Waves and Optics
Introduction
The energy that generated the wave that moves
this surfer was transmitted to the water many
miles away. But the mass of water that originally
received that energy did not move all those miles
to the shore. Rather, it transmitted that energy to
the next mass of water, which
transmitted it to the next.
In waves, energy is
transmitted, not matter.
There are many different kinds
of waves. They include:
waves on the surface of
water,…
…waves on strings,…
…waves within the earth,…
…sound waves,…
…light waves,…
…radio waves,…
…and x-rays.
All of these types of waves transfer energy
without transferring matter. This chapter will
explore aspects of these waves that are
similar to and different from each other.
Mechanical & Electromagnetic Waves
There are two fundamental types of waves:
Mechanical waves and Electromagnetic
waves. What makes these two types of
waves different are their sources and
mediums.
Mechanical Waves have matter as their
source and matter as their medium. As we
have already seen, a guitar string wave has a
finger as its source and the string as its
medium. The wave is propagated as one
particle puts a force on the next particle,
causing it to move and continue the wave.
Sound is a Mechanical Wave. Sound
waves have as their source vibrating
matter. In the case of the horn players,
their vibrating lips start the sound wave.
For a singer, vibrating vocal cords are
the source of the sound wave.
Sound also travels through solids and
liquids. In fact, it travels faster and with less
loss the denser the material. Particles are
closer together allowing for greater elastic
qualities than air.
Scratch on your desk or table so
that you can just barely hear it.
Then put your ear to the desk.
You should be able to
hear the scratching loud and
clear.
The Medium of sound is also matter, in this
case, air. As one air molecule collides with
the next, which collides with the next, sound
is propagated through space.
On the other hand, sound,
like all mechanical waves,
cannot travel in a vacuum.
Sound needs matter to be
transmitted, particles
exerting forces on other
particles. In a vacuum,
there are no particles, so
sound cannot be
transmitted.
Experiments have shown that if you have a
bell in a jar, the bell can be heard. But if
you evacuate the jar of all air, the bell can
no longer be heard.
http://www.youtu
be.com/watch?v=
ce7AMJdq0Gw
On the other hand, the bell itself can
clearly be seen, air or no air. This means
that light is not a Mechanical Wave – it is
instead an Electromagnetic Wave.
Electromagnetic waves include light, radio
waves, microwaves, infrared radiation,
ultraviolet light, x-rays and gamma rays.
The source of Electromagnetic waves is
vibrating charge. In most cases, that
means vibrating electrons.
Because they do not rely on matter for
propagation, Electromagnetic Waves can
travel through vacuum. All Electromagnetic
Waves travel through a vacuum at the speed
of light, ‘c’, where …
c = 3.0 x 108 m/s
When light does travel through matter,
such as water or glass, it travels more
slowly. And different waves travel at
different speeds. The speed of light,
‘c’, is for Electromagnetic Waves moving
through a vacuum only.
A WAVE is a vibration or disturbance in
space that propagates from one place
to another.
A MEDIUM is the substance that all SOUND
WAVES travel through and need to have in order
to move.
Two Types of Waves
The first type of wave is called Longitudinal.
Longitudinal Wave – the displacement of
individual particles is in the same direction
as the direction of the wave propagation
Two Areas of the Wave:
Compression- an area of high molecular
density and pressure.
Rarefaction - an area of low molecular
density and pressure.
Longitudinal waves are mechanical waves.
The second type of wave is called
Transverse.
Transverse Wave – the displacement is at
right angles to the direction of the
propagation.
Light and Radio Waves are examples.
In the animation of the motion of a particle in
a transverse wave, the wave moved left to
right and the particle moved up and down.
Together, they form a plane – the plane of
the screen you are viewing this animation
on.
Direction of Wave Velocity
Direction of Motion of the Medium
Comparing Longitudinal and Transverse Waves
http://www.youtube.com/watch?v=Rbuhdo0AZDU
What about a water wave?
A cork in water
travels both
Horizontally and
Vertically. In that
sense, water has
characteristics of
both longitudinal
and transverse
waves.
λ = wavelength – distance between wave
repeats.
T = Period – time required for 1 one wave to
pass a fixed point.
Units = seconds
f = frequency – waves passing a fixed point
in a given time period. f = 1/T
s-1
1
s
Units = Hertz
The frequency of a sound wave is perceived
as pitch. The higher the frequency, the
higher the pitch. For instance a trombone
has a lower frequency and pitch than a
flute.
