Download Chapter 6 – Optical Methods - Introduction

Chapter 6 – Optical Methods - Introduction
Solutions of the Maxwell equations and energy relationships
Sample Problems
6-S1

To define an electromagnetic field it is necessary to define two vectors, the vector E that

represents the electrical field, and the vector B that defines the magnetic field, also called
magnetic induction. The propagating electromagnetic field has an effect on material objects. The
effect of the electromagnetic field on material objects is defined by a group of vectors, the vector


J f , that represents the electric current density, the vector electric displacement D and the

magnetic vector H .
Write the Gauss equations in the differential form that relate these quantities in Cartesian
coordinates.
Hint: Relate the vectors taking into consideration the constitutive equations of vacuum.
Solution to 6-S1
The answer to these questions is provided by,

B
E 
t

(Farady’s law of electromagnetism)
(6.8)
Expanding the vectorial equations in Cartesian coordinates,

E z E y
 Bx


y
z
t

 By
E x E z


z x
t

E y
E
 Bz
 x 
x y
t
And

E
  B   0 J f   0 0
t
One gets


(Ampère’s law of electromagnetism)
(6.9)
Bz By
 Ex

  0J fx  0 0
y z
t
 Ey
Bx Bz

  0 J fx  0 0
z
x
t
By
x

Bx
 Ez
  0 J fz   0 0
y
t



To relate the vectors J f , D and H it is necessary to consider in the case of vacuum the
constitutive laws,


D 0 E
Hence
(6.10)

D 
E
0

Dz Dy
 Bx

 0
y
z
t

 By
D x D z

 0
z
x
t

D y
D x
 Bz

 0
x
y
t


The above equations relate the vector electric displacement D to the vector B that defines the
magnetic field.
Taking into consideration
H z H y
 Dx

 J fx 
y
z
t
 Dy
H x H z

 J fx 
z
x
t
H y
x

H x
 Dz
 J fz 
y
t
B
H
0
(6.11) and (6.9)

D
The above term J T  J f 
represents the ordinary electrical current plus the so called
t
displacement current, defined by Maxwell that introduced it to include transient fields. The
above set of equations relates the magnetic field to the total current density.


6-S2
The propagation of an electromagnetic field obeys the Maxwell equations. Assume there is a
field propagating in the vacuum that is characterized by the fact that the components of the field
are contained in one plane and remain in this plane throughout the propagation. It is necessary to
utilize a Cartesian coordinate system such that the normal to the plane is the z-coordinate. It is
assumed that the field propagates in the z-direction,


Show that the E and B components are perpendicular to the z-axis.
Solution to 6-S2
It was assumed that the field propagates only in the z-direction hence the field must be
independent of x,y, hence it will change only as a function of z, and t, the time. We can write,


E (z.t) and B( z, t ) .
Utilizing

E 

0
(Gauss Law for the electrical field)
(6.6)
In view that the wave propagates in vacuum and there are no charges,

  E  0 in the expanded form,
E x E y  E z


0
x y
z
The field propagates in the z-direction and hence must be independent of x and y, then the
corresponding derivatives must be zero. From the above equation,
 Ez
 0 . This means that Ez is constant with respect to z and according to our initial
z
assumption that the components are in a plane, we can assume this constant to be zero. From
this result it follows that the field can have only transversal components Ex and Ey. Let us


assume now that E  E x ( z, t ) i .
The field only propagates in the z-direction hence the field will change only as a function of z,
and t, the time. From

B
E  
t

(Farady’s law of electromagnetism)
(6.8)
Utilizing the expanded form given in sample problem 6-S2, and since E y  E z  0 and
E x  E(z, t )
B
Bx
Bz
 Ex
 0 .The above results indicate that Bx and Bz are constant in time,
0,
 y ,
t
t
z
t
the only variable term is By and hence the magnetic field must be of the form,



B  By ( z, t ) j , the vector B is perpendicular to the direction of propagation. We can therefore
conclude that the vectors are perpendicular to the direction of propagation and perpendicular to
each other. The plane that contains both vectors is the plane formed by the two vectors and has a

normal that is of the direction of k . The direction of propagation is constant in space.
6-S3
In section 6.3.1.1. the harmonic solutions of the Maxwell equations were derived. It was shown
that the argument of the sinusoidal solution is of the form,
  sin k ( x  vt )
x
Show that the argument can be written as   ( t  )
c
Solution to 6-S3
Taking into consideration   2f  2 ( 6.22) and c 
T

T
(6.21) multiplying and dividing by T,
2

x
T ( x  ct )  ( x  ct )   (t  ) . Assuming that t>0,
T
c
c
that is we take time from an assumed time origin t=0, then the negative sign indicates a wave that
propagates in the +x direction and the positive sign indicates a wave that propagates in the – x
direction. Hence the solution becomes,
and recalling that in vacuum v=c gives
x
  ( t  )
c
6-S4
Assume that there is a sinusoidal solution of the Maxwell equations (see Section 6.3.1.1), and
from the solution of sample problem 6-S1 we have a propagating wave of the form,

