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The Wave Theory
Not everyone agreed with Newton’s particle (corpuscular) theory of light. There was a
competing theory which obtained dominance after
Newton’s time. It was called the wave theory of light
and it was championed by a man named Christian
Huygens.
The wave theory compared the propagation of
light from a point source to the dropping of a rock in a
pond. The waves move out in concentric circles. They
get smaller as they spread out over larger areas. This
corresponds to the decreasing brightness of a lamp as
you move away from it.
Part of the theory was the assumption that each
point on a wave front can be looked at as a point source
for new waves.
Go to the following site
http://www.microscopy.fsu.edu/primer/java/reflection/huygens/
to see examples of how Huygens was able to explain refraction and reflection.
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There were several predictions made by the wave theory that completely
contradicted those of the corpuscular theory. The first was the concept of
interference, what happens when light waves overlap and the second was diffraction,
the ability of light to bend around corners.
The
problem is that we do not often see light wrapping around corners. Supporters of
the wave theory had to find a way to demonstrate that light actually does wrap
around corners.
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To study diffraction and interference we need to define a couple of characteristics of transverse
waves.
Wave length and frequency are related to the speed at which the wave travels. They are given
by the following equation:
Vel = Frequency x Wavelength
v =
f

Remember that the speed of light “c” in a vacuum is 3 x 108 m/sec and it is the same for light,
radio waves, UV, gamma rays, Cell phones, and microwaves. Find the missing values
Type of radiation
Frequency
AM Radio
640 Khz
_______________________
FM Radio
100 Mhz
_______________________
A computer
1 Ghz
Microwaves
Wavelength
_______________________
_________________
1.2 cm
Red light
_________________
6.5 x 10-7 m
Blue light
_________________
4.0 x 10-7 m
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Standing Waves on a Spring
When a wave is sent along a spring and it hits a solid object it reflects half a wavelength out o
phase that is a crest comes back as a trough.
If you send a series of waves (wave train) down the spring at the right frequency, the
returning waves will overlap (superposition) in such a way that the resulting wave form will
contain points of constant cancellation (nodes) and others with constant maximum
disturbance. These are called standing waves.
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Velocity of Waves on a Spring
Find velocity of waves traveling along a spring. We will set up standing waves along a spring
of constant tension and constant length. First we will set up the fundamental waveform and
then we will go through several harmonics. Count the time it takes to make 10 cycles and
divide by 10 to get the period of one cycle.
Length of spring is ___________
We will be using the following equations
Freq = 1 / Period
vel = ( freq ) (  )
and
period
Fundamental
______
Frequency
________
Wavelength
_________
Velocity
________
1st Harmonic
______
________
_________
________
2nd Harmonic
______
________
_________
________
3rd Harmonic
______
________
_________
________
4th Harmonic
______
________
_________
________
How do the velocities compare? Does this make sense?
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.
Stringed Instrument
Interference is what makes a note from a guitar (or any other stringed instrument
sound better than the pure note generated by a signal generator. Imagine we have a
string upon which waves travel at 200 meters per second. How long must the
string be for the fundamental to resonate at 200 cycles per second?
Hint: the
fundamental is 1/2 of a wavelength long.
String length
_____________________________
Given the string in the problem above, lets look at the harmonics (other natural
frequencies) at which it simultaneously vibrates. Draw each harmonic and calculate:
Fundamental
wavelength ()
___________ m/cy
frequency (f)
___________ hz
1st harmonic
___________ m/cy
___________ hz
___________ m/cy
___________ hz
2nd harmonic
3rd harmonic
___________ m/cy
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___________ hz
Stringed Instrument
In the previous problem we set up a stringed instrument that had strings that were
0.5 meters long. We want a second string on this instrument that plays a higher
frequency note which has a fundamental of 300 cy / sec and is the same length.
How fast must the waves move on this string?
Wave vel.
_____________________________
Given the string in the problem above, lets look at the harmonics (other natural
frequencies) at which it simultaneously vibrates. Draw each harmonic and calculate:
Fundamental
wavelength ()
___________ m/cy
frequency (f)
___________ hz
1st harmonic
___________ m/cy
___________ hz
___________ m/cy
___________ hz
2nd harmonic
3rd harmonic
___________ m/cy
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___________ hz
Waves in Air
Waves on a string are transverse waves (that is their displacement is perpendicular to
their direction of propagation). Sound waves are different. They are longitudinal waves
(displacement is in line with the direction of propagation).. We sometimes call them
compression waves. The sound wave is carried on the back of a mass of forward moving
particles. When these particles hit other particles in front of them they experience an elastic
collision. (remember elastic collisions? Conservation of KE and momentum) The particle
that was initially moving comes to a stop and the originally stationary particle picks up the
first particle’s speed. The sound is now carried forward by this second particle until it hits a
third particle. And so the wave moves forward.
Above is a representation of two wave-fronts moving through air.
The following web site give a very good physical comparison between transverse and
longitudinal waves using animation.
http://paws.kettering.edu/~drussell/Demos/waves-intro/waves-intro.html
Notice how air particles move forward and then backwards to resume their original positions.
The pressure wave moves through the air while the air particles end up in the same place.
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Woodwinds and Brass Instruments
I am going to give you a simplified version of what happens in many musical instruments.
Imagine that below we have a tall narrow drinking glass that is open on the top. We blow
across the open top causing the air at the top to spin around in a circular motion. The
spinning air at the top rubs against stationary air below causing it to start moving but in the
opposite direction. The directions of spin change back and forth from counter clockwise to
clockwise and back again as we move down toward the bottom of the tube
Notice also that the amount of distance each set of particles moves is less and less (smaller
diameter) as one gets closer to the bottom of the tube. This makes sense because the particles
at the bottom of the tube have less distance to move before they hit the rigid bottom.