Low Frequency
And Pitch
High Frequency
And Pitch
d
(m)
d
(m)
t
(s)
t
(s)
The frequency of a light wave is perceived
as color. The higher the frequency, the
closer to the violet end of the spectrum.
The lower the frequency, the closer to the
red end. White light is a mixture of all
frequencies at once.
d
(m)
d
(m)
t
(s)
t
(s)
Low f
Red
High f
Violet
The greater the frequency of a light wave,
the shorter the wavelength.
d
(m)
t
(s)
d
(m)
t
(s)
Amplitude = height of the wave
There are a number of measurements that
can be made on a wave. For mechanical
waves, the Amplitude is the maximum
distance the medium deviates from its
equilibrium position. (click to animate)
Amplitude
Amplitude
The Power/Energy of a wave is
proportional to the square of the
Amplitude. The greater the Amplitude,
the more power in the wave. The greater
the Amplitude of an earthquake wave, the
more destructive the earthquake.
The greater the Amplitude of a sound
wave, the greater the volume.
The greater the Amplitude of a light wave,
the brighter or more intense it is.
Speed of a Wave (m/s)
v=λ
T
v = λf
Sound waves (longitudinal waves)
traveling in air with a speed of
343 m/s. The lowest frequency of
sound we can hear is 20 Hz. The
highest is 20.0 kHz. Find their
wavelengths.
v=λ
f
v=λ
f
343m/s = λ
20 s-1
λ = 17.2 m
343 m/s = λ
20,000 s-1
λ = 0.0172 m
or … λ = 1.72 cm
Factors that influence the speed of sound
are density of solids or liquid, and
temperature.
How long will it take a shout to travel 75.0 m
across a football field on an evening when
the temperature is 18OC?
vsound = 331 m/s + 0.6 m/s (18OC) = 341.8 m/s
OC
t = d
v
t = 75.0
m
341.8 m/s
t = .219 s
Speed in a Solid
v = є
ρ
ε = modulus of elasticity
ρ = density
= N/m2
= kg/m3
Modulus of Elasticity is the mathematical
description of an object or substance's
tendency to be deformed elastically (i.e., nonpermanently) when a force is applied to it.
What is the speed of sound in a sample of
aluminum that has an є of 69 x 109 N/m2
and a density of 2700 kg/m3?
v = є
ρ
v =
69 x 109 N/m2
2700 kg/m3
v = 5055 m/s
Speed in a Liquid
v = β
ρ
β
ρ
= bulk modulus
= density
= N/m2
= kg/m3
Bulk Modulus of a substance measures
the substance's resistance to uniform
compression.
What is the speed of sound in a
sample of ethanol alcohol that has an
β of 1.02 x 109 N/m2 and a density of
789 kg/m3 at 20.0OC?
v = β
ρ
v =
1.02 x 109 N/m2
789 kg/m3
v = 1137 m/s
Miscellaneous Sound Information
Supersonic speed is a rate of travel
of an object that exceeds the speed
of sound (Mach 1). For objects
traveling in dry air of a temperature
of 20°C (68°F) this speed is
approximately 343 m/s (1,125 ft/s,
768 mph or 1,236 km/h). Speeds
greater than five times the speed of
sound (Mach 5) are often referred to
as hypersonic.
A sonic boom is the sound associated
with the shock waves created by the
supersonic flight of an aircraft. Sonic
booms generate enormous amounts of
sound energy, sounding much like an
explosion.
When an object passes through the air, it
creates a series of pressure waves in front of it
and behind it, similar to the bow and stern
waves created by a boat. These waves travel at
the speed of sound, and as the speed of the
object increases, the waves are forced together,
or compressed, because they cannot "get out of
the way" of each other, eventually merging into
a single shock wave at the speed of sound.
This critical speed is known as Mach 1 and is
approximately 1,225 km/h (761 mph) at sea level
and 20 °C (68 °F).
Ultrasound and what we hear …
The sound we normally hear is from
20 to 20 000 cycles per second (Hz).
Ultrasound means sound that has a
higher frequency than our normal
hearing. Ultrasound used for medical
purposes is from one MHz (one million
cycles per second) to 20 MHz.
Sea Navigation and Sonar
Sonar is a device that uses sound energy to
locate objects; measure their distance,
direction, and speed; and produce pictures
of them.
Sonars generally use frequencies in the 20 to
20,000 Hertz range.
Sound Waves
Sound Waves are a common type of
standing wave as they are caused by
RESONANCE.