E( z, t )  E x ( z, t ) .
With the above information determine the B field. Show that Figure 6.6 is correct.
Solution to 6-S4
Since the field is sinusoidal the solution is E x ( x, t )  E0 x Sin where the argument  is of the
z
form,    (t  ) (6.47), where an initial phase that is of no consequence in this case was
c

z
removed. Then, E ( z, t )  E0 x Sin (t  ) , one should remember that c is the speed of light.
c
It is known that Ez=Ey=0. From the expanded form of (6.8) and the result from sample problem
6-S1,
B
 Ex
  y , then it is possible to write,
z
t
By

x
E 0 x cos ( t  )
t
c
c
Integrating with respect to t,

E0x
x 1
sin ( t  )  E x ( z, t )
c
c
c
We can see that the magnitudes of the two vectors are related by the equation E=cB. We know
from the previous problem that the two vectors are orthogonal. From the previous derivations the
two vector fields are in phase. Consequently Figure 6.6 is correct. The above derivations yield
the following result the magnitudes of the magnetic and the electric field in vacuum are related
by the equation E=cB.
By ( z, t ) 
6-S5


We have shown in sample problems 6-S2 and 6-S3 that two vectors E and B , are a solution of
the Maxwell equations. These two vectors define a plane that we call a plane wave front. In these
problems it was assumed that the field propagated normal to one of the coordinate axis in a

Cartesian system. Find the equations of propagation in an arbitrary direction r . Utilizing (6.22),

it is possible to make an expression of the plane as a function of the wave number k .
a harmonic solution, provide an equation of the propagating wave front.
Solution to 6-S5
Assuming

In Figure P6.1there is a plane whose normal is the direction r . In a three dimensional coordinate

system the equation of planes normal to the direction r is given by
Figure P6.1.Plane wave front propagating in space

   
 r  r 0   r  0 . The vector r is normal to the plane, and we can adopt for the normal the


.

k
vector wave number

with the modulus

k
2
. The preceding equation becomes,


   
 r  r 0   k  0 , this equation implies that r  k  C , where C is a constant. This equation


represents a family of planes that are solutions of the Maxwell equations corresponding to the




direction of propagation given by, k  k x i  k y jk z k , that gives the Cartesian equation
C  k x rx  k y ry  k z rz . If the solution is harmonic then assuming that the propagating vector has





amplitude A, A(r.t )  A sin k  r , this function defines a family of planes perpendicular to r .

Moving along r , every time that the argument makes a cycle, that is every time that the distance
d between two points increases by  the vector takes the same amplitude zero. To represent a




 
wave advancing in the positive r direction it should be expressed as A( r.t )  A sin  k  r  t  .


This solution can be represented in the exponential form,


A(r.t )  A e
 

i  k  r  t 


6-S6
 

F ( k  r  t )
Show that the equation A(r , t ) 
is the solution of a 3-D equation representing the
r
propagation of a spherical wave front. In addition indicate the properties of the Poynting vector.

Solution to 6-S6


Either the vector E or B satisfy the wave equation (6.20a) and (6.20b). Recalling (6.24) the
Laplace equation in Cartesian coordinates become,




A A A 1 A



x 2 y 2 z 2 v 2 t 2
By utilizing the spherical coordinates shown in the Figure P.6.1, x=r cos ϕ sinθ, y=r cosθ sin ϕ,
z= r cosθ, the wave equation becomes:







A 2A
1
 
 A
1
2 A 1  A


sin 


r 2 r r r 2 sin   
  r 2 sin 2   2 v 2 t 2


The wave front must have spherical symmetry because it propagates equally in all directions.
Hence the wave equation must be independent of θ and ϕ. Consequently the expression of the


vector field must be A( r, , , t )  A( r.t ) . The consequence of the symmetry is that the derivatives
with respect to θ and ϕ must be zero. Consequently the wave equation should be of the form:



 A( r, t ) 2  A( r, t ) 1  A

 2 2
r 2
r
r
v t
We can use the identity,
 

 2 r A
A 2A 1   1 A


 2 2
r 2 r r r r 2
v t


The solution of the equation must be valid for any r independently of t or for any t independently
of r. Then
 
 
 2 r A
 r A 
 
 
2
2
r
v t 2
With the substitution


r A  A'


 2 A'
1  A'
 2 2
2
r
v t
It is important to recall that in the case of plane wave fronts there are two solutions 1) the
outgoing wave fronts and 2) the converging wave fronts. Hence the solutions are of the form:
 

 

F( k  r  t ) F( k  r  t )
A1 ( r , t )  A 2 ( r , t ) 

r
r


From the solution above it can be concluded that the surfaces of constant phase are spheres.
Indeed the arguments of the vectorial function will be the same for any possible direction of r in
space. Also the Poynting vector will be radial and the corresponding energy will go down with
1/r2. The solution satisfies the principle of conservation of energy.
6-S7
Give arguments that explain that as an approximation in many cases the vectorial functions that
are solutions of the Maxwell equations can be replaced by scalar functions. Thus many problems
in optics can be solved ignoring the vectorial nature of light propagation.
Solution to 6-S7
Starting with the Maxwell equations,

0
(Gauss Law for the electrical field)
(6.6)
B 0
(Gauss Law for the magnetic field)
(6.7)
(Farady’s law of electromagnetism)
(6.8)
(Ampère’s law of electromagnetism)
(6.9)

