Download Normal incidence

Magnetostatics:
Stationary Charges ➾ Constant electric fields: electrostatics.
Steady Currents ➾ Constant magnetic fields: magnetostatics.
Static electric fields characterized by E or D and related
according to D = εE.
Static magnetic fields characterized by H or B and related
according to B = μH.
B=Magnetic flux density,
H = Magnetic field intensity
two major laws governing magnetostatics:
1. Bio-Savart law
2. Ampere’s circuital law
Just as Gauss’s law is special case of Coulombs law, Ampere’s law is a
special case of Bio-Savart’s law and is easily applied in problems
involving symmetrical current distribution.
Biot-Savart Law
• Currents, i.e. moving electric charges, produce
magnetic fields. There are no magnetic charges
Ampere’s Circuital Law** in Integral Form
Ampere’s Circuital Law -“the circulation of the magnetic flux density in
free space is proportional to the total current through the surface
bounding the path over which the circulation is computed.”
 B  dl  0 I encl
C
The line integral is around any
closed contour bounding an open
surface S. I encl is current through S:
I encl 
 J  dS
S
Where J is defined as current density
  B  0 J
Note: ** Ampere’s circuital Law is
Maxwell’s IV equation.
H  J  0
This shows that magnetostatic
field is nonconservative in nature.
Electricity  Magnetism
ELECTRIC FLUX
Electricity  Magnetism
MAGNETIC FLUX
Magnetic lines of force produced in the medium surrounding electric currents or
magnets and is expressed as surface integral of the magnetic flux density.


   B  dS
s
weber
(Tesla m2)
   B dS  0
B
.B  0
Gauss’ law for magnetic fields** says there can not be a net magnetic
flux through the surface since there can be no net magnetic charge
enclosed by the surface.  magnetic monopoles do not exist.
** Maxwell’s II equation
1
If the magnetic flux density in a medium is given by B  cos  aˆr ,
r
what is the flux crossing the surface defined by 
Solution:

4
 

4
,0  z  2m.
1
B  cos  âr ,
r
   B  dS
Where
dS  r d dz âr
s
1
   cos  âr  rd dz âr
sr
2


4
 cos  d dz
0 
4
 = 2.83 wb
Laws of Electromagnetism
1. Gauss Law of electrostatics
Electric Flux   E   E  dS 
S
Qenclosed
o
Over any closed surface S.
2. Gauss Law of magnetostatics
Magnetic Flux   B   B  dS  0
Over any closed surface S.
S
Consider a surface S around one of the poles of a magnet. Magnetic lines of
force travel inside the magnet from one pole and emerges out from the other
pole, so the flux enters through the surface and the same flux emerges out of
the surface. Therefore the net flux over the surface S will be zero.
3. Ampere’s LAw
 B  dl  0 I encl
C
Faraday’s law: Changing magnetic field
induces electrical current
(a) When a magnet is moved
toward a loop of wire connected
to
a
galvanometer,
the
galvanometer
deflects
as
shown, indicating that a current
is induced in the loop.
(b) When the magnet is held
stationary, there is no induced
current in the loop, even when
the magnet is inside the loop.
(c) When the magnet is moved
away from the loop, the
galvanometer deflects in the
opposite direction, indicating that
the induced current is opposite
that shown in part (a).
3. Faraday’s Law **
Changing magnetic field gives rise to electric current.
Induced emf in the loop, due to changing magnetic flux.

  B
t
i.e. rate of change of magnetic flux is the e.m.f. induced
in the circuit.
E
ΦB
If q0 is the charge taken around the loop.Then Force
F  q0 E
Now work done in taking the charge
around the loop will be
E
B (Magnetic field inward)
dB
dt
dW  F  dl  W  q 0 E 2 r
q 0   q 0 E 2 r  
 E  dl  
P
 B
t
 B

t
 E  dl  
P
 E  dl
P
 B


t
t
 B.ds
S
Integral form
Note: ** Faraday’s Law is Maxwell’s III equation.
or   E  
B
t
Differential form
1.
2.
 E  dS 
S
Qenclosed
o
 B  dS  0
S
3.
4.
 B
 E  dl   t
P

