The First Part of the Optics Story:
Light behaving as a wave
1
Diffraction
2
Diffraction
3
through a narrow slit, they too spread out behind the slit, just as the water wave did be
the opening in the barrier. The light is exhibiting diffraction, the sure sign of waviness.
will look at diffraction in more detail later in the chapter. For now, though, we see th
sufficiently narrow slit does act as a source of light waves that spread out behind it.
Interference
0.1-mm-wide
slit in an
opaque screen
Young’s Double-Slit Experiment
In order to observe interference, we need two light sources whose waves can ove
and interfere. FIGURE 17.6a shows an experiment in which a laser beam is aimed a
opaque screen containing two long, narrow slits that are very close together. This
of slits is called a double slit, and in a typical experiment they are
wide
Incident laser beam
spaced
apart. We will assume that the laser beam illuminates both
equally, and any light passing through the slits impinges on a viewing screen. Su
2 The Interference of Light
double-slit experiment was first performed by Thomas Young in 1801, using sunl
SECTION 16.6 you learned that waves of equal wavelength emitted from two
instead of a laser beam. It provided the first definitive evidence that light is a wave
es—such as the sound waves emitted from two loudspeakers—can overlapWhat
and should we expect to see on the viewing screen? FIGURE 17.6b is a view f
ere, leading to a large amplitude where the waves interfere constructively and
above the experiment, looking down on the top ends of the slits and the top edge of
ll or even zero amplitude where they interfere destructively. Interference is
viewing screen. Because the slits are very narrow, light spreads out behind each
ntly a wave phenomenon, so if light acts like a wave we should be able to
as it did in Figure 17.5, and these two spreading waves overlap in the region betw
ve the interference of light waves.
4 screen.
theThis
slits
do so we need two sources of light waves with exactly the same wavelength.
is and the
lt to arrange with conventional light sources. Instead, consider the situation shown
FIGURE 17.6
A double-slit
interference
experiment.
RE 17.5. Laser light passes through
a narrow
opening—a
slit—that is
only 0.1 mm
about twice the width of a human
(a) hair. This situation is similar to Figure 17.2, where
waves spread out after passingThe
through
a narrow opening. When light waves pass
drawing is not to scale: The distance
Creating a
h a narrow slit, they too spread out
behind
slit, just
as the
water wave did behind
to the
screen the
is actually
much
greater
Diffraction
the distance
betweenthe
the sure
slits. sign of waviness. We
ening in
the barrier. The light is than
exhibiting
diffraction,
Viewing
Pattern
ok at diffraction
in more detail later in the chapter. For now, though,
we see that a
screen
ently narrow slit does act as a source of light waves that spread out behind it.
(b)
3. Constructive interference occurs
when r1 and r2 differ by a whole
1. A plane wave is incident
number of wavelengths.
on the double slit.
m=4
2. Waves spread out
l
behind each slit.
The bright fri
m=3
are labeled by
r1
integer m, sta
m=2
at the central
maximum.
r2
m=1
ng’s Double-Slit Experiment
er to observe interference, we need two light sources whose waves can overlap
terfere. FIGURE 17.6a shows an experiment in which a laser beam is aimed at an
r1
e screen containing two long, narrow slits that are very close together. This pair
Theare
two waves overlap
as and
wide
s is called a double slit, and in a typical experiment they
r2
they spread out behind the
apart. We will assume that the laser beam
illuminates
both slits
d
two slits.
The two overlapped
y, and any light passing through the slits impinges onwaves
a viewing
screen.
Such
interfere,
resulting
in a
Top view of
a pattern
of light
and sunlight
dark
e-slit experiment was first performed by Thomas Young
in 1801,
using
the double slit
bands on the screen.
d of a laser beam. It provided the first definitive evidence that light is a wave.
4. Destructive interference occurs when
beam 17.6b is a view from
FIGURE
hat should we expect to see on the viewingIncident
screen?laser
r1 and r2 differ by a whole number of
the experiment, looking down on the top ends of the slits and the top edge of the
wavelengths plus half a wavelength.
ng screen. Because the slits are very narrow, light spreads out behind each slit
5
id in Figure 17.5, and these two spreading waves overlap in the region between
In Figure 17.6b, suppose that for some point P on the screen
ts and the screen.