E 


B
E  
t


E
  B   0 J f  0 0
t


Any solution of an optical problem has to satisfy the above equations. Recall that besides the
field equations of the continuum fields postulated by the Maxwell theory of light, it is necessary
to have constitutive equations that represent the interaction of the field with the medium it is
interacting with. One assumed property is that the media with which the light is interacting is
linear, this property is represented by (6.12) and (6.13), Section 6.3 The Electromagnetic
Theory of Light. The second property is that the media is isotropic, this means that the vectors


E and B are independent of their particular position in space. Another property of the media of
interest is that they are homogeneous indicating that the properties are the same in the region of
interest. The media is non dispersive indicating that the permittivity ε of the material is
independent of the light frequency (wave length) in the region under analysis. Finally the media
under analysis is non magnetic, then the permeability of the media is always the permeability of
the vacuum μ0. Under all these assumptions it is possible to derive the equations of wave
propagation of the electromagnetic field (6.20), Section 6.3 The Electromagnetic Theory of


Light. The wave equation is satisfied by both vectors E and B and also by all the components of
these vectors in any system of coordinates. The equation (6.20) is also satisfied by any scalar
function of the form U( P, t ) , where P indicates a position in space and the propagating field must
satisfy the equation of propagation,
 2 U( P, t ) 
1  2 U( P.t )
0
v 2 t 2
Therefore under the assumed conditions the solutions of optical problems can be described by
the scalar function U ( P, t ) . In view of the linearity of the Maxwell equations more complex
problems can be solved by adding the corresponding scalar functions that are solutions of the
wave equation. This means that the vectorial functions representing solutions of plane wave
propagation and the spherical wave fronts could have been obtained by scalar functions as well.
In many problems the main quantity of interest is the intensity of light I(P), since the
measurements are carried in times that are extremely large compared to the times involved in the
2
light oscillations. Hence (6.60) and (6.61) are replaced by I( P)  U( P) . In one word the final
outcome that can be measured, the modulus of the Poynting vector is independent of the type of
representation of the electromagnetic field (vectorial or scalar). If some of the postulated
conditions are violated, the propagation of the electromagnetic field will no longer be provided
by solutions of (6.20).
Note: A principle similar to the Saint Venant principle in Solid Mechanics is valid for the
Maxwell equations. At some distance of the region where the prescribed conditions are violated
the scalar theory can be applied again. An example is the problem of diffraction caused by an
aperture, at some distance from the aperture the scalar theory can be utilized to obtain the
diffracted field.
6-S8
Prove that (6.61.a) and (6.61b) are correct.
Solution to 6-S8



The Poynting vector was defined by S  E  H
W
(6.56). The magnetic field is given by
m2

1


B
and the magnetic induction is related to the electrical field by B  ( ) 2 E (6.45).
H


1
  2 
Hence B    E .If we consider a harmonic solution of the Maxwell equations of the form
 




E( t, z)  E0 cos( kz  t ) and a similar equation for the magnetic induction field, B , then the
Poynting vector is,

S

 2
E 0 cos 2 ( kz  t ) n

(6.60)
Calculating the average value of the modulus of the vector during an interval of time T,
t 
1
 cos 2 ( kz  t )   cos 2 ( kz  t ) dt .
 t
 cos 2 ( kz  t ) 
1
1
sin 2kz  2( t  ) sin 2(kz  t )

2 4
When    the limit of the above expression is  cos 2 ( kz  t ) 
 cos 2 ( kz  t ) 

1
1
and  S( z, t 
2
2
 2
E0

1
. Then
2
(6.61a)
In the case that the propagation takes place in vacuum (or as a good approximation in air),





B
S  E  H , in vacuum H 
0
0  
S  E H 
E n
0



2



2
Multiplying and dividing by 0 S  E  H  c 0 E n .By utilizing the same steps applied in the

1
preceding derivations,  S ( z, t  c 0 E 02 (6.61b)
2
6-S9
A solid state laser emits a collimated circular beam of 1 mm diameter from the beam conditioner
system, a power meter indicates a 50 mW output. Compute the irradiance of the laser.
Solution to 6-S9
Since the power output is given, irradiance is given by the power of the laser divided by the cross
section of the beam that is a circle.
I
Power 50 103 W

 15,900W / m2
3
2
Area
(5 m)
6-S10
In an optical system the axis of propagation is labeled the z-axis. A plane wave front is
propagating along the optical axis; the amplitude of the electrical field vector is 50 V/m.
Calculate the flux density of the plane wave front.
Solution to 6-S10
Applying (6.61), from section 6.3 The Electromagnetic Theory of Light.
1
I  c0 E 02
2
Rounding the value of c  3  108 m / s , the value of 0  8.854187817...  10-12 F / m
3  108 m / s  8.85  1012 F / m  50V / m 
W
I
 3.32 2
2
m
2
Taking into consideration units relationships F 
Ws
, the resulting units are:
V2
m  Ws  V 2
W
 2
2 2
smV m
m
6-S11
The plane wave front of the preceding problem is converted into a spherical wave front by a
converging lens. We assume that the plane wave front is square in shape and has a side of 7.5
cm. The converging lens is a plane convex lens of diameter 7.5 cm and focal distance f=24.5 cm.
If the source is a helium-neon laser of λ=632.8 nm ,write the equation representing the spherical
wave front.
Solution to 6-S11
From the solution of sample problem 6-S5 and taking into consideration that the wave front is
converging towards the focal point of the lens, the lens the equation of the wave front is


