Waves and Optics

1
Chap. 7 Interference
Optical interference may be termed as
interaction of two or more lightwaves yielding a
resultant irradiance that deviates from the sum
of the component irradiances.
7.1 General consideration
Consider two point sources, and
, emitting monochromatic waves of the same
frequency in a homogeneous medium (Fig. 7.1).
Fig. 7.1 Waves from two point sources overlapping
in space.
2
Locate the point of observation far enough
away from the sources so that at the
wavefronts will be planes, as shown in Fig. 7.2.
For the moment, we will consider only linearly
polarized waves
(7.1a)
and
Fig. 7.2 Two plane waves meet at P.
(7.1b)
The resultant wave at can be given by
The irradiance is defined as
3
Inasmuch as we will be concerned with relative
irradiance within the same medium, we will simply
neglect that constants and set
Obviously,
Taking the time average of both sides, we find
that the irradiance becomes
provided that
4
and
The latter expression is known as the
interference term. To evaluate it in this
specific instance, we form
The interference term is then
and , equal to
is the phase difference arising from a
combined path-length and initial phase-angle
difference. Notice that if and are
perpendicular, and
The most common situation in the work to
follow corresponds to parallel to
. In that case, the irradiance reduces to the
value found in the scalar treatment. Under those
conditions
5
This can be written in a more convenient way by
noticing that
and
The interference term becomes
So, the total irradiance is
At various points in space, the resultant
irradiance can be greater, less than, or equal to
, depending on the value of
, that is, depending on . A maximum in the
irradiance is obtained when so
that
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when
In this case the phase difference between the two
waves is an integer multiple of , and the
disturbances are said to be in phase. One speaks
of this total constructive interference. When
the waves are out of phase,
, and the result is
known as constructive interference.
At , the
disturbances are said to be out of phase,
and For
we have the condition of destructive
interference, When

a minimum in the irradiance is
obtained
It is referred to as total destructive
interference. Another somewhat special yet
very important case arises when the amplitudes of
both waves reaching in Fig. 7.1 are equal.
Let . Eq. 7.12 can now be
written as
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from which it follows that and
Equation 7.12 holds equally well for the
spherical waves emitted by and .
Such waves can be expressed as
and
The terms and are the radii of the
spherical wavefronts overlapping at . In
this case
Equation 7.15 becomes
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The maximum irradiance occurs when
and minimum when
where and

