Topic 7 wave interference

Topic 4.1 Waves, Interference and Optics
1
UEEP1033 Oscillations and Waves
Topic 7:
Interference and Diffraction

2
• Interference
e.g. rainbow colours produced by a thin film of oil on a
wet road, where the light reflected off the surface of the
oil interferes with the light reflected off the water surface
underneath
• Diffraction
e.g. the waves spread out in a semicircular fashion after
passing through the narrow mouth of a harbour
• Both result from the overlap and superposition of waves
Interference and Diffraction of Waves

3
Interference
two waves are
out of phase
destructive
interference
two waves are
in phase
constructive
interference
amplitude of their
superposition is zero
amplitude of the
superposition
(ψ1 + ψ2) = 2A
A is the amplitude of the
individual waves

4
Figure (a)
• Two monochromatic waves ψ1 and ψ2 at a particular point
in space where the path difference from their common
source is equal to an integral number of wavelengths
• There is constructive interference and their superposition
(ψ1 + ψ2) has an amplitude that is equal to 2A where A is
the amplitude of the individual waves.
Figure (b)
• The two waves ψ1 and ψ2 where the path difference is
equal to an odd number of half wavelengths
• There is destructive interference and the amplitude of
their superposition is zero
Interference

5
Point source of light is
illuminating an opaque object,
casting a shadow where the
edge of the shadow fades
gradually over a short distance
and made up of bright and dark
bands, the diffraction fringes.
Shadow fades gradually
>> Bright and Dark Bands
= Diffraction Fringes
Diffraction

6
Francesco Grimaldi
in 1665 first accurate report
description of deviation of light from
rectilinear propagation (diffraction)
The effect is a general characteristics of wave phenomena
occurring whenever a portion of a wavefront is
obstructed in some way
Diffraction

7
Plane wavefronts approach a barrier with an opening or an
obstruction, which both the opening and the obstruction are
large compared to the wavelength
Opening
(size = d)
Obstruction
(size = d)
wavelength,  d >> 

8
• If the size of the opening or obstruction becomes comparable to the
wavelength
• The waves is not allowed to propagate freely through the opening or past the
obstruction
• But experiences some retardation of some parts of the wavefront
• The wave proceed to "bend through" or around the opening or obstruction
• The wave experiences significant curvature upon emerging from the opening
or the obstruction
curvatured  

9
As the barrier or opening size gets smaller,
the wavefront experiences more and more curvature
More curvature
Diffraction
d  

10

11
Historical Background

12
Light source
Aperture
Observation
plane
Screen
Arrangement used for observing
diffraction of light
Corpuscular Theory
shadow behind the screen
should be well defined, with
sharp borders
Observations
• The transition from light
to shadow was gradual
rather than abrupt
• Presence of bright and
dark fringes extending far
into the geometrical
shadow of the screen

13
Christian Huygens
Huygens’s Principle
Each point on the wavefront of a disturbance were considered to be a
new source of a “secondary” spherical disturbance, then the
wavefront at a later instant could be found by constructing the
“envelope” of the secondary wavelets”

14
Every point on a propagation wavefront serves as the source of
spherical secondary wavelets, such that the wavefront at some
later time is the envelope of these wavelets
Plane wave Spherical wave

15

16
Plane wave
Spherical wave
Every point on a propagation wavefront serves as the
source of spherical secondary wavelets
the wavefront at some
later time is the
envelope of these
wavelets

17
• When a wavefront encounters an
aperture in an opaque barrier, the
barrier suppresses all propagation of
the wave except through the aperture
• Following Huygen’s principle, the
points on the wavefront across the
aperture act as sources of secondary
wavelets
• When the width of the aperture is
comparable with the wavelength, the
aperture acts like a point source and
the outgoing wavefronts are
semicircular
Huygen’s Principle

18
18
• Ignores most of each secondary wavelet and only retaining the
portions common to the envelope
• As a result, Huygens’s principle by itself is unable to account
for the details of the diffraction process
• The difficulty was resolved by Fresnel with his addition of the
concept of interference

19
Augustin Jean Fresnel
• 1818, Fresnel brought together the ideas of Huygens and Young
and by making some arbitrary assumptions about the amplitude
and phases of Huygens’ secondary sources
• Fresnel able to calculate the distribution of light in diffraction
patterns with excellent accuracy by allowing the various
wavelet to mutually interfere
Huygens-Fresnel Principle

20
Huygens-Fresnel Principle
Every unobstructed point of a wavefront, at given instant, serves
as a source of spherical secondary wavelets
(with the same frequency as that of the primary wave)
The amplitude of the optical field at any point beyond is the
superposition of all these wavelets
(considering their amplitudes and relative phases)

21
James Clerk Maxwell
Electromagnetic Wave
t
H
E





t
E
H





0 E

0 H


22
Gustav Kirchhoff
• Kirchhoff's diffraction formula can be used to model the
propagation of light in a wide range of configurations, either
analytically or using numerical modelling
• It gives an expression for the wave disturbance when
a monochromatic spherical wave passes through an opening in
an opaque screen
• The equation is derived by making several approximations to the
Kirchhoff integral theorem which uses Green’s theorem to derive
the solution to the homogeneous wave equation
• In 1882, Kirchhoff developed a more rigorous theory
based directly on the solution of differential wave
equation
Kirchhoff's diffraction formula