Resonance – when a FORCED vibration
matches an object’s natural frequency
thus producing vibration, sound, or even
damage.
One example of this involves shattering a
wine glass by hitting a musical note that is on
the same frequency as the natural frequency
of the glass.
Natural frequency depends on the size, shape,
and composition of the object in question.
Because the frequencies resonate, or are in
sync with one another, maximum energy
transfer is possible.
Electromagnetic
Spectrum
Our eyes can only
see certain wavelengths.
c=λf
f=c
λ
•What is the frequency of a
radio wave that has a length
of 1.67 X 1012nm?
f=c
λ
= 3.00 X
8
10
m/s
1670 m
f = 1.80 X 105/s
Note: changed nm to m:
(1 m = 1 x 109nm)
(Hz)
Find the frequency of a
gamma wave that has a
length of 0.00300 nm.
=
f=c
λ
3.00 X 108 m/s
3.00 X 10-12 m
f
= 1.00 X 1020/s (Hz)
•Comparing ...
long radio wave = 1.80 X 105/s
20
short gamma
= 1.00 X 10 /s
•What are the wavelengths of these
two radio stations?
KNYE broadcasts at 95.1 MHz (FM)
KDKA broadcasts at 1020 KHz (AM)
KNYE = 3.00 X 108 m/s = 3.15 m
6
95.1 X 10 Hz
MHz = 106 Hz
KDKA = 3.00 X 108 m/s = 294 m
1020 X 103 Hz
KHz = 103 Hz
•
•
•
•
AM broadcast 500 KHz - 1600 KHz
FM broadcast 88 MHZ - 108 MHz
Cordless phones 46 MHz - 49 MHz
TV remote operates in the Infrared
region of the EM (electromagnetic)
spectra.
Wave
Hz
Meters
Water
5 x 10-2
120
7 x 10-7
Red Light
4.3 x 1014
Violet Light
7.5 x 1014
4 x 10-7
• As the frequency goes up:
–wavelength decreases.
–wave becomes more energetic.
Doppler Shift
When you hear the sound of an object
moving toward you—such as a siren from
and emergency vehicle—the pitch of that
sound is higher than if the vehicle was
standing still.
(f is greater and a shorter λ)
Likewise, as the vehicle moves away from
you, the siren would have a lower pitch.
(f is less and a greater λ)
Object is moving towards the right.
What would happen if the source moved quicker?
f would increase and λ would decrease.
Doppler Effect on a Moving Observer
f’ = ( 1 +/- u ) f
v
f = normal frequency
v = speed of wave
f ‘= new frequency
u = speed of observer
+ = observer moving closer to source
- = observer moving away from source
A street musician sounds the A string of his violin
producing a tone of 440 Hz. What frequency does a
bicyclist hears as he (A) approaches and (B)
recedes from the musician at 11.0 m/s?
f’ = ( 1 +/- u ) f
v
A=+
B=-
f’A = 454 Hz
f’B = 426 Hz
Doppler Effect on a Moving Source
1
1 -/+ u
v
(
(
f’ =
f
v = speed of wave
u = speed of source
- = source moving towards observer
+ = source moving away from observer
A police car’s siren produces a frequency
of 1600 Hz when at rest. (air temperature
20OC) What is the frequency detected if
the car is moving …
A) toward an observer at 25 m/s?
B) away from an observer at 25 m/s?
v = 331 m/s + 0.6 m/s (20OC)
OC
v = 343 m/s
1 -/+ u
v
f’ =
(
1 -
f
1_______
25 m/s
343 m/s
f’ = 1726 Hz
(
1
(
(
f’ =
1600 Hz
(
1 +
1_______
25 m/s
343 m/s
f’ = 1491 Hz
(
f’ =
1600 Hz
Nodes and Antinodes
Superposition and Interference
Superposition: Combination of two or
more waves to form a resultant wave
y = y1 + y2
Constructive Interference: When waves
combine and the resulting pulse has an
amplitude equal to the sum of the amplitude
of the individual waves.
The net displacement is the sum of the two
individual displacements.
They pass through each other without
being disturbed.
Destructive Interference: When the positive
displacement of one waves combines with the
negative displacement of another to create a
net displacement of zero.
In and Out of Phase
These two waves are in phase, meaning that
every point on the top wave is at the same
part of the wave cycle as the same point on
the bottom wave.
The crests line up with crests and the
troughs line up with troughs.
These two waves are out of phase with
each other. None of the points on the top
wave are at the same part of the wave
cycle as the same point on the bottom
wave.
