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Chapter 35
The Wave Nature of Light ― Interference
by C.-R. Hu
1. Conceptual remark
Light can be treated as rays only if the sizes of all objects, gaps, and holes
involved are much bigger than the light wavelength. Otherwise the wave
nature of light becomes important, the ray method can no longer give correct
results, and new properties of light can be revealed.
2. Interference and the concept of coherence
According to the principle of superposition, two (or more) waves can exist
in the same spatial location at the same time, and therefore overlap each other. Then one should add up their amplitudes at each spatial location and
time moment. For water waves, sound waves, or waves on a rope or membrane, amplitudes are just displacements. Light waves are electromagnetic
waves, so they have both electric field and magnetic field as their (dual)
amplitudes. For transverse waves such as waves on a rope, or light waves,
the amplitudes are vectors perpendicular to the direction of propagation of
the wave, so they must be added up vectorially. This is related to the
concept of polarization. When two (or more) light waves exist in the same
spatial location, one must add up two (or more) electric-field vectors and
two (or more) magnetic-field vectors separately. But the two additions are
very similar, so one usually does just the electric field explicitly. Plane
waves belong to an idealistic situation: They have infinite length. Actual
waves have finite length, and are sometimes called wave trains (for long
ones) or wave packets or pulses (for short ones). Unlike ordinary trains,
which must move on separate tracks, two (or more) wave trains can move on
the same “track”. (There are really no tracks involved. The two (or more)
light waves just propagate in the same region of space.) The situation then
leads to the concept of interference when their amplitudes are added up.
Non-trivial results are obtained in such additions only when the two (or
more) wave trains overlap in space. Locally, the situation then appears like
the addition of two (or more) plane waves. This is what we consider below:
For waves, 1 + 1 is not always equal to 2.
When two plane waves of the same wave
length, frequency, and direction of propa-
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gation are shifted with respected to each other, as shown above, they are
said to have a phase difference. Clearly, the phase difference becomes zero
again if two plane waves are shifted with respect to each other by an integer
number of wavelengths. Thus we care about shifts of lengths less than a
wavelength only. Phase differences are measured in degrees. A shift of a
whole wavelength is called a 360° (or 2π radians) phase shift, which is
equivalent to no phase shift. When one wave is flipped over with respect to
another, they are exactly 180° (or π radians)
out of phase. They are shifted with respect
to each other by half of a wavelength. When such two waves are added together, they will cancel each other at all time, because a positive amplitude
and a negative amplitude of the same magnitude exist at the same spatial
location at the same time. So one gets no wave at the end as a result! Thus
in this case, 1 + 1 = 0! This is called destructive interference. On the other
hand, if two waves are in phase, or having no phase difference, then they
add up to a wave of twice the amplitude, or four times the intensity, since the
inten-sity of a wave is proportional to the square of the amplitude. So now
one has 1 + 1 = 4! This is called constructive interference. You can also
get any intensity between 0 and 4 times the intensity of each wave, when
two waves of the same wavelength, frequency, and amplitude are added up
together. It all depends on the relative phase of the two waves. This phenomenon is called interference. Thus for waves, 1 + 1 can equal to any
number between 0 and 4! This is, of course, because waves are not simply
numbers, but have amplitudes and phases, and several other properties.
A very, very long sinusoidal wave train, containing some 1010 oscillations or
more, is said to be coherent, because two such wave trains, shifted by great
distances relative to each other, can still produce constructive or destructive interference. A laser beam, for example, has such good coherence. If
you imagine chopping such a long sinusoidal wave train into many short
segments, each containing some 107 oscillations or less, ― For visible light,
it means each segment is a few meters long or less, ― separate them by nowave regions, and then shift the position of each segment forward or
backward by a small random amount, then the resulting wave is said to be
incoherent. Two such incoherent waves, generated from two independent
such incoherent sources, can no longer give any clear interference phenomenon. One might get constructive or destructive interference at some instant, but this situation can last only about ten to twenty nanoseconds, and
then two new light wave segments arrive, and their phase relation changes!
Since any detector will examine the light intensity for at least, say, a micro-2-
second, it will detect time-averaged behavior, which obeys 1 + 1 = 2 ― no
interference! Most light sources are incoherent, including sunlight, light
from all sorts of lamps, light from white-hot objects, etc. The key concept is
that two coherent waves maintain a definite phase relation with respect to
each other for an observable duration of time, whereas two incoherent waves
do not maintain such a definite relation for so long.
