Unformatted text preview: Physics 227: Lecture 24
Light • Lecture 23 review: •
• RLC circuits with impedance: • ZR = R, ZL = jωL, ZC = 1/jωC = -j/ωC. Transformers: •
• Monday, December 5, 2011 V1/N1 = V2/N2
Current reduced by same factor voltage increased.
Load resistance also transformed. Light iclicker
Which of the following is true about the nature of light.
There may be more than one correct answer.
A. Light is an electromagnetic wave.
B. Light is made of particles called photons.
C. The speed of light in vacuum is c ≈ 3x108 m/s.
D. The direction of motion of light is ⊥ its E and B ﬁelds.
E. Light can be produced by accelerating charged particles. All answers are correct!
Monday, December 5, 2011 Some Math •
• We want to ﬁnd a solution to Maxwell’s Equations in a region
far, far away from any sources.
Following our usual technique - knowing the answer in
a dvance - I propose the following as a possible solution:
E = E0 cos(kz − ω t)ˆ
= B0 cos(kz − ω t)ˆ
y Recall (Physics 124) that this is a transverse traveling wave •
• Fields are in x and y directions.
Movement is to the +z direction.
Wave speed is ω/k. Let’s check that this solution is consistent with Maxwell’s
Equations. Monday, December 5, 2011 Math I • Divergence of E = 0? (No charges.)
∇·E = •
z · E0 cos(kz − ω t)ˆ = 0
∂z Yes! The derivatives are zero except for the d/dz, and the dot
products are 0 except for the x-component, so the divergence
Similarly, the divergence of the B-ﬁeld is zero. Monday, December 5, 2011 Math II • Curl of E = -dB/dt?
z × E0 cos(kz − ω t)ˆ
∇ × E = −kE0 sin(kz − ω t)ˆ
= − B0 cos(kz − ω t)ˆ = −ω B0 sin(kz − ω t)ˆ
∂t • Yes, as long as kE0 = ωB0! Monday, December 5, 2011 Math III • Curl of B = μ0ε0dE/dt? (No currents.)
∇ × B = µ0 0
z × B0 cos(kz − ω t)ˆ
∇ × B = kB0 sin(kz − ω t)ˆ
= µ0 0 E0 cos(kz − ω t)ˆ = µ0 0 ω E0 sin(kz − ω t)ˆ
∂t • Yes, as long as kB0 = μ0ε0ωE0! Monday, December 5, 2011 Math IV: Summary • The proposed solution to Maxwell’s Equation far away from
charges is a transverse E+B traveling wave:
E = E0 cos(kz − ω t)ˆ B = B0 cos(kz − ω t)ˆ
• The divergences of E & B are 0.
The curl of E = -dB/dt as long as kE0 = ωB0! • Denoting the traveling wave velocity as c = ω/k, E0 = cB0. •
• Using E0 = cB0: kB0 = μ0ε0ωcB0 → 1 = μ0ε0c2 → c2 = 1/μ0ε0. Since there are no currents, the curl of B = μ0ε0dE/dt holds
as long as kB0 = μ0ε0ωE0!
c = 1/√(μ0ε0) = 1/√(4πx10-7 x 8.854x10-12) = 3.0x108 m/s Conclusion: one solution to M.E. is a plane wave of E & B ﬁelds
traveling at speed c = 1/√(μ0ε0). This is light. Monday, December 5, 2011 Key points about light: •
• A transverse E+B traveling wave - there is no oscillating
medium like for a mechanical wave.
Deﬁnite ratio between E & B: E0 = cB0.
Fixed speed: c = 1/√(μ0ε0) = 3.0x108 m/s.
E ⊥ B, E & B ⊥ direction of motion.
While our “demonstration” has been for plane waves, other
solutions are possible that have these features. Monday, December 5, 2011 A traveling EM Wave
my x = their y y = their
= my z Monday, December 5, 2011 Backward Wave iclicker
What would change in the picture if the
wave moved to the left - “kz+ωt” instead
There may be more than one correct
answer. A. E would change sign.
B. B would change sign
C. E and B would change sign.
D. k would change sign.
E. The wave cannot travel to the left. Monday, December 5, 2011 With a +ωt, in the evaluation
of the curls of E and B,
there is an additional “-” sign
that indicates E and B have
o pposite signs. See slides 5
and 6. A traveling EM Wave
If we go back to
o ur solution to
and change (kz-ωt)
to (kz+ωt) -assume k and ω > 0
-- you will see that
for M.E. to hold, one
of E0 and B0 has to
change sign, to be
< 0. Monday, December 5, 2011 How do you generate such a traveling wave? •
• Monday, December 5, 2011 Consider a line of charges along
the y axis, at x = z = 0.
Let them oscillate up and down
along the y axis.
As you look from large x, there
w ill be kinks in the electric ﬁeld
in the y direction resulting from
With the current moving in the
y direction, you see B-ﬁeld kinks
in the z direction. Methods of Generating Light •
• Classically: accelerate / oscillate particles along a line, light
moves out in a dipole pattern ⊥ the line. • dipole antennas, synchrotron radiation Classical thermodynamics: black body radiation (really
quantum mechanical) results from interactions (and
accelerations) of moving particles. (Incandescent lights.)
Quantum mechanical: atomic electrons moving from highenergy atomic states to lower energy ones emit photons. Also
occurs in nuclear and particle systems. (Fluorescent lights,
Interaction of charged particles with matter: bremsstrahlung
radiation, Cerenkov radiation, transition radiation.
Subatomic physics: particle and anti-particle annihilate
generating a photon. Monday, December 5, 2011 Methods of Generating Light
lots of light
a⊥ = asinθ
• a little light in
this direction For a dipole antenna, the component of acceleration a you see
Expect E-ﬁeld to vary as sinθ, and power to vary as sin2θ. Monday, December 5, 2011 The EM Spectrum •
• Why is visible light visible?
About peak in solar power generation, and
About peak in atmospheric transmission of light
Recall from earlier wave physics: λf = c. Monday, December 5, 2011 Light iclicker
In vacuum, red light has a
wavelength of 700 nm and violet
light has a wavelength of 400 nm.
That means that in vacuum, red
light The speed of light in
vacuum is independent
of the wavelength.
The longer wavelength
red light has to have a
lower frequency. A. has a higher frequency and moves faster than violet light.
B. has a higher frequency and moves slower than violet light.
C. has a lower frequency and moves slower than violet light.
D. has a higher frequency and the same speed as violet light.
E. has a lower frequency and the same speed as violet light.
Monday, December 5, 2011 Thursday: More light, last lecture of the term except for the review
Some summer job opportunites - internships
with Federal government:
http://www.fossil.energy.gov/education/lelandfellowships/ http://www.nsf.gov/crssprgm/reu Monday, December 5, 2011 ...
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