These two waves are exactly out of phase
with each other. Notice that crests all line
up with troughs. This is called being 180º
out of phase.
Photoelectric Effect
The phenomenon of emission of electrons
from mainly metal surfaces exposed to light
energy (X – rays, UV rays, Visible light and
even IR rays) of suitable frequency is
known as photoelectric effect.
Visible light
Photoelectrons
Alkali Metals
The electrons emitted by this effect are
called photoelectrons.
The current constituted by photoelectrons
is known as photoelectric current.
The photoelectric effect demonstrates
that light has particle properties.
The photoelectric effect demonstrates
that light has particle properties.
… on the other hand
Diffraction
When light OR sound is produced by
TWO sources a pattern results as a result
of interference.
Diffraction of Light
http://www.youtube.com/watch?v=-mNQW5OShMA&NR=1
Interference Patterns
Diffraction … occurs 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 a
source emitting a
wave through a small
opening or slit.
The distance between the slits = d.
The distance from the slit spacing to the
screen = L.
If two waves go through the slit and act
constructively, their amplitudes will build
and leave a very bright intense spot on the
screen. We call this the CENTRAL
MAXIMUM.
Figure 1
Figure 2
In figure 2 there are several bright spots and
dark areas in between. The spot in the middle
is the BRIGHTEST and thus the CENTRAL
MAXIMUM.
These spots are called FRINGES.
Each additional fringe is a little less intense.
These additional bright spots are known as
ORDERS.
The first bright spot on either side of the
central maximum is called the FIRST ORDER
BRIGHT FRINGE.
Figure shows the intensity of
the orders as we move farther
from the bright central
maximum.
Orders, m
Second Order
Bright Fringe
First Order
m=2
Bright Fringe
m=1
m, represents
the ORDER of
the fringe from
the bright
central.
First Order
Central
Bright Fringe
Maximum
m=1
Second Order
Bright Fringe
m=2
Bright Fringes = CONSTRUCTIVE INTERFERENCE.
Dark Fringes
In the pattern there are dark the DARK
REGIONS. These are areas where
DESTRUCTIVE INTERFERENCE has
occurred.
These areas DARK FRINGES or MINIMUMS.
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
Dark Fringes = DESTRUCTIVE INTERFERENCE.
d
If the fringe is further
out, Ө changes.
X
X
X
X
X Ө
d
X
Two slits with a separation of (d) 8.5 x 10-5 m
create an interference pattern on a screen (L)
2.3 m away. If the 10th bright fringe above the
central fringe is a linear distance of 12 cm from it,
what is the wavelength of the light used in the
experiment?
TanӨ = 0.12 m
2.3 m
Ө = 3.0O
10th bright fringe …
λ = d sinӨ
m
λ = (8.5 x 10-5 m) sin3.0
10
λ = 4.4 x 10-7 m or 440 nm
Light, Reflection, & Mirrors
The Law of “REFLECTION”
Law of Reflection … the angle of
incidence (incoming ray) equals the
angle of reflection (outgoing ray)
NORMAL
Angles are
measured from a
perpendicular line
to the surface
called a NORMAL.
The Law of Reflection is always
observed (regardless of the orientation
of the surface)
Two Types of Reflection
Specular Reflection –
reflection off a
smooth shiny
surface.
Notice the symmetry.
Diffuse Reflection –
reflection off a rough
surface.
Notice the
Light
reflected in
sXymmetry.
many directions
Plane Mirror
A flat , plane mirror is 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?
Same side as
the object?
Behind the mirror?
Object Distance, Do = 10 cm
Behind the mirror.
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 images that are formed in
locations where light does not actually reach.
Light does not actually pass through the
location on the other side of the mirror; it
only appears to an observer as though the
light is coming from this location.
VIRTUAL IMAGES are ALWAYS on the
OPPOSITE side of the mirror relative to the
object.
Real Image
The point of convergence of the bundle of
rays emergent from the lens is a real
image of that object.
For MIRRORS
real images
always appear
on the SAME
SIDE of the
mirror as the
object.
Object
Image
X
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)
Spherical Mirrors – Concave &
Convex
X
X
X
… also called
X
DIVERGING
X
X mirror
X
X
X
X
… also called
X
CONVERGING
X
mirror
X
X
X
Looking
into a
X
cave
X
X
X
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. This
point is called the FOCAL POINT.