 
E
E( r, t )  0 cos( k  r  t )
r



Since k is of the same direction than r, k  r  k  r . The value of k 
frequency of the 632.8 wavelength is  
  2 4.74  1014
c


2
. The
632.8  10  9 m
3  108 m / s
1
 4.74  1014 then
9
632.8  10 m
s
1
s
The values of the flux density of the wave front must be computed. The power density in the
wave front is 3.32 W/m2. Figure P.6.2 shows the front view and the side view of the optical
system. The area of the wave front entering the lens is a circle of radius 3.5 cm. This area is
converted in a spherical sector. Assuming very small loses in the lens, the density of energy has
to change according to the ratio of the area of the circle of 3.5 cm that enters the lens and the
spherical sector that emerges and is shown in Figure P.6.3.The tangent of the angle θ is equal to
3.5
tg 
 0.14286,   8.130 . Then h=24.5(1-cos 8.13) = 24.50.010=0.245 cm.
24.5
The area of the spherical sector Figure P.6.3, is A=2πrxh and r=(h2+r12)/2h=25.12 cm2. Then
A=38.67 cm2. The area of the entering wave front is, A=πx3.52=38.5 cm2.The energy density of
the spherical wave front is the: 3.3238.5/38.67=3.305 W/m2. The equation of the spherical
wave front is

3.305 W m 2
2
1 

E( r, t ) 
cos 
 r  2  4.74 1014  t 
9
r
s 
 632.8 10 m
Figure P6.2. Front and side view of the optical system.
Figure P6.3.Spherical sector.
6-S12
A plane polarized beam of light of =632.8 nm propagating in air enters in the direction of the
normal of the surface of separation between air and glass. The index of refraction of the glass is
na=1.55. What is the speed of propagation of the light in the glass? Give the wavelength of the
light in glass and the state of polarization.
Solution to 6-S12
According to (6.26), v 
c 3  108 m s

 1.935  108 m s
na
1.55
According to (6.24),   vT , we can also write  0  cT , where  0 is the wavelength in air
(vacuum).

c
Then since the frequency of light is a constant we arrive at 0   n a . Then
. v

632.8nm
 0 
nm  408.25 nm . The state of polarization remains unchanged upon entering
na
1.55
the glass.
6-S13
Compute the optical path of a light beam of wavelength λ=540 nm going through a glass plate of
thickness 20 mm. The index of refraction of the glass is 1.52. Compute the optical path of a
parallel beam going through air. Compute the difference of path of both beams. Convert the
difference of optical path into a phase difference.
Solution to 6-S13
It is possible to represent the electrical and magnetic fields in exponential form as shown in
(6.38) and (6.39), see section 6.3 Then taking into consideration sample problem 6-S4 they can
be written as,


 
 i  k  r  ωt 


E e


and B   e

 
i  k  r  t 


,


where  and  are constant vectors. Make the following consideration; in most of our optical
developments of interest measurements are made on periods of times that are extremely long
when compared to the frequencies of the propagating light beams. Utilizing this argument it is
possible to ignore the time argument in the preceding equations. The spatial argument

 
2  
k r 
n  r can be analyzed further. The modulus of the vector k was derived analyzing the
0
propagation in vacuum and the wavelength in vacuum is given as λ0 ; this expression is also a
good approximation for propagation in air. To generalize the argument for other linear media
write λi, calling i=1,2, 3…the wavelength of the different media that the beam of light goes
2

through, k i 
and taking into consideration the result of sample problem 6-S4,  i  0 ,we
na
i




2n ia
2n ia
get k i 
then the scalar product is k i  r 
n  r . Looking at Figure 6.5 it is possible to
0
0

see that in the case of a plane wave front the vector k i is of the same direction of the normal to
the wave front. Since the modulus of the normal vector is unity it is possible to write,
 
2n ia
ki  r 
r.
0
The above product is called the optical path of the light beam. In simple terms the above result
tell us that when the light enters a linear medium of index of refraction nia the path is increased of
the amount of the index of refraction with respect to the same path in vacuum and as a good
approximation in air. The above property has been derived for a plane wave front. However it
can be generalized for linear media for an arbitrary wave front propagating along a general

