Download Ch 35 Interference

Ch 35
Interference
The colors in many of a hummingbird’s
feathers are not due to pigment. The
iridescence (虹彩) that makes the brilliant
colors that often appear on the throat and
belly (腹部) is due to an interference effect
caused by structures in the feathers. The
colors will vary with the viewing angle.
Examples of Interference
The bright colors of peacock
feathers are also due to
interference. In both types of
birds, structures in the feathers
split and recombine visible light
so that interference occurs for
certain colors.
The blue of the top surface of a
Morpho butterfly wing is due
to optical interference and
shifts in color as your viewing
perspective changes.
Examples of Interference
Governments worldwide scurry to stay ahead of
counterfeiters (偽造者) who are quick to use the
latest technology to duplicate paper currencies.
Some of the security measures now used to thwart
counterfeiters are security threads and special
watermarks (both of which can be seen if the
currency is held up against a light) and
microprinting (which consists of dots too small to be
picked up by a scanner).
The feature that is probably the most difficult for a
counterfeiter to duplicate is the variable tint that
results from color-shifting inks. For example, the
“100” in the lower right of the front face of a U.S.
$100 bill contains color-shifting ink. If you look
directly down on the number, it is red or red-yellow.
If you then tilt the bill and look at it obliquely, the
color shifts to green. A copy machine can duplicate
color from only one perspective and therefore
cannot duplicate this shift in color you see when you
change your perspective.
 One of the major goals of physics is to understand
the nature of light. This goal has been difficult to
achieve simply because light is complicated (Einstein had
ever said that).
 This complication means that light offers many
opportunities for applications, and some of the
richest opportunities involve the interference of light
waves — optical interference.
Huygens
The first person to advance a convincing wave
theory for light was Dutch physicist Christian
Huygens,
in
1678.
Although
much
less
comprehensive than the later electromagnetic
theory of Maxwell, Huygens’ theory was simpler
mathematically and remains useful today.
Its great advantages are that it accounts for the
laws of reflection and refraction in terms of waves
and gives physical meaning to the index of refraction.
Huygens’ wave theory: where a given wavefront of
optical wave will be at any time in the future if we
know its present position. This construction is based
on Huygens’ principle:
惠更斯原理: 光波波前上的每一點都可視為一新的波源，
以 波 速 向 四 面 八 方 擴 散 出 球 面 次 級 子 波 (secondary
wavelets)，經過一段時間後，這些次級子波的公切面[即
包跡(envelope)]將形成新的波前。
A given wavefront (plane ab) at t=0, where will
the wavefront be at time Δt later ?
Sources of spherical
secondary wavelets
(次級子波波源)
At time Δt, the radius
of all these spherical
wavelets will have
grown to cΔt.
¨ New wavefront (各
次級子波的公切面)
Fig. 35-2
We can use Huygens’ principle to derive the law of
refraction: n1sinθ1=n2sinθ2.
λ1 and λ2: wavelength in medium 1 and 2
v1 and v2: velocity of light in air and in glass (v1>v2)
位於波峰之波前
Fig. 35-3
<由惠更斯原理出發推導光波折射定律>
[Fig. 35-3a] θ1: angle between the
incident wavefront and the interface =
angle of incidence (亦即入射角).
Self-study
[Fig. 35-3b] As the wave moves into the
glass, point “e” (由惠更斯原理知波前上
每一點都可當成一新的波源) will expand
to pass through point “c”, at a distance
of λ1 from point “e”. ¨ Δt1=λ1/v1.
Now note that in this same Δt, point
“h” (另一新的波源) will expand to pass
through point “g”, at the reduced speed
v2 and with λ2. ¨ Δt2=λ2/v2.
By Δt1 = Δt2, we obtain
Fig. 35-3
θ 2: angle between the refracted wavefront and the
interface, is actually the angle of refraction (亦即折射角).