To find this point
you MUST use
light from
“infinity”
Light from an
“infinite” distance,
most likely the sun.
http://www.youtube.com/watch?v=np8lE
Nrge0Q&feature=player_embedded
Converging (Concave) Mirror
The mirror is
spherical. It has a
CENTER OF
CURVATURE, C.
The focal point
happens to be HALF
this distance.
Draw a line through the center of the mirror
and call it the PRINCIPAL AXIS.
f =
C
C= f
Ray Diagram
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.
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.
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.
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.
http://www.youtube.com/watch?v=KVpSCICCD9A&feature=player_embedded
2) Is the image ENLARGED or REDUCED?
3) Is the image INVERTED or RIGHT SIDE
UP?
The Mirror/Lens Equation
There is ANOTHER way to predict image
characteristics besides the ray diagram.
Use the MIRROR/LENS equation to
CALCULATE the position of the image.
1
1 1
=
+
f do di
A 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 do di
10cm 25cm d i
di =
16.67 cm
What does this tell us?
First we know the image is BETWEEN C
and 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?
Magnification Equation
To calculate the orientation and size of the
image we use the MAGNIFICATION
EQUATION.
d i hi
M =− =
d o ho
A POSITIVE magnification, the image is
UPRIGHT.
A NEGATIVE magnification, the image is
INVERTED
The magnification value is GREATER than 1,
the image is ENLARGED.
The magnification value is LESS than 1, the
image is REDUCED.
The magnification value is EQUAL to 1, the
image is the SAME SIZE as the object.
Using our previous data …
16.67cm
M =−
25cm
M = −0.67 x
…the image is INVERTED, and REDUCED.
The +/- Sign Conventions
The sign conventions for the given
quantities in the mirror equation and
magnification equations are as follows:
f is + if the mirror is a concave mirror
f is - if the mirror is a convex mirror
di is + if the image is a real image and located
on the object's side of the mirror.
di is - if the image is a virtual image and
located behind the mirror.
hi is + if the image is an upright image (and
therefore, also virtual)
hi is - if the image an inverted image (and
therefore, also real)
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 do di
10cm 5cm d i
d i = -10 cm
di
M =−
=
5cm
2
Characteristics?
Upright
di = -10 cm
M=+2
•VIRTUAL (opposite side)
Enlarged
Convex Mirrors
OBJECT
X
X
X
X
A convex mirror is sometimes referred to
as a diverging mirror due to the fact that
incident light originating from the same
point and will reflect off the mirror surface
and diverge.
The diagram shows an object placed in front of
a convex mirror. Light rays originating at the
object location are approaching and then
reflecting from the mirror. Observers must
sight along the line of a reflected ray to view
the image of the object.
Each ray is extended
backwards to a point of
intersection - this point of
intersection of all extended
reflected rays is the image
location of the object.
The image is virtual. Light does not pass through
to the image location.
It only appears to observers as though all the
reflected light from each part of the object is
diverging from this virtual image location.
Convex Mirrors – Ray Diagrams
The revised rules can be stated as follows:
Any incident ray traveling parallel to the
principal axis on the way to a convex mirror
will reflect in such a manner that its extension
will pass through the focal point.
Any incident ray traveling towards a convex
mirror such that its extension passes through
the focal point will reflect and travel parallel
to the principal axis.
1) Pick a point on the top of the object and draw two
incident rays traveling towards the mirror.
Draw a ray such that it travels exactly parallel to the
principal axis.
Draw one ray so that it travels towards the focal
point on the opposite side of the mirror.
2) Once these incident rays strike the mirror,
reflect them according to the two rules of
reflection for convex mirrors.
The ray that travels towards the focal point will
reflect and travel parallel to the principal axis. The
ray that traveled parallel to the principal axis on
the way to the mirror will reflect and travel in a
direction such that its extension passes through
the focal point.
3) Locate and mark the top of the image.
The image point of the top of the object is the
point where the two reflected rays intersect. Since
the two reflected rays are diverging, they must be
extended behind the mirror in order to intersect.
Draw the extensions until they intersect. The point
of intersection is the image point of the top of the
object. Both reflected rays would appear to
diverge from this point.
Characteristics of Convex Mirrors .
The diagrams show that in each case, the
image is located behind the convex mirror.
… a virtual image
… an upright image
reduced in size (i.e., smaller than the object)
Another characteristic of the images
of objects formed by convex mirrors
pertains to how a variation in object
distance affects the image distance
and size.
