Lecture26 Double Slit Interference

8/3/2019 Lecture26 Double Slit Interference

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Announcements

y Final exam day events:50-point multiple choice end-material test (coveringassigned sections from chapters 33-36). You get a free 8-point question to make up for my incompetence.*

200 point cumulative final exam, all problems (no multiplechoice), with 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 1:45 pm!

You may spend your two hours however you see fit (all onend-material, all on final exam, some mix).

*Assuming I exhibit some incompetence! Missing boardwork can wreck your grade!



K now the exam time!

Find your room ahead of time!

If at 1:30 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.

y Physics 24 Final Room Assignments, Fall 2011:

Coming soon.

Final Exam is1:30 pm,Tuesday,

December 13.

No oneadmitted after

1:45 pm!



Todays agenda:

Review of Waves. You are expected to recall facts about waves from Physics 23.

Youngs 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 interferencein 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 yourprevious physics course (probably Physics 23).

Consider a wave described by

y(x,t) A sin (kx t ) .!

The phase of this wave is

( x , t ) k x t .!

Alsod d x

k .

dt dt !

y

x



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

dx.dt k

[

!

The phase velocity, vp, is given by

p

v .k!

Imagine yourself riding on any point on thiswave. 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 thesame time, the amplitude at that point is the sum of theamplitudes of all the waves

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

In Physics 23 you may have learned that power (or intensity)is proportional to amplitude squared. The intensity of thesuperposed waves is proportional to the square of theamplitude of the resulting sum of waves.

Superpositiona Characteristic of All Waves



Constructive Interference: If the waves are in phase, theyreinforce 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.

Interferencea Result of the Superposition of Waves



Todays agenda:







This experiment demonstrates thewave nature of light.

Consider a single light source, andtwo slits. Each slit acts as a

secondary source of light.

Light waves from secondary slitsinterfere to produce alternatingmaxima and minima in the

intensity.

Reference and toys: fsu magnet lab, coloradolight cannon.

Youngs Double Slit Experiment

Interesting reading: the double slit experiment and quantum mechanics.



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.

In phase

constructive.

Out of phase

destructive.



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.



Heres a video, with some teasers about quantum mechanics.

Disclaimer! The video ends up heading towards shaky ground.

See, for example, Schrödingers Cat:

One can even set up quite ridiculous cases. A cat is penned up in a steel chamber, along with thefollowing device (which must be secured against direct interference by the cat): in a Geiger counter thereis a tiny bit of radioactive substance, so small, that perhaps in the course of the hour one of the atomsdecays, 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 thisentire system to itself for an hour, one would say that the cat still lives if meanwhile no atom hasdecayed. 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, isa concept that does not scale to large systems (like cats), which are not indeterminably probabilistic innature.Wikipedia

To avoid streaming that video, Ill play this in class.



A note on course grades:sample.xls

More examples next Monday,including examples crucial for borderline grades!



Todays agenda:



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




Conditions for Interference

Sources must be monochromatic-of a single wavelength.

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

Heres the geometry I will usein succeeding diagrams.



U

d

L2

L1

(L = L2 ± L

1= d sin U

For an infinitely distant* screen:

U

P

S2

S1

(L

R

y

L1

L2

U

d

tan y

R U !

U

*so that all the angles labeled

U are approximately equal



Destructive Interference:

Constructive Interference:

The parameter m is called the order of the interference

fringe. The central bright fringe at U = 0 (m = 0) is knownas the zeroth-order maximum. The first maximum on eitherside (m = ±1) is called the first-order maximum.

U

d L

2

L1

(L = L2 ± L

1= d sin U

U

, , ,( ! U ! P s sL d sin m m=0 1 2...

, , ,¨ ¸( ! U ! P s s© ¹ª º

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

2



U

P

S2

S1

(L

R

y

L1

L2

U

d

tan y

R U !

! U } Uy R tan R sin

Bright fringes:

P ! Um d sin

P !y

m d

R

P!

Ry m

d

This is not a startingequation!

Do not use the small-angleapproximation unless it is valid!



U

P

S2

S1

(L

R

y

L1

L2

U

d

tan y

R U !


Dark fringes:

This is not a startingequation!

¨ ¸ P ! U© ¹

ª º

1m d sin

2

¨ ¸ P !© ¹ª º

1 ym d2 R

P ¨ ¸! © ¹ª º

R 1y m

d 2

Do not use the small-angleapproximation unless it is valid!



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.T

he second-order bright fringe (m = 2) is 4.5 cm fromthe center line. Determine the wavelength of the light.


Bright fringes:P ! Um d sin

P !y

m dR

P !yd

Rm

v v

P ! ! v !

-2 -5

74.5 10 m 3.0 10 m

5.6 10 m 560 nm1.2 m 2

U

P

S2

S1

(L

R

y L

1

L2

U

tan y

R U !

d




mm.T

he second-order bright fringe (m = 2) is 4.5 cm fromthe center line. Find the distance between adjacent bright fringes.


Bright fringes:

P ! Um d sin

P !y

m d

RP

!R

y md

vP P P! ! ! ! v !

v

7

2m+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

U

P

S2

S1

(L

R

y L

1

L2

U

tan y

R U !

d




mm.T

he second-order bright fringe (m = 2) is 4.5 cm fromthe center line. Find the width of the bright fringes.

Define the bright fringe width to bethe distance between two adjacent

destructive minima.¨ ¸

P ! U !© ¹ª º

darky1m d sin d

2 R

P ¨ ¸! © ¹

ª º

dark

R 1y m

d 2

v! !

v

7

dark,m+1 dark,m -5

5.6 10 m 1.2 my -y 2.2 cm

3.0 10 m

U

P

S2

S1

(L

R

y L

1

L2

U

tan y

R U !

P P P¨ ¸ ¨ ¸! !© ¹ © ¹

ª º ª ºdark,m+1 dark,m

R 1 R 1 Ry -y m 1 m

d 2 d 2 d

d



Todays agenda:







Intensity in the Double Slit Experiment

Our equations for the minimaand maxima intensity positionsare for the centers of thefringes.

In this section, we calculatedistribution of light intensity inthe double-slit interference

pattern.



The derivation of the double-slit intensity equation is not particularly difficult, so study it if you find derivations helpfulfor your understanding.

A path length difference(L=P corresponds to aphase difference of J=2T.

A path length difference(L=mP corresponds to aphase difference of J=2Tm.

In general, for non-integral m, the phase difference at Pbetween the waves from S1 and S2 is

( U U

L d sin 2= = = d sin

2 x x x

(

2= L is also "official"



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

the previous slide.

Your starting equation for theintensity is

¨ ¸© ¹ª º

20I=I cos

where I0 is 4 times the peakintensity of either of the two

interfering waves:

0 single waveI = 4I

Why did my previous diagrams show this?

Lecture26 Double Slit Interference

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