STOP TO THINK 17.2
where is the wavelength of the light. The interference at point P is
A. Constructive.
B. Destructive.
C. Something in between.
(b)
3. Constructive interference occurs
when r1 and r2 differ by a whole
number of wavelengths.
1. A plane wave is incident
on the double slit.
l
m=4
2. Waves spread out
behind each slit.
The bright fringes
are labeled by the
integer m, starting
m=2
at the central
maximum.
m=1
Central
m = 0 maximum
m=3
KNIG9721_03_chap_17.indd 540
r1
r2
r1
m=1
r2
m=2
d
Top view of
the double slit
4. Destructive interference occurs when
r1 and r2 differ by a whole number of
wavelengths plus half a wavelength.
m=3
m=4
Front view
of screen
6
Central
m = 0 maximum
m=1
m=2
m=3
m=4
Front view
of screen
l and
remains dark.
FIGURE 17.12 The intensity on the screen due to three diffraction gratings. Notice that the
More
Slits
Means
a Sharper Pattern
intensity axes
have different
scales.
n
N=2
y
N = 10
N = 50
The microscopic pits
that store information
on the DVD act as a
diffraction grating.
ptical perspective,
sion of the reflecction gratings can
or grooves
into a
Light
100
0
2500
intensity 4 0
velty items.
As theRainnumber of slits in the grating increases, the fringes get narrower and brighter.
is diffracted at a
740 nm
0
Spectroscopy
As we’ll see in Chapter 29, each atomic element in the periodic table, if appropriately excited by light, electricity, or collisions with other atoms, emits light at only
certain well-defined wavelengths. By accurately measuring these wavelengths, we
can deduce the various elements in a sample of unknown composition. Molecules
also emit light that is characteristic of their composition. The science of measuring
spectroscopy.
the wavelengths
of atomic
is called
diffraction
gratings
are and
themolecular
same: emissions
A bright
fringe
occurs on
tion gratings
nteger number
times the wavelength .
Diffraction gratings
_03_chap_17.indd 545
b
7
lL
d
d sin um = ml
ym = L tan um
um is the angle of the
mth bright fringe.
of
ge.
s.
l
Grating
u1
u2
d is the distance
between slits.
L
23/08/13 10:45 AM
ym is the position of
the mth bright fringe.
y
Screen
y2
m=2
y1
0
m=1
m=0
m=1
m=2
“Order number” m
8
For a diffraction grating, the small-angle approximation is generally not valid. To find the brightfringe positions ym, you must first find the angle
um from d sin um = ml, then use ym = L tan um.
A grating with N slits or lines per mm has slit
spacing d = (1 mm)/N.
grating and forms
w,
How Many Orders?
The diffraction grating glasses we use in class are rated at
6,750 lines per inch. The green laser pointer I use has a
ide wavelength
is farthest of 530 nm.
When I shine the laser through a pair of glasses, how
et side
is orders
closestwill show on the screen?
many
9
Ordering the Rainbow
Think about the angles at which light is bent by the
rainbow glasses.
When you use the rainbow glasses to look at white light,
which of the following is the correct order?
red on the inside (closer to the zeroth
order) and blue on the outside
blue on the inside (closer to the
zeroth order) and red on the outside
10
20 microns
Reflection Grating in Nature:
Ground Beetle Colors
11
0
1
2
3
20 microns
Reflection Grating in Nature:
Ground Beetle Colors
12
straight-ahead position at
but you saw in Figures 17.5 and 17.25b that
is the central maximum, not a minimum.
ANOTE
specimen
of ground
beetle
▶ Equations
17.17 and
17.18isare mathematically the same as the condition
illuminated
with a beam
yellow interference pattern. But the physical
for the mth maximum
of theofdouble-slit
meaning
is quite
different.
Equation 17.18 locates the minima (dark fringes)
light
withhere
!=570
nm,
produces
◀
of
the
single-slit
diffraction
pattern.
clear maxima at 13° and 25°. What
isIt the
spacingalthough
of the beyond
lines inthe
thescope of this textbook, to calculate the entire
is possible,
light
intensity
pattern.
The
results
of
exocuticle that produces this such a calculation are shown graphically in
FIGURE
17.26 . You
can see the bright central maximum at
the weaker secondary
diffraction
effect?
maxima, and the dark points of destructive interference at the angles given by Equation 17.18. Compare this graph to the photograph of Figure 17.23 and make sure you
see the agreement between the two.