A 4.0-cm tall light bulb is placed a distance
of 35.5 cm from a convex mirror having a
focal length of -12.2 cm. Determine the
image distance and the image size.
ho = 4.0 cm
di = ???
do = 35.5 cm
hi = ???
f = -12.2 cm
1 1 1
=
+
f do di
1/(-12.2 cm) = 1/(35.5 cm) + 1/di
-0.0820 cm-1 = 0.0282 cm-1 + 1/di
-0.110 cm-1 = 1/di
di = -9.08 cm
To determine the image height (hi), the
magnification equation is needed.
hi/ho = - di/do
hi /(4.0 cm) = - (-9.08 cm)/(35.5 cm)
hi = (4.0 cm) • (9.08 cm)/(35.5 cm)
hi = 1.02 cm
The negative values for image distance
indicate that the image is located behind
the mirror.
In the case of the image distance, a
negative value always indicates the
existence of a virtual image located behind
the mirror.
In the case of the image height, a positive
value indicates an upright image.
From the calculations in this problem it
can be concluded that if a 4.0 cm tall
object is placed 35.5 cm from a convex
mirror having a focal length of -12.2 cm …
… The image will be upright and
… 1.02 cm tall and
… located 9.08 cm behind the mirror.
Convex mirrors always produce images
that are upright, virtual, reduced in size,
and located behind the mirror.
Lenses – The Mirror/Lens Equation
CALCULATE the image’s position and
characteristics using the same equations
used for mirrors.
An object is placed 35 cm in front of a
converging lens with focal length of
20 cm. Calculate the image’s position
relative to the lens as well as the image’s
characteristics.
1
1 1
=
+
f do di
di
M =−
do
1
1 1
=
+
20 35 d i
di
M =−
35
d i = 46.7 cm
M = -1.33x
This image is REAL (since the object
distance is positive) and on the OTHER side
of the lens. The image is INVERTED and
ENLARGED.
Refraction
Refraction is based on the idea that
LIGHT is passing through one
MEDIUM into another. The
question is, WHAT HAPPENS?
Suppose you are running on the beach
with a certain velocity when you suddenly
need to run into the water. What happens
to your velocity?
IT CHANGES!
Refraction Fact #1: As light goes
from one medium to another, the
velocity CHANGES!
Suppose light comes from air (vacuum),
and strikes a boundary at some angle of
incidence measured from a normal line
and goes into water.
The ratio of the two
speeds can be
compared.
The denominator will ALWAYS be smaller
and produce a unitless value greater or
equal to 1.
This value is called the new medium’s
INDEX OF REFRACTION, n.
All substances have an index of refraction
which can be used to identify the material.
Suppose you decide to go spear fishing,
but unfortunately you aren’t having much
luck catching any fish.
WHY?
Light BENDS when it reaches a new medium.
The object is NOT directly in a straight line
path, but rather it’s image appears that way.
The actual object is on either side of the
image you are viewing.
Refraction Fact #2: As light goes from
one medium to another, the path
CHANGES!
Speed of Light Outside a Vacuum
What is the speed of light in a sample of
water that has an index of refraction of
1.33?
v = c
n
v =
3.0 x 108 m/s
1.33
v = 2.26 x 108 m/s
What ALL happens when light reaches a
new medium?
Some of the light
REFLECTS off the
boundary and some of
the light REFRACTS
through the boundary.
Angle of incidence =
Angle of Reflection
Angle of Incidence > or < the Angle of
refraction depending on the direction of the
light
Going from Air to Water
The index of
refraction, n, for
air (vacuum) = 1.
The index of
refraction for
water = 1.33.
If you are going from a LOW “n” to a HIGH
“n”, the speed DECREASES and the angle
BENDS TOWARDS the normal.
Going from Water into Air
If you are going from
a HIGH “n” to a LOW
“n”, the speed
INCREASES and the
angle BENDS AWAY
the normal
Note: If the angles are
EQUAL, then the “n”
must be equal for
each. The ray will
pass straight through.
Snell’s Law
Snell discovered that the ratios of the
index’s and the ratio of the sine of the
angles WRT the normal are equal!
Snell ' s Law
n1 sin θ1 = n2 Sinθ 2
The refractive index of the gemstone,
Aquamarine, is 1.577. A ray of light strikes a
horizontal boundary of the gemstone with
an angle of incidence of 23O from air.
Calculate the SPEED of light in Aquamarine.