If the waves are in phase at the emitter,
and Eqs. 7.19a and 7.19b
can be simplified to
for maximum and minimum irradiance, respectively.
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7.2 Youngs double-slit interference experiment
In 1801, Thomas Young experimentally proved
that light is a wave. He did so by demonstrating
that light undergoes interference, as do water
waves, sound waves, and waves of all other types.
We shall here examine Youngs historic experiment
as an example of the interference of light waves.
Fig. 7.3 gives the basic arrangement as an
example of the interference of light waves.
Fig. 7.3 Youngs interference experiment.
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Light form a distant monochromatic source
illuminating slit in screen . The
emerging light spreads via diffraction to
illuminate two slits and in screen
The light waves traveling from the two
slits overlap and undergo interference, forming
an interference pattern of maxima and minima on
viewing screen as shown in Fig. 7.4.
Fig. 7.4 A photograph of an interference pattern
produced by the arrangement shown in Fig. 7.3.
The interference pattern consists of
alternating bright and dark bands which are
called bright fringes and dark fringes,
respectively. What exactly determines the
locations of the fringes? To answer, we shall use
the arrangement in Fig. 7.5(a). There, a plane
wave of monochromatic light is incident on two
slits and in
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screen B the light diffracts through the slits
and produces an interference pattern on screen C.
We draw a central axis from the point halfway
between the slits to screen C as a reference. We
then pick, for discussion, an arbitrary point P
on the screen, at angle to the central
axis. This point intercepts the wave of ray
from the bottom slit and the wave of from the
top slit. Notice that these waves are in
phase when they pass through the two slits
because there they are just portions of the same
incident wave. But once they have
Fig. 7.5 (a) Waves from and combine at
. (b) For we can approximate
rays and as being parallel.
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passed the slits, the two waves must travel
different distances to reach . The path
length distance can be given
provided the distance from the slits to
the screen is much grater than the slit
separation .
For bright fringes, we must have
For dark fringes, the requirement is
With Eq. 7.22 and 7.23, we can find the angle
to any fringe and thus locate that fringe.
Further we can use the values of and
to label the fringes. For , Eq. 7.22
tells us that a bright fringe is at ,
that is on the central axis. Now let us
consider the irradiance. According to Eq. 7.18,
the resultant irradiance
13
is given by
where
Fig. 7.6 shows the irradiance of a
double-slit interference pattern as a function
of the phase difference between the waves from
the two slits.
Fig. 7.6 Idealized irradiance versus the phase
difference.
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7.3 Double beam interference interference from
thin films
Interference effects are observable in
sheet transparent materials for a given
wavelength of electromagnetic radiation when
their thickness is of the order of that
wavelength. Consider a thin transparent film with
a thickness illuminated by a monochromatic
light of wavelength (Fig. 7.7).
The optical path-length difference for the
first two reflected beams is given by
and since
Fig. 7.7 Ray representation of thin-film
interference.
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Now, to find an expression for
write
if we make use of Snells law, this becomes
where
The expression for now becomes
16
or finally
The corresponding phase difference associated
with the optical path-length difference is then
just the product of the free-space propagation
number and that is, Notice that
in either case there will be an additional phase
shift arising from the reflection themselves. For
incident angles up to , regardless of the
polarization of the incoming light, the two
beams, one internally and one externally
reflected, will experience a relative phase shift
of radians. Accordingly,
and more explicitly
When in other words, an even
multiple of , an interference maximum, a
bright spot, appears at . In that case Eq.
7.26 can be rearranged to
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where use has been made of the fact that
.
Interference minima in reflected light result
when that is, odd
multiples of . For such cases Eq. 7.26 yields
The appearance of odd and even multiples of
in Eqs. 7.27 and 7.28 is rather
significant, as we will see presently. For an
extended source, light will reach the lens from
various direction, and the fringe pattern will
spread out over a large area of the film (Fig
7.8). The angle or equivalently ,
determined by the position P, will in turn
control . The fringes appearing at
position P1 and P2 in Fig 7.8 are, accordingly,
known as fringes of equal inclination. Each
source point on the extended source is incoherent
with respect to the others.
Fig 7.8 All rays inclined at the same angle
arrive at the same point
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7.4 Multiple-beam interference
Fig 7.9
Figure 7.9 shows a multiple beam interference on
a transparent film. As before, r and t denote
reflection and transmission amplitude
coefficients when the incident beam is out of the
film while r and t when the incident beam is
inside the film. Since all beams are almost
parallel to each other so their electric field
directions are also almost parallel to each
other. As a consequence, we can use scalar to
describe the electric field. The amplitudes of
the reflected beams are
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E0 is the incident electric field.
is the phase
arising from an optical path-length difference
between adjacent rays. d and are the film
thickness and the refraction angle inside the
film respectively. The relative phase shift
undergone by the first ray as a result of the
reflection is embodied in the quantity r. The
resultant reflected scalar wave is then
When and the number of terms in
the series approaches infinity, we have
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In the case of zero absorption, the Stocks
relations hold r-r and tt1-r2. We have
The reflected irradiance at point P is
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Similarly the amplitudes of the transmitted waves
given by
Multiplying this by its complex conjugate, we
obtain the irradiance of the transmitted beam
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For the non-adsorbing film, Eqs (7. 32) and
(7.35) show that Ir / Ii It / Ii 1
(7.36) Figure
7.10 shows It/Ii as a function of the phase .