When two incoherent light sources shine light on the same spot, the total
intensity of light reaching that spot will simply be thr sum of the intensities
due to the separate sources. When two coherent light sources of the same
intensity shine light on the same spot from equal distance, the intensity of
light reaching that spot can range between 0 and 4 times as strong as the
intensity of only one source, depending on the relative phase of the two
coherent light waves, because one must first add up their wave amplitudes
before squaring the sum amplitude to get the final intensity.
3. Young’s double-slit experiment
The simplest way to demonstrate interference
A
is Young’s double-slit experiment. Let a
plane light wave coming from the left impinge
B
on a wall that contains two parallel slits, denoP
ted A and B. (The slits are perpendicular
R
to the paper). Then cylindrical waves will be
emitted to the right from these two slits (Huygen’s principle). If there is a
screen at a distance R to the right of the double slits, and P is a point on this
screen, then waves from both slits will reach this point. The two waves
reaching P will have traveled different distances: The upper one has traveled
distance AP, and the lower one has traveled distance BP. Thus the waves
arriving at P will in general be not in phase because one wave takes a longer
time to reach P. For the interference phenomenon to be prominent, the separation d between the two slits should not be way bigger than the light
wavelength λ. (Actually the condition is that the length difference AP-BP
should be much shorter than the average length of the wave trains involved,
so the two arriving wave trains can still overlap. So how long can d be to
still get interference will depend on the average length of a wave train in the
light source. The longer is this length, the easier it is to see interference.
And the light source is said to be more coherent.)
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For simplicity we assume that each slit is so narrow that we can neglect its
width. That is, we will not worry about the difference in distance between
measuring AP from the upper edge or the lower edge of the same slit. We
also assume that the distance R is very large in comparison with d. Then the
distances AP and BP will only differ by an amount much smaller than R.
But it can still be larger or smaller than
d sin θ
λ. In such a situation the calculation
A
R
can be done approximately. We have
O
d
θ
B
then:
y
θ
AP− BP ≈ AB sin θ = d sin θ.
P
[If θ is small, that is, if y << R , then we can in addition use
sin θ ≈ θ (in radians) ≈ tan θ = y/R. But don’t use it unless the
condition is met.] When this difference in distance is an integer number
of the wavelength, the two waves will add up constructively (meaning
adding up in phase), producing highest intensity. This is when
d sin θ = mλ, with m = 0, ≤1, ≤2, ···. (constructive interference)
This formula gives a sequence of values of θ, corresponding to a sequence
of line segments on the screen parallel to the slits, (on either side of O,)
where the EM wave intensity will be peaked. On the other hand, if this
difference in distance is equal to half of a light wavelength plus any integer
number of light wavelengths, then the two waves arriving at P will cancel
each other, producing zero intensity. This is called destructive interference.
So we have
d sin θ = ≤(m + 1/2)λ, with m = 0, 1, 2, ···. (destructive interference)
Again this formula gives a sequence of values of θ, corresponding to a sequence of line segments on the screen parallel to the slits, where the light
wave intensity will vanish. Clearly, between every two neighboring constructive points on the screen, there is one destructive point, and vice versa.
These points are actually line segments aligned parallel to the slits because
the picture shown above is only a cut view. The intensity variation on the
screen, on a line perpendicular to the forward direction and the slit direction,
will appear as:
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Aside from the (rise-fall-rise-fall ...) pattern, there is also a gradual drop of
the peak heights on the two sides of the central peak because of a general
increase of distance between the point on the screen and the center of the
two slits when θ increases on either side of zero. (The central peak is located at the point O on the screen that makes equal distance with the two slits.
It is also the point on the screen that is closest to the center of the two slits.)
4. Graphic method to determine the intensity in the doublesplit interference pattern
If d is much larger than light wavelength λ, many intensity peaks will appear
in a narrow θ range centered at zero, in which the graduate drop of the peak
heights can be neglected. In this approximation, the electric field arriving at
the point P on the screen from the lower slit may be written as:
E1 = E0 cos(ω t + η ), and the electric field arriving at P from the upper slit
may be written as E2 = E0 cos(ω t + η + φ ), where φ = (2π / λ) d sinθ is the
phase difference in radians between the two light rays when they both arrive
at P. The two electric fields are from the same coherent source, so their
polarization directions are clearly the same. This is why we can forget about
their polarization directions and treat them as if they are not vectors. They
can be added up like ordinary numbers, so the total electric field at P is just
the sum E P = E1 + E 2 (polarized in the same direction). This sum is the vertical projection of the sum of two vectors (see figure): (These vectors are
called phasors, and the diagram, the phasor diagram. Note that these
vectors have nothing to do with polarization. They are introduced in order
for their pojections to give the two cosine factors.)
