Transcript Six

1
Today’s agendum:
Electromagnetic Waves.
Energy Carried by Electromagnetic Waves.
Momentum and Radiation Pressure of an Electromagnetic Wave.
2
We began this course by studying fields that didn’t vary with
time—the electric field due to static charges, and the magnetic
field due to a constant current.
In case you didn’t notice—about five lectures ago things
started moving!
We found that changing magnetic
field gives rise to an electric field.
Also a changing electric field gives
rise to a magnetic field.
These time-varying electric and magnetic fields can propagate
through space.
3
Electromagnetic Waves
Maxwell’s Equations
  q
enclosed
E

d
A


o
  d

B
E

d
s



dt
 
BdA0

d

E
B

d
s


I



oencl
o

dt
These four equations provide a complete description of
electromagnetism.

 E 
0
 B  0
dB
×E=dt
1d
E


B
=2 +
μ
J
0
cd
t
4
Production of Electromagnetic Waves
Apply a sinusoidal voltage to an antenna.
Charged particles in the antenna oscillate sinusoidally.
The accelerated charges produce sinusoidally varying electric
and magnetic fields, which extend throughout space.
The fields do not instantaneously permeate all space, but
propagate at the speed of light.
y
x
direction of
propagation
z
5
y
x
direction of
propagation
z
This static image doesn’t show how the wave propagates.
Here are a couple of animations, available on-line:
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=35
6
Electromagnetic waves are transverse waves, but are not
mechanical waves (they need no medium to vibrate in).
Therefore, electromagnetic waves can propagate in free space.
At any point, the magnitudes of E and B (of the wave shown)
depend only upon x and t, and not on y or z. A collection of
such waves is called a plane wave.
y
x
direction of
propagation
z
7
Manipulation of Maxwell’s equations leads to the following
plane wave equations for E and B:
2
2

E

E
(
x
,t
)
y
y
=


00
2
2

x

t
2
2

B

B
(
x
,t
)
z
z
=


00
2
2

x

t
These equations have solutions:
E
E
s
in
x
-
t
k
y=
m
a
x
B
B
s
in
x
-
t
k
z=
m
a
x
where
2


k
=
,
=
2

f
,a
n
d
f

=
=
c
.

k
You can verify this by direct substitution.
Emax and Bmax in these notes are sometimes written by others as E0 and B0.
8
You can also show that
Ey
Bz
=x
t
E
k
c
o
s
k
x

t
=
B

c
o
s
k
x

t




m
a
x
m
a
x
E
1
m
a
x E
=
=
=
c
= .
B


m
a
x Bk
0
0
At every instant, the ratio of the magnitude of the electric field
to the magnitude of the magnetic field in an electromagnetic
wave equals the speed of light.
9
Summary of Important Properties of Electromagnetic Waves
The solutions of Maxwell’s equations are wave-like with both E
and B satisfying a wave equation.
E
E
s
in
x
-
t
k
y=
m
a
x
B
B
s
in
x
-
t
k
z=
m
a
x
Electromagnetic waves travel through empty space with the
speed of light c = 1/(00)½.
Emax and Bmax are the electric and magnetic field amplitudes.
10
Summary of Important Properties of Electromagnetic Waves
The components of the electric and magnetic fields of plane EM
waves are perpendicular to each other and perpendicular to the
direction of wave propagation. The latter property says that EM
waves are transverse waves.
y
x
direction of
propagation
z
The magnitudes of E and B in empty space are related by
E/B = c.
E
E 
m
a
x
== =
c
B
B k
m
a
x
11
Today’s agendum:
Electromagnetic Waves.
Energy Carried by Electromagnetic Waves.
Momentum and Radiation Pressure of an Electromagnetic Wave.
12
Energy Carried by Electromagnetic Waves
Electromagnetic waves carry energy, and as they propagate
through space they can transfer energy to objects in their path.
The rate of flow of energy in an electromagnetic wave is
described by a vector S, called the Poynting vector.*
1
S= EB
0
The magnitude S represents the rate at which energy flows
through a unit surface area perpendicular to the direction of
wave propagation.
Thus, S represents power per unit area. The direction of S is
along the direction of wave propagation. The units of S are
J/(s·m2) =W/m2.
13
*J. H. Poynting, 1884.
y
E
B
z
For an EM wave EB=EB
EB
so S = .
0
1
S= EB
0
S
c
x
Because B = E/c we can write
2
2
E
cB
S= = .


0c
0
These equations for S apply at any instant of time and
represent the instantaneous rate at which energy is passing
through a unit area.
14
2
2
E
B E
c
B
S
= = =

c 
0 
0
0
E
s
in
x
-
t
EM waves are sinusoidal. E
k
y=
m
a
x
B
B
s
in
x
-
t
k
z=
m
a
x
The average of S over one or more cycles is called the wave
intensity I.
The time average of sin2(kx - t) is ½, so
2
2
E
B
Ec
B
m
a
x
m
a
x m
a
x
m
a
x
I
=
S
=
S
=
=
=
a
v
e
r
a
g
e
2


c
2

0 2
0
0
15
Energy Density
The energy densities (energy per unit volume) associated
with electric field and magnetic fields are:
1B2
uB =
2 0
1 2
uE = 0E
2
Using B = E/c and c = 1/(00)½ we can write

2
E
2
2


E
1
B1
1
12
c
0
0
u
== = =

E
B
0
2



2
02
0 2
0
2
1 2 1
B
u
=
u
=

E
=
B
E
0
2
2

0
16
2
1 2 1
B
u
=
u
=

E
=
B
E
0
2
2

0
For an electromagnetic wave, the instantaneous energy density
associated with the magnetic field equals the instantaneous
energy density associated with the electric field.
Hence, in a given volume the energy is equally shared by the
two fields. The total energy density is equal to the sum of the
energy densities associated with the electric and magnetic
fields:
2
2 B
u
=
u
+
u
=

E
=
B
E
0

0
17
2
B
2
u
=
u
+
u
=

E
B
E
0 =

0
When we average this instantaneous energy density over one
or more cycles of an electromagnetic wave, we again get a
factor of ½ from the time average of sin2(kx - t).
2
1 2
1B
m
ax
u
=

E
,
u
=
, and
E
0m
a
x
B
4 0
4
2
B
1 2 1
m
a
x
u
=

E
=
0m
a
x
2
2
0
2
2
E
c
B
1
1
m
a
x
m
a
x
Recall S
=
S
=
=
so we see that S =c u .
a
v
e
r
a
g
e
2

c2

0
0
The intensity of an electromagnetic wave equals the average
energy density multiplied by the speed of light.
18
Example: a radio station on the surface of the earth radiates a
sinusoidal wave with an average total power of 50 kW.
Assuming the wave is radiated equally in all directions above
the ground, find the amplitude of the electric and magnetic
fields detected by a satellite 100 km from the antenna.
All the radiated power passes
through the hemispherical
surface* so the average power
per unit area (the intensity) is
Satellite
R
Station
5
.
0
0

1
0
W


p
o
w
e
r
P


I
=
=
=
7
.
9
6

1
0
W
m
 =
a
r
e
a

R

 2
2

1
.
0
0

1
0
m


4
7 2
a
v
e
r
a
g
e
2
2
5
*In problems like this you need to ask whether the power
is radiated into all space or into just part of space.
19
2
1E
a
x
I=S= m
2
c
0
Satellite
R
Emax= 2

I
0c
Station
7
8
7
=
2
4


1
0
3

1
0
7
.
9
6

1
0


 


-2V
=
2
.4
51
0
m


2
V
2
.
4
5

1
0
E
m
1
1
m
a
x
B
=
=
=
8
.
1
7

1
0
T
m
a
x
8
c 
3

1
0
m
s

20
Example: for the radio station in the example on the previous
two slides, calculate the average energy densities associated
with the electric and magnetic field.
1 2
u
E
E = 
0
m
ax
4
2
1Bm
ax
uB =
4 0
2
1
1
2
2
u
=
8
.
8
5