B  dl   0 i
P
Asymmetry in the above laws
1. In equation s1 and 2, the R.H.S. of equation 2 contains no value, because
of the non existence of the magnetic monopole.
2. In equation 4 R.H.S. contains μ0 i → μ0(dq/dt), but there is no such term
in equation 3 corresponding to μ0 i. This asymmetry is also because of
nonexistence of magnetic monopole that is no magnetic currents.
3. In equation 3 changing magnetic flux gives rise to electric field. There is
no such corresponding term relating to changing electric flux producing
magnetic field in equation 4. This led Maxwell to introduce a new term
corresponding to changing electric flux in equation 4.
The corresponding term
 E
 B  dl  00
t
P
ΦE
B
 E
 0 ( i  id )
 B  dl  0i  00
t
P
id  displacement curent  0
 E
t
Maxwell’s Equations in Integral Form
1.
Qenclosed
E

d
S


o
3.
 B  dS  0
S
 E  dl  
P
S
2.
E
B
4.
 B
t

 E

B

d
l


i


0

0
t

P



Modification to Ampere’s Law:
-Ampere’s law must be wrong!
 B  dl   I
0 enc
- it depends on what “enclosed” means
Surface S1 encloses a current
Surface S2 does not!
What if we moved S1 into the gap?
How can we modify the rule
to handle all situations?
Ampere’s Law (constant currents):
 B.dl   I
Ampere’s Law for
constant currents.
0 enc
What about currents which are not continuous?
Displacement current in a capacitor
E-field increasing as Q increases!
I
I
+Q charge
deposits on
the plate
The capacitor holds a charge Q
over the two plates. How can
there be a current emerging
from the capacitor plates?
-Q charge induced
by E-field
Modified Ampere’s Law




d
B
.
dl

I

I



E
.
dA
B
.
dl


I
0
enc
d
0
0
0
enc


dt
To see how the displacement current comes about, one has
to consider the electric flux through the capacitor’s plate
(Gauss’s Law).
 E.dA   E 
Qenc
0
 Qenc   0  E.dA
Q increases on the capacitor, the electric flux also increases
at the same rate.
dQenc
d
d
 0 E  0
dt
dt
dt
 E.dA  I
d
CONTINUITY EQUATION
From the principle of charge conservation, the time rate of decrease of
charge with in a given volume must be equal to the net outward current
flow through the surface of the volume. Thus current Iout coming out of
the closed surface is
Qin is the total charge
 dQin enclosed
by the closed
I out   J  dS 
dt
surface.
Using the divergence theorem
 J  dS     Jdv
S
v
dQin
d
v
and 
   v dv   
dv
dt
dt v
t
v
v
dv
   Jdv   
v
v t
v
J  
t
This equation is called continuity equation. It is derived from principle of
conservation of charge and shows that there can be no accumulation of charge at
any point.
v
For steady current
0
t
 J  0
Hence the total charge leaving a volume is the same as the total charge entering it.
Maxwell’s modification of Ampere’s Law
 B  dl   I
From Ampere's Law
0 encl
where I encl 
P
 J  dS
S
Applying Stroke's theorem to left-hand side of above equation
0 I encl   B  dl     B  dS
L
S
   B  dS  0  J  dS
S
Differential form of Ampere’s Law
S
  B  0J
Taking divergence of above equation
    B   0   0   J 
Since R.H.S. of above equation is not zero**, so let us apply
Continuity equation and Gauss’s law (Electrostatics) we get
 v

 E 


J  
   0  E      0

t
t
 t 
 E 
  J    0
0
 t 
Hence
  B   0 J   0 0
E 

or .  J   0
0
t 

E
t
E
E
The term  0
is called as displacement current density . J d   0
t
t
Hence modified Ampere’s Law
(differential form)
  B   0 J  J d 
**Note: Fundamental theorem of vector analysis;
    B  0
Exercise:
Write Maxwell’s equations in integral form and
obtain their differential form by using vector
analysis. Write statement and physical
significance of every equation.
Maxwell’s Equations in Integral Form
1.  E  dS 
Qenclosed

S
2.  B  dS  0
S
o
 “Gauss Law in Electrostatics ” : Electric flux

 coming out from the surface of the body is
 equivalent to the charge enclosed by the body

 “Gauss Law in Magnetostatics ” : Magneticflux

 coming out from the closed surface of the body
is zero as no magnetic monopole exist .


