Download Chapter 25: Interference and Diffraction

Chapter 25: Interference and
Diffraction
•Constructive and Destructive Interference
•The Michelson Interferometer
•Thin Films
•Young’s Double Slit Experiment
•Gratings
•Diffraction
•Resolution of Optical Instruments
1
Interference and Diffraction
Certain physical phenomena are most
easily explained by invoking the wave
nature of light, rather than the particle
nature.
These include:
interference, diffraction, polarization
2
Coherence
If the phase of a light wave is well defined at all times
(oscillates in a simple pattern with time and varies in a
smooth way in space at any instant), then the light is
said to be coherent.
If, on the other hand, the phase of a light wave varies
randomly from point to point, or from moment to
moment (on scales coarser than the wavelength or
period of the light) then the light is said to be
incoherent.
For example, a laser produces highly coherent light. In
a laser, all of the atoms radiate in phase.
An incandescent or fluorescent light bulb produces
incoherent light. All of the atoms in the phosphor of the
bulb radiate with random phase. Each atom oscillates
for about 1ns, and produces a wave about 1 million
wavelengths long.
3
Interference
Just like sound waves, light waves also display
constructive and destructive interference.
For incoherent light, the interference is hard to
observe because it is “washed out” by the very
rapid phase jumps of the light.
Soap films are one example where we can see
interference effects even with incoherent light.
4
Soap Film Interference
(White light)
5
Conditions for Interference
The following four conditions must be true in order
for an interference pattern to be observed.
9 The source must be coherent – has a constant
phase relationship
9 Wavelengths must be the same –
monochromatic
9 The Principle of Superposition must apply.
9 The waves have the same polarization state.
6
Constructive and Destructive Interference
(the same for all waves!)
Two waves (top
and middle) arrive
at the same point in
space.
The total wave
amplitude is the
sum of the two
waves.
The waves can add
constructively or
destructively
7
Constructive interference occurs when two waves are in
phase. To be in phase, the points on the wave must have
Δφ=(2π)m, where m is an integer.
When coherent waves are in phase, the resulting amplitude
is just the sum of the individual amplitudes. The energy
content of a wave depends on A2. Thus, I∝A2.
8
The resulting amplitude and intensity are:
A = A1 + A2
I = I1 + I 2 + 2 I1 I 2
9
Destructive interference occurs when two waves are a half
cycle out of phase. To be out of phase the points on the
wave must have Δφ=(2π)(m+½), where m is an integer.
10
The resulting amplitude and intensity are:
A = A1 − A2
I = I1 + I 2 − 2 I1 I 2
11
Coherent waves can become out of phase if they travel
different distances to the point of observation.
P
S1
θ
d
S2
This represents the extra path length
(Δl) that the wave from S2 must
travel to reach point P.
Δl = d sin θ
12
When both waves travel in the same medium the
interference conditions are:
For constructive interference
Δl = mλ
where m = an integer.
1⎞
⎛
For destructive interference Δl = ⎜ m + ⎟λ
2⎠
⎝
where m = an integer.
13
Fringes
If at point P the path difference yields a
phase difference of 180 degrees between
the two beams a “dark fringe” will appear.
If the two waves are in phase, a “bright
fringe” will appear.
14
Example A 60.0 kHz transmitter sends an EM wave to a
receiver 21 km away. The signal also travels to the
receiver by another path where it reflects from a helicopter.
Assume that there is a 180° phase shift when the wave is
reflected.
(a) What is the
wavelength of
this EM wave?
c 3.0 ×105 km/sec
λ= =
= 5.0 km
3
60 ×10 Hz
f
(b) Will this situation give constructive interference,
destructive inference, or something in between?
The path length difference is Δl = 10 km = 2λ, a whole
number of wavelengths. Since there is also a 180° phase
15
shift there will be destructive interference.
Michelson Interferometer
In the Michelson interferometer, a beam of coherent light is
incident on a beam splitter. Half of the light is transmitted
to mirror M1 and half is reflected to mirror M2.
16
The beams of light are reflected by the mirrors, combined
together, and observed on the screen.
If the arms are of different lengths, a phase difference
between the beams can be introduced.
17
Example A Michelson interferometer is adjusted so that a
bright fringe appears on the screen. As one of the mirrors
is moved 25.8 μm, 92 bright fringes are counted on the
screen. What is the wavelength of the light used in the
interferometer?
Moving the mirror a distance d introduces a path length
difference of 2d. The number of bright fringes (N)
corresponds to the number of wavelengths in the extra
path length.
Nλ = 2 d
2d
λ=
= 0.561 μm
N
18
Thin Films
When an incident light ray reflects from a boundary with a
higher index of refraction, the reflected wave is inverted (a
180° phase shift is introduced).
