summer12 lectures lect03

8/13/2019 Summer12 Lectures Lect03

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Diffraction&

Spectroscopy(Lecture 4 in Notes)

y

L

d

q

Spectra of atoms reveal the

quantum nature of matter


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My office number is

390X(not 390V)


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Phasors Finding the resultant amplitudes of

two or more waves with differentamplitudes, using phasors

A1

A

A1 coscos 11 AAA +

cos2 1A

Now /2 2cos2 1

AA

To get the intensity, we simply square this amplitude:

2cos4 2

2

1

AI

2cos4 21

II where I1= A1

2is the intensitywhen only one slit is open

This is identical to our previous result !

Represent a wave by avector with magnitude(A1) and direction ().One wave has = 0.

More generally, if the phasors havedifferentamplitudes Aand B,

C2

= A2

+ B2

+ 2AB cos Here isthe externalangle.

A

C

B

See text, Secs 35.3, 36.3, 36.4and appendix to lecture 3Phasors make it easier to solveother problems later on

h f l


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Phasors for 2-Slits Plot the phasor diagram for different :

A1

A1

=458/

A1

A1

=904/

A1

A1

=1358/3

A1

A1

=180

2/

A1 A1

A

=360

A1 A1

A

=0

0

A1

A1

=2258/5

A1

A1

=2704/3

A1

A1

=

315

8/7

I

0

4I1

0 2p-2p

q*/d-/dy/d)L-(/d)L

Slits Demo

(Small-angle approx.assumed here)

l l f


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Multi-Slit InterferenceWhat changes if we increase the numberof slits (e.g., N = 3, 4, 1000, . . .) ?

First look at the principle maxima:

The positions of principle interference maximaare the same for any number of slits!

S3

S2

P

Incident wave(wavelength )

y

L

d

S1

If slit 1 and 2 are in phase

with each other, then slit 3will also be in phase, etc.

d sinq= m Slits DemoLaser Demo

M l l f


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Multi-Slit Interference - II

Atot= 3 A1 I

tot= 9 I

1

For N slitsIN= N

2I1 What is the intensity at

the principle maxima?

0 2p-2p

I

0

16I1N=4

0 2p-2p

I

0

25I1N=5

0 2p-2p

I

0

9I1N=3

-/d 0 /d

q

q-/d 0 /d -/d 0 /d

q


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General properties of N-Slit Interference

The positions of the principal maxima of the intensity patterns alwaysoccur at= 0, 2p, 4p, ...[is the phase between adjacentslits]

i.e., dsinq= m, m = 0, 1, 2,).The principal maxima become taller and narrower as N increases.

The intensity of a principal maximumis equal to N2times the maximumintensity from one slit. The width of a principal maximum goes as 1/N.

The # of zeroes between adjacent principal maxima is equal to N-1.The # of secondary maximabetween adjacent principal maxima is N-2.

0 2p-2p

I

0

16I1N=4

0 2p-2p

I

0

25I1N=5

0 2p-2p

I

0

9I1N=3

-/d 0 /d

q

q-/d 0 /d -/d 0 /d

q

A 1


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Act 1Light interfering from 10 equally spaced slits initially illuminates

a screen. Now we double the number of slits, keeping thespacing constant.

1. What happens to the intensity I at the principal maxima?

a. stays same (I) b. doubles (2I) c. quadruples (4I)

2. What happens to the net power on the screen?

a. stays same b. doubles c. quadruples

A 1 S l i


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Act 1 - SolutionLight interfering from 10 equally spaced slits initially illuminates

a screen. Now we double the number of slits, keeping thespacing constant.

1. What happens to the intensity I at the principal maxima?

a. stays same (I) b. doubles (2I) c. quadruples (4I)

2. What happens to the net power on the screen?

a. stays same b. doubles c. quadruples

If we doublethe number of slits, we expect the netpower on the screen to double. How does it do this

The location and number of the principle maxima (whichhave most of the power) does notchange.

The principle maxima become 4x brighter.

