Clicker Question
Two traveling waves 1 and 2 are described by the equations.
General functional form of a traveling wave:
y ( x, t ) = A sin( kx − ωt )
⎡ ⎛ x t ⎞⎤
y ( x, t ) = A sin ⎢2π ⎜ − ⎟⎥
⎣ ⎝ λ T ⎠⎦
k=
2π
λ
ω = 2πf =
2π
T
y1 ( x, t ) = 2 sin(2 x − t )
y 2 ( x, t ) = 4 sin( x − 2 t )
v=
λ
T
= λf =
ω
k
All the numbers are in the appropriate SI (mks) units.
Which wave has the higher speed?
A) Wave 1
B) Wave 2
⎡ ⎛ x t ⎞⎤
y ( x, t ) = A sin ⎢2π ⎜ − ⎟⎥
C) Both have the same speed.
⎣ ⎝ λ T ⎠⎦
Answers: Wave 2 has higher speed. Using the
equations below one can see that v1 = ½ and v2 = 2.
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Clicker Question
Two traveling waves 1 and 2 are described by the equations.
y1 ( x, t ) = 2 sin(2 x − t )
y 2 ( x, t ) = 4 sin( x − 2 t )
The wavelength λ of wave 1 is most nearly
A) 1m
B) 2m
C) 3m
D) 4m
y ( x, t ) =
E) Impossible to tell
⎡ ⎛ x t ⎞⎤
A sin ⎢2π ⎜ − ⎟⎥
⎣ ⎝ λ T ⎠⎦
⎡ ⎛ x t ⎞⎤
y ( x, t ) = A sin ⎢2π ⎜ − ⎟⎥
⎣ ⎝ λ T ⎠⎦
v=
λ
T
= λf =
ω
k
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Electromagnetic Wave:
v
E ( x, t ) = E peak sin( kx − ωt ) yˆ
v
B ( x, t ) = B peak sin( kx − ωt ) zˆ
Electric and Magnetic Fields have the same wavelength,
frequency and thus speed. They are both in a direction
perpendicular to the motion (x direction) and perpendicular to
each other as well.
In fact, direction of wave propagation v = direction of E x B
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Clicker Question
1. E and B fields are perpendicular to each other and to the
direction of propagation (i.e. EM waves are transverse)
An electromagnetic wave is propagating
in the positive x-direction. At this instant
of time, what is the direction of at the
center of the rectangle?
2. E and B fields are “in phase” (i.e. that is E reaches the
maximum at the same time and place as B reaches
maximum, etc.)
A. In the positive x-direction
B. In the negative x-direction
C. In the positive y-direction
D. In the positive z-direction
E. In the negative z-direction
3. B(peak) = E(peak) / c
Note that E field is much larger than the B field !
4. EM waves carry energy.
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1
v v
v | E || B |
1 v 2
| S |=
=
|E|
μ0
cμ 0
Poynting Vector
v
S=
v v
E×B
S
μ0
v
v
| E |= c | B |
Now if we substitute in our traveling wave equation:
• Vector S points in the direction that the wave travels
• |S| has magnitude equal to the energy carried by the wave
v
1
1 1 2
<| S |>=
E p2 sin 2 (kx − ωt ) =
Ep
2 cμ 0
cμ 0
S has units of Watts/meter2 , so the electromagnetic wave
delivers energy per time over a spatial area.
v
1 2
<| S |>=
Erms = Intensity (Watts/m 2 )
cμ 0
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Clicker Question
Average Intensity of Sunlight on a clear day at noon…
How many solar collectors would you need to replace a
4.8 kWatt electric water heater?
Assume each collector is 2 meters2 and has an efficiency of
40% for converting light energy to usable energy.
Intensity ≈ 1 kW/m 2
How large are the E and B fields associated with sunlight?
v
1 2
I =<| S |>=
Erms
cμ 0
Erms = 614 N/C
Brms =
48
Erms
= 3 × 10 −6 Tesla
c
A)
B)
C)
D)
1 panel
2 panels
6 panels
20 panels
4.8kW = (1kW / m 2 ) × ( Area) × (0.4)
A = 12m 2 ⇒ 6 panels
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Light not only carries energy, but can exert pressure (i.e. forces) !
Energy is stored in the EM fields.
v
1
u E = ε 0 | E |2
2
1 1 v 2
uB =
|B|
2 μ0
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energy absorbed
c
Δp Power
F=
=
c
Δt
Δp =
(Energy per volume)
radiation pressure =
Despite the fact that the B field is much smaller than the E
field, there is equal energy in each. Energy is then
transported as the wave moves.