1
02
.
4
5

1
0




E
4
8
.
1
7

1
0

1
u=
B
7
4 4


1
0

J
u=
.
3
3

1
0 3
E 1
m
J
u=
.
3
3

1
0 3
B 1
m
1
5
1
12
1
5
21
Today’s agendum:
Electromagnetic Waves.
Energy Carried by Electromagnetic Waves.
Momentum and Radiation Pressure of an
Electromagnetic Wave.
22
Momentum and Radiation Pressure
EM waves carry linear momentum as well as energy. When
this momentum is absorbed at a surface pressure is exerted on
that surface.
If we assume that EM radiation is incident on an object for a
time t and that the radiation is entirely absorbed by the
object, then the object gains energy U in time t.
Maxwell showed that the momentum
incident
change of the object is then:

U

p
= (
t
o
t
a
la
b
s
o
r
p
t
i
o
n
)
c
The direction of the momentum change of the object is in the
direction of the incident radiation.
23
If instead of being totally absorbed the radiation is totally
reflected by the object, and the reflection is along the incident
path, then the magnitude of the momentum change of the
object is twice that for total absorption.
incident
reflected
2

U

p
=
(
t
o
t
a
l
r
e
f
l
e
c
t
i
o
n
a
l
o
n
g
i
n
c
i
d
e
n
t
p
a
t
h
)
c
The direction of the momentum change of the object is again
in the direction of the incident radiation.
24
Radiation Pressure
The radiation pressure on the object is defined as the force per
unit area:
F
P=
A
From Newton’s
2nd
F 1dp
P
=
=
Law (F = dp/dt) we have:
A Adt
For total absorption, p=
U
c
d
U

S
1
d
p1
d
U1

t
d

= = 
=
=
So P
A
d
tA
d
tc
A
c
c
 
incident
(Equations on this slide involve magnitudes of vector quantities.)
25
This is the instantaneous radiation pressure in the case of total
absorption:
S
P=
c
For the average radiation pressure, replace S by <S>=Savg=I:
S
I
a
v
e
ra
g
e
P
=
ra
d=
c
c
26
I
P
=(
t
o
t
a
la
b
s
o
r
p
t
i
o
n
)
r
a
d
c
incident
Following similar arguments it can be shown that:
2
I
P
=(
t
o
t
a
lr
e
f
l
e
c
t
i
o
n
)
r
a
d
c
incident
reflected
27
Example: a satellite orbiting the earth has solar energy
collection panels with a total area of 4.0 m2. If the sun’s
radiation is incident perpendicular to the panels and is
completely absorbed find the average solar power absorbed
and the average force associated with the radiation pressure.
The intensity (I or Saverage) of sunlight prior to passing
through the earth’s atmosphere is 1.4 kW/m2.


3
2
3
W
P
o
w
e
r
=
I
A
=
1
.
4

1
04
.
0
m
=
5
.
6

1
0
W
=
5
.
6
k
W


2
m
Assuming total absorption of the radiation:


3
W
Caution! The letter P
1
.
4

1
0
2
S
a
v
e
r
a
g
eI
6
m
(or p) has been used
P
=
=
=
=
4
.
7

1
0
P
a
r
a
d
in this lecture for
8
m
c c 3

1
0
power, pressure, and
s




momentum!
6
2
5
N
F
=
P
A
=
4
.
7

1
0
4
.
0
m
=
1
.
9

1
0
N


2
r
a
d
m
28
Today’s agendum:
Introduction to Light.
You must develop a general understanding of what light is and how it behaves.
Reflection and Refraction (Snell’s “Law”).
You must be able to determine the path of light rays using the laws of reflection and refraction.
Total Internal Reflection and Fiber Optics.
You must be able to determine the conditions under which total internal reflection occurs, and
apply total internal reflection to fiber optic and similar materials.
Dispersion.
You must understand that the index of refraction of a material is wavelength-dependent.
29
Light
Normally, “light” refers to the type of electromagnetic wave
that stimulates the retina of our eyes.
Light acts like a wave except when it acts like particles.
30
*Light—Waves or Particles?
http://www.nearingzero.net (quantum007.jpg)
*Both! Take Physics 203 for further enlightenment!
31
The Speed of Light
Light is a type of electromagnetic wave and travels with the
speed c = 2.9979x108 m/s in a vacuum. (Just use 3x108!)
How many physicists does it take to change a light bulb?
Eleven. One to do it and ten to co-author the paper.
32
Visible light is a small part of the electromagnetic spectrum.
33
Geometric Optics
Although light is actually an electromagnetic wave, it generally
travels in straight lines (like particles do!).
We can describe many properties of light by assuming that it
travels in straight-line paths in the form of rays.
A ray is a straight line along which light is propagated. In
other contexts, the definition of ray might be extended to
include bent or curved lines.
http://www.nearingzero.net (ray.jpg)
34
A light ray is an infinitely thin beam of light. Of course, there
really isn’t such a thing, but the concept helps us visualize
properties of light.
there really isn’t such a thing
http://www.nearingzero.net (rays.jpg)
35
Light rays from some
external source strike an
object and reflect off it in all
directions.
We only see those light rays
that reflect in the direction
of our eyes.
If you can see something, it
must be reflecting light!
Zillions* of rays are simultaneously reflected in all directions
from any point of an object. Later, when we study mirrors and
lenses, we won’t try do draw them all! Just enough
representative ones to understand what the light is doing.
*one zillion = 10a big number
36
Today’s agendum:
Introduction to Light.
You must develop a general understanding of what light is and how it behaves.
Reflection and Refraction (Snell’s “Law”).
You must be able to determine the path of light rays using the laws of reflection and refraction.
Total Internal Reflection and Fiber Optics.
You must be able to determine the conditions under which total internal reflection occurs, and
apply total internal reflection to fiber optic and similar materials.
Dispersion.
You must understand that the index of refraction of a material is wavelength-dependent.
37
Reflection
Light striking a surface may be reflected, transmitted, or
absorbed. Reflected light leaves the surface at the same angle
it was incident on the surface:
i r
Real Important Note: the angles are measured relative to the surface normal.
38
Reflection from a
smooth surface is
specular (mirrorlike). Reflection
from a rough
surface is diffuse
(not mirror-like).
http://acept.la.asu.edu/PiN/rdg/reflection/reflection.shtml
http://www.mic-d.com/java/specular/
39
Refraction
Light travels in a straight line
except when it is reflected or
when it moves from one
medium to another.
http://id.mind.net/~zona/mstm/physics/light/rayOptics/refraction/refraction1.html
Refraction—the “bending” of light rays when light moves from
one medium to a different one—takes place because light
travels with different speeds in different media.
40
The speed of light in a vacuum is c = 3x108 m/s. The index of
refraction of a material is defined by
c
n=
,
v
If you study light in advanced classes,
you’ll find it is more complex than this.
where c is the speed of light in a vacuum and v is the speed of
light in the material.
The speed and wavelength of light change when it passes
from one medium to another, but not the frequency, so
c

v
=a
n
d
=.
n
n
n
41
Because light never travels faster than c, n  1. For water, n =
1.33 and for glass, n 1.5. Indices of refraction for several
materials are listed in your text.
Example: calculate the speed of light in diamond (n = 2.42).
c
v =
n
8
3
×
1
0
m
/
s
v=
2
.4
2
8
v=
1
.
2
4
×
1
0
m
/
s
42
Snell’s “Law”
When light moves from one medium into another, some is
reflected at the boundary, and some is transmitted.
The transmitted light is refracted (“bent”).
a is the angle of incidence, and b is the angle of refraction.
a
b
air (na)
water (nb)
water (na)
b
nb>na
air (nb)
a
na>nb
43
Light passing from air (n  1) into water (n  1.33).
Light “bends” towards the normal to the surface as it slows
down in water.
a
air (na)
water (nb)
b
nb>na
44
Light passing from water (n  1.33) into air (n  1).
Light “bends” away from the normal to the surface as it
speeds up in air.
b
air (nb)
water (na)
a
na>nb
45
Snell’s “Law”, also called the law of refraction, gives the
relationship between angles and indices of refraction:
n
s
i
n
θ
n
s
i
n
θ

