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Announcements 3/23/11 Prayer Two labs this week (telescope, interferometer)
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
Fourier Transforms?
From last time: what did our two-slit analysis have 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 1! (Could have an overall reference phase…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 wave on a string moves from a fast
velocity section to a slow velocity section, the reflected wave is phase-shifted by 180. When a wave moves from a slow section to a fast section, however, the reflected wave has no phase shift.
How does the phase shift of a light wave moving from one medium to another compare to that? (close to normal incidence)
a. The phase shift obeys the same rulesb. The phase shift is the same for fast-to-slow,
but reversed for slow-to-fastc. The phase shift is the same for slow-to-fast,
but reversed for fast-to-slow d. The phase shift is reversed for both cases
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
Just the 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!
Back to 1D case
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
What changes if rays really are at an angle?
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?
air
thick glass,n = 1.5
thin coating, n = 1.3
thickness t
Pretty pictures
What’s going on here?
http://twilit.wordpress.com/2008/03/15/bubbles-and-interference/
http://superphysics.netfirms.com/pp_optics.html