552
C H A P T E R 17
Wave Optics
The Width of a Single-Slit Diffraction Pattern
We’ll find it useful, as we did for the double slit, to measure positions17.5
on the screen
Single-Slit Diffraction
rather than angles. The position of the pth dark fringe, at angle up is p
up
We
opened
this
where L is the distance from the slit to the viewing screen. Using Equation 17.18 for chapter with a photograph of a water wave passing through a
a barrier, then spreading out on the other side. The preview to this chapter s
up up we find that the dark fringes
the
up and the small-angle approximation
13 inafter
how light,
passing a narrow needle, also spreads out on the other side. Th
single-slit diffraction pattern are located at
nomenon is called diffraction. We’re now ready to look at the details of diffra
FIGURE 17.23 again shows the experimental arrangement for observing the d
tion
of light through a narrow slit of width a. Diffraction through a tall, narrow
l
p = 1, 2, 3, c
(17.19)
yp =
width
a is known as single-slit diffraction. A viewing screen is placed a dist
Secondary maxima
The Single-Slit Pattern
behind the slit, and we will assume that
The light pattern on the v
Positions of dark fringes for
screen consists of a central maximum flanked by a series of weaker seco
w
screen
N.B.: Same for
an
single-slit
diffractionViewing
with screen
distance L
maxima and dark fringes. Notice that the central maximum is significantly b
obstruction! So can
than the secondary maxima. It is also significantly brighter than the secondar
use same formula
ima, although that is hard to tell here because this photograph has been overe
Central maximum
Again,
is explicitly excluded because the midpoint on the viewing
is the
toscreen
make the
secondary maxima show up better.
for diffraction
Distance L
FIGURE 17.23 A single-slit diffraction
experiment.
central
maximum,
pattern
from a not
hair.a dark fringe.
A diffraction pattern is dominated by the central maximum, which
is muchPrinciple
Huygens’
Single
slit maximum, shown in
brighter than the secondary maxima. The width w of the
central
Our analysis of the superposition of waves from distinct sources, such as two
Figure 17.26, is defined as the distance between the twoof width a minima on either side
speakers or the two slits in a double-slit experiment, has tacitly assumed t
of the central maximum. Because the pattern is symmetrical, the width
is simply
sources
are point sources, with no measurable extent. To understand diffracti
This is
Incident light of
need to think about the propagation of an extended wave front. This problem w
wavelength l
w=
2lL
a
Width of the central maximum for single-slit diffraction
considered by the Dutch scientist Christiaan Huygens, a contemporary of New
In ◀◀ SECTION 15.5 you learned how wave fronts—the “crests” of a wave—
with(17.20)
time for plane and spherical waves. Huygens developed a geometrical
to visualize how any wave, such as a wave passing through a narrow slit, e
Huygens’ principle has two parts:
14
1. Each point on a wave front is the source of a spherical wavelet that spre
at the wave speed.
An important implication of Equation 17.20 and one contrary to common
2. At a The
later time, the shape of the wave front is the curve that is tangent to
sense, is that a narrower slit (smaller a) causes a wider diffraction pattern.
smaller the opening a wave squeezes through, the more it spreads outwavelets.
on the
other
side. Has Thicker Hair?
Who
A
FIGURE 17.24 illustrates Huygens’ principle for a plane wave and a spherical wa
you can see, the curve tangent to the wavelets of a plane wave is a plane that has
gated to the right. The curve tangent to the wavelets of a spherical wave is a larger
FIGURE 17.24 Huygens’ principle applied to the propagation of plane waves and
spherical waves.
(a) Plane wave 23/08/13 10:45 AM
B
1.6 cm
If this experiment is done with light of 670 nm, the hair
1.0 m from the screen and a central width of 1.6 cm,
what is the hair thickness?
Initial
wave
front
Initial
wave
front
Each point on the
initial wave front
is the source of a
spherical wavelet.
15
(b) Spherical wave
The wave front
at a later time
is tangent to all
the wavelets.
