Download In our everyday life there are many examples of

Waves
™In our everyday life there are many examples
of waves for example, Sound, ocean waves,
strings of musical instruments, organ pipes…
™These are examples of waves which need a
medium to travel through and the general
definition of such waves is
A disturbance which propagates through a
medium but the medium itself only moves
locally as the wave passes.
™A special case of a wave is an
electromagnetic wave which can propagate
through vacuum, e.g. radio, light, x rays ….
™Waves transport energy and momentum and
energy is required to generate waves
™The medium must have elastic properties
Week 13
Physics 214
Spring 2010
1
Wave properties
There are two types of waves
Longitudinal – consists of the
propagation of a series of
compressions and rarefactions and
the local movement of the medium
is an oscillation back and forward
along the direction of the wave
Sound is one example
Transverse – where the movement
of the medium is at right angles to
the velocity of the wave. The strings
of musical instruments are an
example.
Week 13
Physics 214
Spring 2010
anim0019.mov
2
Wave properties
Any material medium supports
Longitudinal (compression) waves:
solid, liquid, or gas.
Fluids do NOT support transverse
waves in their interior– because the
restoring force needed would be
shear restoring force, which clearly
fluids don’t posess. You can stir
water (you can’t stir set jello, and
jello CAN wiggle sideways
elastically.
Seismic waves in the earth tell us
about the intermediate liquid layer!
Week 13
Physics 214
Spring 2010
anim0019.mov
3
Wave properties
Water surface waves are a
third type, not purely
transverse nor longitudinal.
The water molecules dance in
place in “elliptical” orbits.
Remember, ALL matter waves
consist of matter dancing in
place under restoring forces
while the wave PATTERN
carries energy.
Week 13
Physics 214
Spring 2010
anim0019.mov
4
Periodic waves
We will focus on regular repetitive waves
λ
These waves have a pattern which repeats. The length of
one repeat is called the wavelength λ
The number of repeats (wavelengths) which pass a point, per
second, is called the frequency f
T is the time for one pattern to pass.
Then f = 1/T
v = λ/T = fλ
The wave velocity, v, is then
Week 13
Physics 214
Spring 2010
5
Waves on a string
If we shake the end of a
rope we can send a wave
along the rope. The rope
must be under tension in
order for the wave to
propagate
v = √(F/μ)
F = TENSION
μ = MASS/UNIT LENGTH
Tighter = faster wave
Lighter = faster wave
Week 13
Physics 214
Spring 2010
6
Standing waves
If two identical waves exist on the
same string but traveling in
opposite directions the result can
be standing waves in which some
points never have a deflection.
…These are called nodes
Halfway between nodes, the string
oscillates with the maximum
amplitude.
…These are called antinodes.
Standing waves provide the notes
on musical instruments. When a
string is secured at both ends and
plucked or hit the generated waves
will travel along the string and be
reflected and set up standing
waves.
Week 13
Physics 214
Spring 2010
7
Waves on a Spring
The spring in the demonstration
shows many nifty simplifications as
you stretch it. Mass is fixed, but if you
double the length, μ is halved.
And by Hooke’s law, the Tension
(F) doubles because of the stretch.
F = TENSION = kL;
μ = MASS/ L
v = √(F/μ) = √(kL/(M/L)) = √(kL2/M) = L√(k/M)
If you double L, you double v.
Round-trip travel time of a pulse does not
change, because it has to travel twice as
far, but at twice the speed!
Neat. But rather a special case.
Week 13
Physics 214
Spring 2010
8
Musical notes
Each end of the string must be a node
so the possible standing waves must
be multiples of λ/2
Fundamental f = v/λ =
v/2L
2nd Harmonic f = v/λ = v/L=2v/2L
3v/2L
3rd Harmonic f = v/λ =
Musical sound is a mixture of
harmonics modified by the body of the
instrument or where the string is bowed
or plucked (near an end enhances high
frequency overtones, relative to low)
v = √(F/μ) so a piano or a violin is
tuned by changing the tension in
the string. (playing the violin
you change the length of the
string with your finger.)
Week 13
Physics 214
Spring 2010
9
“Closed” (at one end) organ pipe
Node at one end and an antinode at
the other
Fundamental f = v/4L = 1st harmonic
2nd Harmonic f = 3v/4L = 3xFund.
3rd Harmonic f = 5v/4L = 5xFund.
higher overtones 7x, 9x, 11x, etc.