trajectory s in the linear space. Since according (6.61a) this is also the direction of the average
Poynting vector, this implies that the energy propagates in the same direction. It follows that the
average Poynting vector is normal to the geometrical wave front. Geometrical light rays can be
defined as orthogonal trajectories to the wave fronts defined by S=constant. The direction of the
rays coincides with the direction of the Poyting vector and the velocity of the vector is v=c/n ia.
Then the concept of optical path can be extended to a general light ray.
S(s)   n(s) ds
Applying the above developments to the present problem, then ST1  20mm 1.52  30.4 mm , the
trajectory of the other beam is ST 2  20mm . The path difference is S  30.4  20  10.4 mm .
The difference of path can be transformed in difference of phase as shown here and is given in
2
 10.4 103 m  2  19.21  103 radians.
radians.  
9
540  10 m
6-S14
A light beam of λ= 630 nm enters a pipe that is filled with a liquid of index of refraction n 1=1.65
There is a second pipe filled with a liquid of index of refraction n2=1.12.Both pipes are 50 m
long. The two beams are fired simultaneously and the arrival of the beams is recorded at the end
of the pipes. Compute the differences of the arrival times.
Solution to 6-S14
For the first pipe the time to arrival is: t1=50 m  1.65/3108m/s = 27.510-8s.
For the second pipe the time of arrival is:t2=50m  1.12/3108m/s = 18.6610-8s
The difference of the arrival times, Δt = t1-t2 = 8.8410-8 s or 88.4 nanoseconds.
Polarized Light
6-S15
Consider two orthogonal plane polarized beams propagating along the same direction, compute
the resulting beam arising from the superposition of the two beams if the phase difference is β=0
or β= nx2π
Solution to 6-S15
We assume that one of the beams is horizontally polarized and the other is vertically plane
polarized. The horizontally polarized beam is represented by (6.70)
E x eix 
 0 
JH  
 and the vertically polarized is J V  E eiy 
 0 
 y 
If the two beams are superimposed we get, (6.76)
 E x 0 e i  x   0   E x 0 e i x 
JR  
i y   
i y 

 0  E y 0 e  E y 0 e 
If β= nx2π, including n=0, the two vectors are in phase and we can write,
 E e i . 
E x 0 
J R   x 0 i .   e i .  . 

E y 0 
E y 0 e 
Applying (6.72) and (6.73) the amplitude of the vector is E 2 R  E 2x 0  E 2yo , and E R  E 2x 0  E 2yo .
We have a vector of the above amplitude that is inclined of the tg 
E0 y
ER
.
cos 
The Jones vector is J R  E R e i.. 
.
 sin  
The resulting beam is a plane polarized beam inclined of the angle θ with respect to the x-axis
see Figure P6.4.
Figure P6.4
6-S16
 E x1 
E x 2 
Two waves corresponding to plane polarized light have equations ei..   and ei..  
E y1 
E y 2 
compute the scalar product, indicate what condition is required for the scalar product to be zero.
Solution to 6-S16
The superposition of the two waves gives:
 E x1  E x 2   i E x1  E x 2  i
i 
J R  e        e 
 e
 E y1  E y 2  
 E y1  E y 2 
E xR 
E 
 yR 
The length of the resultant vector is:
E 2 T  E 2xR  E 2yR and E R  E 2xR  E 2yR
The resultant vector is inclined with respect to the x-axis of tg 
E yR
ER
.
cos 
The Jones vector is J R  E R ei.. 
.
 sin  
We have plane polarized light. The scalar product of the two vectors is:
E x 2 
J1  J 2  ei E x1 E x 2 ei    E x1 E x 2  E y1 E y 2 if this product is zero, the resultant vector is zero.
E y 2 
This result implies that E x1 E x 2   E y1 E y 2 . This means that the two vectors are orthogonal,
consequently orthogonal plane polarized beams that have the same phase have a scalar product
that equals zero.
6-S17
Assume that we have two orthogonal beams but they have a difference of phase β. Compute the
scalar product of the two beams.
Solution to 6-S17
From the preceding problem,
E x 2 
J1  J 2  ei E x1 E x 2 ei    ei E x1 E x 2 E y1E y 2
E y 2 




If the vectors are orthogonal, the vectors satisfy the condition E x1 E x 2   E y1 E y 2 and the scalar
product is zero. Consequently orthogonal vectors whether they are in phase or have different
phase have a scalar product equal to zero.
6-S18
Represent in Jones notation circularly polarized light and indicate all possible cases of wave
superposition that generate circular polarization.
Solution to 6-S18
The circular polarization implies that the amplitude of the two interacting beams that produce the wave
front must be equal and the relative phase must be π/2. It is assumed that the wave front propagates in the
+z direction. It is assumed that at time t=0 the field is also zero for one of the components of the
propagating wave. Furthermore, assuming that this component is in the x-direction, it must be a sine
function. Hence the component in the y-direction is retarded in phase of π/2. In the Jones notation
J CR
 E 0ei  
  i     
E 0e  2  


i

In the above expression, in the unit complex circle e 2  i consequently one can write.
1
J CR  E 0e  
i 
The above expression is a Jones vector of right circularly polarized light. Adopting a graphical

representation of the polarization vector in 2-D, (Figure P6.4 the vector E rotates in the plane with
constant angular speed ω and the components Ex and Ey are the projections of the vector in the xy axes. The vector can rotate in the clockwise and in the counter-clockwise direction. According
to the adopted assumptions at the origin of times the Ex is zero and the Ey will be maximum.