8
c
c 3 x10 m / s
n=
→ vm = =
1.577
vm
n
vm =
1.90 x 108 m/s
Calculate the angle of refraction within
Aquamarine.
n1 sin θ1 = n2 sin θ 2
(1)(sin 23) = 1.577 sin θ 2
sin 23
θ = sin (
)=
1.577
−1
14.34 degrees
Total Internal Reflection
There is a special type of refraction that can
occur ONLY when traveling from a HIGH “n”
medium to a LOW “n” medium.
Suppose we are traveling FROM water and
going into air. Should the ANGLE OF
INCIDENCE get TOO LARGE and the Angle
of Refraction EQUALS 90 DEGREES.
This special angle of incidence is called
the CRITICAL ANGLE, θc, for that material.
If the critical angle is EXCEEDED, for that
material, the ray will reflect INTERNALLY
within the material. We call this idea TOTAL
INTERNAL REFLECTION.
The angle of
incidence EXCEEDS
the critical angle for
water and the ray
reflects according to
the law of reflection
at the boundary.
The Critical Angle
How can you calculate the critical angle?
… the refracted ray is equal to 90O.
n1 sin θ c = n2 sin θ 2 ,
Sin90 = 1
θc
n2
sin θ c =
n1
A light ray is traveling in flint glass (n = 1.65)
upon striking the boundary, enters air.
Calculate the critical angle (Өc) for flint glass.
n1 sin θ c = n2 sin θ 2 , Sin90 = 1
n2
sin θ c =
n1
1
1
−1
→ θ c = sin (
sin θ c =
)
1.65
1.65
θc =
O
37.3
Lenses – Application of Refraction
There are 2 basic types of lenses.
A converging lens (Convex) takes light rays
and brings them to a point.
A diverging lens (concave) takes light rays
and spreads them outward.
Converging (Convex) Lens
Much like a mirror, lenses also take light
rays from infinity and converge them to a
FOCAL POINT, f. The difference, however,
is that a lens does not have a center of
curvature, C, but rather has a focal point
on EACH side of the lens.
Converging Lenses - Ray Diagrams
The rules for ray diagrams are the SAME for
lenses as they were for mirrors except you
go THROUGH the lens after refraction and
instead of going through, C (center of
curvature) you go through the actual center
of the lens.
f
f
1. Pick a point on the top of the object and draw
three incident rays traveling towards the lens.
Draw a ray such that it travels exactly parallel to
the principal axis.
Draw a second ray so that it passes exactly
through the focal point on the way to the lens.
Draw the third incident ray
such that it travels directly
to the exact center of the
lens.
2. The ray that traveled parallel to the principal
axis on the way to the lens will refract and travel
through the focal point.
The ray that passes through the focal point on the
way to the lens will refract and travel parallel to the
principal axis.
The ray that traveled to
the exact center of the
lens will continue in the
same direction.
3. The image point of the top of the object is the
point where the three refracted rays intersect. All
three rays should intersect at exactly the same
point. This point is merely the point where all light
from the top of the object would intersect upon
refracting through the lens.
Since this image could be projected on to a
screen it is a REAL IMAGE and real images
ALWAYS are found on the OPPOSITE side of
the lens from the object.
Ray Rules For Concave Lenses
Any incident ray traveling parallel to the principal
axis of a diverging lens will refract through the lens
and travel in line with the focal point (… in a
direction such that its extension will pass through
the focal point).
Any incident ray traveling towards the focal point on
the way to the lens will refract through the lens and
travel parallel to the principal axis.
An incident ray that passes through the center of the
lens will in affect continue in the same direction that
it had when it entered the lens.
Diverging Lenses - Ray Diagrams
1. Pick a point on the top of the object and
draw three incident rays traveling towards the
lens.
Draw one ray so that it travels towards the focal
point on the opposite side of the lens; this ray will
strike the lens before reaching the focal point;
stop the ray at the point of incidence with the lens.
Draw the second ray such that
it travels exactly parallel to the
principal axis.
Draw the third ray to the
exact center of the lens.
2. These incident rays strike the lens and refract.
The ray that travels towards the focal point will
refract through the lens and travel parallel to the
principal axis.
The ray that traveled parallel to the principal axis on
the way to the lens will refract and travel in a
direction such that its extension passes through the
focal point on the object's side of the lens.
The ray that traveled to the
exact center of the lens will
continue to travel in the
same direction.
The three rays should be diverging upon
refraction.
3. Locate and mark the image of the top of the
object.
The image point of the top of the object is the point
where the three refracted rays intersect.
Since the three refracted rays are diverging, they
must be extended behind the lens in order to
intersect. Using a straight edge, extend each of the
rays using dashed lines. Draw the extensions until
they intersect.
