Announcements 11/12/12 Prayer Labs 8 & 9 due Saturday Progress Reports – replies sent In the...

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Announcements 11/12/12 Prayer Labs 8 & 9 due Saturday Progress Reports – replies sent 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 In the Bleache rs Review

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Fourier Transforms How can our two-slit analysis possibly have anything to do with Fourier transforms? (this is the y-coordinate on the slits, not the y-coordinate on the screen) compare to:

Transcript of Announcements 11/12/12 Prayer Labs 8 & 9 due Saturday Progress Reports – replies sent In the...

Page 1: Announcements 11/12/12 Prayer Labs 8 & 9 due Saturday Progress Reports – replies sent In the Bleachers Review.

Announcements 11/12/12

Prayer Labs 8 & 9 due Saturday Progress Reports – replies sent

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

In the Bleachers

Review

Page 2: Announcements 11/12/12 Prayer Labs 8 & 9 due Saturday Progress Reports – replies sent In the Bleachers Review.

From warmup Extra time on?

a. How do beam splitters work?

Other comments?a. Is there a demo on Newton's rings?b. I worked on a research group at my previous

university. The leading professor in our group is/was a researcher at VIRGO and and LIGO. For LIGO (in the US), he said they could detect vibrations in the earth's crust that varied due to the rush hour(s) in nearby large cities. I thought that was pretty cool.

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Fourier Transforms How can our two-slit analysis possibly

have anything to do with Fourier transforms?

~ 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( )cosL

nL

nxa f x dxL L

compare to:

2 2i iE stuff e e

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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.)

slitsscreen

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 sin yPL d dL

2I E

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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

slitsscreen

2 for eachslitPL

2I Etwo-slit sin yPL d dL

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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

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Clicker question: How many “sub” peaks are there

between the “main” peaks in a 5-slit interference pattern?

a. 1b. 2c. 3d. 4e. 5

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Five Slit Problem: Scanning ThetaFive Slit Problem: Scanning Theta

Note: for some reason he picked the overall reference phase to be about 20-30

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Clicker question: 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

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From warmup In Fig. 37.9 (8th edition) the textbook compares

the reflection of light from a surface to the reflection of a wave on a rope. How does the analogy work?

a. I don't have the 8th editionb. Light reflecting off a higher n substance is like a

wave reflecting off a rigid support, and reflecting off a lower n substance is like reflecting off a free support. The n value correlates to how easily the light can move through the substance, which is similar to a rope being free vs rigid.

c. Colton: CAUTION! this analogy only works for close to normal incidence, i.e. 0 deg. (this fact not mentioned in book).

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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 21T r

For arbitrary angle

1 1 2 2.

1 1 2 2

cos coscos coss polar

n nrn n

1 2 2 1.

1 2 2 1

cos coscos cosp polar

n nrn n

1 1.

1 1 2 2

2 coscos coss polarnt

n n

1 1.

1 2 2 1

2 coscos cosp polarnt

n 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 don’t need this

much truth!

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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)

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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

Quick writing: 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).

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From warmup The concept of "optical path length" (OPL) is used

in many places to analyze optical problems. (Unfortunately your textbook doesn't use it.) See for example Wikipedia, http://en.wikipedia.org/wiki/Optical_path_length. In materials with a constant index of refraction n, the OPL is just the path length times n. In the context of today's reading, why might that be a helpful concept?

a. I think it would help when we analyze reflections off of thin films because the destructive interference depends on the index of refraction of the film times the thickness times two. That would be the optical path length.

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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)

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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

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Pretty pictures What’s going on here?

http://twilit.wordpress.com/2008/03/15/bubbles-and-interference/

http://superphysics.netfirms.com/pp_optics.html

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Demo Demo: Soap film

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Interferometer From lab 9:

Interference!

How does this disprove the ether?

changing optical path length, yields ngas