23
Although the problem was physically somewhat unrealistic , i.e. it
involved an infinitely thin yet opaque, perfectly conducting plane
screen, the result was nonetheless extremely valuable, providing a good
deal of insight into the fundamental processes involved
Arnold Johannes Wilhelm Sommerfeld
In 1896, Sommerfeld published the first exact solution for a particular
diffracting configuration , utilizing the electromagnetic theory of light
Rigorous solutions of this sort do no exist even today for many of the
configurations of practical interest
The approximation treatments of Huygens-Fresnel and Kirchhoff
are adequate for many purposes

24
Interference
Young’s Double-Slit Experiment

25
L >> a
a = slits separation

26
• A monochromatic plane wave of wavelength λ is incident upon
an opaque barrier containing two slits S1 and S2
• Each of these slits acts as a source of secondary wavelets
according to Huygen’s Principle and the disturbance beyond
the barrier is the superposition of all the wavelets spreading out
from the two slits
• These slits are very narrow but have a long length in the
direction normal to the page, making this a two-dimensional
problem
• The resultant amplitude at point P is due to the superposition
of secondary wavelets from the two slits

27
• Since these secondary wavelets are driven by the same incident
wave there is a well defined phase relationship between them
• This condition is called coherence and implies a systematic
phase relationship between the secondary wavelets when they
are superposed at some distant point P
• It is this phase relationship that gives rise to the interference
pattern, which is observed on a screen a distance L beyond the
barrier

28
The secondary wavelets from S1 and S2 arriving at an arbitrary
point P on the screen, at a distance x from the point O that
coincides with the mid-point of the two slits
Distances: S1P = l1 S2P = l2
Since L >> a it can be assumed that the secondary wavelets
arriving at P have the same amplitude A
The superposition of the wavelets at P gives the resultant
amplitude:
 )cos()cos( 21 kltkltAR 
ω = angular frequency
k = wave number
(5)

29
This result can be rewritten as:
Since L >> a, the lines from S1 and S2 to P can be assumed to be
parallel and also to make the same angle θ with respect to the
horizontal axis
 2/)(cos[]2/)(cos2 1212 llkllktAR 
The line joining P to the mid-point of the slits makes an angle θ
with respect to the horizontal axis
21 cos/ lLl 
 cos/212 Lll
(6)

30
When the two slits are separated by many wavelengths, θ is very
small and cos θ  1. Hence, we can write the resultant amplitude
as:
)2/cos()cos(2 lkkLtAR 
= path difference of the secondary wavelets
The intensity I at point P = R2
12 lll 
)2/(cos)(cos4 222
lkkLtAI 
This equation describes the instantaneous intensity at P
The variation of the intensity with time is described by the
cos2(ωt − kL) term
(7)
(8)

31
• The frequency of oscillation of visible light is of the order of
1015 Hz, which is far too high for the human eye and any
laboratory apparatus to follow.
• What we observe is a time average of the intensity
• Since the time average of cos2(ωt − kL) over many cycles = 1/2
 the time average of the intensity is given by:
)2/(cos2
0 lkII 
2
0 2AI  = intensity observed at a maximum of the interference pattern
described how the intensity varies with l)2/(cos2
lk
(9)

32
I = maximum whenever l = n (n = 0,±1, ±2, …)
I = 0 whenever l = (n + ½) 
From figure on slide-25: l  a sin θ
Substituting for l in Equation (9), we obtain:
(10))2/sin(cos)( 2
0  kaII
When θ is small so that sinθ  θ, we can write:
)/(cos)(
)2/(cos)(
2
0
2
0


aII
kaII
(11)
 /2where k

33
separation of the
bright fringes
If there were no
interference, the
intensity would be
uniform and equal
to Io/2 as indicated
by the horizontal
dashed line
Light intensity I (θ) vs angle θ

34
Intensity maxima: .....,2,1,0, 

 n
a
n
.....,2,1,0, 

 n
a
L
nLx
(12)
(13)
(14)
(15)
The bright fringes occur at distances from the point O given by:
Minimum intensity occur when:
The distance between adjacent bright fringes is:
.....,2,1,0,
2
1








 n
a
L
nx
a
L
xx nn

1

35
Fraunhofer and Fresnel
Diffraction

36
Observation
screen 
Fraunhofer and Fresnel Diffraction
S
Lens
Plane
waves
Opaque shield , with a single
small aperture of width a is
being illuminated by plane wave
of wavelength  from a distant
point source S
Case-1
observation screen is very
close to 
Image of aperture is projected
onto the screen

37
Observation
screen 
S
Lens
Plane
waves
Case-2
observation screen is moved farther
away from 
Image of aperture become
increasingly more structured as the
fringes become prominent
Fresnel or Near-Field
Diffraction