“Faraday'sLaw”:Changing Magnetic flux produce

 B
 electric current  or field  in a closed loop.where
3. E  dl  

t
L
 Magnetic flux φB =  B.ds
S

“Modified Ampere’s Law”or “Maxwell-Ampere’s
 Law”: Changing electric flux can produce magnetic

 E  
4.  B  dl    i   0
  field in a discontinuous circuit to hold Ampere’s

t

 
L
 circuital Law.where Electric flux φE =  E.ds
S


0
Proof: Differential form
Use equation 1 and apply Gauss divergence theorem
 E  dS   (  E) dv
V
S
 (  E) dv 
Qenclosed
o
V
 dv


V 0

E 
0
Use equation 2 and apply Gauss divergence theorem
 B  dS   (  B) dv  0
V
S
B  0
Use equation 3 and apply Stoke’s theorem
 E  dl   (   E )  dS
P
S

B
 (  E)  dS  - t    t  dS
S
S
E  -
B
t
Use equation 4 and apply Stoke’s theorem
 B  dl   (   B )  dS
P
S
 E 

(


B
)

d
S


JA




0
0

t


S
( EA ) 


 (   B )  dS   0  JA   0
t 

S
E 

(


B
)

d
S


J


  dS

 0
0
t 

S
S
E 

  B  0  J  0

t 

Since D   0E
Where A   dS
D 

  B  0  J 

t 

Curl of B is due to current flow and a changing electric field.
Maxwell's Equations with Physical Interpretation
1. Relates net electric flux to net enclosed electric charge.
Qenclosed
 E  dS 
o
E 
S
v
0
2. Relates net magnetic flux to net enclosed magnetic charge.
 B  dS  0
B  0
S
3. Relates induced electric field to changing magnetic flux.
 B
 E  dl    t
P
E  -
B
t
4. Relates induced magnetic field to changing electric flux to and to current.

 E
 B  dl   0  i   0 t

P



D 

  B  0  J 

t 

Case 1: Maxwell equations in free space*
(no free charges and no currents)
q  0    0, i  0  J  0,
  E  0 /  0
B  0
B
E  t
E E
  B  00i0 0 0
t t
* Helpful to understand Electromagnetic waves in free space.
Maxwell’s Equations in linear media (Perfect dielectric)
Inside matter, but in regions where there is no free charge or free current,
1.
D  0
2.
B  0
If the medium is linear
3.
4.
1
B
E  t
H 
D
t
D  E, H  B

That is homogeneous (  and  do not vary from point to point), and
isotropic (in which μ and ε are invariant with field orientation )Maxwell’s
equation reduces to
B
1.   E  0
E  3.
t
2.
B  0
4.
  B  
E
t
•Derivation of Electromagnetic wave equation
•Properties of E.M. Waves.
•E is prependicular to k (direction of propagation); k.E=0
•B is prependicular to k (direction of propagation); k.B=0
•E is prependicular to B too; k  E=?
•Wave Impedance E0/B0=c
•Poynting Theorem.
Electromagnetic waves in free space or vacuum
Take curl of equation 3
B
t
     E       E    2E
E  -
 B 
 2 E     
 t 

 2E     B 
t
 2E
  o  o 2
t
 2E
2
 E  oo 2
t
Take curl of equation 4
E
  B  00
t
     B       B    2B
E 