A light ray can be reflected many times within a medium.
19
Phase of wave reflected by interface
between two media
20
21
Interference in Thin Films
We have all seen the colorful patterns which appear in
soap bubbles. The patterns result from an interference
of light reflected from both surfaces of the film. Some
colors undergo constructive some destructive
interference:
180o phase change
0o phase change
λ / n= λn
n>1
t
2t = mλn = mλ / n (m = 0, 1, 2…) Destructive (invisible)
2t = (m+1/2)λn = (m + ½)λ / n (m = 0, 1, 2…) Constructive
(bright color)
22
Air Wedge
Monochromated Light
23
Newton’s Rings
24
Example A thin film of oil (n=1.50) of thickness 0.40 μm is
spread over a puddle of water (n=1.33). For which
wavelength in the visible spectrum do you expect
constructive interference for reflection at normal incidence?
Water
Oil
Air
Incident
wave
Consider the first two
reflected rays. r1 is
from the air-oil
boundary and r2 is from
the oil-water boundary.
r1 has a 180° phase shift (noil >nair),
but r2 does not (noil<nwater).
25
To get constructive interference, the reflected waves must
be in phase. For this situation, this means that the wave
that travels in oil must travel an extra path equal to multiples
of half the wavelength of light in oil.
The extra path distance traveled is 2d, where d is the
thickness of the film. The condition for constructive
interference here is:
1⎞
1 ⎞⎛ λair ⎞
⎛
⎛
⎟⎟
2d = ⎜ m + ⎟λoil = ⎜ m + ⎟⎜⎜
2⎠
2 ⎠⎝ noil ⎠
⎝
⎝
2dnoil
∴ λair =
Only the wavelengths that
1⎞
⎛
⎜m + ⎟
satisfy this condition will have
2
⎝
⎠
constructive interference.
26
Example continued:
Make a table:
m λair(μm)
0
1
2
3
4
2.40
0.80
0.48
0.34
0.27
All of these wavelengths will show
constructive interference, but it is
only this one that is in the visible
portion of the spectrum.
27
White light is incident on a soap film (n = 1.30) in
air. The reflected light looks bluish because the
red light (λ = 670 nm) is absent in the reflection.
What is the minimum thickness of the soap film?
2t = mλ / n (m = 1, 2…)
t = 1λ
Destructive interference
/ 2n = 258nm
28
Young’s Double-Slit Experiment
Place a source of coherent light
behind a mask that has two vertical
slits cut into it. The slits are L tall,
their centers are separated by d,
and their widths are a.
29
The slits become sources of
waves that, as they travel
outward, can interfere with
each other.
30
The pattern seen on the screen
There are
alternating
bright/dark
spots.
An
intensity
trace
31
If the two slits are
separated by a
distance d and the
screen is far away
then the path
difference at point P
is Δl = dsinθ
If Δl = λ, 2λ, 3λ, etc,
then the waves will
arrive in phase and
there will be a bright
spot on the screen.
32
Interference Conditions
For constructive interference,
the path difference must be
zero or an integral multiple of
the wavelength:
d sinθ = nλ , (n = 0, ± 1, ± 2...)
For destructive interference, the
path difference must be an odd
multiple of half wavelengths:
dsinθ = (n + 1/2)λ
(n = 0, ± 1, ± 2...)
n is called the order number
33
Example: Show that the interference fringes in a double-slit
experiment are equally spaced on a distant screen near the
center of the interference pattern.
The condition for constructive
interference is
Δl = d sin θ = mλ
mλ
∴ sin θ =
d
From the geometry of the
problem,
h
tan θ =
D
34
Example continued:
The screen is far away compared to the distance between
the slits (D>>d) so tanθ ≈ sinθ ≈ θ. Here,
mλ
h
sin θ ≈ θ =
and tan θ ≈ θ =
d
D
mλ h
∴
=
d
D
mλD
h=
d
The distance between two adjacent minima is:
h2 − h1 =
λD
d
(m2 − m1 ) =
λD
d
35
Example
If the distance between two slits is 0.050 mm and
the distance to a screen is 2.50 m, find the spacing
between the first- and second-order bright fringes
for yellow light of 600 nm wavelength.
x2 − x1 ≈ 2λL / d − λL / d
2 ⋅ 600 ⋅10 m ⋅ 2.5m
= 0.06m = 6cm
=
−3
0.05 ⋅10 m
−9
36
When green light (λ = 505 nm) passes through a pair of double slits,
the interference pattern shown in (a) is observed. When light of a
different color passes through the same pair of slits, the pattern shown
in (b) is observed. (a) Is the wavelength of the second color greater
than or less than 505 nm? Explain. (b) Find the wavelength of the
second color. (Assume that the angles involved are small enough to
set sinθ = tanθ.)
xgreen , 4.5 = x?,5 ⇒ 4.5λgreen L / d = 5λ? L / d
4 .5
⇒ λ? =
λgreen = 455nm
5
37
Gratings
A grating has a large
number of evenly spaced,
parallel slits cut into it.
d sin θ = mλ
38
39
Example Red light with λ=650 nm can be seen in three
orders in a particular grating. About how many rulings per cm
does this grating have?