Each point
is the source
of a spherical
wavelet.
is tangent to all the wavele
maximum
p=2
p=3
A Circular OpeningLight
Makes a Diffraction Pattern Too
intensity
FIGURE 17.27 The diffraction of light by a circular opening.
y
LWD
Screen
Circular
aperture
0
p=3
Angle u locates the first minimump in
= 2the intensity, where there is perfect destructive interference.uA
mathematical analysis
p = 1 of circular diffraction finds that
1
l
u1 =Width w
Diameter D
(17.21)
p=1
p=2
Central
maximum
where D is the diameter of the circular opening. Equation 17.21 has assumed the
=3
small-angle approximation, which is palmost
always valid for the diffraction of light.
Light
Within the small-angle
approximation,
the
width of the central maximum on a
intensity
0 is
screen a distance L from the aperture
LWD
w = 2y1 = 2 L tan u1 ≈
2.44lL
(17.22)
D where there is perfect destrucAngle u locates the first minimum in the intensity,
tive interference. A mathematical
analysis
of circular
diffraction finds that
Width of central
maximum
for diffraction
from a circular aperture lof diameter D
u1 =
(17.21)
16
The diameter of the diffraction pattern increases with distance L, showing that light
behind aofcircular
aperture,
but it decreases
the sizehas
D ofassumed
the aperture
where Dspreads
is theout
diameter
the circular
opening.
Equationif 17.21
the is
increased.
small-angle
approximation, which is almost always valid for the diffraction of light.
Within the small-angle approximation, the width of the central maximum on a
EXAMPLE 17.9
Finding
theisright viewing distance
Resolution
screen
a distance
L from the
aperture
Light
from a helium-neon laser l
passes through a 0.50-mm-diameter hole.
OneHow
light
far away should a viewing screen be placed to observe a diffraction pattern whose
or two?
central maximum
mm in diameter?
w = is2y3.0
(17.22)
1 = 2 L tan u1 ≈
SOLVE
Equation 17.22 gives us the appropriate screen distance:
Width of central
for diffraction
(3.0 10
m)(5.0 10 m)
wD maximum
=
= 0.97 m
L=
-9
from a circular
2.44l aperture of diameter D
The diameter of the diffraction pattern increases with distance L, showing that light
we’veaseen,
we need
to usebut
theitwave
model ifofthe
light
to understand
the passage
spreads outAsbehind
circular
aperture,
decreases
size
D of the aperture
is
of light through narrow apertures, where “narrow” means comparable in size to the
increased.
wavelength of the light. In the next two chapters, we’ll consider the interaction of
light with objects much larger than the wavelength. There, the ray model of light will
EXAMPLE
17.9appropriate
Findingforthe
right viewing
be more
describing
how lightdistance
reflects from mirrors and refracts from
lenses.
But
the
wave
model
will
reappear
in
Chaptera19
when we study hole.
telescopes
Light from a helium-neon laser l
passes through
0.50-mm-diameter
microscopes.
We’ll find
that be
theplaced
resolution
of these
instruments
has awhose
fundamenHow and
far away
should a viewing
screen
to observe
a diffraction
pattern
talmaximum
limit set by
the mm
wave
of light.
central
is 3.0
in nature
diameter?
Change
aperture
of optical system
SOLVE Equation 17.22 gives us the appropriate screen distance:
L=
(3.0
wD
=
2.44l
10
m)(5.0
10
-9
m)
= 0.97 m
17
As we’ve seen, we need to use the wave model of light to understand the passage
of For
light through
narrow apertures,
the Human
Eyewhere “narrow” means comparable in size to the
wavelength of the light. In the next two chapters, we’ll consider the interaction of
light
with objects
L=.017
m much larger than the wavelength. There, the ray model of light will
be more appropriate for describing how light reflects from mirrors and refracts from
!=450
nmwave model will reappear in Chapter 19 when we study telescopes
lenses.
But the
and microscopes. We’ll find that the resolution of these instruments has a fundamental D=.003
limit set bym
the wave nature of light.
L
23/08/13 10:45 AM
D
2.4 " L
w!
D
23/08/13 10:45 AM
18
Warming Up
19
20
The Index of Refraction
21
Changing Speed Makes Light Bend
n1 sin !1 = n2 sin ! 2
22
Sample Problem
n1 sin !1 = n2 sin ! 2
23
Apparent Depth
A fish is 10
cm below
the surface
of the water.
Does the
fish appear
to be at a
greater or a
shallower
depth than
this?
24
Apparent Depth
25
Just Checking.
A ray of light is passing from under the surface of the
water (n=1.33) into the air as shown. What is the angle
in the air?
