Download 5. INTERFERENCE. Introduction.

Electromagnetic Waves & Optics: Lecture Notes
5.
©Kevin Donovan
INTERFERENCE.
Introduction.
Previously, in these notes the amplitude of an electromagnetic wave has been considered
and its vector quality when discussing polarization. The vector nature of the electric field is an
important property because it is frequently the case that there are several electromagnetic
waves from different sources or more frequently different parts of the same source, and the
need is to find the resultant field in a particular region of space by invoking the superposition
of fields. In that superposition it is essential to take account of the vector nature of the field by
formally superposing those fields using vector addition.
In the previous analysis of polarisation it was noted that the change in direction of the electric
field vector as a result of superposition leads to different polarization states, left and right
handed elliptically and circularly polarised light and plane polarised light. In this section
interest is in the change in amplitude of the electric field vector (and consequently light
intensity) as a result of superposition or addition of waves from multiple sources or from one
source first split and then later recombined at some new location. This brings about a variety
of effects grouped collectively under the title of interference; effects including, two slit
interference (Young’s Slits), thin film interference and various interferometers.
Diffraction phenomena are closely related to interference phenomena and are dealt with
separately using the tools that are developed for interference in what follows.
When considering interference it is convenient to continue using plane waves to describe the
electromagnetic waves as these are more easily manipulated mathematically. It will later be
important to consider how useful this plane wave approach may be in describing reality and
this will lead to the concept of coherence.
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Interference.
To establish the mathematical background required to describe interference the transverse
plane wave remains useful;


 z
 
E ( z, t )  E0 cos2  t   E0 coskz  t 

 
(5.1a)


E(z, t )  E0 exp j kz  t 
(5.1b)
or equivalently
as the most simple solution to the electromagnetic wave equation yet containing all essential
features.
As before the electric field in a region of space resulting from two separate sources of field
can be found by adding the two fields
 

E  E1  E2
(5.2)
Noting that when discussing optics these fields oscillate at typically 5  1014Hz and any
measurement made will not be of the instantaneous field but of the light intensity I which is
related to the time averaged square of the electric field as previously demonstrated
I
E2
(5.3)

where  is the impedance of the medium (see previous notes) and the triangular brackets
represent a time average over several cycles.




 








E 2  E  E  E1  E 2  E1  E 2  E12  E 22  2E1  E 2
(5.4)
Taking the time average of each side and assume that everything takes place in the same
medium and that therefore  is common to all fields, this will lead to the cancellation of  in
much of what follows.
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NB. It is important to note here that when taking the dot product in 5.4 the value of the
projection of one field on the other is being found. This automatically takes care of the
possibility that the two fields are of different polarization when adding them to
discover interference effects.
I  I1  I 2  I12
I1 

E12

,
(5.5)
I2 

E 22
(5.6)

The third term is the interference term and is of interest in this section
I 12 


2 E1  E 2
(5.7)

NB Hecht chooses
(i)
To omit the impedance in his derivations, a perfectly reasonable course to
take as the medium doesn’t change and the impedance will cancel when
converting from fields to intensity. I choose to keep the impedance in as this
is formally correct.
(ii)
To use the symbol  to represent the additional phase of the plane wave
whereas I will use the symbol  to represent the additional phase and  to
represent the total phase.
(iii)
To use  to represent the phase difference between two waves whereas I will
use  to represent the phase difference between two waves.
In order to evaluate the interference term the two electric fields are written as two plane
waves.
One of the most important properties of the electromagnetic wave is its phase, and in the
case of plane waves travelling in the z direction the phase is simply the argument of the
cosinusoid or exponential,   kz  t .
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This phase does not depend on x or y once z has been fixed, ie. the phase is the same at
any value of x and y for a given value of z and t, or otherwise stated the phase is
constant over a plane perpendicular to the direction of travel hence the name plane
wave.
With two or more plane waves it is necessary to specify the intrinsic phase of each wave as
their peaks and troughs do not generally coincide in time and space. To make things more
general in order to account for this the phase of a wave traveling in the z direction becomes
  kz  t   .
For a plane wave traveling in an arbitrary direction the information concerning this direction is

represented by the fact that k is a vector (a fact up until now ignored when the light was
assumed propagating in the z direction) and the phase is more precisely written as
 
  k  r  t  
The two electric fields are then;


 
E1  E01 cos k1  r  t  1


 
E2  E02 cos k 2  r  t  2



(5.8a)

(5.8b)
NB Whilst the wavevectors k are identified by a subscript 1 or 2 there is no subscript
on . This is because the fields are oscillating at the same frequency, the magnitude of
the k vectors are the same for both fields and it is the direction of propagation of the
wave (or equivalently wavevector) and the electric field vectors (polarisation and
amplitude) that are the difference between the two fields.
The interference term can then be found as follows;



 


 
 
E1  E2  E01  E02 cos k1  r  t  1  cos k 2  r  t  2

(5.9)
 
It is the time average of E1  E2 that is required and recognizing this the equation may be rewritten with the time dependence separated out using the trigonometric identity
cosA  B   cos A cos B  sin A sin B
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 

©Kevin Donovan





 
 
