announcements final exam day events (wednesday, december 16, 12:30 pm) 50-point multiple choice...

Announcements

Final exam day events (Wednesday, December 16, 12:30 pm)50-point multiple choice end-material test (covering

assigned sections from chapters 33-36). You get a free 8-point question to make up for my incompetence.*200 point comprehensive final exam, all problems (no multiple choice), about 50% emphasis on chapters 33-36.You may take neither, one, or both of these tests. Your choice. Your grade. No one admitted after 12:45 pm!You may spend your two hours however you see fit (all on end-material, all on final exam, some mix).

*Assuming I exhibit some incompetence soon!

Missing boardwork can wreck your grade!

Know the exam time!

Find your room ahead of time!

If at 5:00 on test day you are lost, go to 104 Physics and check the exam room schedule, then go to the appropriate room and take

the exam there.

Exam is from 5:00-6:00 pm!

Physics 2135 Final Exam Rooms, Fall 2015:

Instructor Sections RoomDr. Hagen J, R TBADr. Hale K, P TBADr. Hor B, D TBADr. Kurter A, C TBADr. Madison G, L TBADr. Parris M, Q TBAMr. Upshaw N, H TBADr. Vojta E, F TBASee notes on next slide for information about room locations.

Special Accommodations Testing Center

Physics 2135 Final Room Assignments, 12:30 PM, DEC 16

Instructor Sections RoomDr. Hagen J, R TBA

room information will be provided here

Dr. Hale K, P TBAroom information will be provided here

Dr. Hor B, D TBAroom information will be provided here

Dr. Kurter A, C TBAroom information will be provided here

Dr. Madison G, L TBAroom information will be provided here

Dr. Parris M, Q TBAroom information will be provided here

Mr. Upshaw N, H TBAroom information will be provided here

Special Accommodations Testing Center

Today’s agenda:

Review of Waves.You are expected to recall facts about waves from Physics 1135.

Young’s Double Slit Experiment.You must understand how the double slit experiment produces an interference pattern.

Conditions for Interference in the Double Slit Experiment.You must be able to calculate the conditions for constructive and destructive interference in the double slit experiment.

Intensity in the Double Slit Experiment.You must be able to calculate intensities in the double slit experiment.

Interference

Review of Waves

This section is a review of material you learned in your previous physics course (probably Physics 1135).Consider a wave described by

y(x,t) A sin (kx ωt) .

The phase of this wave is

θ(x,t) kx ωt .

Also, the phase changes with time:dθ dx

k ω .dt dt

y

x

If is constant with time (i.e., d/dt=0), then we are moving with the wave, and

dx .

dt k

The phase velocity, vp, is given by

p

ωv .

k

Imagine yourself riding on any point on this wave. The point you are riding moves to the right. The velocity it moves at is vp.

If the wave is moving from left to right then /k must be positive.

y

x

When waves of the same nature travel past some point at the same time, the amplitude at that point is the sum of the amplitudes of all the waves

The amplitude of the electric field at a point is found by adding the instantaneous amplitudes, including the phase, of all electric waves at that point.

In Physics 1135 you may have learned that power (or intensity) is proportional to amplitude squared. The intensity of the superposed waves is proportional to the square of theamplitude of the resulting sum of waves.

Superposition—a Characteristic of All Waves

Constructive Interference: If the waves are in phase, they reinforce to produce a wave of greater amplitude.

Destructive Interference: If the waves are out of phase, they reinforce to produce a wave of reduced amplitude.

Interference—a Result of the Superposition of Waves

Today’s agenda:





This experiment demonstrates the wave nature of light.

Consider a single light source, and two slits. Each slit acts as a secondary source of light.

Light waves from secondary slits interfere to produce alternating maxima and minima in the intensity.

Reference and “toys:” fsu magnet lab, Colorado light cannon, wave interference, double slit.

Young’s Double Slit Experiment

Interesting reading: the double slit experiment and quantum mechanics.But don’t take it too literally.

How does this work?

Light waves from the two slits arriving at the detection screen in phase will interfere constructively

In phase—constructive.

Out of phase—destructive.

5

4

Light waves from the two slits arriving at the detection screen in phase will interfere constructively and light waves arriving out of phase will interfere destructively. Another applet (useful).

How does this work?

Light waves from the two slits arriving at the detection screen in phase will interfere constructively and light waves arriving out of phase will interfere destructively.

Video, with some teasers about quantum mechanics.

Disclaimer! The video is part of an advertisement for a New Age religion, and ends up heading towards shaky ground.