!=?
Air
Water
50°
n1 sin !1 = n2 sin ! 2
26
Total Internal Reflection
27
(b) half full of water, and (c) completely full of water?
55. ||| You are standing in a 1.5-m-deep swimming pool at night. The
Going
way.
waterthe
is veryother
still. You
hold a laser pointer just above the water’s
surface and shine it nearly parallel to the surface, but tilted slightly
A ray
of light is passing from the air into a tank of
down so that the beam enters the water 5.0 m from you. How far
water(n=1.33)
What
is the of
angle
in the water?
from you doesas
theshown.
beam strike
the bottom
the pool?
56. ||| What is the exit angle from the glass prism in Figure P18.56?
80°
Air
u
Water
45°
FIGURE P18.56
a
b
b
n
FIGURE P18.57
57. ||| There is just one angle of incidence b onto a prism for which
the light inside an isosceles prism travels parallel to the base and
1 b as shown
2
2
emerges at that1same angle
in Figure
P18.57.
a. Find an expression for b in terms of the prism’s apex angle
and index of refraction n.
b. A laboratory measurement finds that b
for a prism
that is shaped as an equilateral triangle. What is the prism’s
index of refraction?
58. |||| What is the smallest angle u1 for which
a laser beam will undergo total internal
30°
reflection on the hypotenuse of the glass u1
prism in Figure P18.58?
59. ||| A 1.0-cm-thick layer of water stands
on a horizontal slab of glass. Light from
60°
within the glass is incident on the glasswater boundary. What is the maximum FIGURE P18.58
angle of incidence for which a light ray can emerge into the air
above the water?
60. ||||| The glass core of an optical fiber has index of refraction 1.60.
The index of refraction of the cladding is 1.48. What is the maximum angle between a light ray and the wall of the core if the
ray is to remain inside the core?
61. |||| A 150-cm-tall diver is standing completely submerged on the
bottom of a swimming pool full of water. You are sitting on the
end of the diving board, almost directly over her. How tall does
the diver appear to be?
62. || To a fish, the 4.00-mm-thick aquarium walls appear only
3.50 mm thick. What is the index of refraction of the walls?
63.
|| A microscope
is focused
on an amoeba. When a 0.15-mmBending
Light with
a Prism
thick cover glass
is placed over the amoeba, by how
far must
the microscope
objective
be moved
to bring
the what
organA beam
of light
enters a glass
prism
as shown.
Use
ism
back
into
focus?
Must
it
be
raised
or
lowered?
you know about the bending of light at the boundary
64. || A ray
can(and,
be used
to find
the and
location
object if
between
air diagram
and glass
then,
glass
air)oftoantrace
you are given the location of its image and the focal length of
the path
of the light as it traverses the prism.
the mirror. Draw a ray diagram to find the height and position
of an object that makes a 2.0-cm-high upright virtual image that
appears 8.0 cm behind a convex mirror of focal length 20 cm.
n sin ! = n sin !
ing his glasses, he looks not at actual objects but at the virtual
images of those objects formed by his glasses. Suppose he looks
at a 12-cm-long pencil held vertically 2.0 m from his glasses.
Use ray tracing to determine the location and height of the
image.
68. | A 1.0-cm-tall object is 20 cm in front of a converging lens
that has a 10 cm focal length. Use ray tracing to find the position and height of the image. To do this accurately, use a ruler
or paper with a grid. Determine the image distance and image
height by making measurements on your diagram.
69. || A 2.0-cm-tall object is 20 cm in front of a converging lens
that has a 60 cm focal length. Use ray tracing to find the position and height of the image. To do this accurately, use a ruler
or paper with a grid. Determine the image distance and image
height by making measurements on your diagram.
70. || A 1.0-cm-tall object is 7.5 cm in front of a diverging lens that
has a cm focal length. Use ray tracing to find the position and
height of the image. To do this accurately, use a ruler or paper
28 with a grid. Determine the image distance and image height by
making measurements on your diagram.
71. || A 1.5-cm-tall object is 90 cm in front of a diverging lens that
has a cm focal length. Use ray tracing to find the position and
height of the image. To do this accurately, use a ruler or paper
with a grid. Determine the image distance and image height by
making measurements on your diagram.