 
E1  E2  E01  E02 cos k1  r  1 cost   sin k1  r  1 sint 

cosk2  r  2 cost   sink2  r  2 sint 
(5.10)
The time dependent variable (represented in the cosinusoids) has now been separated


making the time average E1  E 2 easier to evaluate. The time average only applies to the
sin2 t  terms and the cos2 t  terms and averaging over a time period, T, much greater
that the period of the cosinusoids,  
2
, they both average to:

cos2 t  sin2 t 
1
.
2
Also the time average of the product of the sin and cosine is zero;
cost sint   0 .
This leads to a tremendous simplification in the time average of 5.10 to
 
 


 



 
 
 
1
E1  E 2  E01  E02 cos k1  r  1 cos k 2  r   2  sin k1  r  1 sin k 2  r   2
2

(5.11)
Using the same trigonometric identity used previously;
cos A cos B  sin A sin B  cosA  B 
further simplification may be achieved and 5.11 becomes;




 


1
E1  E 2  E 01  E 02 cos k1  r  1  k 2  r   2
2
I 12 


2 E1  E 2



E 01  E 02

cos 

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
(5.12)
(5.13)
Electromagnetic Waves & Optics: Lecture Notes
©Kevin Donovan
where
   
  k1  r  k2  r  1  2
(5.14)

is the total phase difference between the two plane waves at the position, r , where the
interference is to be determined. This phase difference is the sum of the phase difference
due to different optical path lengths traversed by each wave and due to the initial phases of
the two waves, 1 and 2.


NB The term E01  E02 account for the case where there are different plane polarization
directions and the projection of field 1 onto field 2 is taken.
Commonly the polarisations of the two interfering waves are identical, ie. the electric fields
are parallel;
E E
I12  01 02 cos 

(5.15)
I12 may be written in terms of intensities by using the earlier expressions relating fields and
intensities
I1 
2
E 01

2
E 01

2
I2 
2
E 02


2
E 02
2
I12  2 I1I 2 cos 
(5.16)
(5.17)
And the total irradiance (power per unit area or intensity) is then
I  I1  I 2  2 I1I 2 cos 
(5.18)
The total irradiance then varies from point to point in space as cos varies between +1 and 1
The maximum irradiance
IMax  I1  I 2  2 I1I 2
(5.19a)
Is obtained when the phase difference is   2m
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m  0,1,2,3
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In this case of total constructive interference the two waves are in phase.
And the minimum irradiance
IMin  I1  I 2  2 I1I 2
(5.19b)
is obtained when   2m  1
m  0,1,2,3
In this case of total destructive interference the two waves are 1800 out of phase.
Another commonly encountered situation is where the electric fields have not only the same
polarization but also the same intensity, I1 = I2 = I0 . In this case the total irradiance may be
written
I  2I 0 1  cos   4I 0 cos2

2
Where the trigonometric identity 1  cos   2 cos2
(5.20)

has been used.
2
From this;
Imin  0 ,
I max  4I 0
Identical arguments and results apply to the interference of two spherical waves emanating
from two point sources S1 and S2 that overlap at a point P.
P
r1
r2
S2
S1
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Previously spherical waves have been introduced propagating in free space and written as


A01
E1(r1, t ) 
exp j k 0 r1  t  1
r1
(5.21a)


A02
E 2 ( r2 , t ) 
exp j k 0 r2  t   2 
r2
(5.21b)
r1 and r2 are the distances of point sources 1 and 2 from the point of overlap, P, or
equivalently the radii of curvature of the two spherical waves.
The phase in this case depends only on the distance r from the point sources and the surface
of any sphere centered on a point source is a surface of constant phase.
The phase difference between the two waves may be easily identified as
  k 0 r1  r2   1   2 
(5.22)
Using the expression for total irradiance as found previously for plane waves, 5.20, and the
new phase difference
1

I  4I0 cos2  k 0 r1  r2   1   2 
2

(5.23)
The condition for maxima and minima are as before ie.  = 2m for maxima and  = (2m
+ 1) for minima where m = 0, 1, 2, 3, ……..
Using the expression for  we obtain constructive interference when
r1  r2    2m  2  1
(5.24a)
k0
And destructive interference when
r1  r2    2m  1  2  1
k0
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(5.24b)
Electromagnetic Waves & Optics: Lecture Notes
©Kevin Donovan
P
r1
r2
S1
S2
Both equations 5.24a and b describe a family of hyperboloids as depicted in the above figure
where the hyperboloid surfaces represent the points P where constructive/destructive
interference occurs. If the intrinsic phases from the two sources, S1 and S2 , are equal, ie, 1
= 2 and chosen to be zero then 5.24 may be rewritten as
r1  r2    2m  m0
(5.25a)
k0
r1  r2    2m  1   m  1  0
k0