For the right triangles “hce” and “hcg” in Fig. 35-3b
we may write
Fig. 35-3
Dividing the first of these two equations by the second
and using Eq. 35-1, we find
We can define the index of refraction n for each
medium as the ratio of the speed of light in vacuum c
to the speed of light v in the medium. Thus,
In particular, for our two media, we have
If we combine Eqs. 35-2 and 35-4, we find
¨
1 The wavelength of light changes when the speed of the light
changes, as happens when light crosses an interface from
one medium into another [Eq. (35-1): v1/v2=λ1/λ2].
2 The speed of light in any medium depends on the index
of refraction of the medium (n=c/v)(進入介質，波速變慢
n倍).
By 12, λ of light in any medium depends on “n” of the medium.
Let a certain monochromatic light have wavelength λ and
speed c in vacuum and wavelength λn and speed v in a
medium with an index of refraction “n”. From Eq. 35-1
Using Eq. 35-3 yields
This equation tells us that the greater the index of
refraction of a medium, the smaller the wavelength
of light in that medium (進入介質，波長變短n倍)
What about the frequency of the light? Let fn
represent the frequency of the light in a medium with
index of refraction n. Then from the general relation of
Eq. 16-13 (v=λ/T=λf), we can write
Substituting Eqs. 35-8 into fn then gives us
where f is the frequency of the light in vacuum.
Although the speed and wavelength of light in the
medium are different (reduced) from what they are
in vacuum, the frequency of the light in the medium
is the same as it is in vacuum.
Equation 35-8 is important in the interference of light waves. In
Fig. 35-4, waves 1 and 2 have identical wavelengths λ and
are initially in phase in air (n ≈ 1). One of the waves travels
through medium 1 (n1) and length L. The other travels
through medium 2 (n2) and the same length L.
When the waves leave the two media, they will have the
same wavelength — their wavelength λ in air. However,
because their wavelengths differed in the two media, the two
waves may no longer be in phase after exiting the media.
Wave 2
Wave 1
In phase (phase
difference=0)
Not in phase (new
phase difference≠0)
Fig. 35-4
To find their new phase difference in terms of wavelengths,
we count the number N1 of wavelengths there are in the
length L of medium 1. The wavelength in medium 1 is
λn1=λ/n1; so
We count the number N2 of wavelengths there are in
the length L of medium 2, where the wavelength is
λn2=λ/n2:
Assuming n2 > n1, we obtain
Suppose Eq. 35-11 tells us that the waves now have
a phase difference of 45.6λ.
A shift of an integer number of wavelengths (such
as 45λ) would put the waves back in phase; so it is
only the decimal fraction (0.6λ) that is important.
¨ A phase difference of 45.6λ is equivalent to an
effective phase difference of 0.6λ.
1 A phase difference of 0.5λ (半個波長) puts two waves
exactly out of phase (完全反相). If the waves had equal
amplitudes and were to reach some common point, they
would then undergo fully destructive interference (完全破壞
性干涉), producing total darkness at that point.
2 With a phase difference of 0λ or integer×λ, they would
undergo fully constructive interference (完全建設性干涉),
resulting in maximum brightness at the common point.
3 Our phase difference of 0.55λ is an intermediate
situation but closer to fully destructive interference, and the
waves would produce a dimly (灰暗の) illuminated common
point.
We can also express phase difference in terms of
radians and degrees. A phase difference of one λ is
equivalent to phase differences of 2π radius or 360o.
→
Δφ = 2π(N2-N1) = (2π/λ)L(n2-n1)
In Section 33-8, we discussed how
the colors of sunlight are separated
into a rainbow when sunlight travels
through falling raindrops.
Here we can see that different
parts of an incoming wave will travel
different paths within the drop. That
means waves will emerge from the
drop with different phases. Thus, we
can see that at some angles (~42o)
the emerging light will be in phase
and give constructive interference.
