cos 2 (kz  t )kˆ
c

1 o
2
P 
E yo kˆ
t
2 c
Consider a slab of E-M radiation whose
thickness is ct and cross-sectional area A:
If the light is absorbed by an object,
the momentum transfer is given by the
impulse force·time:
Vol.=Act
Area A

F  t  p  P Act
t
ct
F 1 o 2
1
2
Pr t  
E yo  c   o E yo
A 2 c
2
 u E  u B t Thus, the energy/vol. contained in the E-M propagation
also represents the pressure exerted on an object.
For example, if Eyo= 1 V/m, then <Pr> = 4.4 x 10-12 N/m2  10-17 atm.
2
Note also that
1 E yo
S
c 2o
1


c
1
Pr
 o E yo 2 c o o
2
or
I  Pr  c
k