2
(5.25b)
Each value of m is represented by a hyperboloid with the positive values on the right hand
side of the mid-line with r1  r2 and the midline (or in 3D a plane perpendicular to the page)
where r1  r2 and the left hand side hyperboloids for r1  r2 . A screen placed as a plane
intersecting this set of hyperboloids may be imagined where at the points of intersection there
is constructive interference. Any point, P, on this midpoint plane has r1  r2 and therefore
represents 5.25a with m = 0, the zeroth order interference fringe.
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These equations can be used to describe the appearance of interference fringes between
two line or point sources such as Young’s slits. It is now possible to explore some examples
of the ways in which interference effects are manifested.
Interference Effects in Practice.
i)
Interference by Division of Wavefront.
Young’s Slits
Thomas Young, in 1801, was one of the first people to demonstrate the wave nature of light
carrying out what has since become a classic experiment that has been applied to particles
to demonstrate matter waves as well as light. In Young’s original experiment he used a
pinhole with a light source behind it to define a point source of spherical waves with two
further pinholes in a screen at a distance from the source pinhole much greater than the
wavelength of light. The two pinholes in the screen act as two separate but related light
sources that have been derived from an original light source by division of the wavefront of
that original source. That the two secondary sources are related allow interference effects to
be observed by placing a second screen at a similar distance from the two secondary light
sources.
To find the light intensity at the second screen the electric fields of the two light sources are
added taking into account the difference in phase between those two fields as usual. If the
two holes providing the interfering sources are of the same size then the approximation that
the intensities of the two sources are equal may be made and equation 5.20 used to find the
intensity at any point P on the screen;
I  4I 0 cos2

2
(5.20)
NB. Before continuing further, it is necessary, for purposes of simplification, to keep the
observation point P close to the centre of the screen and the diagram below is exaggerated
for demonstration purposes. In fact small values only of  are of interest in what follows.
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P
r2
y
S2
r1

S


O
a
S1
d
s
It remains to establish the phase difference,  , at the observation point, P.
The two point sources act as sources of spherical waves and the phase difference was seen
earlier in 5.22 to be;
  k 0 r1  r2   1   2   k 0 (r1  r2 )
(5.22)
If the screen with the two point sources is far enough from the original point source, S,
(orders of magnitude larger than a wavelength) then the spherical wave at S 1 and S2 will be
approximately plane and there will be no other source of phase difference apart from the path
difference and the term (1 - 2) is zero because light at both S1 and S2 have travelled an
equal distance from the source S. With the aid of the figure above, the path difference which
is the same as the optical path difference, , may be found (as the waves propagate in air
with n = 1) and hence the phase difference   k 0  .
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Electromagnetic Waves & Optics: Lecture Notes
  S1P  S2P  r1  r2
©Kevin Donovan
(5.26)
A simplified expression for r1  r2 may be obtained using the law of cosines,
c 2  a2  b2  2ab cosC
as applied to the triangle S1S2P
P
r2
S2
r1

a
90 - 
S1
r22  a 2  r12  2ar1 cos(90  )  a 2  r12  2ar1 sin 
(5.27)
r22  r12  a 2  2ar1 sin 
(5.28)
r2  r1r2  r1  r2  r12r2  a 2  2ar2 sin 
(5.29)
  r1  r2  a sin   a
(5.30)
Where the fact that a  r2 and r1  r2 has been used to simplify 5.29
And a further simplification

y
s
(5.31)
may be used to obtain
  k 0   k 0a
y 2 y

a
s 0 s
(5.32)
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Equation 5.25a and b gave the conditions for maxima and minima respectively and using this
result and 5.25 the condition for a bright fringe may be found;
r1  r2   ay
s
 m 0
(5.33)
Or in terms of position on the screen, y
s
y m  m 0
a
(5.34)
gives the position of the mth bright fringe on the screen. The fringes are equally spaced with a
separation
y  y m  y m 1 
s
0
a
(5.35)
Having found  the intensity may be written as a function of y by using this in 5.20
 ay 
 a sin  
  
  4I 0 cos2 
  4I 0 cos2 
I  4I 0 cos2    4I 0 cos2 
 2 
  0s 
 0 
(5.36)
With

a
a
sin   k 0 sin 
0
2
(5.37)
y = 0s/a
I
y
0
 is the phase difference between rays emanating from each slit when the screen is at
infinity (ie the two rays are parallel) as shown in the diagram below.
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Electromagnetic Waves & Optics: Lecture Notes
k
  0 S1B
2
S2
©Kevin Donovan
(5.38)

