physics 228 today: interference - department of … we generate actual light - visible light, radio...
TRANSCRIPT
Physics 228 Today: InterferenceWebsite: Sakai 01:750:228 or www.physics.rutgers.edu/ugrad/228
We have now finished 3 lecture on geometrical optics, and will continue with 4 lectures on interference phenomena, which show
the wave nature of light.
We studied interference at some level in 124, so this should be in some respects review.
Sunday, February 24, 2013
InteferenceWhy do we see the colors in the soap
bubble?
Interference.
We will begin to see how this works today,
looking at interference of two
sources.
(Soap bubbles will be on Thursday.)
Sunday, February 24, 2013
EM WavesI asked your recitation instructors to show you that
Maxwell's equations in a source free region:
• Divergence of E (and B) = 0? (No charges.)
• Curl of E = -dB/dt?
• Curl of B = μ0ε0dE/dt? (No currents.) have a traveling wave solution...
�E = E0 cos(kz − ωt)x �B = B0 cos(kz − ωt)y
• As long as:
• kE0 = ωB0 ➭ E0 = cB0, as c = ω/k, and
• 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.
Sunday, February 24, 2013
EM WavesWhat happens when we have a point source emitting light
uniformly in all directions? We get a spherical wave.
The wave expands in r, not z, so the argument
kz becomes kr.
Energy conservation leads to the amplitude decreasing with r, since
the area increases as r2.
�E(r, t) =E0
rcos(kr − ωt)θ �B(r, t) =
B0
rcos(kr − ωt)φ
Implicit in these formulas we have a single frequency / wavelength of light. You can add multiple waves with different amplitudes and
frequencies as needed, because electromagnetism is linear.
Since the θ and φ directions are ⊥ to the r direction, we arbitrarily put the E field in the θ direction and the B field in the φ direction.
Sunday, February 24, 2013
Coherence
When we generate actual light - visible light, radio waves, etc. -the sources can be coherent or incoherent.
E.g., when we oscillate electrons back and forth in an antenna, and at a fixed point look at the light broadcast from it, the
electric and magnetic fields are basically fixed in direction and oscillate back and forth as cos(ωt).
For light bulbs, the photons are emitted incoherently - independently in time and space. So at a fixed observing location, the direction and magnitude of the electric and magnetic fields
vary (somewhat) in time. LEDs are also incoherent, but LASERs are coherent.
For intereference, we need coherent light sources.
�B(r, t) =B0
rcos(kr − ωt)φ�E(r, t) =
E0
rcos(kr − ωt)θ
Sunday, February 24, 2013
Spherical Wave
The two sources emit light ``in phase’’ - ``peaks’’ and ``valleys’’ are emitted at the same time from each source. The waves are also ``in phase’’ at points a and b, but not at point c.The distance to a or b is an even number of wavelengths different.The distance to point c is not.
Sunday, February 24, 2013
Maxima and Minima
At some places the waves add up and you get constructive interference and maxima - for || E fields, the sum is (E1+E2)cos(ωt).At other places the waves add up ``out of phase’’ and you get destructive interference and minima - if the E fields are ||, the sum is always 0!And at most places you are somewhere in between these limits.
Sunday, February 24, 2013
Let's see a demo and a simulation of 2-source wave interference.
iClickerI am going to move the sources apart. Will the
minima / maxima lines move?a) No! Of course not.b) Yes! The angles between them increase and you get fewer of them.c) Yes! The angles between them decrease and you get more of them.d) Yes! You get more minima lines but fewer maxima lines.e) Yes! You get more maxima lines but fewer minima lines.
Sunday, February 24, 2013
iClicker:I am going to shift the wave frequency from red to blue. (Make the wave length shorter.) What happens
to the angles of the minima lines?
a) They stay the same. Nothing changes.b) The angles decrease - the lines move in and there are more of them.c) The angles increase - the lines move out and there are fewer of them.d) The angles decrease, but there are still the same number minima / maxima lines.e) I think answer a) is right or wrong.