Thus:
EP =[2E0 cos(φ/2)] cos(ω t +η +φ /2).
The light intensity at P is equal to the
square of EP multiplied by ε0 averaged
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ω t + η+ φ
φ/2
ωt+η
E2
E1
EP
over time. So we have
I P = 4I0 cos 2(φ /2),
t
2
2
where I 0 ≡ ε 0 E 0 cos (ω t + η + φ / 2)
= (1/ 2)ε 0 E02 is the intensity at P due to either slit alone.
If the two light sources at the two slits are incoherent, then the intensity at P
would be
t
I P′ = ε 0 (| E1 | + | E2 | ) = 2I0 .
2
2
Recall that:
φ = (2π / λ) d sinθ ≈ (2π / λ) dθ (in radians) = (2π / λ) (d y / R)
(iassuming d << R) which varies with y, so I P varies with y, but not I P′ :
Ip
1
0.8
0.6
I P′
0.4
0.2
y
-7.5
-5
-2.5
2.5
5. Interference in thin films and gaps
5
7.5
n1
One application of interference is to suppress
n2
reflection by a thin transparent coating. Light
d
is seen reflected from the two surfaces of the
n3
coating. If n1 > n2 > n3 , or n1 < n2 < n3,
and if light is incident on the coating at 90±,
then the lower reflected beam is shifted in phase relative to the upper
reflected beam by an angle equal to (2d/λ)µ360±. Thus if d = λ/4, the
two reflected waves will have a 180± phase-shift relative to each other, and
cancel each other, leading to very low reflection! (The cancellation is not
complete because the two reflected beams will not have equal amplitudes.)
On the other hand, if n1 > n2 < n3 or n1 < n2 > n3, and if light is still
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incident on the coating at 90±, then one needs d = λ/2 to get the two
reflected waves out of phase and nearly cancel! This is because light
reflected from a denser medium into a lighter medium will encounter an
additional 180± phase-shift.
It is important to realize that here λ means the wavelength of light inside
the coating, which has index of refraction n2. Thus this is the wavelength in
vacuum divided by n2, since f λ = v = c/n2 in a medium, but f remains the
same.
air
If a glass plate (or any other transparent film)
is slightly thicker on one edge than on its opposite edge, then one can see parallel alternan
ting dark and bright stripes in the reflected
glass
light. If the upper surface makes a small angle θ in radians with the lower surface, then
air
the separation between neighboring bright (or
dark) lines will be equal to λ/(2θ ) (for vertical incidence), where λ is the
wavelength in the glass.
Separating two glass plates by a thin spacer
on one edge can produce the same effect,
except that one is now using the interference
between reflections from the lower surface of the upper glass plate and the
upper surface of the lower glass plate, so now λ should be essentially the
wavelength in vacuum, if there is only air between the two glass plates. Since the reflection by the lower surface is by a denser medium, it has already a
180º phase shift, locations where you see destructive interference are where
2d = mλ is satisfied. Here d is the separation between the two glass plates
at where you get destructive interference, and m is an integer. (m = 0 gives
the leftmost dark band. To find the total number of dark bands, find the
largest m without making d larger than the separation on the right edge, and
then add 1, because m begins with zero.)
Small alternating concentric dark and bright
rings can also be observed near the contact
point of a spherical lens placed on top of a
flat glass plate. One is again using the gap
between the two glass objects to produce
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the two interfering light beams, except that here the spacing does not grow
linearly, as the radial distance from the contact point increases. The observed rings are called Newton’s rings, because Newton first noticed the phenomenon. These rings do not have uniform spacing because of the way the
spacing between the two glass objects grows.
These phenomena can be used to check the flatness of a glass plate, if the
lens used is perfectly spherical, or any tiny deviation of a spherical lens from
having a truly spherical surface, if the glass plate used is perfectly flat. The
accuracy of these checks is a quarter of a wavelength of the light used.