72. || The moon is
in diameter and
from
the earth’s surface. The 1.2-m-focal-length concave mirror of a
telescope focuses an image of the moon onto a detector. What is
the diameter of the moon’s image?
73. ||| A 2.0-cm-tall candle flame is 2.0 m from a wall. You happen
to have a lens with a focal length of 32 cm. How many places
can you put the lens to form a well-focused image of the candle
flame on the wall? For each location, what are the height and
orientation of the image?
74. || A 2.0-cm-diameter spider is 2.0 m from a wall. Determine the
focal length and position (measured from the wall) of a lens that
will make a half-size image of the spider on the wall.
75. ||| Figure P18.75 shows a meter stick held lengthwise along the
optical axis of a concave mirror. How long is the image of the
29 meter stick?
f = 40 cm
Meter stick
100 cm
FIGURE P18.75
60 cm
76. || A slide projector needs to create a 98-cm-high image of a
2.0-cm-tall slide. The screen is 300 cm from the slide.
a. What focal length does the lens need? Assume that it is a
thin lens.
b. How far should you place the lens from the slide?
NIG9721_03_chap_18.indd 598
30
Dispersion
Blue bends more
31
Red and blue light
enters a fiber. The
light is transmitted
to the end. Which
gets there first?
A. The red
B. The blue
32
How Many Colors?
Pictures show about 4.
Knight, Jones, Field: 6
Color sensors in the eye: 3
33
Primary / Complementary Colors
Magenta
“Not Green”
Green
Red
Cyan
“Not Red”
Blue
Primary
Yellow
“Not Blue”
Complementary
(Secondary)
34
Recall: The Index of Refraction
v=
c
n
35
Just Checking.
Light of wavelength 500 nm in air enters a glass block
with index of refraction n=1.5. When the light enters the
block, which of the following properties of the light do
not change?
speed
frequency
wavelength
color
36
What Determines Color?
The wavelength of a light wave is 700 nm in air; this light
appears red. If this wave enters a pool of water, its
wavelength becomes:
! = !air 1.33 = 530 nm
If you are swimming underwater, this light still appears
red. Given this, what property of a light wave determines
its color?
37
Total Internal Reflection
38
Critical Angles
What are the critical angles for:
• Water (n=1.33)
• Diamond (n=2.42)
39
Mirages
There is no water in this picture.
Total internal reflection is the cause.
40
Making a Rainbow
41
Seeing a Rainbow
42
Scattering.
43
Cloud droplets are much bigger
than the wavelength of light, so
scatter all wavelengths of light
more or less equally.
Scattering from “Large” Objects
44
Skin is made of transparent
components, with differing
indices of refraction. The size
of the elements is a bit larger
than the wavelength of light.
Net result: No pigment
means white skin.
45
Scattering from Small Objects
Air molecules are very
small compared to the
wavelength of light.
Short wavelengths
scatter more.
46
Scattering from Small Objects
Rayleigh scattering
For small objects:
Scattering !
Blue Skies
1
"4
Red Sunsets
47
No Pigment, But Still Blue
Scattering from
melanocytes
Scattering from
particles in feathers
Scattering from
air molecules
48
Thin Film Interference
548
C H A P T E R 17 Wave Optics
FIGURE 17.16 Two reflections are visible
in the window, one from each surface.
17.4 Thin-Film Interference
In ◀◀ SECTION 16.6 you learned about the interference of sound waves in one dimension.
Depending on whether they are in phase or out of phase, two sound waves of the same
frequency, traveling in the same direction, can undergo constructive or destructive
interference. Light waves can also interfere in this way. Equal-frequency light waves
are produced when partial reflection at a boundary splits a light wave into a reflected
wave and a transmitted wave. The interference of light waves reflected from the two
boundaries of a thin film, such as the thin film of water that makes a soap bubble, is
called thin-film interference.
Thin-film interference has important applications in the optics industry. Thin-film
thick, are used for the antireflection coatings on
coatings, less than m
49 The bright colors of
the lenses in cameras, microscopes, and other optical equipment.
oil slicks and soap bubbles are also due to thin-film interference.
FIGURE 17.17 Reflected waves with and
without a phase change.