 B      o  o

t 


  2B  oo    E 
t
 2B
  o  o 2
t
2

B
 2B  oo 2
t
2
26
2

E
 2E  o o 2
t
 2B
2
 B  o o 2
t
1
v
2
 oo
Standard Wave equation:
2
1

f
2
 f  2 2
v t
1
v
oo
o  4 107 weber / amp  m
o  8.542  1012 Farad / m
v  3 108 m / sec  c
Thus we conclude that light is electromagnetic in nature with
electric vector E and magnetic vector B oscillating as a wave
and propagating with a velocity of light in free space .
Maxwell’s equations also imply that empty space supports the
propagation of electromagnetic waves, traveling with speed of
light.
27
•Derivation of Electromagnetic wave equation
•Properties of E.M. Waves.
•E is prependicular to k (direction of propagation); k.E=0
•B is prependicular to k (direction of propagation); k.B=0
•E is prependicular to B too; k  E=?
•Wave Impedance E0/B0=c
•Poynting Theorem.
Solution of Electromagnetic waves in free space
 2E
 E  o o 2
t
 2B
2
 B  o o 2
t
2
The harmonic
solutions to the wave
equations
 E  Eoeikr t 

i kr t 
B

B
e

o
Where, k=2π/λ is propagation vector,
ω=2πν is angular frequency
where Eo and Bo (amplitudes of wave) space and time independent
vectors but may in general be complex.
E(r, t)  Eo e
ikr t 
E o  E ox aˆ x  E oy aˆ y  E oz aˆ z
k  k x aˆ x  k y aˆ y  k z aˆ z
r  xaˆ x  yaˆ y  yaˆ z
Prove-Transverse Nature
k is perpendicular to E and B.
k  E  0 or k  B  0
and E , B and k are orthogonal
29



.E  (aˆ x
 aˆ y
 aˆ z ) Eoeikr t 
x
x
x
k .r  (k x aˆ x  k y aˆ y  k z aˆ z ).( xaˆ x  yaˆ y  yaˆ z )
 kx x  k y y  kz z
.E  i (k x Eox  k y Eoy  k z Eoz )eikr t 
 i (k x aˆ x  k y aˆ y  k z aˆ z ).( Eox aˆ x  Eoy aˆ y  Eoz aˆ z )e ikr t 
 ik .E
Similarly,
B  0
  B  ik  B o eikr t 
 ik  B =0
Now Prove
E , B and k are orthogonal
Proof:
B
  E  i k  E  and ,
 iB
t
ik  E  iB
aˆ x

 E 
x
Ex
aˆ y

y
Ey
aˆ z

z
Ez
where, E x  E ox e
i ( k .r t )
E y  E oy e
E z  E oz e
 Ez E y 
 Ez Ex 
  E  aˆ x 


  aˆ y 

z 
z 
 x
 y
i ( k .r t )
i ( k .r t )
Similarly from equation 4, we can show that
E
  B   0 0
t
ik  B  i 0 0E
or
k  H  D
i.e. E is perpendicular to the Plane formed by k and B !!
Thus, k, E and B vectors are mutually perpendicular to each other.
k .E  0
(i.e. E is perpendicular to k)
k .B  0
(i.e. B is perpendicular to k)


 (i.e. E, B and k are orthoganal)

k  B   o o E 
k  E  B
we conclude that electromagnetic field vectors E and B are both
perpendicular to the direction of propagation vector k. This
shows that Electromagnetic waves are transverse waves.
35
Plane Waves
Ey
Ey
Bz
kx
x
direction of
propagation
Bz
Plane wave because the fields are uniform over every plane
perpendicular to the direction of propagation (i.e. x = constant plane)
as shown in the figure below.
E y  E o cos kx  t 
E y  Eo ei kx t 
Bz  Bo ei kx t 
or
B z  B o cos kx  t 
E y  E o sinkx  t 
or
B z  B o sinkx  t 
36
Show that E and B of plane wave are in same phase at any time in space
E y  Eoei kxt 
From equation 3
Since
But
E  Ey 

B  Bz 
k  k x 
Bz  Boei kxt 
E  

k x E y   Bz 



E y  Bz 
kx

Since /k is a real number, the electric and
magnetic vectors are in phase; thus if at any
instant, E is zero then B is also zero, similarly