Fig 7.10 Transmitted irradiance
The later condition is the same as Eqn. (7.26)
for two-beam interference. When a thin film
surface is coated with adsorbing material (metal)
to increase the reflection coefficient, Eqn
(7.36) doesnt hold. The transmitted intensity
is modified to
So the maximum and minimum conditions above still
hold.
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Fig 7.12 Transmitted light pattern from a
Fabry-Perot interferometer
Fig 7.11 Fabry-Perot interferometer
A Fabry-Perot interferometer is shown in figure
7.11. Two semi-silvered glass optical flats
form the reflecting boundary surfaces. The
enclosed air gap ranges from millimeters to
centimeters. The un-silvered sides of the plates
are made to have a slight wedge shape (a few
minutes of arc) to reduce the interference
pattern arising from reflections off these sides.
The interferometer is illuminated by a broad
source, which might be a He-Ne laser beam spread
out in diameter to several centimeters. Only
one ray emitted from some point S1 on the source
is traced. Entering by way of the plate, it is
multiply reflected within the gap. The
transmitted rays are collected by a lens and
brought to a focus on a screen. Consider this
particular plane of incidence, which contains all
the reflected rays. Any other ray emitted from a
different point S2, parallel to the original ray
will form a spot at the same point P on the
screen. Eq. (7.37) determines the transmitted
irradiance. The multiple waves generated in the
cavity, arriving at P from either S1 or S2, are
coherent among themselves.
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But the rays arising from S1 are completely
incoherent with respect to those from S2. The
contribution to the irradiance It at P is just
the sum of the two irradiance contribution. All
the rays incident on the gap at a given angle
will result in a single circular fringe of
uniform irradiance (Fig 7.12). With a broad
diffuse source, the interference bands will be
narrow concentric rings. A Fabry-Perot
interferometer can be used to measure the
wavelength difference of two closed spectral
lines.
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7.5 Anti_reflection and reflection-enhance
coatings
A substrate-air system can be modified to
substrate-films-air system to either reduce or
enhance the light reflection back to air.
Consider the linearly polarized wave shown in
Fig. 7.13, impinging on a thin dielectric film
between two semi-infinite transparent media.
Each wave and so forth
represents the resultant of all possible waves
traveling in that direction, at that point in the
medium. The summation process is therefore built
in. The boundary conditions require that the
tangential components of both the electric (
) and magnetic ( ) Fields be
continuous across the boundaries. At boundary I
7.13 Fields at the boundaries
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Where use is made of the fact that and
are related through the index of refraction
and the unit propagation vector
. At boundary II
In accord with Eq. (7.25), a wave that traverses
the film once undergoes a phase shift of
So,
The solution of (7.44) is
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(No Transcript)
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are the refractive index, the
thickness and the beam incident angle in the jth
film, respectively. With the help of equations
(7.38) and (7.41), equation (7.48) is expanded
to
For electric field parallel to the incident
plane, equations (7.48) through (7.53) are still
valid except for the Y expression in (7.50). It
should be replaced by
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Lets calculate the reflectance when one layer of
film presents. With Y0, Y1 and Ys defined in
equation (7.50) and the matrix MM1 defined in
equation (7.49), equation (7.52) becomes

So (7.58) becomes the same as (7.32) noting that
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Now consider the extremely important case of
normal incidence, where all

From the single layer formula (7.57), we have
(p1)
To make the film important to the reflectance,
one choose
Generally, the thickness d1 is chosen to satisfy
(7.63) for yellow-green portion of the visible
spectrum, where the eye is most sensitive. MgF2
(n11.38) on a glass substrate (ns1.5), can
reduce the reflection coefficient from 4 to 1
in the visible region.
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For a double layer, quarter-wavelength
antireflection coating as shown on the left panel
in Fig 7.14, MM1M2, or more specifically
Put (7.65) into Eq. (7.52), we have
Fig 7.14 A periodic structure
Its easier to satisfy (7.77) than that (7.63).
Since nsgtn0, from (7.77) n2gtn1. Accordingly, it
is common to design a (glass)-(high index)-(low
index)-(air) system as gHLa, as shown in Fig.
7.14. Zirconium dioxide (n2.1), titanium oxide
(n2.40) and zinc sulfide (n2.32) are commonly
used for H-layers. Magnesium fluoride (n1.38)
often serves as L-layers
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Eq. (7.67) can be used in the opposite way. If
one choose n2/n1gtgt1, one will have a
reflection-enhanced system. To further enhance
the reflection, one can use a periodic
quarter-wave stack system g(HL)ma as shown in
the right panel of figure (7.14) where m3. The
corresponding matrix for the system is
Fig 7.15 Reflectance and transmittance for
several periodic systems
When nH/nLgtgt1 and m is large the reflectance can
be very high. More improvement can be achieved
by adding one more high-index layer to the air
side of the periodic system to become g(HL)mHa
system. The reflectance and transmittance are
shown in figure 7.15 for several periodic
systems.