a
S1
B
This analysis has demonstrated that;
(i)
to have well separated fringes we need a small separation, a, between pinholes or slits
and that
(ii)
longer wavelengths will give rise to broader fringes.
Young had some stringent requirements on the geometry of his experiment in order to see
the interference fringes;
(i)
The slits cannot be too far apart and generally a << s
(ii)
The original source must be at a distance, d, from the screen such that the spherical
wave approximates a plane wave in order that 1   2
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(iii)
©Kevin Donovan
Fringes will only be observed near the centre of the screen where r1  r2 is not too
large.
(iv)
The value of s
a
must be large if fringe separation is to be large enough to observe
due to the small size of the wavelength of visible light.
All of the above apart from (iv) result from the lack of coherence in the sources available to
Young.
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Electromagnetic Waves & Optics: Lecture Notes
ii)
©Kevin Donovan
Interference by Division of Amplitude.
An important way of achieving the conditions for exhibiting interference effects, ie. obtaining
two or more oscillating electric fields and then sending them on different paths before
recombining them to observe interference is to split a primary wave into two separate
secondary waves by partial reflection of the primary and using the reflected and transmitted
waves as the secondary waves to be recombined. This may be achieved in many ways,
sometimes coming about in a quite natural manner eg. thin film interference while often
achieved with a particular optical arrangement known as an interferometer of which there are
several examples.
Dielectric Thin Film Interference.
One of the most commonly observed examples of interference through division of amplitude
occurs where light incident on a thin layer of dielectric undergoes reflections from the top and
bottom surface of the layer and under the right conditions constructive or destructive
interference occurs between the two reflections. Examples of this are the colour effects seen
when a thin layer of oil is floating on water and the colours seen in a soap bubble. The
interference effects caused by multiple reflections in the thin scales of a butterflies wings give
rise to their spectacular iridescence. The effect is also of use in many technological
applications where thin layers are designed and constructed with the intention of creating thin
dielectric film interference eg anti-reflection coatings.
The diagram below indicates schematically how interference effects are produced as a result
of partial reflection/transmission at a thin dielectric film. For purposes of this analysis the
dielectric film of refractive index, nF , is standing on a substrate of refractive index, n S, and
light is incident from a medium of refractive index, n0 (typically air). There will in general be
multiple reflections at the air/film interface and at the film/substrate interface. Transmission
will also occur into the substrate. Equations 5.19 may be used to discover the conditions for
constructive or destructive interference by first calculating the phase difference between the
two rays, 1 and 2, shown propagating upwards from the top of the film in the figure.
The optical path is the real space distance traveled multiplied by refractive index of the
medium and therefore the optical path difference, , between rays 1 and 2 after their
separation on arrival at A is
  nF AB  BC  n0 AD
(5.39)
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Electromagnetic Waves & Optics: Lecture Notes
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And the phase difference due to the optical path difference is the optical path difference
multiplied by the magnitude of the wavevector in free space (vacuum).
OPD  k0   k 0 nF AB  BC  n0 AD 
2
nF AB  BC  n0 AD (5.40)
0
1
2
I
n0
D
A
d
C
nF
T
B
nS
NB. For the total phase change the phase change that occurs on reflection which may
be different for each ray, as we recall from the Fresnel equations, needs to be
included. Thus,
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Electromagnetic Waves & Optics: Lecture Notes
  k0 nF AB  BC  n0 AD   2  1 
©Kevin Donovan
2
nF AB  BC  n0 AD  2  1
0
(5.41)
From geometry and Snell’s law we obtain everything in terms of T,
AB  BC 
d
cos T
AC  2d tan T
n
AD  AC sin I  2d tan T F sin T
n0
2d
AB  BC 
cos T
nF sin2 T
AD  2d
n0 cos T
whence
 



4nFd
2  2nFd
1  sin2 T    2  1 
cos T   2  1 (5.42)

 0  cos T
0

It is often convenient to have the phase difference in terms of the angle of incidence rather
than the internal angle and Snell’s law may be used to achieve this
 
4nFd
4d
2  n2 sin2      
1  sin2 T   2  1 
nF
I
2
1
0
0
0
(5.43)
Before addressing the outstanding question of the reflection phase shifts 1 and 2 note that
nF may be greater than or less than n0 and nS, eg a freestanding soap film in air or an air gap
between two parallel separated glass slides respectively. Recall also that there are two types
of reflection namely internal reflection where the refractive index of the sourced region is
greater than that of the unsourced region and external reflection where the refractive index
of the unsourced region is greater than that of the sourced region. With these possible types
of reflection in mind several possibilities may be identified that apply at near normal
incidence,   300;
(i)
External reflection plus internal reflection
An example is the soap film/bubble free standing in air, nF  n0  nS  n Air . In this case for
near normal incidence,   300 ,  2  1     and the total phase difference is
129
Electromagnetic Waves & Optics: Lecture Notes
 
(ii)
4nFd
cos T  
0
©Kevin Donovan
(5.44)
Internal reflection plus external reflection.
An example of this is an air gap between two glass slides. Again for near normal incidence
2  1    
and the total phase difference is
 
(iii)
4nFd
cos T  
0
(5.45)
Both reflections are internal (or external)
If the reflection at the top and bottom interfaces are both internal reflections (or external
reflections) the factor  2  1   0 and
 
(iv)
4nFd
cos T
0
(5.46)
Reflections from partially metallised surfaces
Examples are a thin metallic film or two partially metallised surfaces separated by a dielectric
(eg. air gap between two partial mirrors. In this circumstance then 1   2  0 as there is no
phase shift upon reflection at a metal where boundary conditions require that the light electric
field is zero and again
 