=

.
a
a
b
b
air (na)
air (na)
a
a
b
water (nb)
b
water (nb)
You are free to choose which is “a” and which is “b.”
 is the angle the ray makes with the normal!
46
Today’s agendum:
Introduction to Light.
You must develop a general understanding of what light is and how it behaves.
Reflection and Refraction (Snell’s “Law”).
You must be able to determine the path of light rays using the laws of reflection and refraction.
Total Internal Reflection and Fiber Optics.
You must be able to determine the conditions under which total internal reflection occurs, and
apply total internal reflection to fiber optic and similar materials.
Dispersion.
You must understand that the index of refraction of a material is wavelength-dependent.
47
Total Internal Reflection; Fiber Optics
n
s
in
n
s
in
θ
θ
=


1
1
2
2
n
2
s
in
=
s
in
θ

θ

1
2
n
1
Suppose n2<n1. The largest possible value of sin(2) is 1
(when 2 = 90). The largest possible value of sin(1) is
n
2
s
in
=
.
θ

1
,m
a
x
n
1
For 1 larger than this, Snell’s
Law cannot be satisfied!
This value of  is called the critical angle, C. For any angle of
incidence larger than C, all of the light incident at an interface
is reflected, and none is transmitted.
48
1 < C
1 close to C
1
1 > C
Another
visualization
here.
49
n2

n1>n2
Ray incident normal to surface is not “bent.” Some is reflected,
some is transmitted.
50
n2

n1>n2
Increasing angle of incidence…
51
n2

n1>n2
Increasing angle of incidence…more…
52
n2

n1>n2
Increasing angle of incidence…more…critical angle reached…
some of incident energy is reflected, some is “transmitted along
the boundary layer.
53
n2

n1>n2
Light incident at any angle beyond C is totally internally
reflected.
54
application: fiber optics
http://laser.physics.sunysb.edu/~wise/wise187/janfeb2001/reports/andr
55
ea/report.html
Example: determine the incident angle i for which light strikes
the inner surface of a fiber optic cable at the critical angle.
Light is incident at
an angle i on a
transparent fiber.
i
f
nf>1
ni=1 (air)
The light refracts at an angle f.
n
in
n
in
θ
θ
is
i=
fs
f
s
in
n
in
θ
θ
i=
fs
f
56
Light strikes the fiber
wall an an angle of
90-f normal to the
surface.
i
f
90-f
90
nf>1
ni=1 (air)
At the critical angle, instead of exiting the fiber, the refracted
light travels along the fiber-air boundary. In this case, 90-f is
the critical angle.
n
s
i
n
9
0
θ
=
n
s
i
n
θ
=
n
s
i
n
9
0

1






f
f
f
c
a
1
s
in
0
-θ
9
f=
n
f
Solve the above for f and use s
in
n
in
θ
θ
i=
fs
fto solve for
i.
57
application: swimming underwater
If you are looking up from underwater, if your angle of sight
(relative to the normal to the surface) is too large, you see an
underwater reflection instead of what’s above the water.
58
application: perfect mirrors
(used in binoculars)
application: diamonds
59
Today’s agendum:
Introduction to Light.
You must develop a general understanding of what light is and how it behaves.
Reflection and Refraction (Snell’s “Law”).
You must be able to determine the path of light rays using the laws of reflection and refraction.
Total Internal Reflection and Fiber Optics.
You must be able to determine the conditions under which total internal reflection occurs, and
apply total internal reflection to fiber optic and similar materials.
Dispersion.
You must understand that the index of refraction of a material is wavelength-dependent.
60
Dispersion
We’ve treated the index of refraction of a material as if it had a
single value for all wavelengths of light.
In fact, the index of refraction is generally wavelength- (or
color-) dependent. When white light passes from air
into glass, the different colours are refracted by different
angles, and therefore spread out, or are dispersed.
It is observed that the shorter the wavelength of the light, the
greater is the refraction.
61
62
Picture from the Exploratorium (http://www.exploratorium.edu/).
Today’s agendum:
Plane Mirrors.
You must be able to draw ray diagrams for plane mirrors, and be able to calculate image and
object heights, distances, and magnifications.
Spherical Mirrors: concave and convex mirrors.
You must understand the differences between these two kinds of mirrors, be able to draw ray
diagrams for both kinds of mirrors, and be able to solve the mirror equation for both kinds of
mirrors.
63
Mirrors
Images Formed by Plane Mirrors
Plane mirrors form virtual images; no light actually comes
from the image. The solid red rays show the actual light path
after reflection; the dashed black rays show the perceived
64
light path.
y
y’
s
s’
*The object distance and image distance are equal: s=-s’.
The object height and image height are equal: y=y’. The
magnification of a plane mirror is therefore one.
The image is upright and virtual.
The image is reversed front-to-back relative to the object.
*The – sign is needed because of sign conventions—see later.
65
Example: how tall must a full-length mirror be?
s
s’
A light ray from the top of your head reflects directly back from
the top of the mirror.
66
y/2


y/2
s

s’
To reach your eye, a light ray from your foot must reflect
halfway up the mirror (because I = R = ).
67
y/2