But they also become only half as wide. (w/2)

Therefore, the net power (integrating over all the peaks)

increases two-fold, as we would expect. (P goes as Iw)


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N-Slit Interference(see Appendix for derivation)

The Intensity for N equally spaced slits is given by:

Lyanddsind q

q

q

p2

2

1)2/sin(

)2/sin(

NIIN

*

*Your calculator can probably plot this up. Give it a try.

** Note: You will not be able to use the small angle approximations if d ~ .

y

L

d

q

is the phase difference between adjacent slits.

As usual, to determine the pattern at the screen(detector plane), we need to relate to qory = Ltanq:

**


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Example Problem #1

0 qmin? /d q

In an N-slit interference pattern, at what angle qmindoes theintensity first go to zero? (In terms of , d and N.)


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Example Problem #1

0 qmin? /d q

In an N-slit interference pattern, at what angle qmindoes theintensity first go to zero? (In terms of , d and N.)

But 2p(d sinq)/2pd q/=2p/N. Therefore, qmin /Nd.

As the illuminated number of slits increases, the peak widths decrease!General feature:

Wider slit features

narrower patterns in the far field.(obviously doesnt hold just past slits!)

2

1

sin( / 2)

sin( / 2)N

NI I

has a zero when Nmin/2 p, or min =2p/N.


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Interference GratingsExamples around us.

CD disk grooves spaced by ~ wavelength of visible light

The color of a butterfly wing!

The material in the wing in not colored!

The color comes from interference ofthe reflected light from the pattern ofscales on the winga grating with spacingof order the wavelength of visible light!

(See the text Young & Freeman, 36.5)


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Optical spectroscopy a major window on the world

Quantum mechanics definite energy levels, e.g.,of electrons in atoms or molecules. When an atom transitions between energy levelsemits light of a very particular frequency.

Every substance has its own signature of whatcolors it can emit. By measuring the colors, we can determine thesubstance, as well as things about its surroundings(e.g., temperature, magnetic fields), whether its

moving (via the Doppler effect), etc.

Optical spectroscopy is invaluable in materialsresearch, engineering, chemistry, biology, medicine

But how do we precisely measure wavelengths???

I t f G ti


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Interference GratingsThe basis for optical spectroscopy

Interference gratings (usually called Diffraction gratings)allow us to resolve sharp spectral signals.

0 2/d sinq

N-slit Interference:

IN = N2I1

0 1/d sin q1

2Shift ofthe peak:

Demo: Red and green light from lasers

S t D t ti


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Spectroscopy DemonstrationWe have set up some discharge tubes with various gases. Notice that

the colors of the various discharges are quite different. In fact thelight emitted from the highly excited gases is composed of manydiscrete wavelengths. You can see this with the plastic grating wsupplied. Your eye

Hold grating less than 1 inch from your eye.

Light source

Your view throughthe grating:

Put light sourceat left side ofgrating.

View spectral lines bylooking at about 45o.If you dont seeanything, rotate

grating 90o.

Room lights mustbe turned off!

At i S t


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Atomic SpectroscopyFigure showing examples of atomic spectra for H, Hg, and Ne.

Can you see these lines in the demonstration?

You can take the grating. See if you can see the atomiclines in Hg or Na street lights or neon signs.

Source for figure:http://www.physics.uc.edu/~sitko/CollegePhysicsIII/28-AtomicPhysics/AtomicPhysics.htm

The lines are explained by Quantum Mechanics!


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Diffraction from gratings anda single slit in more detail

The following slides describe diffraction gratings inmore depth Resolving power of a N-slit grating

The Rayleigh criterion

Example of the requirements to resolve the sodiumdoublet

Diffraction from a single slit

Resolution for a single slitThe Rayleigh criterion again

Combined diffraction of a grating composed of slitswith finite width

Diff ti G ti (1)


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Diffraction Gratings (1) Diffraction gratings rely on N-slit interference.

They simply consist of a large number of evenly spaced parallel slits.