4nF d
cos T
0
(5.47)
So far nothing has been said about the intensities of the two combining beams, I1 and I2 and
so nothing is known about the actual irradiance, IR , achieved in the reflected beam. Neither
has the possibility of multiple reflections been considered nor the intensity of the final
transmitted beam, IT.
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Electromagnetic Waves & Optics: Lecture Notes
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Multiple Beam Interference.
When considering the thin dielectric film interference it was recognized that multiple
reflections could occur and this possibility is now explored.
r/
r
t/
t
d
n1
nF
n1
Consider the above system of a thin dielectric film of thickness d and refractive index nF
embedded in a medium of refractive index n1 eg. air. Further, the film is considered to be nonabsorbing at wavelengths of interest. An electromagnetic wave impinges from the left hand
side and is partially transmitted and partially reflected undergoing a series of subsequent
reflections and transmissions at the left hand and right hand interface. To find the total
reflected or transmitted fields all of the reflected or transmitted waves must be summed
taking into account phase shifts between the waves as usual. The fraction of the electric field
amplitude transmitted on entering the film is denoted as t and on leaving the film as t/ and the
fraction reflected at the n1/nF interface as r and at the nF/n1 interface as r/. These quantities
were discussed earlier when the Fresnel equations were established and they depend, in
general on the angle of incidence and the difference in refractive indices at the interfaces. In
particular, using an argument due to Stokes, it was demonstrated that r I   r T  and
t I t T   1 r 2 where I  and T  come as a pair of angles related by Snell’s law.
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Electromagnetic Waves & Optics: Lecture Notes
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tt/r/9E0
tt/r/8E0expj(t-4)
tt/r/7E0
tt/r/6E0expj(t-3)
tt/r/5E0
tt/r/4E0expj(t-2)
/ /3
tt r E0
tt/r/2E0expj(t-)
/ /
tt r E0
tr/2E0
rE0
tt/E0expj(t)
tr/E0
T
tE0
I
d
E0
The above diagram establishes the sequence of reflected partial amplitudes and of
transmitted partial amplitudes.
The sum of the partial transmitted amplitudes may be established giving the transmitted
electric field amplitude ET.
ET  tt E0e jt  tt r / 2E0e j (t  )  tt r  4E0e j (t  2)  ...  tt r (2N 1)E0e j (t (N 1))
(5.48)


ET  tt E0 e jt 1  r  2e  j  r  4 e  j 2  r  6 e  j 3  r 8 e  j 4  ........
NB.  is the phase difference between two adjacent secondary waves.
Recalling that an infinite geometric progression is
132
(5.49)
Electromagnetic Waves & Optics: Lecture Notes

a
n  1  a  a 2  a 3  a 4  ..... 
n 0
©Kevin Donovan
1
1 a
The geometric progression in square brackets on the RHS has a  r 2e  j  r 2e  j again
and 5.72 may be compactly rewritten as


tt 
ET  E 0 e jt 
2  j 
1  r e

(5.50)
Taking the time average of ET multiplied by it’s complex conjugate and dividing by  gives the
intensity as usual and for the transmitted intensity;
IT 
E02
tt 2


2 1  r 2e  j 1  r 2e  j
IT  I0
(1  r 2 )2
 1 r 4  2r 2 cos 
 I0
1 R 2
(5.51)
1  R 2  2R cos 
The Stoke’s identity r 2  1 tt  has been used.
Finally, using the trigonometric identity, cos   1  2 sin2

an expression for the fractional
2
transmitted intensity, T, is obtained.
I
T  T 
I0
1  R 2

I
T T 
I0
(5.52)
 

1  R 2  2R 1  2 sin2

1  R 2


1  R 2  2R  4R sin2
2
2 
1  R 2
1  R 
2

 4R sin2
1
T 
1
4R
1  R 2

sin2
(5.53)
2
(5.54)
2
A quantity that will appear frequently called the coefficient of finesse, F is now introduced
which is a commonly used figure of merit for a Fabry Perot device
2
 2r 
4R
 
F  
2
 1 r 
1  R 2
and obtain a compact form for the fractional transmitted intensity;
133
(5.55)
Electromagnetic Waves & Optics: Lecture Notes
T 
©Kevin Donovan
1
(5.56)

1  F sin2
2
It is now a simple task to find the fractional reflected intensity, R, as in the absence of
absorption R + T = 1 and thus;

2
R  1




1  F sin2
1  F sin2
2
2
F sin2
1
(5.57)
Maximum transmitted intensity will obtain under the same conditions that give minimum
reflected intensity. It is clear that to maximize the transmitted intensity the denominator in
5.56 needs to be minimized, ie. the sinusoid in the denominator needs to be zero and
therefore

 m
2
(5.58)
for maximum transmission (minimum reflection). Bearing in mind the relation 5.42 between
phase difference and wavelength,
 
4
nF d cos T
0
,this is also a condition on
wavelength.
By rearrangement of 5.42 with the condition 5.58 a new condition is obtained;



2nF d cos T  m
2 0
m
0

 m F  d cos T
2nF
2
where 1   2  0 has been used for simplicity.
When the condition is fulfilled
T=1
The physical meaning becomes clear when normal incidence on the film is considered and
the equation becomes

m F d
2
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Electromagnetic Waves & Optics: Lecture Notes
©Kevin Donovan
This is the requirement that half integer wavelengths, of the wave in the film, fit between the
two inner surfaces of the film, ie. that when standing waves are supported there is maximum
transmission and minimum reflection.
m=1
m= 3
m=2
d
Unsurprisingly the maximum reflection is obtained where transmission is minimum.
Minimum transmitted intensity occurs where the denominator of 5.56 is as large as
possible which is when sin2