y/2
s

s’
The mirror needs to be only half as tall as you.
This calculation assumed your eyes are at the top of your head.
68
Today’s agendum:
Plane Mirrors.
You must be able to draw ray diagrams for plane mirrors, and be able to calculate image and
object heights, distances, and magnifications.
Spherical Mirrors: concave and convex mirrors.
You must understand the differences between these two kinds of mirrors, be able to draw ray
diagrams for both kinds of mirrors, and be able to solve the mirror equation for both kinds of
mirrors.
69
Images Formed by Spherical Mirrors
Spherical mirrors are made from polished
sections cut from a spherical surface.
The center of curvature, C, is the
center of the sphere, of which the
mirror is a section.
C
Of course, you don’t really make these mirrors by cutting out part of a sphere of glass.
70
The radius of curvature, R, is the radius of the sphere, or
the distance from V to C.
R
C
V
71
The principal axis (or optical axis) is the line that passes
through the center of curvature and the center of the mirror.
Principal or Optical
Axis
R
C
The center of the mirror is often called the vertex of the mirror.
V
72
Paraxial rays are parallel to the principal axis of the mirror
(from an object infinitely far away). Reflected paraxial rays
pass through a common point known as the focal point F.
C
F
V
73
The focal length f is the distance from P to F. Your text shows
that f = R/2.
R
f
C
F
P
74
Reality check: paraxial rays don’t really pass exactly through
the focal point of a spherical mirror (“spherical aberration”).
C
F
V
75
If the mirror is small compared to its radius of curvature, or the
object being imaged is close to the principal axis, then the rays
essentially all focus at a single point.
C
F
V
We will assume mirrors with large radii of curvature and
objects close to the principal axis.
76
In “real life” you would minimize aberration by using a
parabolic mirror.
C
F
V
77
Today’s agendum:
Plane Mirrors.
You must be able to draw ray diagrams for plane mirrors, and be able to calculate image and
object heights, distances, and magnifications.
Spherical Mirrors: concave and convex mirrors.
You must understand the differences between these two kinds of mirrors, be able to draw ray
diagrams for both kinds of mirrors, and be able to solve the lens equation for both kinds of
mirrors.
78
Concave and Convex Mirrors
There are two kinds of spherical mirrors: concave and convex.
concave
F
convex
79
Ray Diagrams for Mirrors
We can use three “principal rays” to construct images. In
this example, the object is “outside” of F.
Ray 1 is parallel to the axis
and reflects through F.
Ray 2 passes through F
before reflecting parallel to
the axis.
C
F
Ray 3 passes through
C and reflects back on
itself.
80 2.
We’ll also use three for convex mirrors, but there will be a different version of ray
Ray Diagrams for Concave Mirrors
We use three “principal rays” to construct images.
An image is formed where the rays converge.
The image from a concave
mirror, object outside the
focal point, is real, inverted,
and smaller than the object.
C
F
Two rays would be enough to show us
where the image is. We include the third
ray for “safety.” You don’t have to use
principal rays, but they are easiest to trace.
“Real” image: you could put a camera there and detect the image.
81
The image from a concave mirror, object inside the focal point,
is virtual, upright, and larger than the object.
Ray 1: parallel to the axis
then through F.
Ray 2: “through” F then parallel
to the axis.
Ray 3: “through” C.
C
F
With this size object, there was a bit of spherical
82 C a
aberration present, and I had to “cheat” my
bit to the left to make the diagram look “nice.”
You could show that if an object is placed at the focal point,
reflected rays all emerge parallel, and *no image is formed.
Ray 1: parallel to the axis
then through F.
Ray 2: “through” F then parallel
to the axis. Can’t do!
Ray 3: through C.
no image
C
Worth thinking about: what if the
object is placed between F and C?
F
83
*Actually, the image is formed at infinity.
The Mirror Equation
With a bit of geometry, you can show that
1 1 1
+ =
s s' f
The magnification is the ratio
of the image to the object
y
height:
y'
s'
m= =y
s
s
f
C y’
F
s’
84
1 1 1
+ =
s s' f
y'
s'
m= =y
s
Sign conventions for the mirror equation:
When the object, image, or
focal point is on the reflecting
side of the mirror, the
distance is positive.
When the object, image, or
focal point is “behind” the
mirror, the distance is
negative.
s
f
y
C y’
F
The image height is positive if the image is upright, and
negative if the image is inverted relative to the object.
s’
85
Example: a dime (height is 1.8 cm) is placed 100 cm away
from a concave mirror. The image height is 0.9 cm and the
image is inverted. What is the focal length of the mirror.
1 1 1
+ =
s s' f
y'
s'
m= =y
s
s
f
y
C y’
F
s’
s’, s, or f on reflecting side are +
y is – if image is inverted
86
Applications of concave mirrors.
Shaving mirrors.
Makeup mirrors.
Solar cookers.
Flashlights, headlamps, stove reflectors.
Satellite dishes (when used with electromagnetic radiation).
87
Today’s agendum:
Plane Mirrors.
You must be able to draw ray diagrams for plane mirrors, and be able to calculate image and
object heights, distances, and magnifications.
Spherical Mirrors: concave and convex mirrors.
You must understand the differences between these two kinds of mirrors, be able to draw ray
diagrams for both kinds of mirrors, and be able to solve the lens equation for both kinds of
mirrors.
88
Ray Diagrams for Convex Mirrors
Ray 1: parallel to the axis
then through F.
Ray 2: “through” Vertex.
Ray 3: “through” C.
C
F
The image is virtual, upright, and smaller than the object.
89
Instead of sending ray 2 “through” V, we could have sent it
“through” F. The ray is reflected parallel to the principal axis.
F
Your text talks about all four of the “principal rays” we have
90
used.
C
The mirror equation still works for convex mirrors.
1 1 1
+ =
s s' f
y'
s'
m= =y
s
s
y
s’
f
y’
F
Because they are on the “other” side of the mirror from the
91
object, s’ and f are negative.
C
Example: a convex rearview car mirror has a radius of
curvature of 40 cm. Determine the location of the image and
its magnification for an object 10 m from the mirror.
The ray diagram looks like the one on the previous slide, but
with the object much further away (difficult to draw).
1 1 1
+ =
s s' f
111 1 1
=
= 
s
' f s0
.
2
m
1
0
m
Not on reflecting
sidenegative.
On reflecting
sidepositive.
92
1
1
1
=

s
' 0
.2
m1
0
m
…algebra…
s
'=
0
.
1
9
6
m
=
1
9
.
6
c
m
s
' 0
.
1
9
6
m
1
m
=
-=
=
s
1
0
m5
1
Remind me… what does it say on passenger side rear view
mirrors?
93
Applications of convex mirrors.
Passenger side rear-view mirrors.
Grocery store aisle mirrors.
Railroad crossing mirrors.
Anti-shoplifting (surveillance) mirrors.
Christmas tree ornaments.
94
Sign Conventions Introduced Today
When the object, image, or focal point is on the reflecting
side of the mirror, the distance is positive.
When the object, image, or focal point is “behind” the mirror,
the distance is negative.
The image height is positive if the image is upright, and
negative if the image is inverted relative to the object.
95
Summary of Sign Conventions
Here’s a compact way of expressing mirror and lens
(coming soon) sign conventions all at once.
Object Distance. When the object is on the same side as
the incoming light, the object distance is positive (otherwise
is negative).
Image Distance. When the image is on the same side as
the outgoing light, the image distance is positive (otherwise
is negative).
Radius of Curvature. When the center of curvature C is on
the same side as the outgoing light, R is positive (otherwise
is negative).
96
Today’s agendum:
Refraction at Spherical Surfaces.
You must be able to calculate properties of images formed by refraction at spherical surfaces.
Thin Lenses: Concave and Convex Lenses, Ray Diagrams,
Solving the Lens Equation.
You must understand the differences between these two kinds of lenses, be able to draw ray
diagrams for both kinds of lenses, and be able to solve the lens equation for both kinds of
lenses.
If Time Allows: Lens Combinations, Optical Instruments.
You should be aware of this useful information.
97
Refraction at Spherical Surfaces
Convex surface:
1
R
2
axis
C
F
f
na nb>na
Geometry: a light ray parallel to
the axis passes through F.
 n

b
f=
>
R

R
n
ba
n
98
An extended object will form an image inside the nb medium.
1
Ray 1: parallel to the axis, through F.
Ray 3: through C.
R
2
axis
C
s
F
f
na nb>na
s’
This image is real and inverted.
99
Concave surface:
R
F
C
axis
f
na
Geometry: a light ray parallel to
the axis seems to have come
from F.
nb>na
 n

