Light, Reflection, & Mirrors
AP Physics B
Almost everything we see is
made visible by the light it
reflects. Some materials,
such as air, water, or window
glass, allow light to pass
through. Other materials,
such as thin paper or frosted
glass, allow the passage of
light in diffused directions so
that we can’t see objects
through them.
NOTES 13 : Lights and Optics
Early Concepts of Light and SOUND
1. Scientists now agree that light has a
dual nature, part particle and part wave.
27.1 Early Concepts of Light
Light has been studied for thousands of years.
Some ancient Greek philosophers thought that light consists of tiny particles,
which enter the eye to create the sensation of vision.
Others thought that vision resulted from streamers or filaments emitted by the
eye making contact with an object.
27.1 Early Concepts of Light
Up until the time of Newton and beyond, most philosophers and scientists
thought that light consisted of particles.
However, one Greek, Empedocles, thought that light traveled in waves.
One of Newton’s contemporaries, the Dutch scientist Christian Huygens, also
argued that light was a wave.
27.1 Early Concepts of Light
The particle theory was supported by the fact
that light seemed to move in straight lines instead of
spreading out as waves do.
2.
3. Wave Theory -Huygens showed that under
some circumstances light does spread out and
other scientists found evidence to support the wave
theory.
The wave theory became the accepted theory in the
nineteenth century.
27.1 Early Concepts of Light
4. In 1905, Einstein published a theory explaining
the
photoelectric effect.
According to this theory, light consists of particles
called photons, massless bundles of concentrated
electromagnetic energy.
Scientists now agree that light has a dual nature,
part particle and part wave.
27.1 Early Concepts of Light
What is the nature of light?
27.2 The Speed of Light
5. Michelson’s experimental value for the
speed of light was 299,920 km/s, which is
usually rounded to 300,000 km/s or 3X 108 m/s
27.2 The Speed of Light
It was not known whether light travels instantaneously or
with finite speed until the late 1600s.
Galileo tried to measure the time a light beam takes to
travel to a distant mirror, but it was so short he couldn’t
begin to measure it.
Others tried the experiment at longer distances with
lanterns they flashed on and off between distant
mountaintops. All they succeeded in doing was measuring
their own reaction times.
27.2 The Speed of Light
Olaus Roemer
The first demonstration that light travels at a finite speed was
supplied by the Danish astronomer Olaus Roemer about 1675.
Roemer carefully measured the periods of Jupiter’s moons.
• The innermost moon, Io, revolves around Jupiter
in 42.5 hours.
• The Io disappears periodically into Jupiter’s shadow, so
this period could be measured with great accuracy.
27.2 The Speed of Light
• Roemer found that while Earth was moving away from
Jupiter, the periods of Io were all somewhat longer than
average.
• When Earth was moving toward Jupiter, the measured
periods were shorter than average.
• Roemer estimated that the cumulative discrepancy
amounted to about 22 minutes.
27.2 The Speed of Light
Light coming from Io takes
longer to reach Earth at position
D than at position A. The extra
distance that the light travels
divided by the extra time it takes
gives the speed of light.
27.2 The Speed of Light
Christian Huygens
Christian Huygens correctly interpreted this discrepancy.
• The Io passed into Jupiter’s shadow at the
predicted time.
• The light did not arrive until it had traveled the extra
distance across the diameter of Earth’s orbit.
• This distance is now known to be 300,000,000 km.
27.2 The Speed of Light
Using the travel time of 1000 s for light to move across Earth’s orbit makes
the calculation of the speed of light quite simple:
The speed of light is 300,000 km/s.
27.2 The Speed of Light
Albert Michelson
The most famous experiment measuring the speed of light was performed
by the American physicist Albert Michelson in 1880.
• Light was directed by a lens to an octagonal mirror.
• A beam of light was reflected to a stationary mirror on a mountain 35
km away and then reflected back.
• The distance was known, so Michelson had to find only the time it
took to make a round trip.
27.2 The Speed of Light
• When the mirror was spun, short bursts of light reached the stationary
mirror and were reflected back to the spinning octagonal mirror.
• If the rotating mirror made one-eighth rotation while the light made the
trip, the mirror reflected light to the observer.
• If the mirror was rotated too slowly or too quickly, it would not be in a
position to reflect light.
27.2 The Speed of Light
a.
Light is reflected back to the eyepiece when the mirror is at rest.
27.2 The Speed of Light
a.
b.
Light is reflected back to the eyepiece when the mirror is at rest.
Reflected light fails to enter the eyepiece when the mirror spins too slowly . . .
27.2 The Speed of Light
a.
b.
c.
Light is reflected back to the eyepiece when the mirror is at rest.
Reflected light fails to enter the eyepiece when the mirror spins too slowly . . .
. . . or too fast.
27.2 The Speed of Light
a.
b.
c.
d.
Light is reflected back to the eyepiece when the mirror is at rest.
Reflected light fails to enter the eyepiece when the mirror spins too slowly . . .
. . . or too fast.
When the mirror rotates at the correct speed, light reaches the eyepiece.
27.2 The Speed of Light
When the light entered the eyepiece, the time for the light to make the trip
and the time for the mirror to make one eighth of a rotation were the same.
Michelson divided the 70-km round trip distance by this time and found the
speed of light was 299,920 km/s, which is usually rounded to 300,000 km/s.
Michelson received the 1907 Nobel Prize in physics for this experiment.
Facts about Light
The speed of light, c, is constant in a vacuum.
Light can be:
•REFLECTED
•ABSORBED
•REFRACTED
Light is an electromagnetic wave in that it has wave like properties
which can be influenced by electric and magnetic fields.
27.2 The Speed of Light
The speed of light in a vacuum is a universal constant.
Light is so fast that if a beam of light could travel around Earth, it would make 7.5
trips in one second.
Light takes 8 minutes to travel from the sun to Earth and 4 years from the next
nearest star, Alpha Centauri.
6. The distance light travels in one year is called a light-year.
27.2 The Speed of Light
think!
Light entered the eyepiece when Michelson’s octagonal
mirror made exactly one eighth of a rotation during the time
light traveled to the distant mountain and back. Would light
enter the eyepiece if the mirror turned one quarter of a
rotation in this time?
27.2 The Speed of Light
think!
Light entered the eyepiece when Michelson’s octagonal
mirror made exactly one eighth of a rotation during the time
light traveled to the distant mountain and back. Would light
enter the eyepiece if the mirror turned one quarter of a
rotation in this time?
Answer:
Yes, light would enter the eyepiece whenever the octagonal
mirror turned in multiples of 1/8 rotation— ¼, ½, 1, etc.—in
the time the light made its round trip.
27.2 The Speed of Light
What was Michelson’s experimental value for the speed of
light?
Facts about Light


It is a form of Electromagnetic Energy
It is a part of the Electromagnetic Spectrum and the only part we
can really see
Web Group Research- 5 minutes
Electromagnetic waves
Group 1- Radio waves