Examples – getting a sun burn, LASERs, …
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Change in momentum.
Note that it is doubled if light is reflected.
F Power / A Intensity
=
=
A
c
c
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2
Suppose a source of radiation (EM waves) emits power P0 in
all directions uniformly.
The Sun “beating down” on my head…
F=
Power 1kW / m 2 × (π (0.1m) 2 )
=
= 1.0 × 10 −7 Newtons
c
3 × 108
What is the flux (Power/area) at a distance R from the source?
STAR Trek Trivia Spacecraft used by ancient Bajorans, propelled by light
pressure from Bajor's sun. Solair-sail vessels, also
known as light-ships, had no impulse reaction or warp
propulsion system, but rather had enormous reflective
sails that caught the tenuous solar wind. Crew
accommodations were minimal in order to conserve
mass.
R
Energy is conserved.
All energy emitted must go through the
surface (assume no absorption).
Flux at R =
P
P
= 0
A 4πR 2
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Clicker Question
A point source of radiation emits power Po isotropically (in all
directions uniformly). A detector of area ad is located a
distance R away from the source. What is the power P
received by the detector?
Po
ad
4 πR 2
A:
B: P a
o
D:
2
d
2
Po
ad
πR 2
E: None of these
R
C: Po a d
R
R
Clicker Question
Two radio dishes are receiving signals from a radio station
which is sending out radio waves in all directions with power P.
Dish 2 is twice as far away as Dish 1, but has twice the
diameter. Which dish receives more power?
A: Dish 1
B: Dish 2
C: Both receive the same power
detector
ad
Po
Dish 1
Answer: At a distance R from the source, the power Po is
now spread out uniformly over a sphere of radius R, area
P
a 2
44πR
πR
. So at distance
R, the intensity or brightness, which
power
P
⋅ area of detector =
a
4 πRis P /4πR2. The power received by the
isareapower/area,
o
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detector is given by answer A.
o
2
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d
o
2
d
Dish 2
Answer: Both receive the same power. Dish 2 has twice the diameter, but 4
times the area (area of disk = πd2/4). The energy flux = power/area at any
distance R is Po/4πR2. So at Disk 2, the flux is 1/4 the flux at Disk 1. Power
received = (power/area) × (detector area) (see concept test above) The
factor 4 increase in disk area is just cancelled by the factor of 1/4 decreased
in flux.
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A few notable things…
1. No MP homework due this week
2. No tutorial pre-test this week
3. Tuesday tutorials Æ section FCQ and BEMA (good final review)
Attendance will be taken!
4. Extra final review session Tuesday night 7 pm G1B30
5. Final exam covers everything – including this week !
Must bring Student ID to final !
Seating chart for final will be on the web !
6. Wednesday – finish new material and class FCQ
7. Friday – Professor Zhong will have another review in class
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3
Student Question
Polarization of Light
EM waves have a direction of the Electric field vector.
∫
v
v
dΦ
E ⋅ dl = −
dt
v
E ( x, t ) = E peak sin( kx − ωt ) yˆ
v
B( x, t ) = B peak sin( kx − ωt ) zˆ
B
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Ordinary Light (from the Sun or a lightbulb) is unpolarized,
meaning it is a mixture of waves with Electric field vectors in
random directions within the plane perpendicular to the wave
direction.
c
E
c
B
B
Unpolarized
As viewed from an observer:
E
The error in the student’s reasoning is that E is not proportional
to dB/dt! It is the spatial integral of E, and thus they are in
phase. Consider the traveling wave equations below.
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Polarizer = Polaroid filter = filter that passes light with the
E-field along the “pass axis” of the filter only.
Filter
E
Polarized
Polarized
Light Passes Through…
c
c
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Who are these people and what do they have in common?
Polarizer = Polaroid filter = filter that passes light with the
E-field along the “pass axis” of the filter only.
Filter
E
Polarized
Light Does Not Pass Through…
c
All the EM energy is used up as
work in moving charged
particles in the filter.
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4
Clicker Question
If the E-field is not all parallel to the pass axis, only the
component of E along the pass axis gets through.
θ
An unpolarized beam of light passes through 2 Polaroid
filters oriented at 45o with respect to each other. The
intensity of the original beam is Io. What is the intensity of
the light coming through both filters? I
Only Ecosθ is transmitted.