b
f=
>
R

R
n
ba
n
100
An extended object will form an image inside the na medium.
Ray 1: parallel to the axis, through F.
Ray 3: through C.
R
F
C
axis
f
na
nb>na
The image is virtual and upright.
There are three different places to put the object. The different images
101
formed are always virtual and upright.
We can use geometry to derive an equation relating the image
and source positions, and an equation for the magnification.
axis
C
R
s
F
f
na
nb
n
n-n
a n
+b=b a
s s
'
R
s’
y' ns
a'
m
= =
y ns
b
102
The same equations work for concave surfaces.
R
F
C
s
f
s’
na
n
n-n
a n
+b=b a
s s
'
R
axis
nb
y' ns
a'
m
= =
y ns
b
103
Approximations Were Used!
The equations in this section are excellent approximations if
both the angles of incidence and refraction are small.
104
Sign Conventions
 R is positive when it is in the medium into which the light
propagates. R is negative when it is in the medium from which
the light radiates.
 The image distance is positive when the image is in the
medium into which the light propagates, and negative if it is in
the medium from which the light radiates (virtual image).
 The object distance is positive when the object is in the
medium from which the light radiates (the usual case—a real
object), and negative if on the side opposite to the light
source (a virtual object).
These are really “the same” as for mirrors.
105
Example: a Jurassic mosquito is discovered embedded in an
amber sphere which has an index of refraction of 1.6. The
radius of curvature of the sphere is 3.0 mm. The mosquito is
located on the principal axis and appears to be imbedded 5.0
mm into the amber. How deep is the mosquito really?
n
na n
b
b n
+ = a
s s
'
R
The object is in the
amber, so na=1.6 and
nb=1.
The image is in the medium
from which the light radiates
so s’=-5.0 mm.
nb=1
na=1.6
R
s
s’
Notice the reversed orientation. . .
106
R is negative because it is in the medium from which the
light radiates. R=-3.0 mm.
1
.6 1 1
-1
.6
+ =
s 5
.0 3
s=4 mm
nb=1
na=1.6
R
s
s’
107
Today’s agendum:
Refraction at Spherical Surfaces.
You must be able to calculate properties of images formed by refraction at spherical surfaces.
Thin Lenses: Concave and Convex Lenses, Ray Diagrams,
Solving the Lens Equation.
You must understand the differences between these two kinds of lenses, be able to draw ray
diagrams for both kinds of lenses, and be able to solve the lens equation for both kinds of
lenses.
If Time Allows: Lens Combinations, Optical Instruments.
You should be aware of this useful information.
108
Thin Lenses
A lens in this section is taken to be a single object made of
transparent material of refractive material n>1.
There are two surface boundaries.
Light from an object incident on
the first surface forms an image,
which becomes the object for the
second surface.
A thin lens is one for which the
distance from the object to each of
the two surfaces is the “same”
(and the distance from the image
to each surface is the “same”).
This would NOT
qualify as a thin lens.
109
Until I figure out how to use Powerpoint to
fill in the lens color, I will make my lenses
look “hollow,” like this.
There are several surface combinations from
which we can make lenses. Here are three
(there are more).
110
Converging and Diverging Lenses
Thin lenses can be converging or diverging.
The converging lens is thicker in the center. The diverging
lens is thicker at the edges.
There are focal points on both sides of each lens. The focal
length is the same whether light passes from left to right or
right to left.
111
There are two surfaces at which light
refracts. Our equations (provided later)
“automatically” take care of this.
In your diagrams, simply draw the incident
ray up to the center of the lens, then draw
the refracted ray in its final direction.
112
Today’s agendum:
Refraction at Spherical Surfaces.
You must be able to calculate properties of images formed by refraction at spherical surfaces.
Thin Lenses: Concave and Convex Lenses, Ray Diagrams,
Solving the Lens Equation.
You must understand the differences between these two kinds of lenses, be able to draw ray
diagrams for both kinds of lenses, and be able to solve the lens equation for both kinds of
lenses.
If Time Allows: Lens Combinations, Optical Instruments.
You should be aware of this useful information.
113
Ray Diagrams for Converging Lenses
Ray 1 is parallel to the axis and refracts through F.
Ray 2 passes through F’ before refracting parallel to the axis.
Ray 3 passes straight through the center of the lens.
O
F’
F
I
The image is real and inverted. In this case, it is larger than
the object.
114
Ray Diagrams for Diverging Lenses
Ray 1 is parallel to the axis and refracts as if through F.
Ray 2 heads towards F’ before refracting parallel to the axis.
Ray 3 passes straight through the center of the lens.
O
F
I
F’
The image is virtual and upright. It is smaller than the object.
115
Converging and Diverging Lenses
The image formed by a converging lens may be real, inverted,
and either smaller or larger than the object. It may also be
virtual, upright, and larger than the object. See this web page.
The image formed by a diverging lens is always virtual,
upright, and smaller than the object. See this web page.
Do these lens properties remind you of anything you’ve
studied recently?
116
Today’s agendum:
Refraction at Spherical Surfaces.
You must be able to calculate properties of images formed by refraction at spherical surfaces.
Thin Lenses: Concave and Convex Lenses, Ray Diagrams,
Solving the Lens Equation.
You must understand the differences between these two kinds of lenses, be able to draw ray
diagrams for both kinds of lenses, and be able to solve the lens equation for both kinds of
lenses.
If Time Allows: Lens Combinations, Optical Instruments.
You should be aware of this useful information.
117
The Lensmaker’s Equation
s’
s
s’
1 1

11
+
=
n
1

  
ss
'
RR
a
b


The Lensmaker’s Equation
s
1 1
1
=
1
n
  
f
a R
b
R
1 1 1
+ =
s s' f
y'
s'
M= =y 118s
Sign Conventions for The Lens Equation
1 1 1
+ =
s s' f
y'
s'
M= =y
s
The focal length f is positive for converging lenses and
negative for diverging lenses.
The object distance s is positive if the object is on the side of
the lens from which the light is coming; otherwise s is
negative.
The image distance s’ is positive if the image is on the
opposite side of the lens from where the light is coming;
otherwise s’ is negative.
The image height y’ is positive if the image is upright and
negative if the image is inverted relative to the object.
119
Example: an object is located 5 cm in front of a converging
lens of 10 cm focal length. Find the image distance and
magnification. Is the image real or virtual?
It’s just a coincidence that
the image is located at F’.
O
F’
11
111 1
=
-= - =
s
' fs+
1
0
+
51
0
F
s
' 1
0
M
=
- =
- =
2
s
5
Image distance is 10 cm, image is on side of lens light is
coming from, so image is virtual. M=2 so image is upright. 120
Today’s agendum:
Refraction at Spherical Surfaces.
You must be able to calculate properties of images formed by refraction at spherical surfaces.
Thin Lenses: Concave and Convex Lenses, Ray Diagrams,
Solving the Lens Equation.
You must understand the differences between these two kinds of lenses, be able to draw ray
diagrams for both kinds of lenses, and be able to solve the lens equation for both kinds of
lenses.
If Time Allows: Lens Combinations, Optical Instruments.
You should be aware of this useful information.
121
Lens Combinations
To determine the image formed by a combination of two
lenses, simply...
…calculate the image formed by the first lens…
…then use the first lens image as the source (object) for the
second lens.
There is no homework on lens combinations.
122
Optical Instruments
A Simple Magnifier
h

O
 25 cm (near point)
´
h
F O
I
p
q
Magnifier
123
Refracting Telescope
For viewing very far objects. Object distance taken as infinity.

h
tan
o 
o
fo
h
tan
fe

h
fe
fo

M
 


h
o
fe
fo
124
Terrestrial Telescopes
For producing upright images:
Galilean telescope
Field-lens telescope
125
Reflecting Telescope
Newtonian-focus reflecting telescope
126
Compound Microscope
Again has objective and eyepiece, but because it is for viewing very
near objects it is very different from the telescope.
Objective magnification:
h
f
q l
i
m



 e
o
h
p
p
o
Eyepiece magnification:
.25
Me 
fe
Overall magnification:


.
lf

.
2
5
2
5
l
e
M

M
m





 
e
o
f
p
f
 f
e
e
o

127
Summary of Sign Conventions
Mirrors
Lenses
The focal length f is positive for
converging lenses and negative for
diverging lenses.
When the object, image, or focal
point is on the reflecting side of the
mirror, the distance is positive.
The object distance s is positive if the
object is on the side of the lens from
which the light is coming; otherwise s is
negative (and the object is virtual).
When the object, image, or focal
point is “behind” the mirror, the
distance is negative.
The image distance s’ and radius of
curvature R are positive if the image is
on the side of the lens into which the
light is going; otherwise negative.
The image height is positive if the
image is upright, and negative if the
image is inverted relative to the object.
The image height is positive if the
image is upright, and negative if the
image is inverted relative to the object.
128
Summary of Sign Conventions
Here’s a more compact way of expressing the sign
conventions all at once.
Object Distance. When the object is on the same side as
the incoming light, the object distance is positive (otherwise
is negative).
Image Distance. When the image is on the same side as
the outgoing light, the image distance is positive (otherwise
is negative).
Radius of Curvature. When the center of curvature C is on
the same side as the outgoing light, R is positive (otherwise
is negative).
129
Today’s agendum:
Review of Waves.
You are expected to recall facts about waves from Physics 103.
Young’s Double Slit Experiment.
You must understand how the double slit experiment produces an interference pattern.
Conditions for Interference in the Double Slit Experiment.
You must be able to calculate the conditions for constructive and destructive interference in the
double slit experiment.
Intensity in the Double Slit Experiment.
You must be able to calculate intensities in the double slit experiment.
130
Interference
Review of Waves
This section is a review of material you learned in your
previous physics course (perhaps Physics 103).
Consider a wave described by
y
y
(
x
,
t
)

