Topic 7 wave_interference(latest)

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

2
• 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

3
3
• 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
Huygens’s Principle

4
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

5
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)

6
Christian Huygens
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”

7
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

8

9
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

10
ri 
Law of Reflection
Law of Refraction (Snell’s law)
ttii nn  sinsin
Interface
Incident
medium ni
Refracting
medium ni
Surface
normal

11
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

12
Huygens’ construction to prove the law of reflection
Narrow,
parallel ray
of light
Plane of
interface XY
Angle of
incidence
Angle of
reflection

13
• 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

14
• 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

15
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

16
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





17
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







18
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

19
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



20

21
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.

22
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

23
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

24
• 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?

25
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 

26
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

27
• 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

28
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

29
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.

30
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

31
Interference
Young’s Double-Slit Experiment

32
• 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
)( 




33
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

34
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

35
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

36

37
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

38
L >> a
d
d = slits separation
d

39
• 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

40
• 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

41
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 >> d 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)
d= slits separation

42
This result can be rewritten as:
Since L >> d, 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)

43
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)

44
• 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)

45
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  kdII 
When θ is small so that sinθ  θ, we can write:
)/(cos)(
)2/(cos)(
2
0
2
0


dII
kdII


(11)
 /2where k

46
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
L/d
separation of the
bright fringes

47
Intensity maxima: .....,2,1,0,  n
d
n

.....,2,1,0,  n
d
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
d
L
nx

d
L
xx nn

1

48
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

49
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

50
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 = a)
Obstruction
(size = a)
wavelength,  a >> 

51
• 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
curvaturea  

52
As the barrier or opening size gets smaller,
the wavefront experiences more and more curvature
More curvature
Diffraction
a  

53
Fraunhofer and Fresnel
Diffraction

54
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

55
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

56
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

57
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

58
If a point source S and the
observation screen  are very far
from 
S
Lens
Plane
waves
Observation
screen 
Fraunhofer Diffraction
If a point source S and the
observation screen  are
too near 
Fresnel Diffraction

59
S
Lens
Plane
waves
Observation
screen 
Fraunhofer Diffractiona

R R
R is the smaller of the two
distances from S to  and to 

2
a
R 
d = slit width

60
Practical realization of the Fraunhofer condition
F1 F2

61
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

62
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

63
• 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

64
One can find the second dark fringes above and below the central axis as the
first dark fringes were found, except that we now divide the slit into four
zones of equal widths a/4, as shown in Fig. 36-6a.
In general,

65
Example, Single Slit Diffraction Pattern with White Light:

66
36.4: Intensity in Single-Slit Diffraction Pattern, Qualitatively:

67
Fig. 36-8 The relative intensity in single-slit diffraction for three values of the
ratio a/. The wider the slit is, the narrower is the central diffraction maximum.
The intensity pattern is: where
For intensity minimum,

68
36.5: Intensity in Single-Slit Diffraction Pattern, Quantitatively:
From the geometry, f is also the angle between the
two radii marked R. The dashed line in the figure,
which bisects f, forms two congruent right
triangles.

69
Example, Intensities of the Maximum in a Single Slit Interference Pattern:

70
36.6: Diffraction by a Circular Aperture:

71
36.6: Diffraction by a Circular Aperture, Resolvability:
Fig. 36-11 At the top, the images of two point sources (stars) formed by a
converging lens. At the bottom, representations of the image intensities. In (a)
the angular separation of
the sources is too small for them to be distinguished, in (b) they can be
marginally distinguished, and in (c) they are clearly distinguished. Rayleigh’s
criterion is satisfied in (b), with the central maximum of one diffraction pattern
coinciding with the first minimum of the other.
Two objects that are
barely resolvable
when the angular
separation is given by:

72

73
Example, Pointillistic paintings use the diffraction of your eye:

74
Example, Rayleigh’s criterion for resolving two distant objects:

75
Fig. 36-15 (a) The intensity plot to be
expected in a double-slit interference
experiment with vanishingly narrow
slits.
(b) The intensity plot for diffraction by
a typical slit of width a (not
vanishingly narrow).
(c) The intensity plot to be expected
for two slits of width a. The curve of
(b) acts as an envelope, limiting the
intensity of the double-slit fringes in
(a). Note that the first minima of the
diffraction pattern of (b) eliminate the
double-slit fringes that would occur
near 12° in (c).
The intensity of a double slit pattern
is:

76
Example, Double slit experiment,
with diffraction of each slit
included:

77
Example, Double slit experiment,
with diffraction of each slit
included, cont. :

78
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

79
Diffraction Grating
Grating Equation: d 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
d = grating spacing (spacing
between rulings or slits)
N rulings occupy a total
width w, then d = w/N

80
36.8: Diffraction Gratings:

81
36.8: Diffraction Gratings, Width of the Lines:

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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

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36.9: Gratings, Dispersion and Resolving Power:
A grating spreads apart the diffraction lines associated with the various
wavelengths.
This spreading, called dispersion, is defined as
Here  is the angular separation of two lines whose wavelengths differ by .
Also,
To resolve lines whose wavelengths are close together, the line should also
be as narrow as possible. The resolving power R, of the grating is defined as
It turns out that

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Gratings, Dispersion and Resolving Power, proofs:
The expression for the locations of the lines in the diffraction pattern of a
grating is:
Also, If  is to be the smallest angle that will permit the two lines to be
resolved, it must (by Rayleigh’s criterion) be equal to the half-width of each
line, which is given by :

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36.9: Gratings, Dispersion and Resolving Power Compared:

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