The rainbow is the result of such
constructive interference.
in phase
(a) By the phase difference between the two waves via the
media:
(b) Here the effective phase difference of 0.84 wavelength is
> 0.5 wavelength (for fully destructive interference)
< 1.0 wavelength (for fully constructive interference),
but ~ 1.0 wavelength. Thus, the waves would produce
intermediate interference that is closer to fully constructive
interference — they would produce a relatively bright spot.
Example of diffraction
At a beach in Israel, plane water waves pass through two
openings in a breakwall. Notice the diffraction effect — the
waves exit the openings with circular wave fronts. Notice also
how the beach has been shaped by the circular wave fronts.
In the next section we shall discuss
the experiment – young’s double-slits
interference – that first proved that light
is a wave. To prepare for that
discussion, we must introduce the
idea of diffraction of waves:
If a wave encounters a barrier that
has an opening of dimensions similar to
the wavelength (a~λ), the part of the
wave that passes through the opening
will spread out — will diffract — into
the region beyond the barrier.
a
λ
Fig. 35-6
An incident plane wave of wavelength λ encounters a
slit that has width a=6.0λ. The part of the wave that
passes through the slit flares out (呈喇叭狀展開) on the
far side. Figures 35-7b (with a=3.0λ) and 35-7c
(a=1.5λ) illustrate the main feature of diffraction: “the
narrower the slit, the greater the diffraction.”
Fig. 35-7 The narrower the slit, the greater the diffraction.
Diffraction limits geometrical optics, in which we
represent an EM wave with a ray:
If we actually try to form a ray by sending light
through a narrow slit, or through a series of narrow
slits, diffraction will always defeat our effort because it
always causes the light to spread.
Indeed, the narrower we make the slits, the greater
the spreading is. Thus, geometrical optics (ray optics)
holds only when slits or other apertures (孔洞) do not
have dimensions comparable to or smaller than the
wavelength of the light (that is, a >> λ) such that the
diffraction effect can be neglected.
In 1801, Thomas Young first experimentally proved
that light is a wave, contrary to (對立於) what most
other scientists then thought ̶ light is particle-like,
not a wave. He did so by demonstrating that light
undergoes interference, as do water waves, sound
waves, and waves of all other types (視為波才會有
干涉現象，視為粒子則不會).
In addition, Young was able to measure the
average wavelength of sunlight; his value, 570 nm,
is impressively close to the modern accepted value
of 555 nm.
Diffraction
Diffraction of the light
by these two slits
sends
overlapping
cylindrical waves into
the region between
screens B and C,
where the waves from
one slit interfere with
the waves from the
other slit.
Interference between two
diffracted waves from S1 & S2
Monochromatic
(單色光)
Fig. 35-8 Basic arrangement of Young’s experiment.
A interference of the overlapping waves shows on
screen C:
1 Bright regions → fully constructive interference →
bright fringes (maxima)
2 Dark regions → fully destructive interference → dark
fringes (minima)
h The pattern of bright and dark fringes on screen C is
called an interference pattern.
bright fringes
Fig. 35-9
Interference pattern
dark fringes
screen C
What exactly determines the locations of
the interference fringes ?
We pick an arbitrary point P on
screen C, at angle θ (方位角) to the
central axis. This point intercepts the
wave of ray r1 from the bottom slit and
the wave of ray r2 from the top slit.
Incident waves are in phase when
they arrive S1 & S2 because there they
are just portions of the same incident
wave. However, once they have
passed the slits, the two waves must
travel different distances to reach P via
r1 & r2, respectively. h 產 生 phase
difference
Central axis
Fig. 35-10a
¨ 到達不同觀察點P，r1 與r2 走的路徑差不同→相位差不同→
亮暗程度不同
Consider two waves initially exactly
in phase, traveling along paths with a
path length difference ΔL, and then
arriving at some common point P.