A
s
i
n
(
k
x

ω
t
)
.
x
The phase of this wave is
θ
(
x
,
t
)k

xω
t
.
d
θ d
x
Also

k 
ω
.
d
t
d
t
131
If  is constant with time (i.e., d/dt=0), then we are moving
with the wave, and
dx 
 .
dt k
The phase velocity, vp, is given by
y
ω
vp  .
k
x
Imagine yourself riding on any point on this
wave. The point you are riding moves to the
right. The velocity it moves at is vp.
If the wave is moving from left to right then /k must be positive.
132
Superposition—a Characteristic of All Waves
When waves of the same nature travel past some point at the
same time, the amplitude at that point is the sum of the
amplitudes of all the waves
The amplitude of the electric field at a point is found by adding
the instantaneous amplitudes, including the phase, of all
electric waves at that point.
In Physics 103 you may have learned that power (or intensity)
is proportional to amplitude squared. The intensity of the
superposed waves is proportional to the square of the
amplitude of the resulting sum of waves.
133
Interference—a Result of the Superposition of Waves
Constructive Interference: If the waves are in phase, they
reinforce to produce a wave of greater amplitude.
Destructive Interference: If the waves are out of phase,
they reinforce to produce a wave of reduced amplitude.
134
Today’s agendum:
Review of Waves.
You are expected to recall facts about waves from Physics 103.
Young’s Double Slit Experiment.
You must understand how the double slit experiment produces an interference pattern.
Conditions for Interference in the Double Slit Experiment.
You must be able to calculate the conditions for constructive and destructive interference in the
double slit experiment.
Intensity in the Double Slit Experiment.
You must be able to calculate intensities in the double slit experiment.
135
Young’s Double Slit Experiment
This experiment demonstrates the
wave nature of light.
Consider a single light source, and
two slits. Each slit acts as a
secondary source of light.
Light waves from secondary slits
interfere to produce alternating
maxima and minima in the
intensity.
Reference and “toys:” fsu magnet lab, colorado.
Interesting reading: the double slit experiment and quantum mechanics.
136
How does this work?
Light waves from the two slits arriving at the detection screen in
phase will interfere constructively and light waves arriving out of
phase will interfere destructively.
In phase—
constructive.
Out of phase—
destructive.
137
Today’s agendum:
Review of Waves.
You are expected to recall facts about waves from Physics 103.
Young’s Double Slit Experiment.
You must understand how the double slit experiment produces an interference pattern.
Conditions for Interference in the Double Slit
Experiment.
You must be able to calculate the conditions for constructive and destructive interference in the
double slit experiment.
Intensity in the Double Slit Experiment.
You must be able to calculate intensities in the double slit experiment.
138
Conditions for Interference
Sources must be monochromaticof a single wavelength.
Sources must be coherent-must maintain a constant phase
with respect to each other.
Here’s the geometry I will use
in succeeding diagrams.
139
For an infinitely distant* screen:
L1

L1
y
S1

L2


d
S2
L
y
tan 
R
R
P
d
L2
L = L2 –L1 = d sin 
*so that all the angles labeled
 are approximately equal
140
Constructive Interference:
L1

L

d
s
i
n


m

,
m
=
0
,
1
,

2
.
.
.


d
L2
L = L2 –L1 = d sin 
Destructive Interference:
1


L

d
s
i
n


m
+

,
m
=
0
,,

1

2
.
.
.
 
2
 
The parameter m is called the order of the interference
fringe. The central bright fringe at  = 0 (m = 0) is known
as the zeroth-order maximum. The first maximum on either
side (m = ±1) is called the first-order maximum.
141
y

R
t
a
n


R
s
i
n

L1

y
S1
L2
Bright fringes:
m


d
s
in


d
S2
L
y
tan 
R
P
m
d
y
y
R
R
m
d
R
Do not use the small-angle
approximation unless it is valid!
142
y

R
t
a
n


R
s
i
n

L1

y
S1
 1

m



d
s
in



 2

L2

d
S2
L
tan 
Dark fringes:
y
R
P
y
 1

m



d


R
 2


R
 1

y
 
m

d 2

R
Do not use the small-angle
approximation unless it is valid!
143
Example: a viewing screen is separated from the double-slit
source by 1.2 m. The distance between the two slits is 0.030
mm. The second-order bright fringe (m = 2) is 4.5 cm from
the center line. Determine the wavelength of the light.
y

R
t
a
n


R
s
i
n

Bright fringes:
S1
m


d
s
in

m
d

L1

y
L2

y
R
S2
yd
Rm
L
tan 
y
R
P
R
4
.
5

1
0
m
3
.
0

1
0
m







5
.
6

1
0
m

5
6
0
n
m
2
5
1
.
2
m
2





7
144
Example: a viewing screen is separated from the double-slit
source by 1.2 m. The distance between the two slits is 0.030
mm. The second-order bright fringe (m = 2) is 4.5 cm from
the center line. Find the distance between adjacent bright
fringes.
y

R
t
a
n


R
s
i
n

Bright fringes:
S1

m


d
s
in

y
m
d
R
y
L1
y
L2

S2
L
R
m
d
tan 
y
R
P
R
 
 
5
.
6

1
0
m
1
.
2
m

 2

R

R

R
y
y

m

1

m

 
2
.
2

1
0
m

2
.
2
c
m


m
+
1
m
5
145
d
d
d
3
.
0

1
0
m

7
Example: a viewing screen is separated from the double-slit
source by 1.2 m. The distance between the two slits is 0.030
mm. The second-order bright fringe (m = 2) is 4.5 cm from
the center line. Find the width of the bright fringes.
Define the bright fringe width to be
the distance between two adjacent
destructive minima.
y
 1

d
a
r
k
m



d
s
i
n


d
 
R
 2


R
 1

y

m



d
a
r
k
d
 2

S1
L1

y
L2

S2
L

R1

R
1

R




y
y

m

1


m





d
a
r
k
,
m
+
1
d
a
r
k
,
m
d
d
2
d
 2



5
.
6

1
0
m
1
.
2
m




yy

2
.
2
c
m
3
.
0

1
0
m


tan 
y
R
P
R

7
d
a
r
k
,
m
+
1
d
a
r
k
,
m
5
146
Today’s agendum:
Review of Waves.
You are expected to recall facts about waves from Physics 103.
Young’s Double Slit Experiment.
You must understand how the double slit experiment produces an interference pattern.
Conditions for Interference in the Double Slit Experiment.
You must be able to calculate the conditions for constructive and destructive interference in the
double slit experiment.
Intensity in the Double Slit Experiment.
You must be able to calculate intensities in the double slit experiment.
147
Intensity in the Double Slit Experiment
Our equations for the minima
and maxima intensity positions
are for the centers of the
fringes.
In this section, we calculate
distribution of light intensity in
the double-slit interference
pattern.
148
The derivation of the double-slit intensity equation is not
particularly difficult, so study it if you find derivations helpful
for your understanding.
A path length difference
L= corresponds to a
phase difference of =2.
A path length difference
L=m corresponds to a
phase difference of =2m.
In general, for non-integral m, the phase difference at P
between the waves from S1 and S2 is
φ
L
d
s
i
n

2
π
=
=
φ
=
d
s
i
n

2
π
x
x
λx
λ
λ
149
Your text writes the equation for the intensity distribution in the
double-slit experiment in terms of the phase difference on the
previous slide.
Your starting equation for the
intensity is

2φ
I=
I0co
s 

where I0 is 4 times the peak
intensity of either of the two
interfering waves:
I0=4Isinglewave
150
Why did my previous diagrams show this?
Today’s agendum: Interference Due to Reflection.
Phase Change Due to Reflection.
You must be able to determine whether or not a phase change occurs when a wave is reflected.
Thin Film Interference.
You must be able to calculate thin film thicknesses for constructive or destructive interference.
Examples.
You must be able to solve problems similar to these examples.
151
Interference from Reflection
Phase Change Due to Reflection
Light undergoes a phase change of 180° ( radians) upon
reflection from a medium that has a higher index of refraction
than the one in which the wave is traveling.
152
Today’s agendum: Interference Due to Reflection.
Phase Change Due to Reflection.
You must be able to determine whether or not a phase change occurs when a wave is reflected.
Thin Film Interference.
You must be able to calculate thin film thicknesses for constructive or destructive interference.
Examples.
You must be able to solve problems similar to these examples.
153
Thin Film Interference
Thin film
interference is
caused by…
…phase difference
of reflected waves
due to reflection
off a higher-n
material, and…
http://www.photographyblog.com/gallery/showphoto.php?photo=5545
…phase difference of reflected waves due to path length
differences.
154
Thin Film Interference
180° phase change
Ray  undergoes a
phase change on
reflection.
Ray  has a phase
change due to the
path difference.