Sunday, February 24, 2013
Maxima and Minima AlgebraYou can see from the demo & simulation that when you shift
the frequency or the distance between the sources the interference pattern does move around. Higher frequencies
reduce the angles between the minima and maxima.
Since the sources emit light in phase, maxima and minima at the same time, there is constructive interference when at a
point these add. This happens when the sources are an integral number of wavelengths different distance from some point.
r2 - r1 = mλ, with m = 0, ±1, ±2, ...For minima, we want destructive interference:
r2 - r1 = (m+½)λ, with m = 0, ±1, ±2, ...
You could also analyze the interference based on the time difference from one source vs. the other to a given point being
an integral (or half-integral) number of periods.
Sunday, February 24, 2013
Maxima and Minima Algebra
Assume that the two sources are in phase.
For maxima, we want constructive interference:
r2 - r1 = mλ, with m = 0, ±1, ±2, ...For minima, we want destructive
interference:r2 - r1 = (m+½)λ, with m = 0, ±1, ±2, ...
Sunday, February 24, 2013
Water Analog
For maxima, we want constructive interference:
r2 - r1 = mλ, with m = 0, ±1, ±2, ...For minima, we want destructive
interference:r2 - r1 = (m+½)λ, with m = 0, ±1, ±2, ...
For water, it is pretty easy to keep the two sources in phase - you drop two small stones at the same time.
For radio, with MHz - GHz frequencies, this can be done electronically. But for visible light, with f ≈ 1015 Hz, this is
difficult.How do we get around this?
Sunday, February 24, 2013
In-phase light sources - Old Technique
How do we get around this?We cheat. We take a
coherent LASER source and divide it into
two.
What do we see from the
interference of the two in-phase light sources?
Two in-phase light sources.Diffraction - more about
this in two lectures.
Sunday, February 24, 2013
In-phase light sources
Standard geometrical approximations for analysis: 0) variations in path length over width of band or slit << wavelength1) the two slits are narrow compared to their separation2) The distance to the screen is large compared to the slit separation.
Sunday, February 24, 2013
Geometry and Algebra
Since the path difference is dsinθ, and we get maxima when the difference is an integral number of wavelengths, when
mλ = dsinθ or sinθ = mλ/d.The vertical positions of the bands on the screen become
y = Rtanθ ≈ Rsinθ = mRλ/d.Sunday, February 24, 2013
In-phase light sources
To the left: a picture of alternating light & dark
bands.
maxima at y ≈ mRλ/d.minima at y ≈ (m+½)Rλ/d.
Sunday, February 24, 2013
Maxima and Minima AlgebraAssume that the two sources are in
phase. For maxima, we want constructive
interference:r2 - r1 = mλ, with m = 0, ±1, ±2, ...For minima, we want destructive
interference:r2 - r1 = (m+½)λ, with m = 0, ±1, ±2, ...
There are alternatives - the two sources could be out of phase.
In this case we obtain minima where there were maxima, and vice verse.
More generally, there could be any phase difference between the two
sources.
Sunday, February 24, 2013
Radio Stations
Consider a radio or TV station with a single vertical dipole
antenna. It will broadcast its signal equally in all directions
along the surface.
But if we add a second antenna,The interference pattern
results in more energy going out in some directions - so the signal can be received further away - and less going out in
other directions.
In the configuration shown, interference reduces the energy
going out in some sideways directions. If we changed the antennas to be out of phase,
interference would reduce the energy going to the left and right.
Sunday, February 24, 2013
iClicker:You are going to use two antennas to broadcast radio mainly east/west from the center of Long Island. The antennas have to be east/west of each other. Should they be in phase or out of phase?How many wave
lengths apart should they be? Choose the best answer
a) 0λ in phase.b) 0λ out of phase.c) ½λ out of phase.d) ½λ in phase. e) 1λ, phase does not matter.
Sunday, February 24, 2013