Announcements 11/14/11 Prayer HW 29 problem Labs 8 & 9 due Saturday Progress Reports… Review: 1 i E stuff e 2 2 i i E stuff e e cos( 2) E stuff 2 0 cos ( 2) I I phaseshift 2( ) PL Approx.1: sin PL d A pprox.2:sin yL Rubes
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Announcements 11/14/11 Prayer HW 29 problem Labs 8 & 9 due Saturday Progress Reports… Review: Rubes.
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Announcements 11/14/11
Prayer HW 29 problem Labs 8 & 9 due Saturday Progress Reports… Review:
1 iE stuff e
2 2i iE stuff e e
cos( 2)E stuff 2
0 cos ( 2)I I
phaseshift 2 ( )PL Approx.1: sinPL d
Approx.2: sin y L
Rubes
Fourier Transforms?
How can our two-slit analysis possibly have anything to do with Fourier transforms? 1 iE stuff e
~ each slitiE e
~ i
openareas
E e dY
~ " " iE aperture function e dY
(this is the y-coordinate on the slits, not the y-coordinate on the screen)
2
2
2 2( )cos
L
n
L
nxa f x dx
L L
compare to:
Adding up phases
…
rel. to ref. slit1 slit 2 final slit0 ...i i i i
totE E e e e e
For an equally-spaced pattern of slits, how do the PLs compare?
Each is a multiple of 2! (Could have an overall reference phase for 1…not too important.)
slits
screen
2 for eachslitPL
In short, we need to add up a bunch of vectors that have the same magnitude (1), but angles (phases)
that go like 0, 20, 40, 60, etc.
For a different position on the screen (measured by y or , we need to add up a different set of phases…
perhaps like 0, 21, 42, 63, etc.
two-slit siny
PL d dL
2
I E
Adding up phases, cont.
…
rel. to ref. slit1 slit 2 final slit0 ...i i i i
totE E e e e e
Quick writing: graphically add these three vectors: 10 + 120 + 140
What about 10 + 190 + 1180
slits
screen
2 for eachslitPL
2I Etwo-slit sin
yPL d d
L
Three Slit Problem: Scanning ThetaThree Slit Problem: Scanning ThetaCredit: this animation and the next one are from Dr. Durfee
Note: for some reason he picked the overall reference phase to be about 20
Thought question
How many “sub” peaks are there between the “main” peaks in a 5-slit interference pattern?
a. 1b. 2c. 3d. 4e. 5
Five Slit Problem: Scanning ThetaFive Slit Problem: Scanning Theta
Note: for some reason he picked the overall reference phase to be about 20-30
Reading Quiz
When a light wave travels from a low index to a high index material at normal incidence (perpendicular to surface), what is the phase shift of the reflected wave?
a. 0b. 45c. 90d. 180e. depends on whether it is s- or p-
polarization
Remember these? “Fresnel Equations”
2 1 1 2
1 2 1 2
v v n nrv v n n
2 1
1 2 1 2
2 2v ntv v n n
If near perpendicular (1-D problem)
2R r 2
1T r
For arbitrary angle
1 1 2 2.
1 1 2 2
cos cos
cos coss polarn n
rn n
1 2 2 1.
1 2 2 1
cos cos
cos cosp polarn n
rn n
1 1.
1 1 2 2
2 cos
cos coss polarn
tn n
1 1.
1 2 2 1
2 cos
cos cosp polarn
tn n
Same as strings
The Truth (overlooked by textbook): you don’t always get a phase shift,
even if going fast to slow. (Brewster marks boundary) More Truth: sometimes phase
shifts not just 180: can have complex n, complex , etc.You can’t handle
this much truth!
Air to glass (n=1 to n=1.5)
p-polarization
20 40 60 80 0.2
0.2
0.4
0.6
0.8
1.0
r
t
field amplitudes vs
Brewster180 phase shift
(close to perpendicular)0 phase shift
(close to glancing)
Back to “near normal incidence”
Rays drawn at an angle to make viewing easier. They’re really
perpendicular to surface.
From low to high index: 180 phase shift From high to low index: no phase shift
What does the thickness of this slab need to be to get constructive interference between the two rays?
air
air
thin glass thickness t
If rays at an angle… determine if above/
below Brewster angle (if p-polarization).
Optical path length OPL = Path Length n
since wavelength inside the material is reduced by a factor of n, the distance “looks” bigger than it actually is
Constructive interference: OPL ( any phase shifts) = m
Destructive interference:
OPL ( any phase shifts) = (m+1/2)
New situation
Rays drawn at an angle to make viewing easier. They’re really
perpendicular to surface.
What does the thickness of the COATING need to be to get constructive interference between the two rays?