38
S
Lens
Plane
waves
Case-3
observation screen is at very
great distance away from 
Projected pattern will have spread
out considerably, bearing a little or
no resemblance to the actual
aperture
Observation
screen 

Thereafter moving the screen
away from the aperture change
only the size of the pattern and not
its shape
Fraunhofer or Far-Field
Diffraction

39
S
Lens
Plane
waves
Case-4
If at that point, the wavelength  of
the incoming radiation is reduce
Observation
screen 
the pattern would revert back
to the Fresnel case
If  were decreased even more, so that  → 0
The fringes would disappear, and the image
would take on the limiting shape of the aperture

40
If a point source S and the
observation screen  are very far
from 
S
Lens
Plane
waves
Observation
screen 
Fraunhofer Diffraction
observation screen  are
too near 
Fresnel Diffraction

41
S
Lens
Plane
waves
Observation
screen 
Fraunhofer Diffractiond

R R
R is the smaller of the two
distances from S to  and to 


2
d
R
d = slit width

42
Practical realization of the Fraunhofer condition
F1 F2

43
Diffraction
• Any obstacle in the path of the wave affects the way it spreads out; the
wave appears to ‘bend’ around the obstacle
• Similarly, the wave spreads out beyond any aperture that it meets. such
bending or spreading of the wave is called diffraction
• The effects of diffraction are evident in the shadow of an object that is
illuminated by a point source. The edges of the shadow are not sharp but
are blurred due to the bending of the light at the edges of the object
• The degree of spreading of a wave after passing through an aperture
depends on the ratio of the wavelength λ of the wave to the size d of the
aperture
• The angular width of the spreading is approximately equal to λ/d; the
bigger this ratio, the greater is the spreading

44
The Mechanism of Diffraction
• Diffraction arises because of the way in which waves propagate as
described by the Huygens-Fresnel Principle
• The propagation of a wave can be visualized by considering every point
on a wavefront as a point source for a secondary radial wave
• The subsequent propagation and addition of all these radial waves form
the new wavefront
• When waves are added together, their sum is determined by the relative
phases as well as the amplitudes of the individual waves, an effect
which is often known as wave interference
• The summed amplitude of the waves can have any value between zero
and the sum of the individual amplitudes
• Hence, diffraction patterns usually have a series of maxima and minima

45
• A monochromatic plane wave is incident
upon an opaque barrier containing a single
slit
• Replace the relatively wide slit by an
increasing number of narrow subslits
• Each point in the subslits acts as a point
source for a secondary radial wave
• When waves are added together, their sum is
determined by the relative phases and the
amplitudes of the individual waves, an effect
which is often known as wave interference
• The summed amplitude of the waves can
have any value between zero and the sum of
the individual amplitudes
• Hence, diffraction patterns usually have a
series of maxima and minima
Single Slit Diffraction

46
Diffraction at a Single Slit
The resultant amplitude at
point P is due to the
superposition of secondary
wavelets from the slit
monochromatic
x = 0
x
each of these strips acts as a
source of secondary wavelets
d = slit width

47
Figure in slide-46:
• A monochromatic plane wave of wavelength λ is incident upon
an opaque barrier containing a single slit
• The slit has a width d and a long length (>> d) in the direction
normal to the page, reducing this to a two-dimensional problem
• The resultant amplitude at point P is due to the superposition
of secondary wavelets from the slit

48
• The centre of the slit is at x = 0
• We divide the slit into infinitely narrow
strips of width dx
• Following Huygen’s principle, each of
these strips acts as a source of secondary
wavelets and the superposition of these
wavelets gives the resultant amplitude at
point P

49
• We consider the case in which P is very distant from the slit
• Consequently, all the wavelets arriving at P can be assumed to
be plane waves and to have the same amplitude
• In addition, we can assume that the lines joining P to all points
on the slit make the same angle θ to the horizontal axis
• The amplitude dR of the wavelet arriving at P from the strip dx
at x is proportional to the width dx of the strip
• its phase depends on the distance of P from the strip
i.e. on (l − x sin θ), where l is the distance of P from the
midpoint of the slit

50
Hence dR is given by:
ω = angular frequency k = wave number α = constant
)]sin(cos[  xlktdxdR
The resultant amplitude at P due to the contributions of the
secondary wavelets from all the strips is


2/
2/
)]sin(cos[
d
d
xlktdxR
)cos(]sin)2/sin[(
sin)2/(
kltkd
kd
d
R 



d = slit width

51
Instantaneous intensity I at P:
2
2
2222
]sin)2/[(
]sin)2/[(sin
)(cos



kd
kd
kltdRI
Since the time average over many cycles of cos2(ωt − kl) = 1/2
the time average of the intensity is given by:
2
2
02
2
0
sin
]sin)2/[(
]sin)2/[(sin
)(





 I
kd
kd
II
2/22
0 dI  = maximum intensity of the diffraction pattern
This equation describes how an incident plane wave of wavelength λ
spreads out from a single slit of width d in terms of the angle θ
2/sin kd

52
The diffraction pattern of a single slit
The zeros of intensity in the diffraction pattern occur at θ = ±nλ/d, where
n = ±1,±2, . . . , under the small angle approximation sin θ  θ
This figure is a plot of
I(θ) against θ for a value
of kd/2 = 10π
2
2
0
sin
)(