Bo 
k  E  B
  2, k 
E
i kx t 


i  kx t 
Bz  Boe

E y  Eoe
B
t
y
 cBz
Eo
c
Eo Eo

o  o c 
H o Bo

2

2
 
   c
2

k

Eo 
1
 c
Bo k
 o o
o
o

 o  376.72 
o
 o o
when E attains its maximum value, B also Here ηo is universal const.and called as
attains its maximum value, etc.
Both Ey and Bz are in same phase.
characterstic or intrinsic waveimpedance
of the freespace.
37
Numerical
In
 free space the Electric field is given as
E  10 sin (2 x  100t ) ˆj.
Determine D, B and H by using Maxwell’s equations.
Sol: Wave is propagating along x direction.
(1)
(2)
(3)
D   0 E  10 0 sin (2 x  100t ) ˆj
B
Using   E  
, or k  E   B
t
B
ˆ
20 Cos (2 x  100t ) k   ,
t
1
B  Sin (2 x  100t ) kˆ
5

 B
1
H

Sin (2 x  100t ) kˆ
 0 5 0
Ey
Bz
kx
Energy in EM Waves:
Poynting Theorem and
Poynting Vector
Energy in EM Waves
The energy densities (energy per unit volume) associated with electric
field and magnetic fields are:
Electric field Energy Density
Magnetic field Energy Density
UE 
UB
1
0 E 2
2
1 B2

2 0
J/m 3
J/m 3
In vacuum at any moment of time
UB
1 B2
1 E2
1
2


=

E
=U E
0
2
2 0
2 0 c
2
Therefore, total energy density
UTot = 0 E
2
 0 EBc
J/m3
How much energy passes through 1 m2/sec???
Like energy flux!!!!!
Energy Carried by Electromagnetic
Waves
Total energy passes from volume c m3 through a closed surface of 1m2 per
unit time
Utot c =0EBc2
=EB/0 J/m2 sec
C m
Which is poynting vector
1m2
1
S=EXB/0
Thus, S represents power per unit area. The direction of S is along the
direction of wave propagation. The units of S are J/(s·m2) =W/m2.
The magnitude S represents the rate at which energy flows
from the volume through a unit surface area perpendicular to
the direction of wave propagation.
Energy Carried by Electromagnetic
Waves
Electromagnetic waves carry energy, and as they propagate through space
they can transfer energy to objects in their path. The rate of flow of energy in
an electromagnetic wave is described by a vector S, called the Poynting
vector.*
1
S = E B
0
EB
or S =
0
E2 cB2
or S =
=
0 c 0
y
E
B
EM waves are sinusoidal. E y  E o cos(kx  ωt)
B z  B o cos(kx  ωt)
S
c
x
z
*J. H. Poynting, 1884.
Energy Carried by Electromagnetic
Waves
The average of S over one or more cycles is called the
I  Saverage
wave intensity I.
E o Bo
E o2
cBo2 cμ o H o2 c o E o2
 S 




2μ o
2μ o c 2μ o
2
2
The time average of cos2(kx - t) is ½
The magnitude of S is the rate at which energy is transported by a wave across
a unit area at any instant:
Thus,
 energy

 power 
time


S=
=



area
area

instantaneous


instantaneous
 power 
I= S =

 area average
Relation between S and U
When we average this instantaneous energy density over one or
more cycles of an electromagnetic wave, we again get a factor of
from the time average of cos2(kx - t).
1
2
uE = 0Emax
,
4
2
max
1B
uB =
, and
4 0
2
2
1 Emax
1 cBmax
Saverage = S =
=
2 0 c 2 0
uTotal
2
1
1 Bmax
2
= 0Emax =
2
2 0
S =c u .
The intensity of an electromagnetic wave equals to the ‘c’
times of the average energy density.
Momentum Transport- Radiation Pressure
Relativistic energy and Momentum
2
E=
Energy
2
(pc)
momentum
+
m c 
2 2
o
rest mass energy
For light mo=0
E=U=pc
When em wave strikes a surface , its momentum changes.
Rate of change of momentum is equal to Force. This force
acting per unit area of the surface exert a pressure, called
RADIATION
1 dp 1 d  U