Group 4 – Light
waves

Group 5- Ultraviolet
rays

Group 6- Gamma
rays
Group 2- Microwaves
Group 3- Infrared
waves
CW : Electromagnetic Waves p 408
Define Electromagnetic waves .
Define the Electromagnetic Spectrum
Draw and label the electromagnetic spectrum
Give examples of these waves.
Define each frequency in scientific notation format and standard decimal
format
7. This energy travels in a wave that is partly electric and
partly magnetic. Such a wave is an electromagnetic
wave.
The electromagnetic spectrum consists of radio
waves, microwaves, infrared, light, ultraviolet rays,
X-rays, and gamma rays.
8. Drawing :Electromagnetic Waves
Light is a portion of the family of electromagnetic waves that includes radio waves,
microwaves, and X-rays.
The range of electromagnetic waves is called the electromagnetic spectrum.
Electromagnetic Waves
The lowest frequency of light we can see
appears red. The highest visible light, violet,
has nearly twice the frequency of red light.
9. Electromagnetic waves of frequencies
lower than the red of visible light are called
infrared. Heat lamps give off infrared waves.
10. Electromagnetic waves of frequencies
higher than those of violet are called
ultraviolet. They are responsible for sunburns.
27.3 Electromagnetic Waves
What are the waves of the electromagnetic spectrum?
11.
12.
13.
Light doesn’t travel
through a mirror, but is
returned by the mirror’s
surface. These waves are
reflected.
When waves strike the
surface of a medium at an
angle, their direction
changes. These waves are
refracted. Usually waves
are partly reflected and
partly refracted when they
fall on a transparent
medium.
CW : Comparison of Materials:
Score - # of Information
TRANSPARENT
MATERIALS P409
DEFINITION:
EXAMPLE OF
MATERIALS
WHAT HAPPENS TO
LIGHT WHEN IT HITS
THIS TYPE OF
MATERIAL?
OPAQUE
MATERIALS P 410
Light and Transparent Materials
Light passes through materials whose atoms absorb
the energy and immediately reemit it as light.
27.4 Light and Transparent Materials
Materials that transmit light are
transparent. Glass and water are
transparent.
Materials that are springy (elastic)
respond more to vibrations at some
frequencies than at others.
The natural vibration frequencies of an
electron depend on how strongly it is
attached to a nearby nucleus.
27.4 Light and Transparent Materials
The electrons of atoms in glass can be
imagined to be bound to the atomic nucleus
as if connected by springs.
27.4 Light and Transparent Materials
Electrons in glass have a natural vibration
frequency in the ultraviolet range.
• In ultraviolet light, resonance occurs as the
wave builds a large vibration between the
electron and the nucleus.
• The energy received by the atom can be
either passed on to neighboring atoms by
collisions or reemitted as light.
• If ultraviolet light interacts with an atom
that has the same natural frequency, the
vibration amplitude is unusually large.
27.4 Light and Transparent Materials
The atom typically holds on to this energy
for about 1 million vibrations or 100
millionths of a second.
During this time, the atom makes many
collisions with other atoms and gives up
its energy in the form of heat.
That’s why glass is not transparent to
ultraviolet.
27.4 Light and Transparent Materials
When the electromagnetic wave has a lower
frequency than ultraviolet, as visible light does, the
electrons are forced into vibration with smaller
amplitudes.
• The atom holds the energy for less time, with
less chance of collision with neighboring atoms.
• Less energy is transferred as heat.
• The energy of the vibrating electrons is
reemitted as transmitted light.
27.4 Light and Transparent Materials
Glass is transparent to all the frequencies of
visible light.
The frequency of the reemitted light is identical to
that of the light that produced the vibration to
begin with.
The main difference is a slight time delay between
absorption and reemission.
This time delay results in a lower average speed
of light through a transparent material.
27.4 Light and Transparent Materials
A light wave incident upon a pane of glass sets up
vibrations in the atoms. Because of the time delay
between absorptions and reemissions, the
average speed of light in glass is less than c.
27.4 Light and Transparent Materials
In a vacuum, the speed of light is a constant
300,000 km/s; we call this speed of light c.
• Light travels slightly less than c in the
atmosphere, but the speed is usually rounded
to c.
• In water, light travels at 75% of its speed in a
vacuum, 0.75c.
• In glass, light travels at about 0.67c,
depending on glass type.
• In a diamond, light travels at only 0.40c.
When light emerges from these materials into the
air, it travels at its original speed, c.
Reflection
Materials such as glass and water are not as
rigid to light waves.
• When light shines perpendicularly on the
surface of still water, about 2% of its
energy is reflected and the rest is
transmitted.
• When light strikes glass perpendicularly,
about 4% of its energy is reflected.
• Except for slight losses, the rest is
transmitted.
27.4 Light and Transparent Materials
Infrared waves, with frequencies lower than visible
light, vibrate not only the electrons, but also the
entire structure of the glass.
This vibration of the structure increases the
internal energy of the glass and makes it warmer.
Glass is transparent to visible light, but not to
ultraviolet and infrared light.
27.4 Light and Transparent Materials
What kind of materials does light pass through?
27.5 Opaque Materials
In opaque materials, any coordinated
vibrations given by light to the atoms and
molecules are turned into random kinetic
energy—that is, into internal energy.
27.5 Opaque Materials
Materials that absorb light without
reemission and thus allow no light through
them are opaque.
Wood, stone, and people are opaque.
In opaque materials, any vibrations from
light are turned into random kinetic
energy—that is, into internal energy.
The materials become slightly warmer.
27.5 Opaque Materials
Metals are also opaque.
In metals, the outer electrons of atoms
are not bound to any particular atom.
When light shines on metal and sets
these free electrons into vibration, their
energy does not “spring” from atom to
atom.
It is reemitted as visible light. This
reemitted light is seen as a reflection.
That’s why metals are shiny.
27.5 Opaque Materials
Our atmosphere is transparent to visible light and
some infrared, but almost opaque to high-frequency
ultraviolet waves.
The ultraviolet that gets through is responsible for
sunburns.
Clouds are semitransparent to ultraviolet, so you can
get a sunburn on a cloudy day.
Ultraviolet also reflects from sand and water, so you
can sometimes get a sunburn while in the shade of a
beach umbrella.
27.5 Opaque Materials
think!
Why is glass transparent to visible light but opaque to
ultraviolet and infrared?
27.5 Opaque Materials
think!
Why is glass transparent to visible light but opaque to
ultraviolet and infrared?
Answer:
The natural frequency of vibration for electrons in glass matches the
frequency of ultraviolet light, so resonance in the glass occurs when
ultraviolet waves shine on it. These vibrations generate heat instead of