 II
 sin)2/(kd
θ
d = slit width

53
The first zeros in the intensity occur when  sin)2/(kd
dk /sin/2but 
the degree of spreading depends upon the ratio λ/d
When λ << d, as in the case of light, sin θ  θ, giving the first
zeros in the diffraction pattern at:
...),2,1( 

 n
d
n
d


In general, zeros in intensity occur when:
d = slit width

54
Double slits of finite width
• Consider each of the two slits to be composed of infinitely
narrow strips that act as sources of secondary wavelets
• Then the resultant amplitude R at a point P is the
superposition of the secondary wavelets from both slits








2/2/
2/2/
2/2/
2/2/
)]sin(cos[
)]sin(cos[
da
da
da
da
xlktdx
xlktdxR
d = the width of each slit
a = separation of the slits

55
]sin)2/cos[(
sin)2/(
]sin)2/sin[(
)cos(2 


 ka
kd
kd
kltdR
Double slits of finite width
]sin)2/[(cos
]sin)2/[(
]sin)2/[(sin
)( 2
2
2
0 


 ka
kd
kd
II
Resultant Intensity:
• This result is the product of two functions.
• The first is the square of a sinc function corresponding to
diffraction at a single slit
• The second is the cosine-squared term of the double-slit
interference pattern

56
Diffraction at a Double Slit
)2/sin(cos)( 2
0  kaII
2
2
0
)2/sin(
)2/sin(sin
)(



kd
kd
II
)2/sin(cos
)2/sin(
)2/sin(sin
)( 2
2
2
0 


 ka
kd
kd
II

57
Diffraction at a Double Slit
double-slit interference maxima occur at angles:
...),2,1( 

 n
a
n
zeros in the diffraction pattern occur at angles:
...),2,1( 

 n
d
n

58

59
Observation
screen 
S
Lens
Plane
waves
Opaque shield , with a single
small aperture of width a is
being illuminated by plane
wave of wavelength  from a
distant point source S
Case-1
observation screen is very close
to 
Image of aperture is projected onto
the screen

60
Observation
screen 
S
Lens
Plane
waves
Case-2
observation screen is moved
farther away from 
Image of aperture become increasingly
more structured as the fringes become
prominent
Fresnel or Near-Field
Diffraction

61
S
Lens
Plane
waves
Case-3
observation screen is at very
great distance away from 
Projected pattern will have spread
out considerably, bearing a little or
no resemblance to the actual
aperture
Observation
screen 

Thereafter moving the screen
away from the aperture change
only the size of the pattern and not
its shape
Fraunhofer or Far-Field
Diffraction

62
S
Lens
Plane
waves
Case-4
If at that point, the
wavelength  of the
incoming radiation is reduce
Observation
screen 
the pattern would revert back
to the Fresnel case
If  were decreased even more, so that  → 0
The fringes would disappear, and the image
would take on the limiting shape of the aperture

63
S
Lens
Plane
waves
observation screen  are very far
from 
Observation
screen 
observation screen  are
near to 
Fresnel Diffraction

64
S
Lens
Plane
waves
Observation
screen 
a

R R
R is the smaller of the two
distances from S to  and to 


2
a
R

65
Practical realization of the Fraunhofer condition
F1 F2

66
coherent line source
Introduction
yi
A linear array of N
in-phase coherent
point sources
e.g. secondary sources
of the Huygens-
Fresnel Principle
a long slit whose
width D is much
less than 
Each point emits a spherical
wavelet equal to:
A0 = source strength
)sin(0
krtE r
A


67
Introduction
The sources are very weak and their number N is
tremendously large and the separation between is
vanishing small
Finite segment of array yi contain yi (N/D) sources
Assume that the array is divided into M such segments
(i.e. i goes from 1 to M)
The contribution of electric field intensity at P from the i-
th segment is:
 D
yN
ir
A
i
i
i
krtE

 )sin(0

68
Introduction
The contribution of electric field
intensity at P from the i-th segment
is:
Finite segment of array yi
contain yi (N/D) sources
)sin(0
krtE r
A

 D
yN
ir
A
i
i
i
krtE

 )sin(0
Each point emits a spherical wavelets:

69
Introduction
Net field at P from all M segments is:
AL = source strength per unit length
For continuous line source (yi 0,i.e. M )


M
i
iir
A
ykrtE i
L
1
)sin(
 NA
D
A
N
L 0lim
1


dy
r
krt
AE
D
DL 


2
2
)sin(

70
Introduction
3rd term
Can be ignored as long as its
contribution to the phase is
insignificant
 22
cos)2/(sin RyyRr
For continuous line source (yi 0,i.e. M )
dy
r
krt
AE
D
DL 


2
2
)sin(

71
Single-Slit Fraunhofer Diffraction
Point of observation is very
distant from the coherent line
source and R >> D
 r(y) never deviates
appreciably from its
midpoint value R
 Quantity (AL/R) at P is
essentially constant for all
elements dy