Open pipe overtones (unlike the
string) are only odd multiples of the
Fundamental tone
The velocity of sound in air is
v = 340m/s
Nth harmonic f = (2N-1)v/4L
Week 13
Physics 214
Spring 2010
10
Closed organ pipe
Node at one end and an
antinode at the other
The velocity of sound in air is
v = 340m/s
Nth harmonic f = (2N-1)v/4L
32 foot Organ pipe, L=10m
Fundamental tone has
f = v/4L = 340m/s/40m = 8.5 Hz
Too low to be heard as a tone,
it’s a deep throbbing in your
chest.
Week 13
Physics 214
Spring 2010
11
QUIZ
9 April ‘10
Closed organ pipe
Node at one end and an antinode at the
other
Fundamental f = v/4L = 1st harmonic
2nd Harmonic f = 3v/4L
Velocity of sound in air is v = 340 m/s
Q: ~16 foot pipe (5m) open one end
closed other end. What is the 3d
harmonic, in Hz?
A. 68
D. 51
B. 34
E. 85
C. 17
λ = 20 m, f = v/λ = 340/20 = 17 [1/s]
3d harmonic has f = 5*fund. = 85 Hz,
(E.)
. Week 13
Physics 214
Spring 2010
12
Beats
If two waves have slightly different
frequencies then the sum has a
frequency which is f1 – f2.
The human ear can detect beats
and this is used to tune an
instrument.
For example using a tuning fork at
a known frequency and adjusting a
piano string until no beats heard
Two notes close together in
frequency may sound dissonant
due to the beat tone.
Week 13
Physics 214
Spring 2010
f = f1 – f 2
13
Musical Scales
An Octave occurs when one frequency is twice the
other frequency. There are twelve half-tones in one
octave (corresponding to 12 piano keys in sequence)
Two string tones an octave apart have the higher tone
at the frequency of the first overtone, so the sound is
similar to either of the separate tones.
Week 13
Physics 214
Spring 2010
14
Musical Scales
The notes of the Western musical scale are based on
simple chords, and tones that are simple fractions of
each other.
Take two tones of frequency f and 1.5f. What is the
beat frequency? It’s fbeat = 1.5f - f = 0.5 f
This is one octave below f, and the chord sounds
smooth and mellow.
Take f and 1.25f (ratio is 5/4).
fbeat = 0.25 f
This is two octaves below f and is still a smooth chord
Week 13
Physics 214
Spring 2010
15
Musical Scales
The simple ratios can work fine for one “key signature”, for
example C. But if a piano is tuned with perfect chords for this
key, then chords in another key (say G) might have dissonances
(buzzing beat notes, or so-called discord).
A good compromise is the equal-tempered or well-tempered
scale, invented around the time of J. S. Bach, where the half
tones differ in frequency by a ratio = the twelfth root of 2.
This gives reasonably harmonious chords, and every key has the
same degree of “consonance”, or harmony. This is, however, a
tradeoff since the ratios are no longer perfect 5/4 or 3/2 in any
key.
Week 13
Physics 214
Spring 2010
16
Musical Scales
For example, the chord C-E is two full tones apart,
Which is four half-tones. The frequency ratio is then
24/12 = 21/3 = 1.26 in the equal tempered scale
This is quite close to 5/4, and is still a harmonious
chord.
If we take middle C, 261.5Hz, what frequency is the
“error”? 1.26 differs by 0.01 from 5/4. So E is about
2.6 Hz higher frequency than a perfect chord.
C-G is seven half tones apart. The frequency ratio is
then 27/12 = 1.4983 which is very close to 3/2 – again,
a very harmonious chord. Here, G is 0.0017 x 261.5 Hz
flat (low) compared to a perfect chord. That’s 0.44 Hz.
Week 13
Physics 214
Spring 2010
17
Ch 15 CP4 cont.
21/12 is ~1.059463:
A = 440 Hz
Ab = A/1.0595 = 415.2902 Hz
G = Ab/1.0595 = 391.9681 Hz
Gb= G/1.0595 = 369.9557 Hz
F = Gb/1.0595 = 344.1796 Hz
E = F/1.0595 = 329.5701 Hz
Eb = E/1.0595 = 311.0620 Hz
D = Eb/1.0595 = 293.5932 Hz
Db = D/1.0595 = 277.1054 Hz
C = Db/1.0595 = 261.5435 Hz
A, Bb, B, C → up to scale
A = 440 Hz
Bb = A × 1.0595 = 466.18 Hz
B = Bb × 1.0595 = 493.9177 Hz
C = B × 1.0595 = 523.3058 Hz
middle C = 261.5435 Hz
high C = 523.058 Hz
vs 523.087 = 2x middle C
(roundoff error)
high C/ middle C = 2.00;
Yes, high C is twice the frequency
of middle C!!