1 When ΔL=mλ (m=0, 1, 2…), the
waves arrive at P exactly in phase and they interfere
fully constructively there. → bright fringe
2 When ΔL=(m)λ/2 (m=1, 3, 5…), the waves arrive at P
exactly out of phase and they interfere fully
destructively there. → dark fringe
The path length difference ΔL
(between r1 & r2) = S1b:
As the distance from the slits
to the observed point of the
screen >> the slit separation,
we can approximate rays r1 and
r2 as being parallel to each
other and at angle θ to the
central axis (Fig. 35-10b).
We can also approximate the
triangle formed by S1S2b as
being a right triangle, and
approximate the angle inside
that triangle at S2 as being θ.
Fig. 35-10b
Then, sinθ = ΔL/d and thus
Fig. 35-10b
ΔL
1 For a bright fringe, we saw that ΔL=mλ (m=0, 1,
2…). We can write this requirement as
¨
For m=0, Eq. 35-14 tells us that a bright fringe is at
θ=0 and thus on the central axis. This central
maximum ( 中 央 極 大 ) is the point at which waves
arriving from the two slits have a path length difference
ΔL=0, hence phase difference=0.
For m=2, Eq. 35-14 tells us that bright fringes are at
the angle
Waves from the two slits arrive at these two fringes
with ΔL=2λ and with a phase difference of 2λ. These
fringes are said to be the second-order bright fringes
(m=2) or second maxima.
2 For a dark fringe, ΔL=(m)λ/2 (m=1, 3, 5…). We can
write this requirement as
¨
For m=1, Eq. 35-16 tells us that dark fringes are at
the angle
Waves from the two slits arrive at these two fringes with
ΔL=1.5λ. These fringes are called the second-order dark
fringes or second minima. (The first dark fringes, or first
minima, are at locations for which m=0 in Eq. 35-16.)
第
一
極
大
第
二
極
小
中
央
極
大
第
一
極
小
第
一
極
大
第
一
極
小
第
二
極
小
By Fig. 35-10a (mth
階亮紋發生的位置)
From Eq. 35-14 (mth
階亮紋發生條件)
Then using the approximation sinθ ≈ tanθ
for small θ
P
For the interference pattern
to appear on screen C, the
two light waves from S1 and
S2 reaching any point P on
the screen must have a
phase difference that does
not vary in time.
→ That is the case in Fig. 35-8 because the waves
passing through slits S1 and S2 are portions of the single
light wave that illuminates the slits. Because the phase
difference remains constant (不隨時間變化) at any point
of P on screen C, the light from slits S1 and S2 is said to
be completely coherent (同調/相干).
If we replace the double slits with
two
similar
but
independent
monochromatic light sources, the
phase difference between the two
waves emitted by the two sources
varies rapidly and randomly.
h The two waves are not coherent.
P
φ1(t)
φ2(t)
Δφ(t) at P changes with
time randomly & rapidly
Δφ(t) changes with time
randomly & rapidly
As a result, at any given point P on the viewing screen,
the interference between the waves from the two sources
varies rapidly and randomly【phase difference varies
rapidly and randomly because the phase change of each
wave is as quickly as > 109 Hz】between fully constructive
and fully destructive during the response time of human eyes.
The eye cannot follow such rapid changes, and no (longterm stable) interference pattern can be seen. The fringes
disappear, and the screen is seen as being uniformly
illuminated.
Equations 35-14 and 35-16
show the maxima and minima
of the double-slit interference
pattern as a function of the
angle θ ( 方 位 角 ). Here we
wish to get an expression for
the intensity I of the
interfering
fringes
as
a
function of θ
h I=I(θ)
Assuming that the light
waves from the two slits are
not in phase when they
arrive at point P. The electric
field components of those
waves at point P are not in
phase and vary with time as
Phase difference (由於路徑差造成): φ
Because the phase difference does not vary with time (φ =
const.), the two waves are coherent.
These two waves will combine at P
to produce an intensity I given by
dsinθ = 路徑差
I0: intensity of the light through S1 or S2.