72
Point of observation is very distant from the
coherent line source and R >> D
P
D
r1
rM
R
P
r1  rM  R

73
(AL/R)dy = amplitude of the wave
The field at P due to the differential segment of the source dy
is:

The phase is more sensitive to
variation of r(y) than is the
amplitude
dy
r
krt
AE
D
DL 


2
2
)sin(
dykrt
R
A
dE L
)sin( 

74
Even when y = ± D/2
i.e. (D2/4R)cos2 = negligible
This is true for all values of 
when R is adequately large
  22
cos)2/(sin RyyRr

75
Fraunhofer conditions: The distance r is linear in y
Therefore the phase can be written as a
function of the aperture variable

 sinyRr
dyyRkt
R
A
E
D
D
L


2
2
)]sin(sin[
)sin(  yRkkr

76
Simplify:

where:
dyyRkt
R
A
E
D
D
L


2
2
)]sin(sin[
)sin(
sin)2/(
]sin)2/sin[(
kRt
kD
kD
R
DA
E L




)sin(
sin
kRt
R
DA
E L









 sin)2/(kD

77
22
sin
2
1
)( 














R
DA
I L
Irradiance:
where:
When  = 0, sin/ = 1 
Principle maximum

T
EI 2
)( 
2
1
)sin(  T
kRt
2
2
1
)0( 






R
DA
I L

78
The Irradiance resulting from an idealized coherent line
source in the Fraunhofer approximation is:
When D >> 
the irradiance drop extremely rapidly as  deviates from
zero
Or:
Where:
2
sin
)0()( 







 II
2
2
1
)0( 






R
DA
I L  sin)2/(kD
 sin)/( D

79
An aperture of this sort might typically have a width of
several hundred  and a length of few centimeter

80
l
Each strip is a long coherent line
source and can be replaced by a
point emitter on the z-axis, which
radiates a circular wave in the y =
0 (xz-plane)
Usual Analysis Procedure
Divide the slit into a series of
long differential strips (dz by l)
parallel to the y-axis

81
Our Problem is:
To find the field in the xz-plane due to the infinite
number of point sources extending across the
width of the slit along the z-axis
Only evaluate the integral of the contribution of
dE from each element dz in the Fraunhofer
approximation

82
The complete solution for the slit is:
Here the line source is short, D = b, thus  is not large
Or:Where:
Although the irradiance fall off rapidly, higher-order
subsidiary maxima will be observable
2
sin
)0()( 







 II
 sin)2/(kD  sin)/( D

83
The extreme of I() occur at values of  that cause dI/d
= 0
The irradiance has minima (equal to zero) when sin  = 0
Also when:
0
)sincos(sin2
)0( 3





I
d
dI
......,3,2, 


tani.e.
0sincos

84
The extreme of I() occur at values of  that cause dI/d = 0
The irradiance also has minima when:
0
)sincos(sin2
)0( 3





I
d
dI


tani.e.
0sincos

85
The irradiance also has minima when:
Only one such extremum
exist between adjacent
minima
I() must have subsidiary
maxima at
......,4707.3
,4590.2,4303.1




tani.e.
0sincos

86
Easy Way of Analysis
Every point in the aperture emitting
rays in all direction in the xz-plane
Undiffracted beam arrive on
the viewing screen in phase and
a central bright spot will be
formed by them

87
Rays coming off at an angle 1
Path length difference
between the ray from the
very top and bottom
bsin1 = 
Easy Way of Analysis
bsin1 = 

88
A ray from the middle of the slit will
then lag ½ behind a ray from the
top and exactly cancel it
A ray just below center will cancel a ray
from just below the top, and so on …
All across the aperture ray-pairs will
cancel, yielding a minimum
The irradiance has dropped from its
high central maximum to the first
zero on either size at bsin1 = ±

89
As the angle increases further, some small fraction of the rays will
again interfere constructively, and the irradiance will rise to form
a subsidiary peak.
A further increase in the angle
produces another minimum
bsin2 = 2

90
In general,
the zero irradiance will occur when
bsinm = m
where, m = ±1, ±2, ±3, …
Equivalent to:  = m = (kb/2)sinm
wave number, k = 2/

91
The Fraunhofer diffraction pattern of a single slit
Irradiance drop: 1.0 0.047 0.017 0.008
2.46
3.47
1.43
subsidiary maxima
2
sin
)0()( 







 II

92
Double-Slit Fraunhofer Diffraction
Two long slits of width b
and center-to-center
separation is a
Each aperture by itself
would generate the same
single-slit diffraction
pattern on the viewing
screen
At any point on the
viewing screen, the
contributions from the slits
overlap

93
If the primary plane wave is incident on the double-slit aperture
 at normal incidence, the wavelets are all emitted in-phase
The interference fringe at a
particular point is determined by
the differences in the optical
path lengths traversed by the
overlapping wavelets from the
two slits
The flux-density distribution =
rapidly varying double-slit
interference system modulated
by a single-slit diffraction
pattern