wave reemission, so the glass is opaque to ultraviolet. In the range of
visible light, the forced vibrations of electrons in the glass result in
reemission of light, so the glass is transparent. Lower-frequency infrared
causes entire atomic structures to resonate so heat is generated, and the
glass is opaque.
28.2 Color by Reflection
The color of an opaque object
is the color of the light it
reflects.
28.2 Color by Reflection
The colors of most objects around
you are due to the way the objects
reflect light.
The color of an opaque object is the
color of the light it reflects.
Light reflects from objects similar to
the way sound “reflects” from a
tuning fork when another sets it into
vibration.
28.2 Color by Reflection
If the material is transparent, the reemitted light
passes through it.
If the material is opaque, the light passes back
into the medium from which it came. This is
reflection.
Most materials absorb light of some frequencies
and reflect the rest.
If a material absorbs light of most visible
frequencies and reflects red, for example, the
material appears red.
________________________________________END of CW
Count the information = Score =_________
28.2 Color by Reflection
13. When white light falls on a flower,
light of some frequencies is absorbed
by the cells in the flower and some light
is reflected.
14. Cells that contain chlorophyll absorb
light of most frequencies and reflect the
green part, so they appear green.
The petals of a red rose, on the other
hand, reflect primarily red light, with a
lesser amount of blue.
28.2 Color by Reflection
15. Petals of most yellow flowers, such as daffodils,
reflect red and green as well as yellow.
Yellow daffodils reflect light of a broad band of
frequencies.
16. The reflected colors of most objects are not pure
single-frequency colors, but a spread of frequencies.
So something yellow, for example, may simply be a
mixture of colors without blue and violet—or it can be
red and
green together.
28.4 Sunlight
17. White light from the sun is a composite of all
the visible frequencies. The brightness of solar
frequencies is uneven.
18.. The lowest frequencies of sunlight, in the red
region, are not as bright as those in the middlerange yellow and green region.
19.. Humans evolved in the presence of sunlight
and we are most sensitive to yellow-green.
.The blue portion of sunlight is not as bright, and
the violet portion is even less bright.
28.4 Sunlight
The radiation curve of sunlight is a graph of brightness versus
frequency. Sunlight is brightest in the yellow-green region.
28.5 Mixing Colored Light
When red light, green light, and blue light of equal brightness are projected on
a white screen, the overlapping areas appear different colors.
28.5 Mixing Colored Light
21. Light of all the visible frequencies mixed together produces white.
• White also results from the combination of only red, green, and blue
light.
22. Red and green light alone overlap to form yellow.
23. Red and blue light alone produce the bluish-red color called magenta.
24. Green and blue light alone produce the greenish-blue color called
cyan.
28.6 Complementary Colors
When two of the three additive primary colors are combined:
22. red + green = yellow
23. red + blue = magenta
24. blue + green = cyan
25. When we add in the third color, we get white:
•yellow + blue = white
•magenta + green = white
•cyan + red = white
26. Mixing Colored Pigments
27. Why the Sky Is Blue
Of the visible frequencies, violet light is
scattered the most, followed by blue, green,
yellow, orange, and red, in that order.
• Violet light is scattered more than blue
but our eyes are not very sensitive to
violet light.
• Our eyes are more sensitive to blue, so we
see a blue sky.
28.8 Why the Sky Is Blue
28.10 Why Water Is Greenish Blue
28. Water is greenish blue because water molecules
absorb red.
28.10 Why Water Is Greenish Blue
Ocean water is cyan because it absorbs red. The
froth in the waves is white because its droplets of
many sizes scatter many colors.
Light doesn’t travel through a
mirror, but is returned by the
mirror’s surface. These
waves are reflected. When
waves strike the surface of a
medium at an angle, their
direction changes. These
waves are refracted. Usually
waves are partly reflected
and partly refracted when
they fall on a transparent
medium.
29.1 Reflection
29. The return of a wave back to its original medium is called
reflection.
Fasten a spring to a wall and send a pulse along the
spring’s length.
The wall is a very rigid medium compared with the spring, so
all the wave energy is reflected back along the spring.
Waves that travel along the spring are almost totally
reflected at the wall.
29.2 The Law of Reflection
30. Incident rays and reflected rays make equal angles with
a line perpendicular to the surface, called the normal.
31. The angle between the incident ray and the normal is
the angle of incidence.
32. The angle between the reflected ray and the normal
is the angle of reflection.
33. Angle of incidence = Angle of reflection
The Law of Reflection
33. The
law of reflection states that the angle of incidence
and the angle of reflection are equal to each other.
34. A
virtual image appears to
be in a location where light does
not really reach.
Mirrors produce only virtual
images.
29.3 Mirrors
Your eye cannot ordinarily tell the
difference between an object and its
virtual image.
• The light enters your eye in
exactly the same manner as it
would if there really were an
object where you see the image.
• The image is the same distance
behind the mirror as the object is
in front of it.
• The image and object are the
same size.
Plane Mirror
Suppose we had a flat , plane mirror mounted vertically. A candle is
placed 10 cm in front of the mirror. WHERE IS THE IMAGE OF
THE CANDLE LOCATED?
mirror
On the surface of the mirror?
Behind the mirror?
Object Distance, Do = 10 cm
Same side as the object?
35. Plane Mirror
Suppose we had a flat , plane mirror mounted vertically. A candle is
placed 10 cm in front of the mirror. WHERE IS THE IMAGE OF
THE CANDLE LOCATED?
mirror
Object Distance, Do = 10 cm
Virtual Image
Image Distance, Di = 10 cm
Do=Di, and the heights are equal as well
Virtual Images
Virtual Images are basically images which cannot be
visually projected on a screen.
If this box gave off
light, we could project
an image of this box
on to a screen
provided the screen
was on the SAME
SIDE as the box.
You would not be able to project the image of the
vase or your face in a mirror on a screen, therefore
it is a virtual image.
CONCLUSION: 36. VIRTUAL IMAGES are ALWAYS on the OPPOSITE side
of the mirror relative to the object.
Real Image
Real Images are ones you can project on to a screen.
For MIRRORS they always appear on the SAME SIDE of the mirror as the object.
object
image
37. The characteristics of the
image, however, may be
different from the original object.
These characteristics are:
•SIZE (reduced,enlarged,same
size)
•POSITION (same side,
opposite side)
•ORIENTATION (right side up,
inverted)
What if the mirror isn’t flat?
38. Spherical Mirrors – Concave &
Convex
Also called DIVERGING mirror
Also called CONVERGING mirror
29.3 Mirrors