Week 13
Physics 214
Spring 2010
18
Doppler effect
As the source is approaching
the frequency increases
because the waves are
crowded closer together.
As an observer approaches a
stationary source, she passes
more wave crests per second,
the frequency is increased.
It’s the RELATIVE motion that
matters.
For relative recession, the
wavelength is “stretched”,
which lowers the observed
frequency.
http://www.physics.purdue.edu/class/applets/phe/dopplereff.htm
Week 13
Physics 214
Spring 2010
19
Sonic boom
Sonic boom (a shock wave)
At each point on the path the
sound wave expands radially
and they all combine along a
single wave front which is a
sharp pressure wave
http://www.phy.ntnu.edu.tw/java/airplane/airplane.html
Week 13
Physics 214
Spring 2010
20
Sonic boom
A sonic boom is analogous to
the bow-wake of a boat. In each
case, the emitter is travelling faster
than the wave velocity.
The red & thin black lines are like the
crest of a bow wave, except the sonic
boom is in 3-D so it is a cone
instead of a pair of lines. Arrows point
along the wave velocity.
http://www.phy.ntnu.edu.tw/java/airplane/airplane.html
Week 13
Physics 214
Spring 2010
21
Summary of Chapter 15
v = λ/T = fλ
Transverse or longitudinal
Standing waves
Fundamental f = v/λ = v/2L
2nd Harmonic f = v/λ = v/L
3rd Harmonic f = v/λ = 3v/2L
Fundamental f = v/4L
2nd Harmonic f = 3v/4L
3rd Harmonic f = 5v/4L
Week 13
Physics 214
Spring 2010
22
Sound effects
Doppler effect
f increases as sound
approaches
Beats f = f1 – f2
Sonic boom
Week 13
Physics 214
Spring 2010
23
Electromagnetic waves
v~EXB
(directions)
EM waves consist of oscillating magnetic and electric fields
transmitted through vacuum at a constant speed of c = 3 x
108m/s. Curl you fingers from the E direction to the B
direction: your thumb points in the direction of propagation.
They are produced whenever there is a changing electric or
magnetic field.
The acceleration of electric charge produces EM waves such
as a radio broadcast antenna, AC wiring and lightning. Also
electrons changing orbits in atoms. A simple EM wave has
f = c/λ
and
λ = c/f
Where
c = speed of light = 3 x 108 m/s
Week 13
http://www.physics.purdue.edu/class/applets/phe/emwave.htm
Physics 214
Spring 2010
24
Electromagnetic waves
v~EXB
f = c/λ
and
λ = c/f
(directions)
Visible light with λ = 500 nm (~ green) has
what frequency?
f = c/λ = 3x108 m/s / 5x10-7 m = 6x1014 Hz
If you can discriminate one cycle, you have “measured” one
second to an accuracy of almost one part in 1015 !!!
Week 13
Physics 214
Spring 2010
25
Electromagnetic spectrum
Almost all the information we
receive from outside the earth
is in the form of EM radiation.
Different parts of the
spectrum correspond to
different physical processes
We can understand what is
going on in the Universe and
back in time to near the
beginning of the Universe
using a variety of earth- and
space-based telescopes.
Week 13
Physics 214
Spring 2010
26
←(part of) Electromagnetic spectrum→
Visible light spans 1 “octave”. This figure
spans factor of 1015 This is ≈ 50 “octaves”,
since
210 = 1024 ≈ 103
Week 13
250
≈
1015
Physics 214
Spring 2010
27
A few telescopes
Radio telescope at Arecibo
Space telescopes don’t have to
look through the atmosphere.
Week 13
An array of radio telescopes – can
resolve very fine angular detail due
to large overall size.
Physics 214
Spring 2010
28
How do we see color
An image is formed at the back of
the eye like a camera, on the
retina. There are receptor cells
called cones that respond to
different wavelengths. The brain
interprets the mixture of the three
signals as color.
If we look at an object by reflected
light the color we see is the
wavelengths that were reflected,
not absorbed.