Equations 35-22 and 35-23, which necessarily contain
information about the location of the maxima and
minima.
(1) The intensity maxima will occur when (from Eq. 35-22)
If we put this result into Eq. 35-23, we find
¨
which is exactly Eq. 35-14.
(2) The minima in the fringe pattern occur when (from Eq.
35-22)
If we combine this relation with Eq. 35-23, we are led
to
which is just Eq. 35-16.
Figure 35-12 (a plot of Eq. 35-22) shows the intensity of
the double-slit interference pattern as a function of φ.
The horizontal solid line is I0. Note in Eq. 35-22 and the
graph that the intensity I varies from 0 at the fringe
minima to 4I0 at the fringe maxima.
Fig. 35-12
If the waves from the two sources (slits) were incoherent,
so that no enduring (固定の) phase relation existed between
them (相位差隨時間瞬時任意變化), there would be no fringe
pattern and the intensity would have the uniform value 2I0
for all points on the screen.
Interference cannot create or destroy energy but merely
redistributes ( 重 新 分 佈 ) it over the screen. Thus, the
average intensity on the screen must be the same 2I0
regardless of whether the sources are coherent. This follows
from Eq. 35-22; if we substitute ½, the average value of the
cos2-function, this equation reduces to Iavg=2I0.
→ Self-study (利用相量作圖法)
The phase difference φ between E1 and E2 is associated
with the path length difference S1b. If S1b is λ/2, then φ
is π; if S1b is λ, then φ is 2π, and so on. This suggests
相位差 = (2π/λ)×路徑差
(important!!)
Fig. 35-10b
Fig. 35-10b
The path length difference S1b in Fig. 35-10b is dsinθ ;
so Eq. 35-30 for the phase difference between the two
waves arriving at point P on the screen becomes
which is same as Eq. 35-23.
The colors we see when sunlight illuminates a soap bubble
or an oil slick on the road are caused by the interference of
light waves reflected from the front and back surfaces of a
thin transparent film.
The thickness (L) of the soap or oil film (for causing the
interference fringes) is typically of the order of magnitude
of the wavelength of the (visible) light involved. (L~λ)
Figure 35-15 shows a thin
transparent film (thickness L, index
of refraction n2) illuminated by
bright light of wavelength λ from a
distant point source. Assuming that
n1=n3=nair. For simplicity, θ ≈ 0o.
air
air
θ≈0o
λ
Fig. 35-15
thin film
If reflected waves r1 and r2 are exactly in phase at the
eyes ¨ interference maximum and region ac on the film
is bright to the observer.
If reflected waves r1 and r2 are exactly out of phase
(相差180o), they produce an interference minimum and
region ac is dark to the observer.
If reflected waves r1 and r2 have some intermediate
phase difference, there are intermediate interference
and brightness.
air
air
θ≈0o
λ
Fig. 35-15
thin film
The key to what the observer sees is the phase
difference between reflected waves r1 and r2.
Because θ (入射角) is ~ 0o, we approximate the
path length difference (路徑差) between the waves r1
and r2 as 2L (r2在n2薄膜內多走了2L).
However, to find the phase difference between the
waves, we cannot just find the number of wavelengths
λ that is equivalent to a path length difference of 2L.
This simple approach is impossible for two reasons:
air
(i) The path length difference occurs
in a medium other than air (因為在
介質中折射率與在空氣中不同).
(ii) Reflections are involved, which
can change the phase.
air
θ≈0o
λ
Fig. 35-15
thin film
Refraction at an interface never
causes a phase change — but
reflection can, depending on the
indexes of refraction on the two
sides of the interface.
Figure 35-16 shows what happens
when reflection causes a phase
change, using as an example pulses
on a denser string and a lighter
string.