Returning to the Cartesian notation E( z, t )  E 0 sin ( kz  t ) i  cos ( kz  t ) j . It is possible to


dE x
  cos( kz  t ) . As shown in Figure
compute the rate of change of these two components,
dt
dE y
  sin( kz  t )
P6.5a the component Ex decreases with t, while the Ex component,
dt
increases with t, the vector rotates counter clockwise when looking from +z towards the source.
The light is said to be left circularly polarized light. The opposite occurs when the equation of
propagation is changed switching initial conditions, the x-component is a cosine function and the
y-component is a sine function.
.
a)Left polarized
b)Right polarized
Figure P6.5. Sense of rotation of the polarization vector.
The equation of the left polarized light is,
1
J LR  E 0e  
 i 
6-S19
Compute the magnitude of the right and left circularly polarized beams.
Solution to 6-S19
Applying (6.71) and (6.72), from section 6.5 The Jones vector representation.
i 
2
E CR
 JCR  J CR  E 02  i 1    (1  1) ;hence ECR  2 E0
1
 i 
E 2LR  J LR  J LR  E02 i 1    (1  1) ; hence E LR  2 E0
1
6-S20
 1
Describe the state of polarization of the beam given by J  E 0e  
 i
Solution to 6-S20
In view of the notation the equation stated in the problem represents circularly polarized light.



Going back to the Cartesian notation, E( z, t )  E 0 sin e ( kz  t ) i  cos( kz  t ) j . The


derivatives can be utilized to find out the sense of rotation. An alternative procedure is to utilize

the initial conditions. At z = 0 and t = 0 the field is represented by  E 0 j , calling T the period of

 
 

oscillation of the vector, as such for t=T/4, E0 sin  ( ) i  cos( ) j    E0 i ; for t=T/2,
2
2 






E 0 sin(  ) i  cos(  ) j  E 0 j .


If the vector rotates clockwise then the polarization is right circularly polarized light.
6-S21
1
1
Compute the effect of adding J CR  E 0e   and J LR  E 0 e   
i
i

 
Solution to 6-S21
1  1
1
J T  E 0e 
 2E 0 e   

i  i 
0
The result of the addition is horizontal plane polarized light.
6-S22
In problem 6-S17 it was shown that two plane polarized vectors can be orthogonal. Extending
the concept of orthogonal states of polarization to circular polarization, write the expression of
two circularly polarized beams that are orthogonal.
Solution to 6-S22


Recalling (6.71) and (6.72), the scalar product of two Jones vectors is, R  J x1 J x 2  J y1 J y 2
If we have two circularly polarized beams, R  (i )( i)J x 2  (1)(1)  0 . The component vectors
 i 
i 
can be of the form J CR  E0e   and J LR  E 0e   .
1
1
Hence right circular and left circular states of polarization are orthogonal states of polarization.
6-S23
The Jones vector of a harmonic wave propagating in the z-direction is:
 E 0 x ei  
J
i    
E 0 y e

Characterize the state of polarization of the beam.
Solution to 6-S23
 E0 x 
J  ei 
i  can be written and taking into consideration (6.81),
E 0 y e 
E E
E
 y 2   E x 2  2 x y cos   sin 2 
E0y
E0x
E0x E0y
The beam is elliptically polarized or linearly polarized depending on the value of α.
6-S24
The Jones vector of a harmonic wave is given by the equation,
 E 0 ei  
J   i    Assume that α=π/4
E 0e

Characterize the state of polarization and indicate the sense of rotation of the vector.
Solution to 6-S24
Utilizing equation from problem 6-S23,
E E
E
 y 2   E x 2  2 x 2 y 0.70711  0.5
E0
E0
E0
The above equation is the equation of an ellipse. To get the sense of rotation it is necessary to
return to the time dependence of the wave front

 

E( z, t )  E 0 cos( kz  t ) i  E 0 cos ( kz  t  ) j
4 

Adopting the same procedure of sample problem (6-S23): considering the initial condition z=0,
and taking the successive values of t  0 and t  2n  1 2T 4 where n=0,1.2,3
t
0
T/8
3T/8
5T/8
7T/8
i
E0
E0/ 2
-E0/ 2
- E0/ 2
E0
j
E0/ 2
E0
0
- E0
0
The vector rotates counterclockwise and the elliptical polarization is left handed
Refraction and Reflection of light
6-S25
A light beam enters from air to a stratified medium consisting of m parallel layers of indices of refraction
ni with n=1,2, …m, and exit into air. Show that the outgoing beam is parallel to the incoming beam.
Solution to 6-S25
One can start with the different layers and apply the Snell law at each interface. For the first layer one
gets:
n a sin i1  n1 sin r1
For the intermediate layers one can write utilizing the generic notation layer j
n j1 sin ij  n j sin rj
The following relationship is valid for the intermediate layers. For the layer j
ij   rj1
From the above relationship one can write,
n a sin i1  n1 sin r1 =… n j1 sin ij  n j sin rj …, n m sin im  n a sin ra .
The above equations yield,
n a sin i1  n1 sin r1...  n m sin im n a ra , consequently i1  ra
6-S26
A collimated beam of light is incident in a glass plate with parallel faces. Show that the beams
that are reflected in the upper and bottom faces are themselves parallel.
Figure P6.6.Collimated beam incident in a glass plate with parallel faces.
Compute the difference of optical path for the two beams. Express the retardation as a phase. If
the thickness of the plate is 1.5 mm, n2=1.52, λ=635 nm, θt1=9o, compute the phase retardation in
radians
Solution to 6-S26
From the geometry of parallel lines, corresponding angles are (a) θt1=θi2, (b) θr2=θi3. From the law of
reflection (c) θi2=θr2. From the above relationships it follows that θr1=θi3. Utilizing the Snell law
(d) n1 sin i1  n 2 sin i 2 and n 2 sin i 3  n1 sin t 3 . Taking into consideration (a), (b) and (c) one
concludes that, θr1=θt3.
Calling t the thickness of the plate the difference of optical path is,
t
 n2
cos t1
The second emerging beam is retarded of the above magnitude with respect to the incident beam.
The phase retardation is given by,
2 2 t