Recall that the intensity pattern produced by light of wavelength passing through Nslits with spacing dis given by:2

12

2

)/sin(

)/Nsin(IIN

where:

0 2p2p

I

0

25I1N=5

dsin2

q p

Consider very narrow slits (a


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Diffraction Gratings (2)

How effective are diffraction gratings at resolving lightof different wavelengths (i.e. separating closely-spaced

spectral lines)? Concrete example:Na lamp has a spectrum with two yellow lines

very close together: 1= 589.0 nm, 2= 589.6 nm (D 0.6 nm) Are these two lines distinguishable using a particular grating?

IN = N2I1

0 1/d sin q

0 2/d sin q

1

2

Diff ti n G tin (3)


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Diffraction Gratings (3)

minmin

sind d

d Nd

q q

q

D D

D D

We assume Rayleighs criterion: the minimum wavelength separationwe can resolve Dmin2-1occurs when the maximum of 2overlapswith the first diffraction minimum of

1

.

(Dqmin

=/Nd)

IN = N2I1

N

1m in

D Larger N Smaller Dmin(Higher spectral resolution)

0 1/Nd sin q

q

1/d

Rayleigh Criterion

N = number of illuminatedlines in grating.

Dqmin

2/d

Diffracti n Gratin s (4)


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Diffraction Gratings (4) We can squeeze more resolution out of a given grating by working

in higher order. Remember, the principal maxima occur atsinq= m/d, where m = 1,2,3 designates the order.(Dqmin /Nd still*)You can easily show:

Nm

1m in

D

0 /d 2/d 3/d sin q

First order Second order Third order

m = 1 m = 2 m = 3

* To be precise: Dqmin= /(Nd cosq), (but D/= 1/Nm iscorrect.)

Larger Nm Smaller Dmin(Higher spectral resolution)

Example: Diffraction Gratings


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Example: Diffraction Gratings

Angular splitting of the Sodium doublet:

Consider the two closely spaced spectral (yellow) lines of sodium(Na), 1= 589 nm and2= 589.6 nm. If light from a sodiumlamp fully illuminates a diffraction grating with 4000 slits/cm,what is the angular separation of these two lines in the second-order(m=2)spectrum?

Hint: First find the slit spacing dfrom the number of slits percentimeter.


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ACT 2


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ACT 21. Assuming we fully illuminate the grating from the previousproblem (d = 2.5 m), how big must it be to resolve the Nalines (589 nm, 589.6 nm)?(a)0.13 mm (b)1.3 mm (c) 13 mm

3. Which reducesthe maximum number of interference orders?

(a) Increase wavelength

(b) Increase slit spacing

(c) Increase number of slits

2. How many interference orders can be seen with this grating?(a)2 (b)3 (c) 4

ACT 2 Solution


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ACT 2 - Solution1. Assuming we fully illuminate the grating from the previousproblem (d = 2.5 m), how big must it be to resolve the Nalines (589 nm, 589.6 nm)?(a)0.13 mm (b)1.3 mm (c) 13 mm

3. Which reducesthe maximum number of interference orders?

(a) Increase wavelength

(b) Increase slit spacing

(c) Increase number of slits

We need enough lines to narrow the diffraction peak:

1

Nm

D

589nm491

2(0.6nm)

N

m

D

size = N d

490 (2.5

m)1.2 mm

2. How many interference orders can be seen with this grating?

The diffraction angle can never be more than 90: From sinq= m/d,it must be that m d, or m d/= 2.5 m/0.589m = 4.2 m = 4

Increase , or decreased.Changing the number of slits does

not affect the number of orders.

(a)2 (b)3 (c) 4

Single slit Diffraction


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So far in the N-slit problem we have assumed that each slit is apoint source.

Point sources radiate equally in all directions. Real slits have a non-zero extent - a slit width a. The

transmission pattern depends on the ratio of ato .

In general, the smaller the slit width, the more the wave willdiffract!