A dt A dt  c
PRESSURE
 dU  1
 1
dt   S


 c A  c


Total Absorption
dp 1 dU 1
F= 
 IA
dt c dt c
F 1
P   I (radiation pressure)
A c
Total Reflection
dp 2 dU 2
F= 
 IA
dt c dt c
F 2
1 pascal (Pa) = 1 N/m2 =
P   I (radiation pressure) 1 kg/(m·s2)
A c
Total Absorption
dp 1 dU 1
F= 
 IA
dt c dt c
F 1
P   I (radiation pressure)
A c
Total Reflection
dp 2 dU 2
F= 
 IA
dt c dt c
F 2
1 pascal (Pa) = 1 N/m2 =
P   I (radiation pressure) 1 kg/(m·s2)
A c
Ex: a radio station on the surface of the earth radiates a sinusoidal wave
with an average total power of 50 kW. Assuming the wave is radiated
equally in all directions above the ground, find the amplitude of the electric
and magnetic fields detected by a satellite 100 km from the antenna.
All the radiated power passes through the
hemispherical surface* so the average power per unit
area (the intensity) is
P
 power 
I=
=

2
 area av 2R
= 7.96 10-7 W m2
Satellite
R
Station
2
1 Emax
I= S =
 Emax = 20 cI  2.45 10-2V / m Exercise: calculate the average
2 0 c
energy densities associated with
the electric and magnetic field.
Emax
= 8.17 10-11 T
B max =
-15
3
u
=
u

1.33

10
J
/
m
B
E
c
*In problems like this you need to ask whether the power is radiated into all space or into just part
of space.
If the earth receives 2cal min-1cm-2 solar energy, what are the
amplitudes of electric and magnetic fields of radiation?
E0=1026.8V/m
H0=2.726A-turn/m
The electric field in an em wave is given by E=E0sin(t-x/c),
where E0=100 N/Coul. Find the energy contained in a
cylinder of cross section 10-3m2 and length 100 cm along xaxis.
U=4.42510-11 Joule
Poynting’s Theorem
“Conservation of Energy”
Chapter 8, Page 346, Griffith
Rate of work in the system
System of q and I
Applied Lorentz force
 E  J dv
Power flow

V
 1
1

   E 2  H 2  dv
t  2
2

V
Rate of decrease in stored energy

  E H ds
S
Poynting’s Theorem
“Conservation of Energy”
Chapter 8, Page 346, Griffith
“The work done on the charges by the
electromagnetic force is equal to the decrease in
energy stored in the field, less the energy that flowed
out through the surface”
 1 2 1
2
V  E  J i  dv   t V  2  E  2  H  dv  S  E  H   d s
the total electromagnetic
represent the total
energy generated (Rate of
electromagnetic energy
flow of instantaneous
power out of the volume
stored in the volumeV.
V through the surfaceS.
work done) by the sources
in the volumeV.
Differential form of Poynting’s Theorem
 1 2 1
2
V E  J dv   t V  2  E  2  H  dv 
U mec

V t dv   t V U EM dv 
 E H  ds
S
 S ds
S

V t (U mec  U EM )dv   V .S dv

.S   (U mec  U EM )
t
Shows Conservation of Energy
Refractive index of the medium is defined as
In a dielectric medium,
   r o
Since dielectric is non magnetic,
n  r

n
 r r
0 0
   r o
r  1
v
1
1
c


μo  n μo  o n
That is the speed of electromagnetic waves in an isotropic
dielectric is less than the speed of electromagnetic waves in
free space.
Exercise: The relative permittivity of distilled water is 81.
Calculate refractive index and velocity of light in it .
Boundary conditions, Page 333,
Ch. 7
• If there is no free charge or free current at the
interface of two medium, then
1.  1 E   2 E