E
o
Thus, since intensity is related to
the field strength squared:
Filter Axis
Strans = S 0 cos 2 (θ )
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Clicker Question
A: (1/1.4)Io
B: (1/2)Io
C: (1/4)Io
D: (1/8)Io
E: None
c
Answer: (1/4)Io After the first filter, the intensity is I1 = Io(1/2). Reason:
If the original beam were polarized, then after passing through one filter
the intensity would be I1 = Io cos2θ , but here the original beam is
unpolarized so all angles θ are in the beam. The intensity after one filter
is then I1 = Io (cos2θ)average = Io(1/2). After the second filter the
intensity is I2 = I1 cos245 = I1 (1/2) = Io(1/2)(1/2) = Io(1/4).
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Why do polarizing filter sunglasses work?
If sun light is unpolarized, then it just
makes things darker by removing one
component.
Unpolarized light of equal intensity is
incident on four pairs of polarizing filters.
Rank in order, from largest to smallest,
the intensities Ia to Id transmitted through
the second polarizer of each pair.
A.
B.
C.
D.
E.
Ia = Id > Ib = I c
Ib = Ic > I a = Id
Ib = Ic > I a > Id
Id > Ia > Ib > I c
Id > Ia > Ib = I c
However, scattering can polarize light.
Then the sunglasses work very well
at reducing this light (glare).
67
In many circumstances we ignore the wave nature of light and
assume that light is a stream of particles that travel in straight
lines (rays).
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When a ray of light hits a surface, it can be reflected back,
absorbed, or refracted (bent as it travels into the material).
This type of analysis is
called Geometrical Optics.
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5
Index of Refraction
Rule for Reflection:
θ incident = θ reflected
θi θreflected
Any transparent material (air,
water, glass) can be
characterized by a dimensionless
number (index = n).
Rule for Refraction:
n1 sin(θ incident ) = n2 sin(θ refracted )
Snell’s Law of Refraction
depends on the index of
refraction of a material.
θrefracted
The velocity of light in the
medium v = c / n.
Thus, n is always greater than or
equal to 1.00
Materials
Vacuum
Air
Water
Lucite
Glass
Diamond
Index
1.00
1.00003
1.33
1.51
1.4-1.7
2.4
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Image Formation with Lenses:
Total Internal Reflection:
n1 sin(θ incident ) = n2 sin(θ refracted )
n
sin(θ refracted ) = 1 sin(θ incident ) > 1.0
n2
By shaping a transparent material, one can have the refracted
rays focused to form an image.
Example is a convex
lens = converging lens.
If this condition is met, then
there can be no refraction.
Thus there is total reflection.
Example – fiber optic cable.
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Simple set of equations to calculate images from telescopes,
microscopes, and set of geometrical optic lenses.
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We “see” or measure all things with lenses. In order to
determine the limit (best possible) resolution, we must recall
again that light is a wave !
There is a limit from the diffraction (spreading) of light.
Sometimes there is also a limit from things like atmospheric
blurring (scattering of light with particles).
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6
Ray diagram of geometrical optics indicates that a point
source that is infinitely far away will be imaged at a perfect
point. But light is a wave, not a ray. Thus, due to diffraction,
the image will be a small patch of light, not a perfect point.
Rayleigh Criterion: The minimum angle resolvable:
sin θ min ≈
Δs
λ
= 1 .2
L
D
λ is the wavelength of light and D is the diameter of the lens.
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Your eye has a near point of 25 cm – the nominal distance an
object would have light focused on the retina.
Estimate the minimum separation between two objects such
that the human eye can resolve them.
Key inputs: Pupil Diameter = 2.5 mm, λ = 550 nm
sin θ min ≈
λ
Δs
= 1 .2
L
D
Δs ≈ 0.1mm
Δs
550 × 10 −9 m
= 1 .2
0.25m
0.0025m
Diameter of a thin thread.
Also about separation of retinal cells !
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An asteroid appears on a collision course with Earth and is
currently 20 million km away.
What is the minimum size asteroid the
Hubble Space Telescope can resolve at
this distance?
Note that the HST has a diameter of 2.4 meters and is
diffraction limits (since no atmospheric blurring).
Δs
550 × 10 −9 m
= 1.2
9
20 × 10 m
2.4m
Δs ≈ 5.6km
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Electron Microscope
Serat
It turns out that particles
have wave like properties
as well !
Pointillism Style
We can generate beams
of electrons with
wavelengths almost 1
million times smaller than
visible light.
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7
Thank you for you attention and hard work this semester.
I hope you have learned a good deal of physics and some
appreciation for the sciences.
Good luck on your finals and in the future!
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