94
Double-slit
pattern for
a = 3b
b = individual slit width

95
(a)
(b) (c)
Intensity plot for a
double-slit
interference
Intensity plot for
diffraction by a typical
by a single-slit of width
a
Intensity plot for
diffraction by double-slit
of width a
The curve of (b) acts as an
envelope, limiting the intensity
of the double-slit fringes in (a)

96
The total contribution to the electric field
in the Fraunhofer approximation
where:
constant-amplitude factor C =
secondary source strength per
unit length along the z-axis
divided by R




2/
2/
2/
2/
)()(
ba
ba
b
b
dzzFCdzzFCE
 )sin(sin)(  zRktzF

97
 )2sin()sin(
sin








 kRtkRtbCE
Integration of the equation yield:
where:
Just the sum of the two fields at P, one from each slit
Distance from the first slit to P is R
 phase contribution = -kR
Distance from the second slit to P
is (R - asin ) or (R - 2/k)
 phase contribution = (-kR + 2)
 sin)2/(;sin)2/( kakb

98
2 = phase difference between two nearly parallel rays
arriving P from the edges of one of the slit
2 = phase difference between two ways arriving at P, one
having originated at any point in the first slit, the other
coming from the corresponding point in the second slit
Simplifying the
previous equation 
Irradiance 
)sin(cos
sin
2 







 kRtbCE










 2
2
2
0 cos
sin
4)( II

99
In the  = 0 (i.e. when  =  = 0)
I0 = the flux-density contributing from either slit
I(0) = 4I0 = total flux density
In the case b becomes vanishing small (i.e. kb << 1)
 sin/  1
 I( ) = 4I0cos2 = flux-density for a pair of long line sources
(Young’s double-slit interference
experiment)










 2
2
2
0 cos
sin
4)( II

100
In the case a = 0
 The two slits coalesce into one,  = 0
 I() = 4I0(sin2 )/ 2 = flux-density for single-slit diffraction
with the source strength doubled
The total expression as being generated by a
cos2 interference term
modulated by a
(sin2)/2 diffraction term










 2
2
2
0 cos
sin
4)( II

101
At angular positions where
 = ±, ±2, ±3, …
Diffraction effects are such that no light reaches the viewing
screen, and none is available for interference










 2
2
2
0 cos
sin
4)( II

102
At points on the viewing screen where
 = ±/2, ±3 /2, ±5 /2, …
The various contributions to the electric field will
be completely out-of-phase and will cancel










 2
2
2
0 cos
sin
4)( II

103
Double-slit pattern for a = 3b (i.e.  = 3)
If a = mb, there will be 2m bright fringes
within the central diffraction peak
2  3 = 6 bright fringes within the
central diffraction peak
1
3
5
4
2
# 6: ½ + ½

104
 No light is available at that precise position to partake in the
interference process and the suppressed peak is said to be a
missing-order
An interference maximum and a diffraction minimum may
correspond to the same -value

105
Diffraction by Many Slits
Procedure for obtaining the irradiance
function for diffraction by many slits
is the same as that used when
considering two slits
N long, parallel, narrow slits, each
of width b and center-to-center
separation a
The total optical disturbance at a point
on the viewing screen is given by:




2
2
2
2
)()(
ba
ba
b
b
dzzFCdzzFCE

106
The total optical disturbance at a point on the
viewing screen is given by:
Where:












2)1(
2)1(
22
22
2
2
2
2
)(
.....)(
)()(
baN
baN
ba
ba
ba
ba
b
b
dzzFC
dzzFC
dzzFCdzzFCE
 )sin(sin)(  zRktzF

107
The contribution from the j-th slit:
(by evaluating only that one integral in the previous equation)
After some manipulation:
where:












2)1(
2)1(
22
22
2
2
2
2
)(
.....)(
)()(
baN
baN
ba
ba
ba
ba
b
b
dzzFC
dzzFC
dzzFCdzzFCE
)2sin(
sin
jj kRtbCE 








 sin)2/(;sin)2/( kakb

108
Rj = R - ja sin
 -kR + 2j = -kRj
Total optical disturbance:
)2sin(
sin1
0
1
0
j
N
j
N
j
j
kRtbCE
EE

















109
Total optical disturbance written as the
imaginary part of a complex exponential:
Geometric series
 
















 



1
0
2)(sin
Im
N
j
jikRti
eebCE
  ][
][
1
1
2
21
0
2











 iii
iNiNiN
i
NiN
j
ji
eee
eee
e
e
e
  







 



 sin
sin)1(
1
0
2 N
ee Ni
N
j
ji

110
Total optical disturbance written as the
imaginary part of a complex exponential:

 
















 



1
0
2)(sin
Im
N
j
jikRti
eebCE
])1(sin[
sin
sinsin
















 NkRt
N
bCE

111
Flux-density distribution function:
I0 = flux density in the  = 0 direction emitted by any one of the slits
I(0) = N2I0
The waves arriving at P in the forward direction are all in-
phase, and their fields add constructively yield a multiple-
wave interference system modulated by the single-slit
diffraction envelope
22
0
sin
sinsin
)( 
