See, for example, Schrödinger’s Cat:

“One can even set up quite ridiculous cases. A cat is penned up in a steel chamber, along with the following device (which must be secured against direct interference by the cat): in a Geiger counter there is a tiny bit of radioactive substance, so small, that perhaps in the course of the hour one of the atoms decays, but also, with equal probability, perhaps none; if it happens, the counter tube discharges and through a relay releases a hammer which shatters a small flask of hydrocyanic acid. If one has left this entire system to itself for an hour, one would say that the cat still lives if meanwhile no atom has decayed. The psi-function of the entire system would express this by having in it the living and dead cat (pardon the expression) mixed or smeared out in equal parts.”—Erwin Schrödinger

“The idea of a particle existing in a superposition of possible states, while a fact of quantum mechanics, is a concept that does not scale to large systems (like cats), which are not indeterminably probabilistic in nature.”—Wikipedia

Electron double slit applet.

You can find this video on Youtube. To avoid streaming it, I’ll play this in class.

A note on course grades:sample.xls

More examples next Monday,including examples crucial for borderline

grades!

Today’s agenda:





Conditions for Interference

Sources must be monochromatic-of a single wavelength.

Sources must be coherent-- must maintain a constant phase with respect to each other.

Here’s the geometry I will use in succeeding

diagrams.

dL2

L1

L = L2 –L1 = d sin

For an infinitely distant* screen:

P

S2

S1

L

R

y

L1

L2

d

tany

R

*so that all the angles labeled are approximately equal

Destructive Interference:

Constructive Interference:

The parameter m is called the order of the interference fringe. The central bright fringe at = 0 (m = 0) is known as the zeroth-order maximum. The first maximum on either side (m = ±1) is called the first-order maximum.

dL2

L1

L = L2 –L1 = d sin

, , , L d sin m m=0 1 2...

, , ,

1L d sin m+ m=0 1 2...

2

P

S2

S1

L

R

y

L1

L2

d

tany

R

y R tan R sin

Bright fringes:

m d sin

ym d

R

Ry m

d

This is not a starting equation!

Do not use the small-angle approximation unless it is valid!

For small angles:

P

S2

S1

L

R

y

L1

L2

d

tany

R

y R tan R sin

Dark fringes:

This is not a starting equation!

1m d sin

2

1 ym d

2 R

R 1y m

d 2

Do not use the small-angle approximation unless it is valid!

For small angles:

Example: a viewing screen is separated from the double-slit source by 1.2 m. The distance between the two slits is 0.030 mm. The second-order bright fringe (m = 2) is 4.5 cm from the center line. Determine the wavelength of the light.

y R tan R sin

Bright fringes:

m d sin

ym d

R

yd

Rm

-2 -5

74.5 10 m 3.0 10 m

5.6 10 m 560 nm1.2 m 2

P

S2

S1

L

R

yL1

L2

tany

R

d

Example: a viewing screen is separated from the double-slit source by 1.2 m. The distance between the two slits is 0.030 mm. The second-order bright fringe (m = 2) is 4.5 cm from the center line. Find the distance between adjacent bright fringes.

y R tan R sin

Bright fringes:

m d sin

y

m d R

Ry m

d

72

m+1 m -5

5.6 10 m 1.2 mR R Ry -y m 1 m 2.2 10- m 2.2 cm

d d d 3.0 10 m

P

S2

S1

L

R

yL1

L2

tany

R

d

Example: a viewing screen is separated from the double-slit source by 1.2 m. The distance between the two slits is 0.030 mm. The second-order bright fringe (m = 2) is 4.5 cm from the center line. Find the width of the bright fringes.Define the bright fringe width to be the distance between two adjacent destructive minima.

darky1m d sin d

2 R

dark

R 1y m

d 2

7

dark,m+1 dark,m -5

5.6 10 m 1.2 my -y 2.2 cm

3.0 10 m

P

S2

S1

L

R

yL1

L2

tany

R

dark,m+1 dark,m

R 1 R 1 Ry -y m 1 m

d 2 d 2 d

d

Today’s agenda:





Intensity in the Double Slit Experiment

Our equations for the minima and maxima intensity positions are for the centers of the fringes.

In this section, we calculate distribution of light intensity in the double-slit interference pattern.

The derivation of the double-slit intensity equation is not particularly difficult, so read about it in your text if you find derivations helpful for your understanding.

A path length difference L= corresponds to a phase difference of =2.

A path length difference L=m corresponds to a phase difference of =2m.In general, for non-integral m, the phase difference at P between the waves from S1 and S2 is

φ π φ

π λ λ λ

L d sin 2= = = d sin

2 x x xπ

φλ

2

= L is also "official"

Your text writes the equation for the intensity distribution in the double-slit experiment in terms of the phase difference on the previous slide.

Your starting equation for the intensity is

φ

20I =I cos

where I0 is 4 times the peak intensity of either of the two interfering waves:

0 single waveI =4I

Why did my previous diagrams show this?

announcements final exam day events (wednesday, december 16, 12:30 pm) 50-point multiple choice...

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