No phase change
Air
Film
t
nAir < nFilm
Air
Do the reflected rays interfere destructively or constructively?
Caution! The wavelength in the film is different than in air.
155
Dark lines in drawings are there to help you see the boundaries, and are not a separate medium.
Assume the incident light is nearly perpendicular to the film
surface.
The path length
difference is
approximately 2t.
There is a 180 phase
difference (½ of a
wavelength) due to
the first reflection.
180° phase change

No phase change
Air
Film

film=
nfilm
t
nAir < nFilm
Air
We will get destructive interference when the path
difference is an integral number of wavelengths:

2
t
=
m

=
m

2
n
t
=
m

,



m
=
0
,
1
,
2
.
.
.
f
i
l
m
f
i
l
m
n
f
i
l
m
156
Assume the incident light is nearly perpendicular to the film
surface.
180° phase change
We get constructive
interference when
the path difference
is film/2, 3film/2,
5film/2, etc.

No phase change
Air
Film

film=
nfilm
t
nAir < nFilm
Air
We will get constructive interference when the path
difference is a half-integral number of wavelengths:
1
1

1





2
t
=
m
+

=
m
+2

n
t
=
m
+

,



m
=
0
,
1
,
2
.
.
.
f
i
l
m
f
i
l
m






2
2
n
2





f
i
l
m
157
The equations below are not on your starting equation sheet.
180° phase change
You need to apply
the reasoning used
in deriving them to
each of your thin
film interference
problems.

No phase change
Air
Film

film=
nfilm
t
nAir < nFilm
Air
2
n
t
=
m

,



m
=
0
,,
1
2
.
.
.
f
i
l
m
 1

2
n
t
=
m
+

,



m
=
0
,
1
,
2
.
.
.
f
i
l
m

 2

These are only true when the film is surrounded by a medium with lower
158
index of refraction than the film!
Caution!
These are valid when
the light is incident
almost perpendicular to
the film:
2nfilmt=m

180° phase change

No phase change
Air
Film

film=
nfilm
t
nAir < nFilm
 1

2
n
t
=
m
+
film 
 Air
 2

The incident ray in the diagram clearly does not qualify visually
as “almost perpendicular.” That’s because the angle relative to
the normal is exaggerated for viewing convenience.
159
Caution!
180° phase change

No phase change
Air
Film

film=
nfilm
t
nAir < nFilm
Air
For truly non-perpendicular incidence, you have to take into
account the extra path length of the ray reflected at the air-film
interface (as well as the extra path length inside the film).
160
Thin Film Interference Problem Solving Tips
 Identify the thin film causing the interference.
 Determine the phase relationship between the portion of the
wave reflected at the upper surface and the portion reflected at
the lower surface.
 Phase differences have two causes: (1) path differences and
(2) phase changes upon reflection.
 When the total phase change is an integer multiple of the
wavelength (, 2, 3, etc.) the interference is constructive,
and when it is a half-integer multiple of the wavelength (/2,
3/2, 5/2, etc.) it is destructive.
161
Today’s agendum: Interference Due to Reflection.
Phase Change Due to Reflection.
You must be able to determine whether or not a phase change occurs when a wave is reflected.
Thin Film Interference.
You must be able to calculate thin film thicknesses for constructive or destructive interference.
Examples.
You must be able to solve problems similar to these examples.
162
Example: a glass lens is coated on one side with a thin film of
MgF2 to reduce reflection from the lens surface. The index of
refraction for MgF2 is 1.38 and for glass is 1.50. What is the
minimum thickness of MgF2 that eliminates reflection of light
of wavelength λ = 550 nm? Assume approximately
perpendicular angle of incidence for the light.
180° phase change
Both rays  and 
experience a 180 phase shift
on reflection so the total
phase difference is due to the
path difference of the two
rays.
 
Air
nAir = 1.00
180°
phase
change
MgF2
n= 1.38
t
glass, ng =1.50
163
The reflected light is minimum when the two light rays meet the
condition for destructive interference: the path length difference
is a half-integral multiple of the light wavelength in MgF2.
 1

2
t
=
m
+



m
=
0
,
1
,
2
.
.
.

 ,
n
 2

M
g
F
2
The minimum thickness is for
m=0.

2tmin=
2n
M
gF2
180° phase change
 
Air
nAir = 1.00
180°
phase
change
MgF2
n= 1.38
t
5
5
0
n
m
t=
= =
9
9
.
6
n
m
m
i
n
4
n
1
.
3
8


M
g
F
24
glass, ng =1.50
164
Example: two glass plates 10 cm long are in contact on one
side and separated by a piece of paper 0.02 mm thick on the
other side. What is the spacing between the interference
fringes? Assume monochromatic light with a wavelength in air
of λ = 500 nm incident perpendicular to the slides.
The light that is partly reflected at the bottom of the first glass
surface and partly transmitted is responsible for the interference
fringes.*
Ray  is not phase shifted on reflection.
Ray  is shifted 180 on reflection.
For destructive interference
2
t=
mm


 =
0

1
2
.
.
.


H
t
x
L = 10 cm
H = 2x10-5 m165
*This reference explains why there is no visible interference due to the relatively thick glass plates themselves.
2
t=
mm


 =
0

1
2
.
.
.
t H
H
x
= 
t=
x L
L
0
.
1
m
5
0
0
n
m



H
x
L

2
=
m






x
=
m
=
m
=
m
1
.
2
5
m
m


5
L
2
H
2
2

1
0
m


x is the distance from the contact
point to where destructive
interference takes place.
Successive dark fringes are
separated by 1.25 mm.


H
t
x
L = 10 cm
H = 2x10-5 m166
 1

For constructive interference 2
t
=
m
+





m
=
0

1
2
.
.
.


 2

t H
H
x
= 
t=
x L
L
H
x
L

1

1

2
=
m
+






x
=
m
+




L
2
H
2

2

Successive bright fringes occur for
m+½ and (m+1)+½.
H
x 
1
2 =
+
1+
 
m
L 
2
3L


x=m
 +
22
H



H
t
x
L = 10 cm
H = 2x10-5 m167
Successive bright fringes are
separated by 1.25 mm.


H
t
x
L = 10 cm
H = 2x10-5 m168
Example: suppose the glass plates have ng = 1.50 and the
space between them contains water (nw = 1.33). What
happens now?
Ray  is not phase shifted on reflection. Ray  is shifted 180
on reflection. Both are the same as before.
t=
mm


 =
0

1
2
.
.
.
For destructive interference 2
But the path difference now occurs in
water, where the light will have a
wavelength

nwater
Repeat the calculation, using water.


H
t
x
L = 10 cm
H = 2x10-5 m169
For destructive interference, we now have
0
.
1
m
5
0
0
n
m
/
1
.
3
3




L

w
a
t
e
r
x
=
m
=
m=
m
0
.
9
4
m
m


5
2
H 2
2

1
0
m


Successive dark fringes are separated by 0.94 mm.