N
II

112
If the width of each aperture
were shrunk to zero
i.e.  0 (or N )

Principal maxima occur when (sinN / sin) = N
i.e. when:  = 0, ± , ± 2, …
Or since  = kasin/2 a sinm= m
m = 0, ± 1, ± 2, …
22
0
sin
sinsin
)( 
















N
II
2
sin
)0()( 







 II

113
Minima or zero flux density exist whenever (sinN / sin)2 = 0
i.e. when:
Between consecutive principal maxima, there will be (N-1) minima
Between each pair of minima there will have to be a subsidiary maximum
22
0
sin
sinsin
)( 
















N
II
.....,
)1(
,
)1(
.....,,
3
,
2
,
N
N
N
N
NNN











114
Subsidiary Maximum are located
approximately at point where
(sin N) has its greatest value,
i.e.
The dark regions become wider than the
bright bands as N increase and the
secondary peaks fade out
As N increases, the Principal Maxima
maintain their relating spacing (/a)
while becoming increasingly narrow
.....,
2
5
,
2
3
NN





115
The dark regions become
wider than the bright bands
as N increase and the
secondary peaks fade out
As N increases, the
Principal Maxima maintain
their relating spacing (/a)
while becoming
increasingly narrow

116
Diffraction Grating
Definition
A repetitive array of diffracting elements that has the effect
of producing periodic alterations in the phase, amplitude, or
both of an emergent wave
An idealized grating
consisting of only
five slits
Opaque surface with
narrow parallel grooves
e.g. made by ruling or
scratching parallel notches
into the surface of a flat,
clean glass plate
Each of the scratches
serves as a source of
scattered light, and
together they form a
regular array of parallel
line sources

117
Diffraction Grating
Grating Equation: a sinm = m
m = specify the order of the
various principal maxima
The intensity plot produced by a
diffraction grating consists of narrow
peaks, here label with their order
number m
The corresponding bright fringes seen
on the screen are called lines
The maxima are very narrow and they
separated by relatively wide dark
region
a = grating spacing (spacing
between rulings or slits)
N rulings occupy a total
width w, then a = w/N

118
Diffraction Grating
Grating’s ability to revolve or separate lines of different
wavelengths depends on the width of the lines
Half-width hw of the central line is measured from
the center of that line to the adjacent minimum on a
plot of intensity
The path length difference between the
top and bottom rays is Na sin hw
Na sin hw = 
Na
The first minimum occurs where
Since hw is small, then sin hw = hw
hw =  / Na (Half-width of central line)
Half-width of line at : hw =  / Na cos

119
Diffraction Grating
Application: Grating Spectroscope
collimator
Plane wave
Diffraction grating
telescope
Visible emission lines of cadmium
Visible emission lines from hydrogen
The lines are farther apart at greater angles

120
Reflection and Refraction

121
ri 
Law of Reflection
Law of Refraction (Snell’s law)
ttii nn  sinsin
Interface
Incident
medium ni
Refracting
medium ni
Surface
normal

122
Law of Reflection
When a ray of light is reflected at an interface dividing two
uniform media, the reflected ray remains within the plane of
incidence, and the angle of reflection equals the angle of
incidence. The plane of incidence includes the incident ray and
the normal to the point of incidence
Law of Refraction (Snell’s law)
When a ray of light is refracted at an interface dividing two
uniform media, the transmitted ray remains within the plane of
incidence and the sine of the angle of refraction is directly
proportional to the sine of the angle of incidence

123
Huygens’ construction to prove the law of reflection
Narrow,
parallel ray
of light
Plane of
interface XY
Angle of
incidence
Angle of
reflection

124
• Since points along the plane wavefront do not arrive at the
interface simultaneously, allowance is made for these
differences in constructing the wavelets that determine the
reflected wavefront
• If the interface XY were not present, the Huygens
construction would produce the wavefront GI at the instant
ray CF reached the interface at I
• The intrusion of the reflecting surface, means that during the
same time interval required for ray CF to progress from F to
I, ray BE has progressed from E to J and then a distance
equivalent to JH after reflection

125
• Wavelet of radius JN = JH centered at J is drawn above the
reflecting surface
• Wavelet of radius DG is drawn centered at D to represent the
propagation after reflection of the lower part of the light
• The new wavefront, which must now be tangent to these
wavelets at points M and N, and include the point I, is shown
as KI in the figure
• A representative reflected ray is DL, shown perpendicular to
the reflected wavefront
• The normal PD drawn for this ray is used to define angles of
incidence and reflection for the light

126
The Law of Refraction
Use Huygen’s principle to derive the
law of refraction
The refraction of a plane wave at an
air-glass interface
Figures show three successive stages
of the refraction of several
wavefronts at a plane interface
between air (medium 1) and glass
(medium 2)
1 = wavelength in medium 1
v1 = speed of light in medium 1
v2 = speed of light in medium 2 < v1
1 = angle of incidence

127
As the wave moves into the glass, a
Huygens wavelet at point e will
expand to pass through point c, at a
distance of 1 from point e.
The time interval required for this
expansion is that distance divided by
the speed of the wavelet = 1/v1
In the same time interval, a Huygens
wavelet at point h will expand to pass
through point g, at the reduced speed
v2 and with wavelength 2, i.e. the
time interval = 2/v2
2
2
1
1
vv