The law of reflection holds for curved mirrors. However, the sizes and
distances of object and image are no longer equal.
39. The virtual image formed by a convex mirror (a mirror that curves
outward) is smaller and closer to the mirror than the object is.
29.3 Mirrors
The law of reflection holds for curved mirrors. However, the sizes and
distances of object and image are no longer equal.
40.The virtual image formed by a convex mirror (a mirror that curves
outward) is smaller and closer to the mirror than the object is.
41. When an object is close to a concave mirror (a mirror that curves
inward), the virtual image is larger and farther away than the object is.
Converging (Concave) Mirror
42. A converging mirror is one that is spherical in
nature by which it can FOCUS parallel light rays to
a point directly in front of its surface. Every spherical
mirror can do this and this special point is at a
“fixed” position for every mirror. We call this point
the FOCAL POINT. To find this point you MUST
use light from “infinity”
Light from an “infinite”
distance, most likely the
sun.
Images Formed by Spherical Mirrors
Spherical mirrors are made from polished
sections cut from a spherical surface.
43. 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.
44. The radius of curvature, R, is the radius of the sphere,
or the distance from V to C.
R
C
V
45. The principal axis (or optical axis) is the line that
passes through the center of curvature and the center of the
mirror.
R
Principal Axis
C
The center of the mirror is often called the vertex of the mirror.
V
42. 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
42. The focal length f is the distance from P to F. Your text
shows that f = R/2.
R
f
C
F
P
Reality check: paraxial rays don’t really pass exactly through
the focal point of a spherical mirror (“spherical aberration”).
C
F
V
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.
In “real life” you would minimize aberration by using a
parabolic mirror.
C
F
V
46. Converging (Concave) Mirror
Since the mirror is
spherical it technically
has a CENTER OF
CURVATURE, C. The
focal point happens to
be HALF this distance.
R
f=
2
R=2f
We also draw a line through the
center of the mirror and call it the
PRINCIPAL AXIS.
Ray Diagrams for Mirrors
47. 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.
Ray 3 passes through
C and reflects back on
itself.
A fourth principal ray is the one directed at the vertex V.
V
C
F
Ray Diagrams for Concave Mirrors
We use three “principal rays” to construct images.
An image is formed where the rays converge.
48. 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.
49. 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.
What do you mean, this is a
virtual image? I can see it!
C
F
With this size object, there was a bit of spherical
aberration present, and I had to “cheat” my C a
bit to the left to make the diagram look “nice.”
50. 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
Ray 2: “through” F then parallel
then through F.
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
*Actually, the image is formed at infinity.
Ray Diagram Review
A ray diagram is a pictorial representation of how the
light travels to form an image and can tell you the
characteristics of the image.
object
C
f
Principal axis
Rule One: Draw a ray, starting from the top of the object, parallel to the
principal axis and then through “f” after reflection.
Ray Diagrams
object
C
f
Principal axis
Rule Two: Draw a ray, starting from the top of the object, through the focal
point, then parallel to the principal axis after reflection.
Ray Diagrams
object
C
f
Principal axis
Rule Three: Draw a ray, starting from the top of the object, through C, then
back upon itself.
What do you notice about the three lines?
THEY INTERSECT
The intersection is the location of the image.
Ray Diagram – Image Characteristics
object
C
f
Principal axis
After getting the intersection, draw an arrow down from the principal axis to
the point of intersection. Then ask yourself these questions:
1) Is the image on the SAME or OPPOSITE side of the mirror as the object?
Same, therefore it is a REAL IMAGE.
2) Is the image ENLARGED or REDUCED?
3) Is the image INVERTED or RIGHT SIDE UP?
51. The Mirror Equation
With a bit of geometry, you can show that
1
1
1
=
+
f
so si
f= for a flat mirror.
52. The magnification is the
ratio of the image to the
object height:
M= hi/ho = -si/so
so
f
ho
C hi
F
si
53. Sign Conventions for
Magnification
Orientation of
Sign of M
image with
respect to object
Upright
Inverted
Type of image
+
Virtual
-
Real
1
1
1
=
+
f
so
si
M= hi/ho = -si/so
Sign conventions for the mirror equation:
54. When the object, image,
or focal point is on the
reflecting side of the mirror,
the distance is positive.
55. When the object, image,
or focal point is “behind” the
mirror, the distance is
negative.
s
f
y
C y’
F
56. The image height is positive if the image is upright,
and 57. negative if the image is inverted relative to the
object.
s’
“When the object, image, or focal point is “behind” the mirror,
the distance is negative.”
If the image distance is negative, can the image be real?
The Mirror/Lens Equation
Is there any OTHER way to predict image characteristics besides
the ray diagram? YES!
51. One way is to use the MIRROR/LENS equation to
CALCULATE the position of the image.
1 1 1
= +
f so si
Mirror/Lens Equation
58. Assume that a certain concave spherical mirror
has a focal length of 10.0 cm. Locate the image for
an object distance of 25 cm and describe the
image’s characteristics.
What does this tell us? First we know the image is BETWEEN “C” & “f”. Since
the image distance is POSITIVE the image is a REAL IMAGE.
Real image = positive image distance
Virtual image = negative image distance
What about the size and orientation?
Mirror/Lens Equation
58. Assume that a certain concave spherical mirror
has a focal length of 10.0 cm. Locate the image for
an object distance of 25 cm and describe the
image’s characteristics.
1 1 1
1
1 1
= + ® = +
f so si 10 25 si
si =
16.67 cm
What does this tell us? First we know the image is BETWEEN “C” & “f”. Since
the image distance is POSITIVE the image is a REAL IMAGE.
Real image = positive image distance
Virtual image = negative image distance
What about the size and orientation?
59. Magnification Equation
To calculate the orientation and size of the image we
use the MAGNIFICATION EQUATION.
si hi
M =- =
so ho
16.67
M =25
M = -0.67x
Here is how this works:
•If we get a POSITIVE magnification, the image is
UPRIGHT.
•If we get a NEGATIVE magnification, the image is
INVERTED
•If the magnification value is GREATER than 1, the
image is ENLARGED.
•If the magnification value is LESS than 1, the image
is REDUCED.
•If the magnification value is EQUAL to 1, the image
is the SAME SIZE as the object.
Using our previous data we see that our image was INVERTED, and REDUCED.
60. Example
Assume that a certain concave spherical mirror has a focal
length of 10.0 cm. Locate the image for an object distance of
5 cm and describe the image’s characteristics.
1 1 1
1 1 1
= + ® = +
f so si
10 5 si
si = -10 cm
Characteristics?
si
M = - = 2x
5
•VIRTUAL (opposite side)
•Enlarged
•Upright
CW: Concave Mirrors 1