 1. When this condition holds
2


 2m  1
2
2
  2m  1
And the value of the minimum transmitted intensity using sin2


1
IT Min  I0 1  I0 
4R
1 F
1 
 1  R 2
m  0, 1, 2...

 1 in 5.56 is;
2


2




1  R 2
  I  1  R 

I



0
0

 1  R 2  4R 
1  2R  R 2  4R 


(5.59)
IT Min
 I0
1 R 2
1 R 2
135
(5.60)
Electromagnetic Waves & Optics: Lecture Notes
And the maximum reflected intensity occurs when sin2
IR Max
 I0
F
 I0
1 F

 1 and is
2
 4R 


 1  R 2 


 4R 

1 
 1  R 2 


The fractional transmission and reflection are plotted below.
136
©Kevin Donovan
2
 I0
4R
1  R 2
(5.61)
Electromagnetic Waves & Optics: Lecture Notes
©Kevin Donovan
Transmission through thin film vs
phase shift, r = 0.2, 0.5, 0.9
1
0.6
0.4
0.2
0
-4
-2
0
2
4
Phase shift  ( rads)
 
4nF d
cos T
0
Reflection from thin film vs phase
difference, r = 0.2, 0.5, 0.9
1
0.8
Ir (fraction)
It (fraction)
0.8
0.6
0.4
0.2
0
-4
-2
0
2
Phase shift  ( rads)
137
4
Electromagnetic Waves & Optics: Lecture Notes
©Kevin Donovan
The transmission/reflection intensities from a thin dielectric film as described by equations
5.56 and 5.57 are shown in the two graphs above where three values of r are used, the
sharpest minima/maxima occur for the largest values of r.
There is a maximum in the transmission where  is an integer multiple of 2 and as r is
increased this maximum becomes increasingly sharp around these values. The inverse is
true of the reflection. Such behavior suggests potential use as a tuned bandpass filter.
It is important to pause and to recall once again what it is that  represents physically. It
represents the difference in phase between adjacent reflected (or transmitted) partial waves
and that phase difference is determined by the extra distance traveled by one wave with
respect to the other and so varies with angle of incidence and film thickness. It also depends

inversely on the wavelength of the light within the film,   0
nF
. This means that the
horizontal axis, , in the above graphs could as easily be a plot of either inverse wavelength
or of angle of incidence. This is clearly seen when writing the phase difference explicitly as
  2dnF 2
0
cos T .
In other words there are several variables that could be plotted on the x axis in place of ,
1
namely by holding T constant the inverse wavelength, 01 (or frequency   c
) could be
0
plotted on the x axis with a series of wavelengths/frequencies (or angles) at which sharp
resonant transmission occurs given by;
  2m  2dnF
2
cos T ,
0
0 
2dnF
cos T ,
m
 m 0
T  cos1
 2dnF



NB the resonant frequencies are equally spaced the wavelengths are not!
With modern technology allowing control over very small distances it is also possible to
adjust d and allow this as the variable with maximum transmission when
d m
0
cos T
2nF
Finally, as has been seen in the discussion on birefringence, the refractive index of certain
media may be altered by application of an electric field allowing another way to alter  and
thus the transmission.
In the above equations the integer m is also known as the order of the transmission (or
reflection). An extended white light source would then be split into transmitted spectral
138
Electromagnetic Waves & Optics: Lecture Notes
©Kevin Donovan
components where the wavelength criteria is satisfied. Depending on the order, m, this would
be a wavelength band around some peak that is transmitted whilst the other wavelengths are
reflected and lost until a second order (third order etc) transmission peak allows further
transmission of another band of wavelengths around a second (third etc.) wavelength. The
peak transmission (or reflection) will also depend on the viewing direction and it is this that
gives rise to the colours observed in a thin oil film floating on water or the colours observed in
a soap bubble.
Looking at the above graph for transmission through the thin film an increasingly narrow
range of  (or T or 0 ) over which transmission occurs is seen as r is increased. Ie. the
transmission of the thin film may be highly tuned. This highly defined directionality and
wavelength range, comes about due to the large number of coherent sources (partial waves)
that contribute to the overall beam.
The thin film dielectric with multiple interference clearly demonstrates useful properties and
the potential for constructing a useful device. The Fabry Perot interferometer is an
engineered structure using the basic principles that have just been discussed and finds many
uses in optics from spectroscopy through high resolution optical filters to laser resonators.
For this reason it is worth examining in some detail.
Fabry Perot Interferometer.
One of the simplest realizations of an optical structure, using the principles of the multiple
reflection interference of the thin film previously discussed, is to take a pair of parallel partially
reflecting surfaces separated by a careful engineered spacer.
Spacer
Spacer
Extended
source
d
139
Screen
Electromagnetic Waves & Optics: Lecture Notes
©Kevin Donovan
The above diagram shows such a structure, known as a Fabry Perot interferometer or Fabry
Perot etalon, with a pair of metallised partially reflecting surfaces, in this case evaporated
onto two transparent substrates. The reflecting surfaces are held precisely parallel to one
another and usually the substrates will be slightly wedged on the non-metallised surface in
order to suppress the formation of secondary parallel reflecting systems that would interfere
with the operation of the primary system. The two reflecting surfaces of the primary system
are held at a precise separation, d. There is in the example shown an extended light source
to the left and a screen to the right. Choosing a ray from a given portion of the extended
source traveling at an arbitrary angle, its progress is followed as it undergoes multiple
reflection/transmission events before the transmitted rays from this one coherent point source
are collected by a second lens and brought to a focus at some point P on a viewing screen
(photographic plate, retina etc.). Of course, any ray traveling at the same angle from an
equivalent point on the extended source (at the same distance from the system axis) would
have been brought to a focus on the screen at the same distance from the system axis
resulting in the appearance of a series of concentric rings centered on that axis.
This system has the essential characteristics of the previous situation examined with the
exception that;
(i)
The metallised reflecting surfaces will be a source of dissipation/absorption and this
will mean that the Stokes relations that were used frequently in the previous discussion no
longer hold ie.
I0  I r  It
T R 1
and tt /  r 2  1
Rather these are modified to account for the absorbance, A, in the following way;
I0  I r  It  I A
T R  A 1
(5.62)
Where 5.62 again represents the conservation of energy but with the dissipation (absorption)
term A included .
(ii)
The gap, d, is much larger than in the original thin film scenario and can be from
microns up to centimeters as it is now a factor under control of engineering.
and
(iii)
Metallic films will introduce a phase shift, (), upon reflection which may not be zero
or  and that may depend on the angle of incidence, T.
Now, the phase difference between two successive waves is as usual
140
Electromagnetic Waves & Optics: Lecture Notes
 