2
1
2
1
v
v





128
According to Huygens’ principle, the
refracted wavefront must be tangent
to an arc of radius 2 centered on h,
say at point g
the refracted wavefront must also be
tangent to an arc of radius 1 centered
on e, say at point c
2 = angle of refraction
h c
e
h c
g
hc
1
1sin


hc
2
2sin

 2
1
2
1
2
1
sin
sin
v
v







129
Define: refraction index for a medium
c = speed of light
v = speed of light in the medium
Speed of light in any medium depends on the index of
refraction of the medium
1
1
v
c
n e.g.
2
2
v
c
n 
v
c
n 
1
2
2
1
2
1
2
1
/
/
sin
sin
n
n
nc
nc
v
v




2211 sinsin  nn

130
The wavelength of light in any medium depends on the index
of refraction of the medium
Let a certain monochromatic light:
Medium refraction index wavelength speed
vacuum 1  c
medium n n v
2
1
2
1
v
v



From slide-8:
c
v
n 
The greater the index of refraction of a medium, the smaller the
wavelength of light in that medium
n
n



131

132
Frequency Between Media
• As light travels from one
medium to another, its
frequency does not change.
– Both the wave speed
and the wavelength do
change.
– The wavefronts do not
pile up, nor are they
created or destroyed at
the boundary, so ƒ must
stay the same.

133
n
n
v
f


Frequency of the light in a medium with index of refraction n
 fv
f
c
n
nc
fn 




/
/
f = frequency of the light in vacuum
The frequency of the light in the medium is the same as it is in
vacuum

134
The fact that the wavelength of light depends on the index of
refraction is important in situations involving the interference
of light waves
Example:
Two light rays travel through two media having different
indexes of refraction
• Two light rays have identical wavelength
 and are initially in phase in air (n  1)
• One of the waves travels through medium
1 of index of refraction n1 and length L
• The other travels through medium 2 of
index of refraction n2 and the same
length L

135
• When the waves leave the two media, they will have the same
wavelength – their wavelength  in air
• However, because their wavelengths differed in the two media,
the two waves may no longer be in phase
The phase difference between two light waves can change if
the waves travel through different materials having different
indexes of refraction
How the light waves will interfere if they reach some common
point?

136
Number N1 of wavelengths in the length L of medium 1
11 / nn wavelength in medium 1:



 1
1
1
LnL
N
n
wavelength in medium 2: 22 / nn 



 2
2
2
LnL
N
n
)( 1212 nn
L
NN 


Phase difference
between the waves
21 nn 

137
Example:
phase difference = 45.6 wavelengths
•i.e. taking the initially in-phase waves and shifting one of them
by 45.6 wavelengths
•A shift of an integers number of wavelengths (such as 45)
would put the waves back in phase
•Only the decimal fraction (such as 0.6) that is important
•i.e. phase difference of 45.6 wavelengths  0.6 wavelengths
•Phase difference = 0.5 wavelength puts two waves exactly out
of phase
•If the two waves had equal amplitudes and were to reach some
common point, they would then undergo fully destructive
interference, producing darkness at that point

138
• With the phase difference = 0 or 1wavelengths, they would
undergo fully constructive interference, resulting brightness
at that common point
• In this example, the phase difference = 0.6 wavelengths is an
intermediate situation, but closer to destructive interference,
and the wave would produces a dimly illuminated common
point

139
Example:
 = 550 nm
Two light waves have equal
amplitudes and re in phase before
entering media 1 and 2
Medium 1 = air (n1  1)
Medium 2 = transparent plastic (n2  1.60, L = 2.60 m)
Phase difference of the emerging waves:
o
9
6
1212
1020rad17.8
swavelength84.2
)00.160.1(
10550
1060.2
)(











nn
L
NN

140
Effective phase difference = 0.84 wavelengths = 5.3 rad  300o
• 0.84 wavelengths is between 0.5 wavelength and 1.0
wavelength, but closer to 1.0 wavelength.
• Thus, the waves would produce intermediate interference that is
closer to fully constructive interference,
• i.e. they would produce a relatively bright spot at some
common point.

141
Fermat’s Principle
• The ray of light traveled the
path of least time from A to B
• If light travels more slowly in
the second medium, light
bends at the interface so as to
take a path that favors a
shorter time in the second
medium, thereby minimizing
the overall transit time from
A to B
Construction to prove the law of
refraction from Fermat’s principle

142
• Mathematically, we are required to minimize the total time:
ti v
OB
v
AO
t 
22
xaAO  22
)( xcbOB 
ti v
xcb
v
xa
t
2222
)( 




143
0
)( 2222






xcbv
xc
xav
x
dx
dt
ti
• minimize the total time by setting dt / dx = 0
22
sin
xa
x
i


• From diagram:
22
)(
sin
xcb
xc
t



0
sinsin





t
t
i
i
vvdx
dt
0
/
sin
/
sin





t
t
i
i
ncnc ttii nn  sinsin

Topic 7 wave interference

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