A concave shaving mirror has a focal length of 33
cm. Calculate the image position of a cologne
bottle placed in front of the mirror at a distance
of 93 cm.
CW: Concave Mirrors 1






A concave shaving mirror has a focal length of 33
cm. Calculate the image position of a cologne
bottle placed in front of the mirror at a distance
of 93 cm.
1/f = 1/so + 1/si
1/.33m = 1/.93m + 1/si
1/si = 1/.33m – 1/.93m
1/si = 3.03 – 1.07
si = .51 m
CW : Concave Mirrors 1

A concave shaving mirror has a focal length of 33
cm. Calculate the magnification of the image. Is
the image real or virtual? Is the image inverted or
upright?
CW : Concave Mirrors 1



A concave shaving mirror has a focal length of 33
cm. Calculate the magnification of the image. Is
the image real or virtual? Is the image inverted or
upright?
M = - si / so = - .51m / .93 m = -.55
image is inverted , reduced
CW 2: Concave Mirror

A concave shaving mirror has a focal length of 33
cm. Calculate the image position of a cologne
bottle placed in front of the mirror at a distance
of 93 cm. Draw a ray diagram to confirm your
results.
Draw the diagram
C
The image is inverted
and about half the
height of the object.
f
CW 3 : Concave Mirrors

a.
b.
c.
d.
An object of height 5 cm is placed 50 cm in
front of a concave mirror whose focal length
is 20cm.
where’s the image ?
Is it real or virtual ?
Is it upright or inverted ?
What’s the height of the image ?
1/ so + 1/ si = 1/f
1/ 50cm + 1/si = 1/ 20cm
1/si = .03
si= 33.33 cm
Real image , Inverted,
hi /ho = - si / s o = M
hi = - si ho / so = - 33.33cm (4cm ) / 50 cm
hi = - 2.67 cm inverted
CW:





An object of height of 5 cm is placed 30 cm in
front of a convex mirror whose focal length is
-40cm.
A. Where’s the image ?
B. is it real or virtual ?
C. Is it upright or inverted ?
D. What’s the height of the image ?
1/ so + 1/ si = 1/f
 1/ 30cm + 1/si = 1/ -40cm
 Si = - 1/.058 = -17.24cm
 Image is virtual
 Image is upright
 hi/ ho = - si/ so
 hi= - ho si /so = - 5cm (-17.24cm)/ 30cm
hi = 2.87 cm virtual image is upright

Applications of concave mirrors.
Shaving mirrors.
Makeup mirrors.
Kid scarers.
Solar cookers.
Ant fryers.
Flashlights, headlamps, stove reflectors.
Satellite dishes (when used with electromagnetic radiation).
62. Ray Diagrams for Convex Mirrors
Ray 1: parallel to the axis
then “through” F.
Ray 2: “through” Vertex.
Ray 3: “through” C.
F
The image is virtual, upright, and smaller than the object.
C
63. 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
used.
C
64. The mirror equation still works for convex mirrors.
1 1 1
= +
f so si
M= hi /ho = -si /so
so
ho
f
so
ho
F
Because they are on the “other” side of the mirror from the
object, s’ and f are negative.
C
Convex Mirrors

65. Convex mirrors take objects in a large field of
view and produce a small image

Side-view mirrors on cars are convex mirrors. That’s
why they say “objects are closer than they appear”
Convex Spherical Mirrors

66. A convex spherical mirror (diverging mirror) is
silvered so that light is reflected from the
sphere’s outer, convex surface



The image distance is always negative!
The image is always a virtual image!
The focal length is negative !
Ray diagrams for convex mirrors

67. The focal point and center of curvature are
behind the mirror’s surface


A virtual, upright image is formed behind the mirror
The magnification is always less than 1
f
C
68. Drawing the reference rays

Ray 1 is drawn parallel to the principal axis
beginning at the top of the object. It reflects from
the mirror along a line that intersects the focal
point
f
C
Ray 2

Ray 2 starts from the top of the object and goes as
though its going to intersect the focal point but it
reflects parallel to the principal axis
Ray 1
Ray 2
f
C
Ray 3

Ray 3 starts at the top of the object and goes as
though its going to intersect the center of
curvature
Ray 1
Ray 2
f
C
69. Convex Spherical Image
Formation

The image forms at the intersection of any two of
the three rays behind the mirror.
Ray 1
Ray 2
The rays do not intersect in
front of the mirror!!
f
C
70.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
1 1 1
1
1
=  =

s' f s -0.2 m 10 m
Not on reflecting
sidenegative.
On reflecting
sidepositive.
1
1
1
=

s' -0.2 m 10 m
…algebra…
s' = - 0.196 m= -19.6 cm
s'
-0.196 m 1
m= - = =
s
10 m
51
Remind me… what does it say on passenger side rear view
mirrors?
CW : Convex Mirrors

A candle is 49 cm in front of a convex spherical
mirror that has a focal length of 35 cm. What are
the image distance and magnification? Is the
image virtual or real? Is the image inverted or
upright? Draw a ray diagram to confirm your
results.
CW : Convex Mirrors





A candle is 49 cm in front of a convex spherical
mirror that has a focal length of 35 cm. What are
the image distance and magnification? Is the
image virtual or real? Is the image inverted or
upright? Draw a ray diagram to confirm your
results.
1/f = 1/so + 1/si
1/ si = -2.86-2.04
-1/.35m = 1/.49m + 1/si
= - 4.9
1/si = -1/.35m – 1/.49m
si = - .204 m




M = - Si / S0 = - (- .204 m) / .49 m
M = 0.42
Positive = virtual , upright
Less than 1 = reduced
Ray diagram for the Image
f
C
Remember that you only need to draw two of the three rays to find the image.
71. Applications of convex mirrors.
Passenger side rear-view mirrors.
Grocery store aisle mirrors.
Railroad crossing mirrors.
Anti-shoplifting (surveillance) mirrors.
Christmas tree ornaments.
72. 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.
Hint: download Dr. Hale’s quick reference card.
73. 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).
(negative image distance  virtual image)
(positive image distance  real image)
Radius of Curvature. When the center of curvature C is on
the same side as the outgoing light, R is positive (otherwise
is negative).
I really don’t understand these ray diagrams!
If, like many students, you don’t understand ray diagrams, the
following supplementary (will not be covered in lecture) slides
may help.
For a concave lens, the center of curvature and focal point are
on the same side of the lens as the object.
C
Concave Lens
F
Light coming from the object parallel to the axis will always
reflect through F.
C
Concave Lens
F
Light coming from the object and passing through F before it
hits the mirror will always reflect parallel to the axis.
C
Concave Lens
F
Light coming from the object and passing through C before it
hits the mirror will always reflect back on itself.
C
Concave Lens
F
Light coming from the object and striking the vertex will reflect
off with an outgoing angle equal to the incoming angle. This is
often more difficult to draw (unless you measure the angle).
C
Concave Lens
For a convex lens, the center of curvature and focal point are
on the opposite side of the lens as the object. Light from the
object will never actually pass through C or F.
F
Convex Lens
C
Light coming from the object parallel to the axis will always
reflect back as if it had come from F.
F
Follow the path of the light back “through” the
mirror to see where it appears to have come from.
Convex Lens
C
Light coming from the object and directed at F will always
reflect back parallel to the axis.
F
Follow the path of the light back “through” the
mirror to see where it appears to have come from.
Convex Lens
C
Light coming from the object and directed at C will always
reflect back on itself.
F
Follow the path of the light back “through” the
mirror to see where it appears to have come from.
Convex Lens
C
Light coming from the object and striking the vertex will reflect
back with an outgoing angle equal to the incoming angle. This
is often more difficult to draw (unless you measure the angle).
F
Follow the path of the light back “through” the
mirror to see where it appears to have come from.
Convex Lens
C
CW:
AN OBJECT PLACED 70CM IN FRONT OF A
SPHERICAL MIRROR FORMS A REAL
IMAGE AT A DISTANCE OF 40 CM FROM
THE MIRROR.
.a. Is the mirror concave or convex ?
b. What’s the mirror’s focal length ?
c. Is the image taller or shorter than the object
?