Interference of Reflected Light Waves
17.4 Thin-Film Interference
549
As you know, and as we discussed in Chapter 16, a light wave encountering a piece
1. If neither or both waves have a phase change due to reflection, the net addition to
of glass is partially transmitted and partially reflected. In fact, a light wave is partially reflected from any boundary between two transparent media with different
2. If only
one wave has a phase change due to reflection, the effective path-length
Incident wave
Transmitted
wave
indices of refraction. Thus light is partially reflected not only from the front surface
difference is increased by one half-wavelength to ∆d
2t
l.
of a sheet of glass, but from the back surface as well, as it exits from the glass into
The interference ofthe
the air.
twoThis
reflected
waves
is
then
constructive
if
leads to the two reflections seen in FIGUREl17.16and
.
m+2 l
extra distance
trav- for strings in Figdestructive if ∆deff =Another
important
of wavethereflections
wasisshown
film . Why aspectBecause
eled
so we
needof
to the
compare
2t to the If
wavelength
in the film.
Case
1: inside
n1 7 n2.the film,ure
16.10b
last chapter.
a wave moves
fromFurther,
a string with a higher wave
The
thereflected
film’s wave
index of speed
refraction
n, sowith
the wavelength
in the
filmthe
is lreflected
l wave
where
to a isstring
a lower wave
speed,
is inverted with respect
doesis
notthe
have
a
wavelength
theincoming
light in vacuum
toofthe
wave. Itorisair.
not inverted if the wave moves from a string with a lower
phase change.
Reflected
With this information,
we can
thewith
conditions
forwave
constructive
wave speed
to write
a string
a higher
speed. and destrucwaves
tive2:interference
of theThe
light
waves
reflected
by for
a thin
film:
Case
n1 6 n2.
same
thing
happens
light
waves. When a light wave moves from a medium
Boundary
Index n1
then2path-length difference is zero. The effective path-length difference is
Index
Conditions
The reflected wave
does have a phase
change.
with a higher light speed (lower index of refraction) to a medium with a lower light
l index of refraction), the reflected wave is inverted. This inversion of
speed (higher
2t = m
m = 0, 1, 2, c
(17.14)
the wave, ncalled a phase change, is equivalent to adding an extra half-wavelength
The reflection with the phase change is
to the distance the wave travels. You can see this in FIGURE 17.17 , where a reflected
for constructive interference with either 0 or 2 reflective phase changes
half a wavelength behind, so the Condition
effect
wave withinterference
a phase with
change
is compared to a reflection without a phase change. In
of the phase change is to increase theCondition for destructive
only 1 reflective phase change
summary, we can say that a light wave undergoes a phase change if it reflects from
path length by l/2.
a boundary at which the index of refraction increases. There’s no phase change at
a boundary1where
l the index of refraction decreases.
2t =Consider
am + ba thin, transparent
m = 0, 1, 2,film
c with thickness t and
(17.15)
FIGURE 17.18 In thin-film interference,
index of refraction n coated
2 n
two reflections, one from the film and
onto a piece of glass. FIGURE 17.18 shows a light wave of wavelength approaching
Condition
for destructive interference with either 0 or 2 reflective phase changes
one from the glass, overlap and
interfere.
the film. Most of the light is transmitted into the film, but, as we’ve seen, a bit is
Condition for constructive interference with only 1 reflective phase change
reflected off the first (air-film) boundary. Further, a bit of the wave that continues
Air
50 The two reflected
1. The incident wave is Thin film Glass
into the film is reflected off the second (film-glass) boundary.
Index
n
transmitted through the
waves,
have give
exactly
the same
frequency,
travel
back out into the air where
NOTE ▶ Equations
17.14which
and 17.15
the film
thicknesses
that yield
constructhin film and the glass.
theyinterference.
overlap andAtinterfere.
As we learned
in Chapter
16, the two reflected waves
tive or destructive
other thicknesses,
the waves
will interfere
will interfere
to cause
a strong
reflection
they are in phase (i.e., if
neither fully constructively
norconstructively
fully destructively,
and the
reflected
intensityif will
◀
fall somewhere their
between
these
two extremes.
crests
overlap).
If the two
reflected waves are out of phase, with the crests of
l
lfilm
one wave overlapping the troughs of the other, they will interfere destructively to
These conditions are the basis of a procedure to analyze thin-film interference.
Front
Rear reflection or, if their amplitudes are equal, no reflection at all.
cause a weak
2. Part of the incident 3. Part of the
reflection
wave
wave reflects from the transmitted
Wereflection
found the interference of two sound waves to be constructive if their pathTACTICS
reflects from the Analyzing thin-film interference
first surface.