 n 2 =45,089 radians
 cos  t1
S  2 
6-S27
A collimated beam is incident on a parallel glass plate. Compute the trajectory of the beam and
show that the beam emerges parallel to itself.
Figure P6.7. Beam crossing a parallel plate
Solution to 6-S27
From Figure P6.7, at the emerging face exiting beam forms with the normal to the plate the angle
θi1.=θr1. The distance d can be computed as,

d  AC sin CAB

But CAB  i1   t1 , AC 
d
t sin i1   t1 
cos  t1
t
, then
cos 1t
6-S28
A non polarized beam of light propagating in air impinges at the interface of glass at the
Brewster angle. The glass has an index of refraction n=1.50.What is the state of polarization of
the reflected beam.
Solution to 6-S28
To solve this it is necessary to apply (6.95) i  arctg
n2
and as a good approximation for the air
n1
n=1, therefore i  arctg 1.5  56.30 .
The beam impinging in the glass will be partially transmitted and partially reflected. According
to what was stated in Section 6.7.1 at the Brewster angle Rp becomes zero and no p-polarized
light can be reflected. Upon reflection the light becomes s-polarized.
6-S29
Show that the upon the incidence at the Brewsters’ angle the refraction and reflection angles are
such that the angle of the reflected beam with the surface and the angle made by the refracted
beam with the surface add to 900 , (Figure P6.8) or the refracted and the reflected beams are
perpendicular.
Solution to 6-S29
Let us consider the case that n2>n1, that is the beam comes from the lower index of refraction to
the higher index of refraction (case of external reflection).Utilizing (6.95),one gets, i  56.3
sin 56.3
 055464 giving t  33.7
1.5
The angle between the reflected and the transmitted beams is,
Applying the Snell law n1 sin i  n 2 sin t ; then , sin  t 
90o  56.30  90o  33.7o  90o
FigureP6.8.Beam incident at the Brewster angle, polarization of the reflected beam.
6-S30
A beam is reflected at the interface of a glass plate with air. The angle of incidence is
30o.Compute the degree of polarization of the reflected beam.
Solution to 6-S30
At this point of the example problems, the concept of partially polarized light needs to be
introduced. Polarized light was defined as a light that propagates in a given direction and the two
perpendicular components to this direction have a defined phase difference. For example if the
beam propagates in the z-direction, the Ex and the Ey components have a definite phase
relationship and the beam can be represented by a Jones vector. In partially polarized light the
phase relationship between the two vectors is random. The idea of partial polarization is that in
the propagating beam part of the beam is such that there is a certain proportion that has a definite
relationship between the two orthogonal components. The principle of conservation of energy
tells us that the total energy is the sum of the IP plus the IS where the P and S subscripts of the
light intensity. If one defines the degree of polarization by,
I I
V P s
I P  Is
If I P  Is the degree of polarization is zero, If IS=0 the degree of polarization is 100%. In the case
of reflection I P  R PIiP and IS  R SIiS , where the symbol Ii indicates the incident light. From the
above considerations for reflected light,
R  Rs
V P
R P  Rs
In the particular problem under consideration from the Fresnel relationships, (6.92) and (6.93)
are utilized,
.
2
2
 tg ( t i ) 
 sin(  t  i ) 
Rp  
  0.005766 and R s  
  0.02528
 tg ( t i ) 
 sin(  t  i ) 
V
0.0576  0.0258
%  43.8%
0.0576  0.0258
This result means that from the total energy in the reflected beam 43.8 % is P polarized and the
rest 52.6 % is non polarized indicating that there is a random phase relationship between the two
components of the propagation vector. The light is partially polarized and this effect of reflection
on the light is important in optical systems that are set up for interference of light that should
preserve the state of polarization of the beam.
Lenses and mirrors
6-S31
A point source is located at 50 cm of a plano-convex thin lens and the radius of curvature of the
convex face is 10 cm. The index of refraction of the lens is n=1.5. What is the effect of putting
the planar face towards the source or the convex face in the position of the image?
Solution to 6-S31
Utilizing (6.98) and the convention of signs associated with this equation.
1) R1   , R 2  10
1 1
1 
1
  (1.5  1)  
50 si
   10 
si=14.286 cm, that is the image is to the right of the lens
1 1
 1 1
  (1.5  1)   
50 si
10  
si=14.286 cm, that is the image is to the right of the lens.
Since we are considering a thin lens the position of the image will not be changed by changing
the relative position of the lens faces.
6-S32
A prism of h=10 cm is located at a distance s0=50 cm of a positive lens of focal distance 5 cm.
Give the position and the height of the image of the prism.
Solution to 6-S32
Applying (6.93)
1 1 1
  , si=5.56 cm.
50 si 5
The height of the prism can be computed utilizing (6.100)
s
M T   1  0.111
s2
Then the image size is h=10(-0.111) =-1.11cm. The minus sign indicates that the image is
inverted.
6-S33
An object with h=10 cm must be imaged on a sensor that has h=0.6 cm in the vertical direction.
The focal length of the lens is 5 cm. Give the position of the object with respect to the lens, and
the position of the image.
Solution to 6-S33
We can compute the magnification,
0.6
MT 
 0.06
10
s
hence  i  0.06 =, then si  0.06s0
so
1
1
1