For each of the maxima
d sin θ = mλ
d sin θ 0 = 0
d sin θ1 = 1λ
d sin θ 2 = 2λ
d sin θ 3 = 3λ
d sin θ 4 = 4λ
Third order is observed.
This order is not
observed.
40
Example continued:
Since the m = 4 case is not observed, it must be that
sinθ4>1. We can then assume that θ3≈90°. This gives
d = 3λ = 1.95 × 10 −6 m
and
1
N = = 510,000 lines/m = 5100 lines/cm.
d
41
Diffraction
Using Huygens’s principle: every
point on a wave front is a source
of wavelets; light will spread out
when it passes through a narrow
slit.
Diffraction is appreciable only when
the slit width is nearly the same size
or smaller than the wavelength.
42
Definition and Types of Diffraction
• Diffraction is the bending of a wave around an
object accompanied by an interference pattern
• Fresnel Diffraction - curved (spherical) wave
front is diffracted
• Fraunhofer Diffraction - plane wave is diffracted
Fresnel Diffraction from
a circular obstruction
Fresnel bright spot
43
The
intensity
pattern on
the screen.
44
The minima occur when:
a sin θ = mλ
where m = ±1, ±2,…
45
Example: Light from a red laser passes through a single slit
to form a diffraction pattern on a distant screen. If the width
of the slit is increased by a factor of two, what happens to the
width of the central maximum on the screen?
The central maximum occurs between θ=0 and θ as
determined by the location of the 1st minimum in the
diffraction pattern:
a sin θ = mλ
θ≈
λ
Let m =+1 and assume
that θ is small.
a
From the previous picture, θ only determines the half-width
of the maximum. If a is doubled, the width of the
maximum is halved.
46
Resolution of Optical Instruments
The effect of diffraction is to spread light out. When viewing
two distant objects, it is possible that their light is spread out
to where the images of each object overlap. The objects
become indistinguishable.
47
Resolvability
Rayleigh’s Criterion: two point sources are barely resolvable if their angular
separation θR results in the central maximum of the diffraction pattern of one
source’s image is centered on the first minimum of the diffraction pattern of the
other source’s image.
Fig. 36-10
λ ⎞ θ R small
λ
⎛
θ R = sin ⎜1.22 ⎟ ≈ 1.22
d⎠
d
⎝
−1
(Rayleigh's criterion)
36-48
For a circular aperture, the
Rayleigh criterion is:
Δθ
a sin Δθ ≥ 1.22λ
where a is the aperture size of your instrument, λ is
the wavelength of light used to make the observation,
and Δθ is the angular separation between the two
observed bodies.
49
a sin Δθ ≥ 1.22λ
To resolve a pair of objects, the
angular separation between
them must be greater than the
value of Δθ.
50
Example: The radio telescope at Arecibo, Puerto Rico, has a
reflecting spherical bowl of 305 m diameter. Radio signals
can be received and emitted at various frequencies at the
focal point of the reflecting bowl. At a frequency of 300 MHz,
what is the angle between two stars that can barely be
resolved?
a sin Δθ ≥ 1.22λ
(
8
×
m/s
3
.
0
10
(1.22)
1.22λ
=
sin Δθ ≥
a
sin Δθ ≥ 4.1 × 10−3
)
300 × 10 Hz
6
305 m
Δθ ≥ 0.23 degrees
51
X-Ray Diffraction
X-rays are electromagnetic radiation with wavelength ~1 Å = 10-10 m (visible
light ~5.5x10-7 m)
X-ray generation
X-ray wavelengths to short to be resolved
by a standard optical grating
Fig. 36-27
mλ
−1 (1)( 0.1 nm )
= sin
= 0.0019°
θ = sin
d
3000 nm
−1
36-52
X-Ray Diffraction, cont’d
Diffraction of x-rays by crystal: spacing d of
adjacent crystal planes on the order of 0.1 nm
→ three-dimensional diffraction grating with
diffraction maxima along angles where reflections
from different planes interfere constructively
2d sin θ = mλ for m = 0,1, 2… (Bragg's law)
Fig. 36-28
36-53