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The simplified picture is very represents sound intensity. At the bottom of the tube is a node
(no sound>). At the top of the tube is an anti-node (maximum sound) If you were to lower a
small microphone you would hear this.
Now, lets turn the picture on its side.
With a fundamental, what you have inside the tube is 1/4 of a complete wavelength. We can
work with this the same way we did with vibrations on a string.
If the tube is 0.5 meters long.
How long is one wavelength of this sound pattern? ____________________
Given that the speed of sound in air is approximately 340 m/sec
We can use the equation
v =f 
to find the fundamental frequency of this tube
________ = ( Freq ) __________
( Freq ) = ___________
What frequency would the tube vibrate at if it were each of the following lengths?
Fundamental frequency
Tube is 0.5 meter long (from above)
_______________________
Tube is 1.0 meters long
_______________________
Tube is 1.5 meters long
_______________________
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Like stringed instruments, glass tubes and organ pipes vibrate at several frequencies at the
same time. The higher frequencies are still called harmonics.
Here are the basic rules for drawing wave patterns in a tube.
1. There must be a node at the closed end of the tube.
2. There must be an anti-node at the open end of the tube.
3. Each harmonic adds a node. It also adds one anti-node
I have drawn in the pattern for the fundamental and the first harmonic.
Can you draw the patterns for the second and third harmonics?
Assume that the speed of sound is 340 m/sec and that the tubes above are 1 meter long.
Find the following:
Wavelength of the sound pattern Frequency of the pattern
Fundamental
_________________________
___________________
1st Harmonic
_________________________
___________________
2nd Harmonic
_________________________
___________________
3rd Harmonic
_________________________
___________________
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The speed of sound in air is approximately 340 m/sec. If the fundamental for the top tube
below is 200 hz find the length of the tube and then find the missing information.
(1760 hz)
The tubes above are open at both ends. If they are the same length as the top tubes finish the
drawings and find the missing information.
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Other Forms of Interference
Imagine that wavefronts of monochromatic (one frequency) light is moving toward a wall with
two openings in it. The waves of light bounce of the wall and pass through the openings.
Remember that Christian Huygens considered each wavefront as an infinite series of point
source generators. The result is that the wave spreads out on the other side in a circular
motion. Somewhere on a screen beyond the openings is a point halfway between the two
openings. Light waves hitting this point are in phase with each other. They create an antinode (bright spot)
Screen
The corpuscular theory of Newton would not have predicted this. It would have predicted
two bright spots corresponding to the two slits in the wall. There was a second prediction
that the wave theory made. It predicted that there should be other spots on the wall as well.
These would correspond to other locations where waves were out of phase by a complete
wavelength or multiple wavelengths.
Waves hitting the first order maximum (anti-node) are exactly one wavelength out of phase.
There is a second order maximum with waves that are two wavelengths out of phase.
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Below is a pattern created when a laser is pointed at two vertical slits..
For those who prefer a mathematical equation for determining interference nodes and antinodes below is the derivation for double slits.
On the board your teacher will project an interference pattern. Find the wavelength of the
red light.
X = __________ L = _____________
d = _______________
n = _____
Now we will place a very, very closely spaced grating. Using the wavelength found for red
light found above find the spacing of the grating.
X = __________ L = _____________
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d = _______________
n = _____
Radio transmitters can take advantage of interference.
Below is a top view of waves moving out from a vertical radio transmitting antenna.
A single radio antenna will radiate energy
(in the form of radio waves ) uniformly in all directions. The intensity goes down as 1/r.
Explain why it is 1/r and not 1/r2
From a single tower, soon the signal is too weak to receive easily. If you divide your
transmitter’s energy between two antennas you can create nodes and anti-nodes. Because
there is no measurable energy at the nodal points the missing energy will appear at the antinodes.
Using this property of interference it is possible to direct and intensify a weak signal. This is
something of great value to the owners of radio stations.
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Diffraction
Diffraction is the ability of waves to bend around corners. It happens all the time with ocean
waves.
It is easiest to study diffraction where light moves through a single small vertical opening.
Below is a pattern created when a laser is pointed through a single vertical opening.
The anti-nodes (maximums) fall in broad bands. This makes finding the exact location of the
maximums hard. Points of cancellation (nodes) are much easier to study. The first points of
cancellation on either side of the Central maximum are called 1st order minimums (minima is
the correct turn when there is more than one mimimum). The Second points of cancellation
are called 2nd order minima.
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Below is a diagram designed to explain how it is produced. You will have to fill in the blanks.
Problem: Red light has a wavelength of 6.4 E–9 meters. If we are shining red light through
an opening and showing a diffraction pattern on a wall that is 2 meters away (L) and the
central maximum is 5 centimeters wide ( 2 X ), how wide is the opening of the slit?
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The Second Order Minimum
The equation for first and second order diffraction minimums looks suspiciously like the
equation for the first and second order interference maxima.
Problem; How wide (or narrow) must an opening be so that the first order minimum of
red light (6.4 x 10-7 meters) to be at an angle of 10 degrees?
Hint the sin of  is equal to X/L
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The method we have been using can be used to explain why the central maximum is about
three times brighter than the 1st order maximum.
Important question. What happens as “X” approaches one wavelength?
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Explain.
Review terms
Define and describe
1.
Reflection
2.
Refraction
3.
Diffraction
4.
Interference
5.
Superposition
6.
Amplitude
7.
Focus
8.
Negative focus
9.
Crest
10. Trough
11. Wavelength
12. Period
13. Frequency
14. Snell’s law
15. Node
16. Anti-node
17. Longitudinal
18. Transverse
19. Normal
20. Total Internal Reflection
21. Real Image
22. Virtual Image
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