H
t
x
L = 10 cm
H = 2x10-5 m170
Two lectures ago I showed you these two plots of the intensity
distribution in the double-slit experiment:
Peak intensity varies with angle.
Which is correct?
Peak intensity independent of angle.
171
Diffraction
Light is an electromagnetic wave, and like all waves, “bends”
around obstacles.

d
<<d
d
>>d
This bending, which is most noticeable when the dimension
of the obstacle is close to the wavelength of the light, is
called “diffraction.” Only waves diffract.
172
Diffraction pattern from a penny
positioned halfway between a
light source and a screen.
The shadow of the penny is the
circular dark spot.
Notice the circular bright and
dark fringes.
The central bright spot is not a defect in the picture. It is a
result of light “bending” around the edges of the penny and
interfering constructively in the exact center of the shadow.
173
Single Slit Diffraction
In the previous chapter we calculated the interference pattern
from a pair of slits.
One of the assumptions in
the calculation was that
the slit width was very
small compared with the
wavelength of the light.
a






Now we consider the
effect of finite slit
width. We start with a
single slit.
Each part of the slit acts as a source of light rays, and these
different light rays interfere.
174
Divide the slit in half.
Ray  travels farther*
than ray  by (a/2)sin.
Likewise for rays  and
.
a/2
a/2
If this path difference is exactly half a
wavelength (corresponding to a
phase difference of 180°) then the
two waves will cancel each other and
destructive interference results.
Destructive interference:

a





a
sin 
2
a

sin=
2
2
175
*All rays from the slit are converging at a point P very far to the right and out of the picture.
Destructive
interference:
a

sin=
2
2
asin=

sin =
a
a/2

a





a/2
a
sin 
2
If you divide the slit into 4 equal parts, destructive
2
interference occurs when sin= .
a
If you divide the slit into 6 equal parts, destructive
3
interference occurs when sin= .
a
176
a/2

a





a/2
a
sin 
2
In general, destructive interference occurs when

s
i
n

=
m
,m
=
1
,2
,3
,.
.
.
a
The above equation gives the positions of the dark fringes.
The bright fringes are approximately halfway in between.177
Use this geometry for
tomorrow’s single-slit
homework problems.
y
a
If  is small,* then it is
valid to use the
approximation sin   .
( must be expressed in
radians.)

O
x
*The approximation is quite good for angles of 10
or less, and not bad for even larger angles.
178
Single Slit Diffraction Intensity
I won’t derive the intensity distribution for the single slit.
The general features of that distribution are shown below.
Most of the intensity is in the central maximum. It is twice
the width of the other (secondary) maxima.
179
New starting equations for single-slit intensity:
2
= asin

2
s

in

/2


I=
I0


/
2
  
“Toy”
180
Example: 633 nm laser light is passed through a narrow slit
and a diffraction pattern is observed on a screen 6.0 m away.
The distance on the screen between the centers of the first
minima outside the central bright fringe is 32 mm. What is the
slit width?
y1 = (32 mm)/2
tan = y1/L
tan  sin   for small 

 L

s
i
n

=a

= 
a s
i
ny
1
/
Ly
1
9
.
0
m
6
3
3

1
0
m
6


a
=
3
1
6

1
0
m


-4
a
=
2
.3
71
0
m
181
Resolution of Single Slit (and Circular Aperture)
The ability of optical systems to distinguish closely spaced
objects is limited because of the wave nature of light.
If the sources are far enough apart so that their central
maxima do not overlap, their images can be distinguished and
182
they are said to be resolved.
When the central maximum of one image falls on the first
minimum of the other image the images are said to be just
resolved. This limiting condition of resolution is called
Rayleigh’s criterion.
183
From Rayleigh’s criterion we can determine the minimum
angular separation of the sources at the slit for which the

images are resolved.
a=
These come from
For a slit of width a:  =

a
sin
the small angle approximation,
and geometry.
1.22
For a circular aperture of diameter D: =
D
Resolution is wavelength limited!
184
If a single slit diffracts, what about a double slit?
Remember the double-slit interference pattern from the
chapter on interference?

d
s
in



I
=
I
c
o
s
m
a
x


 
2
If the slit width (not the spacing
between slits) is small (i.e.,
comparable to the wavelength of
the light), you must account for
diffraction.
interference only
185
Double Slit Diffraction
r1
S1
y
a
r2

P
d
S2
L
186
Diffraction Gratings
A diffraction grating consists of a large number of equally
spaced parallel slits.
The path difference between
rays from any two adjacent
slits is  = dsin .

d
 = d sin 
If  is equal to some integer
multiple of the wavelength
then waves from all slits will
arrive in phase at a point on
a distant screen.
Interference maxima occur for d
s
i
n

=
m

,m
=
1
,,,
2
3
.
.
.
187
Ok what’s with this equation monkey business?
d
s
i
n

=
m

,m
=
1
,,,
2
3
.
.
.
double-slit interference
constructive
a
s
i
n

=
m

,m
=
1
,,,
2
3
.
.
.
single-slit diffraction
destructive!
d
s
i
n

=
m

,m
=
1
,,,
2
3
.
.
.
diffraction grating
constructive
188
Diffraction Grating Intensity Distribution
Interference Maxima:
dsin=m


d
 = d sin 
The intensity maxima are
brighter and sharper than for
the two slit case.
189
190
Application: spectroscopy
visible light
hydrogen
helium
mercury
You can view the atomic spectra for each of the elements here.
191
Example: the wavelengths of visible light are from
approximately 400 nm (violet) to 700 nm (red). Find the
angular width of the first-order visible spectrum produced by a
plane grating with 600 slits per millimeter when white light falls
normally on the grating.
Interference Maxima:
dsin=m

1
6
d
=
=
1
.
6
7

1
0
m
6
0
0
s
l
i
t
s
/
m
m
First-order violet:
9
1
4
0
0

1
0
m





V
s
i
n

=
m
=
=
0
.
2
4
0
V
6
d1
.
6
7

1
0
m
V =13.9
192
First-order red:
9
1
7
0
0

1
0
m





R
s
i
n

=
m
=
=
0
.
4
1
9
R
6
d1
.
6
7

1
0
m
R =24.8









=
2
4
.
8
1
3
.
9
=
1
0
.
9
R V
10.9
visible light
193
Example: for this diffraction grating show that the violet end of
the third-order spectrum overlaps the red end of the secondorder spectrum.
9
6
3
4
0
0

1
0
m





1
.
2
0

1
0
m
V
i
n

=
m
=
=
Third-order violet: s
V
d
d
d
9
6
2
7
0
0

1
0
m





1
.
4
0

1
0
m
V
Second-order red: s
i
n

=
m
=
=
R
d
d
d
sin
sin
R
2>
V
3
No matter what the grating spacing, d, the largest angle for
the 2nd order spectrum (for the red end) is always greater
than the smallest angle for the 3rd order spectrum (for the
violet end), so 2nd and 3rd orders always overlap.
194
Diffraction Grating Resolving Power
Diffraction gratings let us measure wavelengths by spreading
apart the diffraction maxima associated with different
wavelengths. In order to distinguish two nearly equal
wavelengths the diffraction must have sufficient resolving
power, R.
Consider two wavelengths λ1 and λ2 that are nearly equal.

2
1+
The average wavelength is avg=
and the difference is
2
=2-1 .
The resolving power is defined as R=
avg

.
195
R=
avg

For a grating with N lines illuminated it can be shown that the
resolving power in the mth order diffraction is
R =Nm.
Dispersion
Spectroscopic instruments need to resolve spectral lines of
nearly the same wavelength.
mercury

 The greater the angular dispersion,
a
n
g
u
l
a
r
d
i
s
p
e
r
s
i
o
n
=

 the better a spectrometer is at
resolving nearby lines.
196
Example: Light from mercury vapor lamps contain several
wavelengths in the visible region of the spectrum including two
yellow lines at 577 and 579 nm. What must be the resolving
power of a grating to distinguish these two lines?
mercury
5
7
7
n
m
+
5
7
9
n
m

=
=
5
7
8
n
m
a
v
g
2


=
5
7
9
n
m
5
7
7
n
m
=
2
n
m

7
8
n
m
a
v
g 5
R
==
=
2
8
9

 2
n
m
197
Example: how many lines of the grating must be illuminated if
these two wavelengths are to be resolved in the first-order
spectrum?
mercury
R =289
R
2
8
9
R
=
N
m

N
=
==
2
8
9
m
1
198
199