If we are looking through a
colored object then the object lets
that color be transmitted and the
other colors are absorbed
Week 13
Physics 214
Spring 2010
S. M and L refer to short, med.
and long wavelengths of light.29
How do we see color
Any color we see excites the S, M,
and L cones each a certain
amount. The color perceived is a
result of the relative proportions of
the three excitations, not the overall
brightness. So the color space
humans see is effectively 2Dimensional. Once you know the
% of L and the % of M, you know
the % of S
[must add to 100%]
Notice
how
but
Week
13 purple is NOT a pure wavelength,Physics
214
rather a mix of the two ends of the visible spectrum.
Spring 2010
30
Two-Slit Interference
When two coherent & equalamplitude monochromatic beams
of light are brought together they
will add just like two waves on a
string.
When two wave peaks coincide
the light will be a maximum.
When a peak coincides with a
valley there will be no light.
At the two slits, the waves are “in
phase”. The waves spread out
from the slits, and at places on
the screen where one slit is an
integer number of wavelengths
farther than the other from the
screen, the waves add up to give
brightness.
Halfway between each bright
stripe is darkness from the waves
cancelling.
Week 13
y is the sideways distance on the
screen, from a point “straight ahead”
Physics 214
Spring 2010
Bright fringes are located
at positions y given by
y/x = n λ/d
where n is an integer
Dark fringes occur when
yd/x = nλ/2
where n is an odd integer
31
TWO-SLIT INTERFERENCE
The picture at the right is for two slits extending in and
out of the picture. Picture shows only the “rays” from
each wavelet that aim at a PARTICULAR POINT ON THE
SCREEN. The screen is far away. The slits have equal
widths, so the waves in the two rays have equal
amplitudes.
The two rays are pointing towards a point on the screen
d
(where the sideways distance is y), for which there is
exactly ONE HALF WAVELENGTH path difference
between the two slits.
These two rays, being exactly out of phase, will cancel
each other. Also for any odd-half-integer path diff.
CONSTRUCTIVE interference gives BRIGHT stripes
straight ahead and for path differences of λ, 2λ, 3λ, etc.
There are intervening DARK fringes at odd half-integer
wavelength path differences.
P/d ≈ y/x (for small angles away from straight ahead)
Similar triangles. Long leg ≈ hypotenuse.
For small angles (in radians) P/d ≈ θ ≈ y/x
Week 13
Physics 214
Spring 2010
almost x
yy
θ
x
P = Path difference
For the first
CANCELLATION, need
a half-wavelenth path
difference λ/2 = d sin(θ)
where θ = deflection
angle.
y/x = tan(θ)
sin(θ) ≈ tan(θ) ≈
for small θ
θ
32
Interference
Where does the energy go
when light waves interfere to
make a dark region?
There MUST also be bright
regions, and the energy is
just REARRANGED in space.
So the “dark bulb” (O. R.
Frisch’s joke) that you turn it
on and the room goes dark,
is an impossibility. Cute
idea, it does make you think.
Week 13
Physics 214
Spring 2010
Bright fringes are located at
positions given by
dy/x = nλ
where n is an integer
Dark fringes occur when
dy/x = nλ/2
where n is an odd integer
33
Thin film interference
Thin film interference occurs when
light is reflected from the top surface
and the underneath surface of film.
This gives two beams of coherent
light, which interfere.
Since we normally observe this with
white light we see color fringes
because the path difference varies with
the angle of observation
So different wavelengths (colors) have
constructive and destructive
interference at different places on the
film.
Week 13
Physics 214
Spring 2010
34
Thin film interference
If the round-trip path in the thin
film is λ / 2 (so the distance
travelled one-way is λ / 4)
the two reflected waves will be
exactly out of step with each
other, and cancel. For different
colors (wavelengths) this
condition will take place at
different angles (if at all).
Redder color with longer
wavelength would need an angle
farther away from the vertical,
and vice versa for bluer color.
Week 13
Physics 214
Spring 2010
35
Anti-reflection Coatings for lenses
As light passes from one transparent medium to another a
few percent of the light will be reflected.
This is a particular problem in optical systems like lenses
where there may be many glass elements.
For example if 96% of the light is transmitted at a surface,
after 18 surfaces only (0.96)18 = 48% of the light is
transmitted overall, and the other 52% scatters everywhere.
Thin coatings are put on lens surfaces with thickness = ¼
wavelength of green light. Then green rays entering
perpendicular to the surface will have the front and rear
reflections from the film λ / 2 out of phase, and cancel.