 , then so  88.33cm and si  0.06  88.33cm  5.3cm
s o 0.06s o 5
6-S34
Two biconvex lenses of focal distance 10 cm, are located apart 100 cm. An object of 10 cm
height is located at 15 cm of the first lens. Give the position and the magnification of the object.
Solution to 6-S34
Utilizing (6.101)
s2 
f 2d  f1f 2s1 / s1  f1  10  100  10  10  15 / 15  10

 11.67
d  f 2  f1s1 / s1  f1 
100  10  10  15 / 15  10
Utilizing (6.103)
f 2s 2
10  11.67
116.7
MT 


 0.33
ds1  f1   s1f1 100  5  15  10
350
As such, the image of the object is at 11.67 cm behind the first lens, the image is upright and the
height of the image is hi=100.33=3.3 cm.
Problems to solve
6.1
We have a wave front propagating in air, the equation of the wave front is,

E x ( z, t )  50

W

 
6 1
14 1
,
E
cos

3
.
16

10

z
(
m
)

9
.
48

10

t
(
s
)
y  0 , Ez  0 .


m2
m
s

The phase of the wave front is defined by the initial condition at t=0, E x  0 .
With the information provided obtain the following information:
The wave length of the propagating radiation, the speed of propagation, the frequency, the time

period, the state of polarization, the amplitude of the vector E x and the initial phase.
6.2

In problem 6.1the electric vector E field of the radiating field was determined. We know that the

luminous radiation requires a vector field B associated with the electrical field. Provide the

equation of the B field associated with the electrical field defined in problem 6.1.
6.3
The same electromagnetic field of problems 6.1 and 6.2 is propagating along the z-axis of a
Cartesian system of axis. The electric vector forms and angle of 30o with the x-axis. Give the
Cartesian equations of the electromagnetic field.
6.4
Given the index of refraction of a glass is 1.556 compute the value of the permittivity in the
glass.
6.5
Assume that the modulus of the k vector in a given medium is 1.6107 rad/m. What is the
wavelength of the light if the index of refraction of the medium is 1.5?
Polarized light
6.6
Write the Jones vectors for + 45o polarized light.
6.7
Write the Jones vectors for - 45o polarized light.
6.8
Write the Jones vector for a harmonic linearly polarized wave propagating along the x-axis. The
plane of vibration is the yz-plane. Also write the Jones vector of the B field.
6.9
 20  V
The Jones vector of a harmonic propagating wave is, J  e 
 .What is the physical
( i)50 m
meaning of the vector length?
6.10
Write the Jones vector of right and left circularly polarized light that will produced a plane
polarized state propagating in the z-direction; the plane of polarization is the yz-plane.
6.11
Characterize the state of polarization of a harmonic beam given by the Jones vector of problem
6.9.
Refraction and reflection of the light
6.12
A beam travels in a parallel face glass plate of index of refraction n=1.54. The glass plate
interfaces with a liquid of index of refraction n=1.52. The beam impinges at 300 with the normal
to the interface. Compute the direction of propagation of the beam in the fluid.
6.13
If the glass plate of the preceding problem is surrounded by air, what is the angle of incidence of
the beam in the glass plate?
6.14
If the glass plate is part of a rectangular container filled with the fluid at what angle will the
beam of light emerge from the container?
6.15
For the beam of problem 6.14 compute the difference of optical path of the propagating beam
with respect to a beam propagating in air. The thickness of the wall of the container is t=4mm,
the dimension of the container in the direction of propagation is 50 mm.
6.16
Utilizing equations (6.92) and (6.93) and Snell’s law for the interface of two media with indices
of refraction n1 and n2 express (6.92) and (6.93) in function of θi, n1 and n2.
6.17
Derive the coefficients RP and RS for normal incidence.
6.18
Compute the coefficients for the interface of air-glass and glass-air. Take the coefficient of the
glass as 1.5.
Lenses and mirrors
6.19
There is an object with h=5 cm, it must be captured by a sensor of h=0.6 cm. Select a lens that
can satisfy the above condition and satisfy the restriction that between the lens and the object the
distance cannot be greater than 70 cm
6.20
One has an optical system formed by a positive lens of focal distance 10 mm. 500 mm behind the
first lens is another lens of focal distance 25 mm. Compute the position, magnification of an
object that is 15 mm from the first lens.
6.21
Add to the system of problem 6.19 a lens system capable to produce an image that can be
captured by a sensor of 0.6 height.