1

1

2
3. E  E
//
2

2
Incident
E
E//
2. B  B
//
1
4.
//
1
B
1

//
2
B
2
E┴
Reflected
Medium-1(ε1 , µ1)
Medium-2(ε2 , µ2)
Refracted
Reflection and Transmission at
Normal incidence
Reference
Introduction to Electrodynamics
By D. J. Griffith
Reflection and Transmission at
Normal incidence
Incident wave
Transmitted wave
1 1
2 2
Reflected wave
Suppose yz plane forms the boundary between two linear media. A
plane wave of frequency ω traveling in The x direction (from left) and
polarized along y direction, approaches the interface from left (see
figure)
Reflection and Transmission at
Normal incidence
E I ( x, t )  E0 I exp  i  k1 x  t   ˆj
B I ( x, t ) 
1
E0 I exp  i  k1 x  t   kˆ
v1
1 1
Reflected wave
Transmitted wave
2 2
Reflection and Transmission at
Normal incidence
E I ( x, t )  E0 I exp  i  k1 x  t   ˆj
B I ( x, t ) 
1
E0 I exp  i  k1 x  t   kˆ
v1
1 1
E R ( x, t )  E0 R exp  i  k1 x  t   ˆj
B R ( x, t )  
1
E0 R exp  i  k1 x  t   kˆ
v1
Transmitted wave
2 2
Reflection and Transmission at
Normal incidence
E I ( x, t )  E0 I exp  i  k1 x  t   ˆj
B I ( x, t ) 
1
E0 I exp  i  k1 x  t   kˆ
v1
1 1
E R ( x, t )  E0 R exp  i  k1 x  t   ˆj
B R ( x, t )  
1
E0 R exp  i  k1 x  t   kˆ
v1
E T ( x, t )  E0T exp  i  k2 x  t   ˆj
BT ( x, t ) 
1
E0T exp  i  k2 x  t   kˆ
v2
2 2
Reflection and Transmission at
Normal incidence
Incident wave
E I ( x, t )  E0 I exp  i  k1 x  t   ˆj
1
B I ( x, t )  E0 I exp  i  k1 x  t   kˆ
v1
E R ( x, t )  E0 R exp  i  k1 x  t   ˆj
Reflected wave
1
B R ( x, t )   E0 R exp  i  k1 x  t   kˆ
v1
E T ( x, t )  E0T exp  i  k2 x  t   ˆj
Transmitted Wave
1
BT ( x, t )  E0T exp  i  k2 x  t   kˆ
v2
Reflection and Transmission at
Normal incidence
At x=o the combined fields to the left
EI+ER and BI+BR, must join the fields to the right ET and BT in
accordance to the boundary condition.
Since there are no components perpendicular to the surface so
boundary conditions (i) and (ii) are trivial. However last two [(iii) &
(iv)] yields:

1
D
B1
E

//
B1
1
//

 x  0   D2  x  0 

(i)
 x  0   B2  x  0 
(ii)
 x  0  E  x  0
(iii)
//
//
B2
 x  0 
 x  0
2
(iv)
E0 R  E0 I  E0T
 1 1

1 1
1
 E0 I  E0 R    E0T 
1  v1
v1
 2  v2

 1 1

1 1
1
 E0 I  E0 R    E0T 
1  v1
v1
 2  v2

E0 R  E0 I  E0T
(1)
or , E0 I  E0 R   E0T
(2)
1v1
where  

2v2
Using (1)
and (2)
 E0 I  E0 R 
 1 

 1 
1v1
E0T
2 v2
1 2 1n2

21 2 n1

 2
 E0 I and E0T  

 1 

E0 R
 E0 I

v1
If  r  1(nonmagnetic media) then  =
v2
thus we have, E0 R
 v2  v1 
 2v2 

 E0 I and E0T  
 E0 I
 v2  v1 
 v2  v1 
The reflected wave is in phase if v2>v1 or n1>n2
and
out of phase if v2<v1.or n1<n2
The real amplitudes are related by
 2v2 
v2  v1
E0 R 
E0 I and E0T  
 E0 I
v2  v1
 v2  v1 
c
in terms of refractive index n=
v
 2n1 
n1  n2
E0 R 
E0 I and E0T  
 E0 I
n2  n1
 n2  n1 
Reflected wave is 180o out of phase when reflected from
a denser medium. This fact was encountered by you during
Last semester optics course. Now you have a proof!!!
Reflectance (R)
and
Transmittance (T)
• Wave Intensity (average power per unit area is given by):
1 B0 E0 1
I
  vE02
2 
2
• If μ1= μ2 = μ0, i.e μr=1 , then the ratio of the reflected
intensity to the incident intensity is
2
2
I R  E0 R   n1  n2 
R