BOX 17.1
and destructive if ∆d = m
l, where m is an intelength difference is
second surface.
ger. The same idea holds true for reflected light waves, for which the path-length
Follow the light wave as it passes through the film. The wave reflecting from
difference is the extra distance
traveled by the wave that reflects from the second
Cells
the second boundary travels an extra distance 2t.
surface. Because this wave travels
twice through a film of thickness t, the path-length
n=1.33
1. Note the indices
of refraction
of the three media: the medium before the
difference
is
4. The two reflected
film, the film itself,
and the
medium
beyond
film. when
The first
andwave
third reflects from a boundary
We noted
above
that the
phasethe
change
a light
waves overlap
and
t
may be the same.
reflective
phase
change
at any boundary
where
Thickness
with There’s
a higheraindex
of refraction
is equivalent
to adding
an extra half-wavelength to
Crystals
interfere.
the index of refraction
increases.
the distance
traveled. This leads
to two situations:
t
n=1.83
2. If neither or both reflected waves undergo a phase change, the phase changes cancel and the effective path-length difference is
. Use Equation
Cells
17.14 for constructive interference and 17.15
for destructive interference.
3. If only one wave undergoes a phase change,n=1.33
the effective path-length difference is d 2t
l. Use Equation 17.14 for destructive interference and
17.15 for constructive interference.
KNIG9721_03_chap_17.indd 548
l
2
Exercises 12, 13
Light of wavelength 600 nm in air passes into the layer of
Iridescent feathers
The gorgeous colors of the hummingbird shown at the
beginning
of thisof
chapter
are due not to pigments
but to interference.
crystals with the noted
index
refraction.
What
is theThis iridescence,
present in some bird feathers and insect shells, arises from biological structures whose
size
is
similar
to
the
wavelength
of
light.
The
sheen
of
an
insect,
for instance, is due to
wavelength of the light in this layer?
▶
thin-film interference from multiple thin layers in its shell. Peacock feathers are also a
layered structure, but each layer itself consists of nearly parallel rods of melanin, as
shown in the micrograph, that act as a diffraction grating. Thus a peacock feather
combines thin-film interference and grating-like diffraction to produce its characteristic
multicolored iridescent hues.
51
* 10,000
23/08/13 10:45 AM
Front
reflection
Rear
reflection
Cells
n=1.33
Thickness
t
Crystals
n=1.83
Cells
n=1.33
If the layer of crystals is 80 nm thick, what is the longest
wavelength for which there is constructive interference?
What is this wavelength in air?
52
Structure of the Scales.
53
598
C H A P T E R
Head Start
18
Ray Optics
Line of sight
54. ||| Figure P18.54 shows a
30°
meter stick lying on the bottom of a 100-cm-long tank
50 cm
with its zero mark against the
Zero
left edge. You look into the
Meter stick
tank at a 30° angle, with your
100 cm
line of sight just grazing the
upper left edge of the tank. FIGURE P18.54
What mark do you see on the meter stick if the tank is (a) empty,
(b) half full of water, and (c) completely full of water?
55. ||| You are standing in a 1.5-m-deep swimming pool at night. The
water is very still. You hold a laser pointer just above the water’s
surface and shine it nearly parallel to the surface, but tilted slightly
down so that the beam enters the water 5.0 m from you. How far
from you does the beam strike the bottom of the pool?
56. ||| What is the exit angle from the glass prism in Figure P18.56?
65. | A 2.0-cm-tall object is located 8.0 cm in front of a converging
lens with a focal length of 10 cm. Use ray tracing to determine
the location and height of the image. Is the image upright or
inverted? Is it real or virtual?
66. || You need to use a 24-cm-focal-length lens to produce an
inverted image twice the height of an object. At what distance
from the object should the lens be placed?
67. |||| A near-sighted person might correct his vision by wearing
diverging lenses with focal length
When wearing his glasses, he looks not at actual objects but at the virtual
images of those objects formed by his glasses. Suppose he looks
at a 12-cm-long pencil held vertically 2.0 m from his glasses.
Use ray tracing to determine the location and height of the
image.
68. | A 1.0-cm-tall object is 20 cm in front of a converging lens
54 that has a 10 cm focal length. Use ray tracing to find the posi-