All three extensions should
intersect at the same location. The
point of intersection is the image
point of the top of the object.
The diagrams above show that in each case, the
image is …
located on the object side of the lens
a virtual image
an upright image
reduced in size (… smaller than the object)
Unlike converging lenses, diverging lenses always
produce images that share these characteristics.
The location of the object does not affect the
characteristics of the image.
Hence, the characteristics of the images formed by
diverging lenses are easily predictable.
The diagram below shows five different object
locations (drawn and labeled in red) and their
corresponding image locations (drawn and labeled
in blue).
The sign convention for lenses is similar to
that for mirrors.
Take the side of the lens where the object is
to be the positive side. Because a lens
transmits light rather than reflecting it like a
mirror does, the other side of the lens is the
positive side for images. In other words, if
the image is on the far side of the lens as the
object, the image distance is positive and the
image is real. If the image and object are on
the same side of the lens, the image distance
is negative and the image is virtual.
For converging mirrors, ______________,
the focal length is positive. Similarly, a
converging lens, ____________________,
always has a positive f, and a diverging
lens has a negative f.
The signs associated with magnification
also work the same way for lenses and
mirrors.
A positive magnification corresponds to an
upright image.
A negative magnification corresponds to an
inverted image.
Upright and inverted are taken relative to
the orientation of the object.
A 4.00 cm tall light bulb is placed a
distance of 45.7 cm from a convex lens
having a focal length of 15.2 cm. Determine
the image distance and the image size.
ho = 4.00 cm
di = ???
do = 45.7 cm
hi = ???
f = 15.2 cm
1 1 1
=
+
f do di
1/(15.2 cm) = 1/(45.7 cm) + 1/di
0.0658 cm-1 = 0.0219 cm-1 + 1/di
0.0439 cm-1 = 1/di
di = 22.8 cm
hi/ho = - di/do
hi /(4.00 cm) = - (22.8 cm)/(45.7 cm)
hi = - (4.00 cm) (22.8 cm)/(45.7 cm)
hi = -1.99 cm
From the calculations in this problem it
can be concluded that if a 4.00 cm tall
object is placed 45.7 cm from a convex
lens having a focal length of 15.2 cm,
then the image will be inverted, 1.99 cm
tall and located 22.8 cm from the lens.
A 4.00 cm tall light bulb is placed a
distance of 8.30 cm from a convex lens
having a focal length of 15.2 cm.
ho = 4.00 cm
di = ???
do = 8.3 cm
hi = ???
f = 15.2 cm
1 1 1
=
+
f do di
1/(15.2 cm) = 1/(8.30 cm) + 1/di
0.0658 cm-1 = 0.120 cm-1 + 1/di
-0.0547 cm-1 = 1/di
di = -18.3 cm
hi/ho = - di/do
hi /(4.00 cm) = - (-18.3 cm)/(8.30 cm)
hi = (4.00 cm) (18.3 cm)/(8.30 cm)
hi = 8.81 cm
From the calculations in the second sample
problem it can be concluded that if a 4.00 cm
tall object is placed 8.30 cm from a convex
lens having a focal length of 15.2 cm, then
the image will be enlarged, upright, 8.81 cm
tall and located 18.3 cm from the lens on the
object's side (virtual).
A 4.00 cm tall light bulb is placed a distance
of 35.5 cm from a diverging lens having a
focal length of -12.2 cm. Determine the
image distance and the image size.
ho = 4.00 cm
di = ???
do = 35.5 cm
hi = ???
f = -12.2 cm
1 1 1
=
+
f do di
1/(-12.2 cm) = 1/(35.5 cm) + 1/di
-0.0820 cm-1 = 0.0282 cm-1 + 1/di
-0.110 cm-1 = 1/di
di = -9.08 cm
hi/ho = - di/do
hi /(4.00 cm) = - (-9.08 cm)/(35.5 cm)
hi = (4.00 cm) (9.08 cm)/(35.5 cm)
hi = 1.02 cm
From the calculations in this problem it
can be concluded that if a 4.00 cm tall
object is placed 35.5 cm from a diverging
lens having a focal length of 12.2 cm, then
the image will be upright, 1.02 cm tall and
located 9.08 cm from the lens on the
object's side.
Converging lenses play an important role
in our lives as our eyes are these types of
lenses. Often times we need additional
corrective lenses to fix our vision.
A camera uses a lens to focus an image
on photographic film.
The End