 

I I  E0 I   n1  n2 
Where as the ratio of transmitted intensity to incident intensity is
2
2
IT  2 v2  E0T 
n2  2n1 
4n1n2
T







I I 1v1  E0 I 
n1  n1  n2 
(n1  n2 ) 2
Use
ε α (n)2
Traveling E and H waves in free space ( region 1) are
normally incident on the interface with a perfect
dielectric (region 2) for which r=3.0. Compare the
magnitudes of the incident, reflected and transmitted
E and H waves at the interface.
Find out Reflectance and Transmittance .
Check the result.
Ans:
NOTE: R+T=1 => conservation of energy
E 0R
E 0T
 0.268;
 0.732
E 0I
E 0I
H 0R
H
 0.268; 0T  1.268
H 0I
H 0I
R  0.072;T  0.928
Chapter 24, Optics by GHATAK 4th Ed.
Oblique Incidence
Case 1:
E parallel to the plane of incidence
The reflection of a plane wave with its electric vector parallel to the plane of
incidence.
The parallel polarization (or the p polarization) is also called the transverse
magnetic (or the TM) polarization as the magnetic field is perpendicular to
the plane of incidence.
Case 1: E parallel to the plane of incidence

1
D
B1
E1
H1

 x  0   D2  x  0 

 x  0   B2

//
 x  0  E2  x  0
//
 x  0  H2
 x  0
//
//
 x  0
The z component of the electric field represents a tangential component which
should be continuous across the surface. Thus at x=0
E1z+E3z=E2z
Further, the normal component of D must also be continuous, and since D = E,
1E1x+ 1E3x= 2E2x
The z component of the electric field represents a tangential component which
should be continuous across the surface. Thus at x=0
Further, the normal component of D must also be continuous, and since D = E,
Substituting for E20 from Eq. (15),
The amplitude reflection coefficient ,
r||=E30/E10
The amplitude refraction coefficient ,
t||=E20/E10
For nonmagnetic media, 120=4x10–7 N A–2, and the expression for the
amplitude reflection coefficient simplifies to
Case 2. E perpendicular to the plane of incidence

1
D
B1
E1
H1

 x  0   D2  x  0 

 x  0   B2

//
 x  0  E2  x  0
//
 x  0  H2
 x  0
//
//
 x  0
The reflection and refraction of a plane wave with the electric vector
lying perpendicular to the plane of incidence. (s-polarized)
Since, the y axis is tangential to the interface, the y component of E must
be continuous across the interface; consequently
E10 + E30 = E20
the z -component of the magnetic field to be continuous,
H10 cos1  H 30 cos1  H 20 cos 2
or
k1
1
( E10  E30 )cos1 
k2
2
E20 cos 2
We summarize the amplitude reflection and transmission coefficients
for the two cases; the results are valid for nonmagnetic media:
Equations are known as the Fresnel equations.
2
2
2
Ex 1: Light is incident at 60o on a boundary separated by
media of refractive index n1 = 1 and n2 = sqrt(3) . Find the
amplitude of reflection coefficient r when electric field is
perpendicular to the plane of incidence.
n1 cosi  n2 cost
r 
n1 cosi  n2 cost
n1 sin i  n2 sin t
Ex 2: Consider a linearly polarized electromagnetic wave
(with its electric field vector along y-direction) of
magnitude 5 V/m propagating in vacuum. It is incident on
a dielectric interface at x = 0 at an angle of incidence of
30o. The frequency associated with the wave is 6×1014 Hz
and refractive index of the dielectric is 1.5.
Show that R+ T=1.
n1 cosi  n2 cost
r 
 0.2404
n1 cosi  n2 cost
2cosi sin t
n2 cos 2 2
t 
 0.7596 ; T 
t
sin(i  t )
n1 cos1
 R  rr '  0.057796
 T 
4n1n2 cos1 cos 2
 n1 cos1  n2 cos2 
Hence Proved,
2
R  T  1
 0.942204