At normal (perpendicular) incidence, with a “quarter-wave
coating”, this is only true for a single wavelength. To
reduce the reflections for a range of wavelengths requires
multiple thin film layers and costs more.
This is why camera lenses often have a purple cast. The
antireflection coating is optimized for green, the middle of
the spectrum and also the color the eye is most sensitive
to. The cancellation is not perfect for red or blue, so the
reflected light is purple = red+blue, the two ends of the
optical spectrum.
Week 13
Physics 214
Spring 2010
9 lenses = 18 surfaces!
36
Anti-reflection Coatings for lenses
At normal (perpendicular) incidence, with a
“quarter-wave coating”, which is λ/4 thick (and
hence the round trip is λ/2), exact cancellation
ALSO requires equal strength of the two reflections
(from front and back surfaces of the coating).
It turns out that this requires the index of refraction
of the coating to be the geometrical mean of air
(n~1.0) and the lens glass (n~1.5) . Then the
“steps-up” in index each have the ratio shown
below. (We’ll define n and its properties on the
next two slides.)
n = Sqrt(1.5) = 1.225 is required, but materials with
this low an index tend to be soft and easily
scratched.
Week 13
Physics 214
Spring 2010
37
REFRACTION, Speed in a Medium
In transparent media, light travels slower than in a
vacuum: v = c/n.
n is called the “index of refraction”.
Since nothing can travel faster than c,
n is always
greater than 1.00 (or equal to 1.00, in vacuum)
For water, n=1.33; for glass, n~1.5; for diamond, ~2.4
Going from air into water or glass, light slows down
and, being a wave, therefore MUST bend its direction
closer to the perpendicular of the interface.
This is the basis of all lenses, prisms, etc.
Week 13
Physics 214
Spring 2010
38
REFRACTION, Speed in a Medium
In transparent media, light travels slower than in a
vacuum: v = c/n.
Charged particles can travel very close to c, even in
transparent media. Thus, they can travel faster than
the speed of light IN THE MEDIUM. Such fast
charged particles will then emit an “optical shock
wave”, which is called Cherenkov radiation after the
Russian physicist who conceived this idea less than
100 years ago.
Cherenkov radiation wavefronts also are coneshaped, and the flow of radiation follows another
cone at right angles to the wavefronts.
Week 13
Physics 214
Spring 2010
39
QUIZ
Two loudspeakers next to each other
broadcast the same tone, same strength.
One speaker is then placed 7 halfwavelengths farther away from you than
the other. You then hear :
A. a pulsating tone
B. a rising tone
C. a falling tone
D. much less sound
E. more intense sound
The answer is D, because there is destructive interference.
The three full wavelenths of shift would not make a difference,
but the extra half-wavelength puts the two waves exactly out
of phase and they cancel each other.
Week 13
Physics 214
Spring 2010
40
Diffraction
Waves travel in straight lines EXCEPT when they
encounter obstacles – then the waves spread out,
and can bend around corners, etc.
That’s why, for example, you can hear sounds
from sources you can’t see because they’re not in
your line of sight.
Water waves are an excellent example.
Photo of wave pool at water park in Cancun,
Mexico.
Week 13
Physics 214
Spring 2010
41
Week 13
Physics 214
Spring 2010
42
Diffraction
Waves travel in straight lines
EXCEPT when they encounter
obstacles – then the waves spread
out, and can bend around corners,
etc.
If it were not for Diffraction, the
two-slit interference could not take
place, since it requires the
coherent waves coming from the
two slits to start spreading out
sideways from each of the slits.
Week 13
Physics 214
Spring 2010
43
INTERFERENCE
Because of diffraction, interference
occurs even for a single aperture.
The pattern shown BELOW is from a
square aperture and the pattern can
be thought of as light from different
parts of the aperture interfering.
As the aperture is made smaller the
pattern actually expands (because a
larger angle of deflection is needed to
generate a full λ path difference
between the edges.)
This effect can limit our ability to see
detail in small objects or to resolve
two stars nearby to one another.
Week 13
Physics 214
Spring 2010
λ/w
≈
y/x
But the wavelets at the
two edges are IN step !?