The fact that the image is real tells us that the
mirror is not convex, since convex mirrors
form only virtual images. The mirror is
Concave
so = 70cm si = 40 cm
1/ so + 1/ si = 1/f
1/70cm + 1/ 40cm = 1/f
1/f = .039 f = 25.64 cm
74. Diffuse Reflection
Diffuse reflection is the reflection of light from a
rough surface.
Each ray obeys the law of reflection.
The many different angles that incident light rays encounter at the surface cause
reflection in many directions.
29.4 Diffuse Reflection
Diffuse reflection allows us to see most things around us.
75. Light is diffusely reflected from paper in many directions.
76. Light incident on a smooth mirror is only reflected in one direction.
29.6 77. Refraction
When a wave that is traveling at an angle
changes its speed upon crossing a boundary
between two media, it bends.
29.6 Refraction
When a wave that is traveling at an angle
changes its speed upon crossing a boundary
between two media, it bends.
Refraction is the bending of a wave as it
crosses the boundary between two media at
an angle.
29.6 Refraction
Water waves travel faster in deep water than
in shallow water.
a. The wave refracts at the boundary where
the depth changes.
29.6 Refraction
Water waves travel faster in deep water than in shallow water.
a. The wave refracts at the boundary where the depth
changes.
b. The sample ray is perpendicular to the wave front it
intersects.
29.6 Refraction
In drawing a diagram of a wave, it is convenient to draw
lines, called wave fronts, that represent the positions of
different crests.
• At each point along a wave front, the wave is moving
perpendicular to the wave front.
• The direction of motion of the wave is represented by
rays that are perpendicular to the wave fronts.
• Sometimes we analyze waves in terms of wave
fronts, and at other times in terms of rays.
29.6 Refraction
What causes a wave to bend?
29.7 Refraction of Sound
Sound waves are refracted when parts of a
wave front travel at different speeds.
29.7 Refraction of Sound
Sound refraction occurs in uneven
winds or when sound is traveling
through air of uneven temperature.
• On a warm day the air near
the ground may be
appreciably warmer than the
air above.
• Sound travels faster in warmer
air, so the speed of sound
near the ground is increased.
• The refraction is not abrupt
but gradual.
• Sound waves tend to bend
away from warm ground,
making it appear that the
sound does not carry well.
29.7 Refraction of Sound
When the layer of air near the ground is colder than the air
above, the speed of sound near the ground is reduced.
The higher speed of the wave fronts above causes a bending
of the sound toward Earth.
Sound can then be heard over considerably longer distances.
29.8 Refraction of Light
Changes in the speed of light as it passes
from one medium to another, or variations in
the temperatures and densities of the same
medium, cause refraction.
29.8 Refraction of Light
Due to the refraction of light:
• swimming pools appear shallower,
• a pencil in a glass of water appears bent,
• the air above a hot stove seems to shimmer, and
• stars twinkle.
The directions of the light rays change because of refraction.
29.8 Refraction of Light
As a light wave passes from air into
water, its speed decreases.
29.8 Refraction of Light
When light rays enter a medium in which their speed
decreases, as when passing from air into water, the rays bend
toward the normal.
When light rays enter a medium in which their speed
increases, such as from water into air, the rays bend away
from the normal.
The light paths are reversible for both reflection and refraction.
If you can see somebody in a reflective or refractive device,
such as a mirror or a prism, then that person can see you by
looking through the device also.
Refraction
Example
Incident
Ray
Less Rigid
Medium
More Rigid
Medium
Refracted
ray bends
towards
the normal.
Refracted Ray
Index of Refraction, n
n=c/v
c : the speed of light in a vacuum,
3 x 108 m/sec
 v : speed of light in the medium.
 n : medium's index of refraction

Indices of Refraction
Vacuum
Air
Water
Ethanol
Crown glass
Quartz
Diamond
1.00
1.0003
1.33
1.36
1.52
1.52
2.42
Note

The speed of light has a lower speed in
a more optically dense medium.
Problem



What is the speed of light in quartz?
V = c/n = 3X108 m/s / 1.52
V = Answer: 1.97 x 10 8 m/s
Snell’s Law

n1 sin q1 = n2 sin q2
Problem





What is the angle of refraction when a
ray from air with an angle of incidence
of 25 o is incident to water?
Draw the ray diagram.
n1 sin θ1 = n2 sin θ2
1.0003 sin 25o = 1.33 sin θ2
Answer: 18.5 o
Total Internal Reflection




Can occur when ray goes from higher n to
lower n.
Above a Critical angle (of incidence) the ray
is reflected, not refracted
http://www.walter-fendt.de/ph14e/refraction.htm
For problems, set the angle of refraction to
90, and solve for critical angle
Problem





To solve for critical angle use the angle
of refraction = 900
Find the critical angle for a light ray
that is incident from water to air.
n1 sin θ1 = n2 sin θ2
1.33 sin θ1 = 1.0003 sin 90
θ1 = 48.80
Dispersion