4nF d
cos T  2
0
©Kevin Donovan
(5.63)
A complete analysis with absorbance included gives
2
A 
1

T  1 

 1  R  1  F sin2 
2
With maximum transmission
A 

TMax  1 

 1 R 
2
Ignoring all these differences and using what has already been found from analysis of
multiple reflections in thin films.
I
T  t 
I0
1

1  F sin 2
(5.64)
2
This expression of 5.64 makes it much easier to understand what is happening with the
transmission as it did with the thin film analysis as maxima are simply obtained when the
sinusoid in the denominator goes to zero.
Two graphs shown below demonstrate the effect of the absorbance. The effect of A is to
reduce the maximum value of the transmission. The graphs show the transmission of the
Fabry Perot with an absorbance, A = 0.1 as a function of phase difference (equivalently,  or
0 ). Also shown is the normalised transmission. Note in the first that the maximum is well
below 1 as a result of the absorbance. In those two graphs the amplitude reflectance, r, is
changed from 0.9 to 0.7 and this has big effects on the transmission with the width of the
transmission getting much larger and the transmission never dropping to zero in the case of a
reflectance of 0.7.
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Electromagnetic Waves & Optics: Lecture Notes
©Kevin Donovan
Transmission of Fabry Peroy
Transmission of a Fabry Perot with r= 0.9, A = 0.1
0.25
0.2
0.15
0.1
0.05
0
-4
-2
0
2
4
Phase Difference  (units of )
Normalised Transmission.
Fabry Perot Transmission Normalised to the
Maximum Transmission, r = 0.9, 0.7
F = 22, 1.66
1
0.75
0.5
0.25
0
-4
-2
0
2
Phase difference,  (units of )
142
4
Electromagnetic Waves & Optics: Lecture Notes
©Kevin Donovan
Properties of the Fabry Perot.
When using the Fabry Perot as a filter or monochromator (wavelength selective device cf
prisms and gratings) several important questions arise concerning the resolution of the
Fabry Perot, ie. how well defined is the transmitted wavelength/frequency?
To answer this a number of things need to be considered.
i)
The free spectral range.
This is the separation of the transmission peaks of adjacent orders which occur at  = 2m
so peaks are separated in phase by 2. The separation of the peaks of adjacent orders in
terms of wavelength is called the free spectral range of the Fabry Perot and indicates the
wavelength range over which the interferometer may be operated without adjacent orders
mixing. It may be seen how this matters by considering operating the interferometer as a
variable plate separation device to find an unknown wavelength 1. As d is altered  will
change with it until at some point  satisfies the condition for maximum transmission at the
specific wavelength of interest and the light is transmitted. Knowing d and  should
therefore allow knowledge of 1. But recalling that maximum transmission occurs for
d m
0
cos T there is a problem as m is unknown. There is a second transmission peak
2nF
in the neighbourhood of this one where the order number is m + 1 and the wavelength may
correspond to m+1 and not m. To avoid any such confusion it is necessary to know how far
apart the two transmission windows are in terms of wavelength (or frequency). The peak
separation or free spectral range is simply found as
FSR  m  m1 
2nF d cos T 2nF d cos T 2nF d cos T


m
m 1
m(m  1)
This may be written in a different way using
2nF d cos T
 m
m
 FSR 
2nF d cos T

 m
m(m  1)
m 1
(5.65)
NB. Remember that in this preceding discussion  m is that wavelength in free space!
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Electromagnetic Waves & Optics: Lecture Notes
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The free spectral range in terms of frequency is
m1  m 
c
m1

c
c(m  1)
cm
c
1




m 2nF d cos T 2nF d cos T 2nF d cos T 
 FSR 
(5.66)
c
c

2nF d cos T m(m  1)FSR
(5.66a)
Generally the device is used with an air gap, nF = 1, and at normal incidence in which case
5.65 simplifies to
FSR 