If the incident light is a mixture of all visible frequencies, called white light,
then one will see colored lines or bands in the reflected light. This is not
because the index of refraction n depends on frequency somewhat, as in the
case of a prism or in a rainbow, but is simply because the predicted locations
of the bright lines or bands are wavelength dependent. (In the case of
Young’s double-slit interference experiment, the separation of color will be
larger for constructive interference involving larger phase differences, i.e.,
mλ for larger integer m.) This is why a thin oil film or soap bubble often
shows beautiful colored bands under bright white lighting. A better demonstration is to look at a CD under bright white lighting. You will notice that
colored bands show up only where the recording tracks are located, because
light reflected from the different recording tracks can cause interference,
much like the Young’s double-slit interference experiment. However, note
that nothing needs to be recorded in the CD. A blank CD is sufficient for
this purpose.
6. Michelson interferometer
Interference can also be used to accurately measure length with an accuracy
of a quarter of a wavelength, using a device called Michelson interferometer. This device has also been used to prove that electromagnetic waves
require no medium to carry them — this imaginary medium was called
ether, which was proven to not exist by using this device), and to confirm
Einstein’s special theory of relativity.
Design of a Michelson interferometer:
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movable mirror
1
light source (single
known frequency)
partially reflecting mirror
2
fixed mirror
bright or dark view depending on path-length difference between paths 1 and 2.
Move the movable mirror forward or backward by a precisely measured
distance d to get two consecutive dark views at the exit end of light. Then
2d must be λ. If the mirror has been moved so much that you see, say, 11
dark views separated by bright views, then 2d must be divided by 11 − 1
= 10 to get λ. Then the error in λ will be a factor of 10 smaller! This is
because a one millimeter error in, say, a ten meter length is a much smaller
percentage error than a one millimeter error in, say, a one meter length! So
when a larger 2d is measured, its percentage error can be smaller. Then the
λ so determined will have a smaller percentage error.
7. Energy Considerations in the Presence of Interference
One plus one (i.e., 1 + 1 ) for the combination of two waves of the same amplitude, wave-length, and frequency can equal to anything between 0 and 4
due to interference, depending on the relative phase of the two waves. [If
the amplitudes and/or wavelengths of the two waves are not the same, interference still occurs but the outcome is more complicated. For example, if
the amplitudes of the two waves are E1 ≠ E2 , but their wavelengths are still
the same, then constructive interference will give the intensity ∝ | E1 + E2 |2 ,
whereas destructive interference will give the intensity ∝ | E1 − E2 |2 , when the
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individual waves alone will give intensities ∝ | E1 |2 and | E2 |2 , with the same
proportionality constant for all four cases.]
An important question to answer about the interference of two or more
waves is:
Is energy still conserved?
The short, generally-true, answer is:
The energy is redistributed in space and time, but is not gained
or lost. The law of conservation of energy is not violated.
To understand why this is true, we shall not give a general abstract proof,
which is actually quite difficult to do. We will simply consider several concrete examples.
Consider first the Young's double-slit interference experiment: In this case,
the redistribution in space is seen by the bright and dark fringes seen on a
screen, which are steady in time. The bright fringes have intensities four
times stronger then those of the individual waves, and the dark fringes have
zero intensity. Their spatial average is two times those of the individual
waves, just what is expected if energy is conserved. Actually, this answer
has cheated somewhat. It has over-simplified the matter. We know that as
the angle gets large, the maximum intensities will not always stay at four
times that of each incident beam, but will gradually drop off. Thus the average intensity along the screen is not two times that of each incident beam, as
seems to be required by energy conservation. The true answer is that the
total energy must be obtained by integrating over the whole space, not just
along the screen. The calculation is not trivial, but it can be done, and the
total energy is equal to that of the two incident beams, so that energy is only
redistributed in space, but not gained or lost.
In the above case, the two waves of the same amplitude and wavelength are
not propagating in the same direction. What if the two waves are exactly
propagating in the same direction? Wouldn't that give zero intensity for destructive interference and four times the individual wave intensity for constructive interference everywhere in space and every moment in time?
Wouldn't that situation necessarily violate the law of conservation of
energy? To answer this tricky question, it is necessary to first consider how
this situation can be achieved in reality. For the sources of the two waves to
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be not in each other's way, some bending of light must take place. The
simplest situation involves only one interface of two different optical media
with indices of refraction n1 and n2, respectively. If one incident beam
shines on the interface, from the n1 side, say, then by choosing the incident
angle properly, one can make exactly half of the energy reflected back into
the n1 side, and the other half transmitted into the n2 side. This is illustrated
by the following figure:
incident light
reflected light
n1
n2
transmitted light
Now consider two incident beams, one on each side, as shown below:
incident beam a
reflected light of beam a and
transmitted light of beam b
n1
n2
incident beam b
transmitted light of beam a
and reflected light of beam b
Could both beams leaving the interface (henceforth called the exit beams)
involve destructive interference or constructive interference, thereby violating energy conservation? The answer is no, because one of the reflected
beam is by a denser medium, so that reflection involves a 180° phase shift.