The index of refraction of glass is different
for the colors that make up white light
because the speed of light is slightly
different in glass for each frequency of
light. (In vacuum all colors have speed
c=3x108m/s.)
CW : Problem
The index of refraction for crown glass
for red light is 1.514.
What is the speed of red light in crown
glass?
CW : Problem
The index of refraction for crown glass
for red light is 1.514.
What is the speed of red light in crown
glass?
Answer: 1.98 x 10
8
m/s
CW: Refraction
A beam of light in air is incident upon a piece of
glass , striking the surface at an angle of 300. If
the index of refraction of the glass is 1.5 , what
are the angles of reflection and refraction ?
CW: Refraction
A beam of light in air is incident upon a piece of
glass , striking the surface at an angle of 300. If
the index of refraction of the glass is 1.5 , what
are the angles of reflection and refraction ?
Law of Reflection : 600
Snell’s Law : 350
n1Sin60o = n2sin θ2
1.0003 (.866) = 1.5 sin θ2
Angle of refraction = 19.2o Angle of reflection
= 30o
Diffraction
AP Physics B
Superposition ..AKA….Interference
One of the characteristics of a WAVE is the ability to
undergo INTERFERENCE. There are TWO types.
We call these waves IN PHASE.
We call these waves OUT OF
PHASE.
Diffraction
When light OR sound is produced by TWO sources a
pattern results as a result of interference.
Interference Patterns
Diffraction is normally taken to refer to various phenomena which occur
when a wave encounters an obstacle. It is described as the apparent
bending of waves around small obstacles and the spreading out of waves
past small openings
Diffraction – The Central Maximum
Suppose you had TWO sources
each being allowed to emit a
wave through a small opening or
slit.
The distance between the slits
denoted by, d.
The distance from the slit spacing
to the screen is denoted by the
letter, L.
If two waves go through the slit and then proceed straight ahead to the
screen, they both cover the SAME DISTANCE and thus will have constructive
interference. Their amplitudes will build and leave a very bright intense spot
on the screen. We call this the CENTRAL MAXIMUM.
Diffraction – The Central Maximum
Figure 2
Figure 1
Here is the pattern you will see. Notice in figure 2 that there are several
bright spots and dark areas in between.
The spot in the middle is the BRIGHTEST and thus the CENTRAL
MAXIMUM. We call these spots FRINGES.
Notice we have additional bright spots, yet the intensity is a bit less. We
denote these additional bright spots as ORDERS.
So the first bright spot on either side of the central maximum is called the
FIRST ORDER BRIGHT FRINGE. Figure 1 represents the intensity of
the orders as we move farther from the bright central maximum.
Diffraction – Orders, m
We use the letter, m, to
represent the ORDER of
the fringe from the bright
central.
Second Order
Bright Fringe
First Order
m=2
Bright Fringe
m=1
First Order
Central
Bright Fringe
Maximum
m=1
Second Order
Bright Fringe
m=2
It is important to understand that we see these bright fringes as a result of
CONSTRUCTIVE INTERFERENCE.
Diffraction – Bright Fringes
The reason you see additional
bright fringes is because the
waves CONSTRUCTIVELY build.
There is a difference however in
the intensity as you saw in the
previous slide.
As you can see in the picture, the
BLUE WAVE has to travel farther
than the RED WAVE to reach the
screen at the position shown.
For the BLUE WAVE and the
RED WAVE to build constructively
they MUST be IN PHASE.
Here is the question: HOW MUCH FARTHER DID THE BLUE WAVE HAVE TO
TRAVEL SO THAT THEY BOTH HIT THE SCREEN IN PHASE?
Diffraction – Path difference
Notice that these 2 waves are IN
PHASE. When they hit they screen
they both hit at the same relative
position, at the bottom of a crest.
How much farther did the red wave
have to travel?
l
Exactly – ONE WAVELENGTH
The call this extra distance the PATH DIFFERENCE. The path difference and
the ORDER of a fringe help to form a pattern.
Diffraction – Path Difference
2l
1l
2l
1l
The bright fringes you see
on either side of the central
maximum are multiple
wavelengths from the bright
central. And it just so
happens that the multiple is
the ORDER.
Therefore, the PATH DIFFERENCE is equal
to the ORDER times the WAVELENGTH
P.D. = ml
(Constructive)
Diffraction – Dark Fringes
We see a definite DECREASE in intensity between the bright fringes. In
the pattern we visible notice the DARK REGION. These are areas where
DESTRUCTIVE INTERFERENCE has occurred. We call these areas
DARK FRINGES or MINIMUMS.
Diffraction – Dark Fringes
First Order
Dark Fringe
m=1
ZERO Order
Dark Fringe
m=0
ZERO Order
Central
Dark Fringe
Maximum
m=0
First Order
Dark Fringe
m=1
It is important to understand that we see these dark fringes as a result of
DESTRUCTIVE INTERFERENCE.
Diffraction – Dark Fringes
On either side of the bright
central maximum we see areas
that are dark or minimum
intensity.
Once again we notice that the
BLUE WAVE had to travel
farther than the RED WAVE to
reach the screen. At this point ,
however, they are said to
destructively build or that they
are OUT OF PHASE.
Here is the question: HOW MUCH FARTHER DID THE BLUE WAVE HAVE TO
TRAVEL SO THAT THEY BOTH HIT THE SCREEN OUT OF PHASE?
Diffraction – Path difference
Notice that these 2 waves are OUT OF
PHASE. When they hit they screen they
both hit at the different relative positions,
one at the bottom of a crest and the other
coming out of a trough. Thus their
amplitudes SUBTRACT.
0.5 l
How much farther did the red wave have to
travel?
Exactly – ONE HALF OF A WAVELENGTH
The call this extra distance the PATH DIFFERENCE. The path difference and
the ORDER of a fringe help to form another pattern.
Diffraction – Path Difference
1.5l
1.5l
0.5l 0.5l
The dark fringes you see on
either side of the central
maximum are multiple
wavelengths from the bright
central. And it just so
happens that the multiple is
the ORDER.
Therefore, the PATH DIFFERENCE is equal
to the ORDER plus a HALF, times A
WAVELENGTH
P.D. = (m+1/2)l
(Destructive)
Path Difference - Summary
For CONSTRUCTIVE INTERFERENCE or
MAXIMUMS use:
P.D. = ml
For DESTRUCTIVE INTERFERENCE or
MINIMUMS use:
P.D. = (m+1/2)l
Where “m”, is the ORDER. Refer to slides 7 and 12.
Young’s Experiment
In 1801, Thomas Young successfully showed that light does produce an
interference pattern. This famous experiment PROVES that light has WAVE
PROPERTIES.
Suppose we have 2 slits separated by a distance, d and a
distance L from a screen. Let point P be a bright fringe.
P
y
d
q
B.C.
We see in the figure that we can
make a right triangle using, L, and
y, which is the distance a fringe is
from the bright central. We will
use an angle, q, from the point in
the middle of the two slits.
We can find this angle using
tangent!
L
Diffraction – Another way to look at
This right triangle is
“path difference”
SIMILAR to the one
made by y & L.
P
q
d
q
B.C.
d
P.D.
P.D. = dsinq
P.D.
Notice the blue wave travels farther. The
difference in distance is the path difference.
Similar Triangles lead to a path difference
y
q
These angles
Are EQUAL.
d
L
dsinq
q
Diffraction – Putting it all together
Path difference is equal to the following:
 ml
 (m+1/2)l
 dsinq
Therefore, we can say:
Will be used to find the
angle!
Example
A viewing screen is separated from a 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 central
maximum. Determine the wavelength of light.
L  1.2m d  3.0 x10 5 m q  tan 1 ( y )  tan 1 ( 0.045 ) 
L
1.2
m  2 y  0.045m
" bright"  Constructi ve
l ?
d sin q  ml
(3x10 5 ) sin( 2.15)  2l
l
5.62x10-7 m
2.15 degrees
Example
A light with wavelength, 450 nm, falls on a diffraction grating (multiple slits).
On a screen 1.80 m away the distance between dark fringes on either side
of the bright central is 4.20 mm. a) What is the separation between a set of
slits? b) How many lines per meter are on the grating?
l  450 x10 9 m L  1.80m
y  0.0021m q  ?
0.0021
q  tan (
)
1.8
1
0.067 degrees
d ?
d sin q  (m  1 )l
2
d sin( 0.067)  (0  1 )450 x10 9
2
d  0.0001924 m
meters
d
line
lines
1
1
 d or  5197.2 lines/m
meter
d
Mirage
http://www.warrenwilson.edu/~physics/PhysPhotOfWeek/2007PPOW/20070921RoadMirage/IMG_
1444MirageCrop500.jpg
Mirage
Total internal reflection occurs because
hot air has a lower n, than cold air.
http://www.warrenwilson.edu/~physics/PhysPhotOfWeek/2007PPOW/20070921RoadMirage/index.html
29.8 Refraction of Light
The laser beam bends toward the normal when it enters the water, and away from
the normal when it leaves.
29.8 Refraction of Light
a.
The apparent depth of the glass block is less than the real depth.
29.8 Refraction of Light
a.
b.
The apparent depth of the glass block is less than the real depth.
The fish appears to be nearer than it actually is.
29.8 Refraction of Light
a.
b.
c.
The apparent depth of the glass block is less than the real depth.
The fish appears to be nearer than it actually is.
The full glass mug appears to hold more root beer than it actually does.
These effects are due to the refraction of light whenever it crosses a
boundary between air and another transparent medium.
29.8 Refraction of Light
What causes the refraction of light?
29.9 Atmospheric Refraction
A mirage is caused by the refraction of light
in Earth’s atmosphere.
29.9 Atmospheric Refraction
The speed of light in air is only 0.03% less than c, but in some situations,
atmospheric refraction is quite noticeable.
A distorted image, called a mirage, is caused by refraction of light in
Earth’s atmosphere.
• A layer of very hot air is in contact with the ground on very hot days.
• Light travels faster through it than through the cooler air above.
• The speeding up of the part of the wave nearest the ground
produces a gradual bending of the light rays.
• Light is refracted.
29.9 Atmospheric Refraction
29.9 Atmospheric Refraction
Wave fronts of light travel faster in the hot air near the ground, thereby bending the
rays of light upward.
29.9 Atmospheric Refraction
A motorist experiences a similar situation when driving along a hot road
that appears to be wet ahead.
The sky appears to be reflected from a wet surface, but, in fact, light from
the sky is being refracted through a layer of hot air.
A mirage is not a “trick of the mind.”
A mirage is formed by real light and can be photographed.
29.9 Atmospheric Refraction
When you watch the sun set, you see the sun for several minutes after it
has really sunk below the horizon.
Since the density of the atmosphere changes gradually, refracted rays
bend gradually to produce a curved path.
The same thing occurs at sunrise, so our daytimes are about 5 minutes
longer because of atmospheric refraction.
29.9 Atmospheric Refraction
When the sun is near the
horizon, the rays from the
lower edge are bent more than
the rays from the upper edge.
This produces a shortening of
the vertical diameter and
makes the sun look elliptical
instead of round.
Atmospheric refraction
produces a “pumpkin” sun.
29.9 Atmospheric Refraction
think!
If the speed of light were the same for the various temperatures
and densities of air, would there still be mirages?
29.9 Atmospheric Refraction
think!
If the speed of light were the same for the various temperatures
and densities of air, would there still be mirages?
Answer:
No! There would be no refraction if light traveled at the same
speed in air of different temperatures and densities.
29.9 Atmospheric Refraction
What causes the appearance of a mirage?
Light’s Nature

