Announcements 11/14/11 Prayer HW 29 problem Labs 8 & 9 due Saturday Progress Reports… Review:...

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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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Transcript of Announcements 11/14/11 Prayer HW 29 problem Labs 8 & 9 due Saturday Progress Reports… Review:...

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?

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

Demo

Demo: Soap film

Interferometer From lab 9:

Interference!

How does this disprove the ether?

changing optical path length, yields ngas