Counter-intuitive
The position of the first
dark fringe is given by
y = λx/w. This means the
central bright fringe has a
width ~ 2y which increases
as w gets smaller
44
INTERFERENCE
This picture is for a single slit of width w,
extending in and out of the figure. Every point
on the wavefront AT the aperture radiates a
semi-circular “wavelet”. The picture shows
only the “rays” from each wavelet that aim at a
PARTICULAR POINT ON THE SCREEN.
We are looking at a DARK point on the screen
(where the sideways distance is y), for which
there is exactly ONE WAVELENGTH difference
between the two edges of the slit.
To see the complete cancellation, pair up the
wavelets that are HALF a wavelength out of
step (bottom edge and mid-slit). Then move a
tiny way up and repeat until you have swept
across half the slit with each wavelet of the
pair. All wavelet pairs CANCEL. And this uses
ALL the wavelets (don’t worry about using the
middle wavelet twice, it contributes an
infinitesimal amount to the overall wave sum.)
Week 13
Physics 214
Spring 2010
The central bright
band is 2y wide on
the screen
45
Diffraction grating
If we have multiple slits very close together, say several
1000 slits per centimeter the diffraction from each slit
adds in a way that produces very narrow and bright
fringes. Call the slit spacing d, x=dist. to screen,
y = deflection
For small angles the location of the bright fringes is given
by y/x ~ angle ~ m(λ /d)
y = mλx/d where m is an odd or even integer
y
Week 13
Physics 214
Spring 2010
46
Diffraction grating
Diffraction gratings are excellent for resolving two nearby
colors (wavelengths). They are particularly useful in
astronomy, and are how we detect atomic spectral color
lines and measure their redshifts.
Week 13
Physics 214
Spring 2010
47
Polarization
“In a normal beam of light all possible
transverse polarizations are present.”
A light wave is made up of zillions of
photons, which randomly have all
possible polarizations of their E fields.
But due to the wave nature of light, this
normal light beam can validly be
represented as 50% vertical
polarization and 50% horizontal
polarization. Strangely, you can use
any two orthogonal transverse
directions to represent normal light.
A polarizing filter lets through one
component and blocks the other, thus
cutting the light intensity in half.
Week 13
Physics 214
Spring 2010
48
Polarization
There are materials like polaroid that
will only transmit light (photons)
with the electric field vector along a
specific direction.
Light reflected by water or other
shiny surfaces is also is polarized
so polaroid sun glasses at the right
orientation will preferentially block
the reflected light.
Transparent objects under stress
can also cause transmitted light to
be polarized. Car windshields are
one example where patterns can be
seen when wearing polaroid sun
glasses
Week 13
Physics 214
Spring 2010
At a special angle,
Brewster’s angle, light
reflected off a transparent
interface is totally
horizontally polarized, and
the other polarization
entirely enters the denser
medium. This condition is
when the reflected and
transmitted rays are
perpendicular to each other.
At other angles, the
polarization is only partial.
49
Polarization
The effects of polarizing filters:
Picture at top is a polarized light
beam. A vertical polarizing filter
allowed through only half the light:
all of the vertical component and
none of the horizontal.
This light hits a second filter which
is rotated 45o from the original filter.
The E field (green) has a vector
component along the new filter axis,
so a reduced amplitude passes
through (the red E field)
Crossed polaroids (second figure)
will kill ALL the light.
Almost magically, an intermediate
polaroid between the crossed
polaroids will let some of the light
through (third figure
Week 13
Physics 214
DEMO
Spring 2010
Filter axis
Filter axis
Blocks all light
E after first filter
E after
intermed. filter
E after crossed
(third) filter
50
QUIZ
Polarization
An unpolarized light beam is passed through a
linear polaroid polarizing fillter, which
removes half of the light intensity.
The polarized beam then passes through a
polaroid filter whose axis is at 45o to the
first filter. The light intensity is:
A. Further reduced B. Reduced to zero
D. Increased
C. Not changed
There is a non-zero component of E that can pass the filter, but
it’s not the full E out of the 1st filter
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Physics 214
Spring 2010
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Summary of Chapter 16
f = c/λ c = 3 x 108m/s in vacuo
In a medium with index n, f DOES NOT
CHANGE, the wave slows down
(relative to in vacuum) and λ changes
(shorter).
Polarization
Thin film interference
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Summary of Chapter 16 contd.