Wave nature (electromagnetic wave)

Particle nature (bundles of energy called
photons)
Past- Separate Theories of Either
Wave or Particle Nature




Corpuscular theory of Newton (1670)
Light corpuscles have mass and travel at
extremely high speeds in straight lines
Huygens (1680)
Wavelets-each point on a wavefront acts as a
source for the next wavefront
Why was it difficult to prove the
wave part of the nature of light?
Proofs of Wave Nature

Thomas Young's Double Slit Experiment (1807)
bright (constructive) and dark (destructive)
fringes seen on screen

Thin Film Interference Patterns

Poisson/Arago Spot (1820)

Diffraction fringes seen within and around a
small obstacle or through a narrow opening
Proof of Particle Nature:
The Photoelectric Effect



Albert Einstein 1905
Light energy is quantized
Photon is a quantum or packet of energy
The Photoelectric Effect


Heinrich Hertz first observed the
photoelectric effect in 1887
Einstein explained it in 1905 and won the
Nobel prize for this.
Thomas Young’s Double Slit
Interference Experiment


Showed an interference
pattern
Measured the wavelength
of the light
Two Waves Interfering
Young’s Double Slit
Interference Pattern
http://galileo.phys.virginia.edu/classes/USEM/SciImg/home_files/introduction_files/doubleslit.jpg
Interference of Waves From
Two Sources

Simulation “Ripple Tank”
http://www3.interscience.wiley.com:8100/lega
cy/college/halliday/0471320005/simulations6
e/index.htm?newwindow=true
Interference
Young’s Double Slit Interference
 http://galileo.phys.virginia.edu/classes
/109N/more_stuff/flashlets/youngexpt
4.htm

For Constructive
Interference:
The waves must arrive to
the point of study in
phase.
So their path difference
must be integral
multiples of the
wavelength:
DL= ml
m=0,1,2,3,………
For destructive interference:
, the waves must arrive
to the point of study
out of phase.
So the path difference
must be an odd
multiple of l/2:
DL= m l
m=1/2,3/2,5/2,….
Typical Question

Where is the first location of constructive or
destructive interference?
Fo Constructive Interference of Waves from
Two Sources
x=Ltanq
sinq DL/d
DL=nl
x
For small angles:
Lsinq~Ltanq
q
d
q
dsinqnl
nl  dx
L
L
n=0,1,2,3,…
Double Slit Interference
dsinqnl
nl  dx
L
Constructive (brights) n=0,1,2,3,…..
Destructive (darks) n=1/2, 3/2, 5/2,…..
Note:
To find maximum # of fringes set q to 90o for n.
Question

How does x change with wavelength?

How does x change with slit distance?
Problem
Two slits are 0.05 m apart. A laser of
wavelength 633nm is incident to the slits.
A screen is placed 2m from the slits.
a) Calculate the position of the first and second
bright fringe.
b) What is the maximum number of destructive
interference spots there can be on either side
of the central maximum?
Diffraction Grating
http://des.memphis.edu/lurbano/vpython/matter_interactions/spectrum/spectrum_02.jpg
Diffraction Grating


Large number of equally spaced parallel slits.
Equations are same as for double slit interference
but first calculate the d (slit separation) from the
grating density, N.
d=1/N , N slits per unit length
dsinqnl
nl  dx
L
Constructive (brights) n=0,1,2,3,…..
Destructive (darks) n=1/2, 3/2, 5/2,…..
Problem
A neon laser of wavelength 633nm is pointed
at a diffraction grating of 3000lines/cm. Find the
angle where the first bright occurs.
(Hint: slit separation d is inverse of grating
density)
Diffraction
Wave bends as it passes an obstacle.
Diffraction through a Narrow Slit
Each part of the slit acts as a point source
that interferes with the others.
(Based on Huygens Principle)
Pattern of Diffraction of Light
through a Narrow Slit
x
w
L
Intensity of the Diffraction
Interference Patterns

Simulation “Interference of Light”

http://www3.interscience.wiley.com:8100/lega
cy/college/halliday/0471320005/simulations6
e/index.htm?newwindow=true
Diffraction from Narrow Slit
wsinqnl
l nw y
L
w: is the width of the slit
Destructive (dark fringes): m=0,1,2,3,….
Questions

How does x change with the width?

How does x change with the wavelength
Diffraction around a Penny and
Poison Spot
Example of Diffraction