Similarly, for light
(光密→光疏)
(光疏→光密)
denser
lighter
同相
同相
denser
lighter
同相
反相
Fig. 35-16
There are 3 ways in which the phase difference between
two waves can be generated:
1. by reflection
2. by the waves traveling along paths of different lengths
(路徑差)
3. by the waves traveling through media of different
indexes of refraction (折射率差)
When light reflects from a thin
film, producing the waves of rays r1
and r2 shown in Fig. 35-15, all 3
ways described above are involved.
air
air
θ≈0o
λ
Fig. 35-15
thin film
<反射造成相差之解析>
反射造成相差之解析
On the 1st interface, the incident wave (in air) reflects
from the medium n2 (n2>nair) ; so the wave r1 has its
phase shifted by  wavelength (π radius 或 180o).
On the 2nd interface, the incident wave reflects from
the medium (air) having the lower of the two indexes of
refraction; so the wave r2 reflected there is not shifted in
phase by the reflection.
1st
So far, as a result of the
reflection phase shifts, the waves
of r1 and r2 have a phase difference
of  wavelength and thus are
exactly out of phase (相差180o).
2nd
r2
r1
incident light
Fig. 35-17
< 光程差(路徑差×折射率)與反射造成相差之解析>
與反射造成相差之解析
n If the reflected waves of r1 and r2 are to be exactly in
phase so that they produce fully constructive interference,
the path length 2L must contribute an additional phase
difference of 0.5, 1.5, 2.5, . . . wavelengths. Then the net
phase difference [ 路徑差×折射率+反射(π或 λ)] is an
integer number of wavelengths. Thus, for a bright film (完
全建設性干涉), we must have
λn2: the light in the medium with n2
Fig. 35-17
o If the waves are to be exactly out of phase so that
there is fully destructive interference, the path length 2L
must contribute an additional phase difference of 0, 1, 2,
3, . . . wavelengths. Then the net phase difference be an
odd number of half-wavelengths. For a dark film (完全破
壞性干涉), we must have
λn2: the light in the medium n2
Fig. 35-17
Use Eq. 35-8 (λn=λ/n) to write the wavelength of the
wave of ray r2 inside the film as
where λ is the wavelength of the incident light in vacuum.
Substituting Eq. 35-35 into Eq. 35-32 and replacing
“odd number/2” with (m+) give us
Similarly, with m replacing “integer,” Eq. 35-34 yields
(干涉極大)
(干涉極小)
(L « λ)
A special situation arises when a film is so thin that L <<
λ (L<λ/10). Then the path length difference 2L can be
neglected, and the phase difference between r1 and r2 is
due only to reflection phase shifts (π). → The film we see
is dark, regardless of λ.
L
λ
λ>>L
Thin
film
Figure 35-18 shows a vertical soap film whose thickness
increases from top to bottom because gravitation has
caused the film to slump (下降).
L<<λ → dark (r1與r2反相，破壞性干涉)
Color interference fringes
(L夠厚，厚度造成之相差不可
忽略 → 隨厚度改變 → 不同波
長反射光在不同厚度位置產生
干涉極大 → 彩色干涉條紋)
Fig. 35-18
air
θ=0
λ=400-690nm
water
air
完全破壞性干涉
θ=0
π
π
Fig. 35-20
θ=0
π
π
Homeworks
Ans:
Ans: (a) 50º (b) 0.14 ps
Ans: (a) 0.833 (b) intermediate but closer to fully constructive
Ans: 140 bright fringes
Ans:
Ans:
Ans: (a) 34 bright rings (b) 46 bright rings
Ans: (a) 22o (b) The angle of incidence is gradually reduced due
to refraction. Eventually after many refractions, θ2 will be
virtually zero. This is why most waves come in normal to a shore
Ans: (a) 1.6
(b) 1.4
Ans:
tzig zag – tdirect
=(n1L/c)-[n12L/(n2c)]
=51.6 ns.
Ans: x= (m+)(D/2a)λ.
Ans: (a)
(b)
Ans: (a)
(b)
Ans: (a)
(b)