The other reflection, on the other hand, is by a lighter medium, which does
not involve any phase shift. Thus if the exit beam in the n1 side is destructive, then the exit beam in the n 2 side will necessarily be constructive, and
vice versa. The intensity of the exit beam on the destructive side is zero, but
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it can be shown that the intensity of the exit beam on the constructive side is
always equal to twice of the intensity of each incident beam, so energy is
still always conserved in this system. In fact, with more sophisticated analysis, it can be shown that energy is always conserved no matter what is the
relative phase difference between the two incident beams. The two incident
beams don’t even have to have the same amplitude. In the general case,
both exit beams will not have zero intensity, and their intensities will not be
equal in general, but the sum of them will still be equal to the sum of the
intensities of the two incident beams, in spite of the fact that each exit beam
has already had the effect of interference, so energy is still conserved.
An even trickier situation to analyze is when use is made of a semi-transparent mirror which reflects half of the wave's energy, and let the other half
through. It is called a beam splitter. Consider a beam splitter which is at
45° with light beams a and b that are at 90° with each other:
detector 1
beam splitter
detector 2
light source a
light source b
We assume that light source a emits a monochromatic (single-wavelength),
plane-wave light beam to the right, which is half reflected by the beam
splitter to go upward, and the other half is transmitted to go rightward. Light
source b emits upward a monochromatic, plane-wave light beam of the same
amplitude and wavelength, which is half reflected by the semi-transparent
mirror to go rightward, and half transmitted to go upward. This time, only
the beam splitter itself involves an index of refraction n, otherwise the whole
instrument is in the air or in vacuum. So the two reflections by the beam
splitter no longer differ by a 180° phase shift. The intensities arriving at the
two detectors are then seemingly the same. But the correct analysis must
take into account the thickness of the beam splitter (a dielectric slab with
some silver coating as one design) and the multiple reflections inside it. The
end result is that the light beams arriving at the two detectors will indeed
have the same intensity, but they will each equal to the intensity of either
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incident beam (assumed equal), so energy is still conserved! We shall not
give the details of the analysis here because it is somewhat involved, but will
just mention the key point that each reflected beam is in phase with the
corresponding incident beam, but the corresponding transmitted beam is at
90° phase difference from the incident beam!
That energy can be redistributed in space due to interference is most obvious
if you go to a poorly constructed music hall to listen to a performance. If
sound can reach you in two or more paths – a direct path and one or more
reflected paths via the walls, then depending on the phase relation between
these different paths, you can hear particularly laud or weak sound at your
seat. Energy is redistributed in the music hall, but the total energy in the
music hall is still equal to the total energy emitted by the music instruments,
if only absorption by the walls is negligible, and no energy leaks outside the
hall. A good music hall therefore makes sure that all walls are made of
strongly absorbing material, so no reflection can take place at the walls.
If one light source actually emits two waves of the same wavelength and
amplitude but 180D out of phase, then their destructive interference will
mean that actually no light will come out of this source. But then the source
will not lose any energy either, so energy is still conserved. Similarly, If one
light source actually emits two waves of the same wavelength, amplitude,
and phase, then their constructive interference will mean that actually four
times the intensity of each wave will be emitted. But then the source will
also lose four times the energy it will lose if it emits only one of the two
waves. Thus energy is again conserved.
So far, all redistributions of energy are in space but not in time. For what
situation will redistribution of energy occur in time? The answer is that
when two (or more) light beams of different frequencies interfere. The
result can then be a ... strong-weak-strong-weak ... sequence of light intensity variation in time. The same phenomenon also occurs with the interference of two sound waves of slightly different frequencies, in which case you
can sense the intensity variation with your ears, and is called beat. (This is
what you hear when you blow at a pair of holes of a harmonica at the same
time.) If the two waves of different frequencies are also at an angle with
respect to each other, then there is redistribution of energy in space as well
as in time, which is then the most general situation with regard to the energy
redistribution in the presence of interference.
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