Bright fringes are located at positions
given by
y/x = nλ/d n is an integer
Dark fringes occur when dy/x = nλ/2
where n is an odd integer 2 slits
Diffraction grating dy/x = mλ
where m is any integer
The position of the first dark fringe is
given by y = λx/w. This means the central
bright fringe has a width ~ 2y and
increases as w gets smaller one slit
Week 13
Physics 214
Spring 2010
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Lightning Tour of Chapt. 17
This chapter deals with refraction of
light, including index of refraction, n,
which I have already lectured about a
little bit. n is >= 1
In the medium,
vlight = c/n
is < c
Refraction is the basis of prisms and
lenses, also of some types of lightguide optical fibers.
When light passes from a less optically
dense to a more dense medium (i.e.
from lower n to higher n) it is bent
CLOSER to the normal (perpendicular)
to the interface.
Week 13
Physics 214
Spring 2010
n1 <
n2
θ2
θ1
54
Lightning Tour of Chapt. 17
NOTE: This chapter is not assigned material
for testing, but you need to have seen just a
little bit about lenses, total internal reflection,
n1
and so on – to help you understand the
simple principles underlying these
phenomena. I also recommend
θ1
reading Ch. 17.
Let’s continue:
The bending towards the slower medium is
because in the slower medium, the wave
crests must be closer together. To match up
crests at the interface, the direction of flow
must change
Remember, f const., λ shrinks as wave slows
Week 13
Physics 214
Spring 2010
<
n2
θ2
55
Lightning Tour of Chapt. 17
To match up crests at the interface,
the direction of flow must change
Remember, f does not change.
Instead, λ shrinks as wave slows
down in the denser medium.
λ = v/f and now v is less than c
n1 <
n2
θ2
θ1
f does not change because the
photon energy is directly related to
its frequency (basic quantum
behavior)
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Chapt. 17
Some lenses cause bundles of light rays to become
more CONVERGENT. This can focus the almost
parallel rays from a very distant obejct into a point
at a certain distance beyond the lens, called the
focal point of the lens.
If you are far-sighted, you need such a lens, of
some strength, to help the lens of your eye to focus
images on your retina.
If you are near-sighted, you need a DIVERGING
lens of some strength to move focused images
further from your eye’s lens and onto your retina.
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Chapt. 17
CONVERGING lenses are fat in the middle and thin
at the edge. So called “convex” shape, or bulging
outward.
DIVERGING lenses are thin at the center and thick
at the edge, “dished-in” which we call concave.
HOWEVER, a “hollow air lens” immersed in water
reverses the effects, fat lenses are diverging and
concave lenses are converging.
This is because of the inversion of dense vs less
dense index of refraction, compared to “normal”
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Chapt. 17
In media such as glass (or water) the
index of refraction depends on
frequency (color) so one color bends
more or less than another color.
This is called DISPERSION
PRISM demonstration
In lenses, this produces unwanted
“chromatic aberration” (different colors
focus at different distances from the
lens, but retina or film is at a fixed
distance.) This must be compensated
for by complicated arrangements of
several lenses using DIFFERENT
glasses having different dispersions.
Week 13
Physics 214
Spring 2010
n1 <
n2
..
θ2
θ1
59
Ch. 17 Total Internal Reflection
Light in a denser medium aimed at an
interface, if it’s oblique enough, will not
be able to pass into the less dense
medium. This follows from Snell’s Law,
which precisely describes: smaller angle
in the “denser” medium, bigger angle in
the less “dense” medium.
n1 sin(θ1) = n2 sin(θ2)
If θ1 is too large, it requires sin(θ2) > 1
which is impossible, so no wave goes
into the less dense medium 2 and all
energy is reflected back into medium 1.
The “critical angle” has sin(θ1) =(n2/n1)*1
Because 1 = sin(90o)
Week 13
Physics 214
Spring 2010
denser
n1 >
less dense
n2
θ2=90o
θ1
The diagram
shows the critical
angle. Bigger
angles away from
the normal (that
is, rays skimming
closer to the
interface) totally
60
reflect
Ch. 17 Total Internal Reflection
Clad thin (1mm) plastic fibers can
transmit light with reasonably small
losses, using total internal reflection. We
use a core of polystyrene with n = 1.59,
the cladding is pmma (plexiglas) with n =
1.49. The light must aim within 20o of the
fiber axis, but this traps enough light to 20
read out the light from scintillating plastic
particle detectors over several meters of
length. Angles bigger than 69.6o (relative to the
o
70o =~ θcritical
axis
cladding
PERP. to the interface) will totally internally reflect
sin 69.6o = 1.49/1.59 = 0.937
Week 13
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