2d
 m
m(m  1) m  1
FSR 
c
c
1


2d m(m  1)FSR 
And 5.66 simplifies to
Where  is the time taken for the wave to travel from one reflecting surface to the other and
back again, also known as the round trip time within the interferometer.
If the free spectral range is made to be larger than the interval over which any emission is
expected then the problem of confusing orders can be avoided.
ii)
Width of the transmission window.
As a measure of the width of the transmission window the value of  where the maximum
transmission has dropped by half,  1 is found and the width of the transmission window is
2
defined as twice this ie Full Width at Half Maximum, FWHM  2   1
, a commonly
2
employed criterion.
The curve is simply T 
1

1  F sin2
and where this has dropped by 50% is given by.
2
1
1  F sin2
 1
2
2
144
 0 .5
(5.67)
Electromagnetic Waves & Optics: Lecture Notes
©Kevin Donovan
Re arranging to make  1 the subject of the equation
2
F sin2
 1
2
2 1
(5.68)
1
F
(5.69)
 1
sin
2
2 
 1  2 sin1
2
1
F
(5.70)
For large coefficient of finesse, F, the approximation
sin1
1
1

F
F
may be used
 1  2
2
1
F
(5.71)
And the width of the peak, at half the maximum is
FWHM  2 1 
2
4
F
It is the case that the phase cannot be measured and only relative phases have any practical
effect. It is of more practical interest to ask questions about wavelength or frequency when
discussing light waves. These are, of course, related to the phase as seen many times,
  2dnF
2
cos T  m2 for maximum transmission of the mth order wavelength m. If the
m
phase is known but it is the wavelength that is required then proceeding as follows;
d 
1
 4nF d cos T
d 0
2
Where the device is used at m
145
(5.72)
Electromagnetic Waves & Optics: Lecture Notes
©Kevin Donovan
d 
1
 4nF d cos T
d 0
2m
FWHM  2m
1
1
FWHM   2m
4nF d cos T
nF d cos T
(5.72a)
1
F
(5.73)
Using the previously established condition for maximum transmission of mth order
m 
2nF d cos T
m
5.73 may be written
FWHM  m
2 1
m F
(5.74)
An important property, the resolution of the Fabry Perot, used in mth order is defined as
m
FWHM

m
F
2
(5.75)
and serves as a dimensionless figure of merit.
This may be required in terms of frequency where the relationship between frequency and
wavelength,  m  m  c , for the mth order may be used. Using a procedure identical to that
used to translate from phase to wavelength translating from wavelength to frequency is done
as follows;
d
d

c
2
At the specific frequency, m where  = m
 FWHM  
c
2m
 FWHM  
c  FWHM
2 1
  m
m
m
m F
(5.76)
Therefore the resolution in mth order is as previously found for this quantity in terms of
wavelength
146
Electromagnetic Waves & Optics: Lecture Notes
m
 FWHM

©Kevin Donovan
m
F
2
(5.77)
NB the minus signs that occur in these equations when differentiating has been
subsequently ignored as the meaning is simply that for example on going from  to  +
 the frequency will go from  to  -  and in this context it carries no significance.
iii)
The ratio of peak separation to peak width or finesse, F
What is a good figure of merit when defining the performance of a Fabry Perot?
Comparing the separation of peaks or free spectral range with the width of a single peak a
figure of merit for the resolution of the Fabry Perot. May be obtained where the former is
usually required to be large and the latter small by taking as a figure of merit.
F 
FSR
2
 F


4
FWHM
2
F
(5.78)
known as the Finesse
NB The finesse and the coefficient of finesse , while closely related are not the same
thing!!
By comparing the frequency resolution expression 5.77 and the definition of finesse, F it is
seen that the frequency resolution of the mth order may also be written as
m
FWHM

m
F  mF
2
(5.79)

m
F  mF
2
(5.79b)
and
m
FWHM
iv)
What is the resolving power, R , of the Fabry Perot?
This is a question that asks, “ if a source has two wavelengths very close together what
is the minimum wavelength separation, Min that a Fabry Perot would be able to
“see”?
147
Electromagnetic Waves & Optics: Lecture Notes
©Kevin Donovan
Min
1.0
1
2
0.5
The criteria for being able to resolve two slightly different wavelengths is that their half
maximum height values as measured by the Fabry Perot just cross as shown in the diagram
above where monochromatic wavelengths 1 and 2 are measured as finite width features
due to the finite width of the transmission window of the Fabry Perot.
1  11   2   2 1
2
2
Therefore
Min   2  1  2 1  FWHM  2 m
2
1 1
m F
The resolving power is simply defined as
R
m

m F


 mF
 Min Min
2
Finally, knowing F or F the minimum resolvable wavelength difference for a given order, m,
may be obtained;
148
Electromagnetic Waves & Optics: Lecture Notes
Min 
2m
m F

m
mF
149
©Kevin Donovan