Laser Light(wavelength )

screen

Diffractionprofile

I1

Small slit:

Laser Light(wavelength )

screen

Diffractionprofile

I1

Large slit:

Lets examine this effect quantitatively.

Single-slit Diffraction Laser Demo

Single Slit Diffraction


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Single-Slit Diffraction Slit of width a. Where are the

minima? The first minimum is at an angle such

that the light from the top and themiddleof the slit destructivelyinterfere:

The second minimum is at an anglesuch that the light from the top and apoint at a/4 destructively interfere:

Location of nth-minimum:

P

Incident Wave(wavelength ) y

L

aqa/2

minsin2 2

a q

minsinaq

min,2sin4 2

a q

min,2

2sin

a

q

min,sin nn

a

q

qa/4

(n= 1, 2, )


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Single-slit Diffraction: Example Problem

Suppose that when we pass red light (= 600 nm) through a slit of

width a, the width of the spot (the distance between the first zeroson each side of the bright peak) is W = 1 cm on a screen that is L = 2mbehind the slit. How wide is the slit?

2 m

a 1 cm = W


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Solution

Suppose that when we pass red light (= 600 nm) through a slit of

width a, the width of the spot (the distance between the first zeroson each side of the bright peak) is W = 1 cm on a screen that is L = 2mbehind the slit. How wide is the slit?

Solution:

The angle to the first zero is: q= /a

1 cm = W

L = 2 m

a

q

Solve for a: a =2L/W= (4m)(610-7m) /(10-2m)

= 2.410-4

m = 0.24 mm

W = 2 Ltan q 2 Lq 2L/a (use tan q q)

ACT 3


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ACT 3

Which of the following would broaden thediffraction peak?

a. reduce the laser wavelength

b. reduce the slit width

c. move the screen further away

2 m

a 1 cm = W

ACT 3 Solution


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ACT 3 - Solution

Which of the following would broaden thediffraction peak?

a. reduce the laser wavelength

b. reduce the slit width

c. move the screen further away

2 m

a 1 cm = W

Single-slit Diffraction Summary


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Single-slit Diffraction Summary(see Appendix for derivation)

The intensity of a single slit has the following form:

Single Slit Diffraction Features:

First zero: b/2= p q /a

(agrees with phasor analysis)

Secondary maximum is quite small.

b

q

-4p -2p 0 2p 4p

-/a 0 /a

0

I

I0

At P, the phase difference bbetween 1st and last source is

given by:

2

01

/a

)/asin(I)(I

qp

qpq

Screen(far away)

a

= a sinq aq

q

a

L

qq

q

p

bLy

asina

2

a

Therefore,

P

2

012/

)2/sin(II bb

b= angle between 1st and last phasor

S Slit I t f Diff ti


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Summary: Slit Interference + DiffractionCombine:

Multi-slit Interference,

and

Single-slit Diffraction,

to obtain

Total Interference Pattern,

Remember:/2p /= (d sinq)/d q/b/2p a/ (a sinq)/ aq/

2

012/

)2/sin(

b

bII

2

1)2/sin(

)2/sin(

NIIN

22

0

)2/sin(

)2/Nsin(

2/

)2/sin(II

b

b

angle between adjacent phasorsb angle between 1st and last phasor

You will explore these concepts in Lab this week!

(for plane-wave sources, I1= constant)

Summary


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Summary

*Derivations in Appendix(also in Young and Freeman, Secs. 36.2 and 36.4)

Multiple-slit Interference*

Principle Maxima given by Bragg Law: d sin q= m Peaks narrower & more intense as the number of slits increases

Diffraction Gratings

Typically very large number of slits small angle approx. valid

Spectral Resolution increases with the number of slits

Optical Spectroscopy

Spectra of atoms reveals the nature of the atomic scale world- Explained by Quantum Mechanics!

Single-Slit Diffraction*

Interference + Diffraction full description


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Home Exercise


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Home Exercise

Light of wavelength is incident on an N-slitsystem with slit width aand slit spacing d.

1. The intensity Ias a function ofyat a viewingscreen located a distance Lfrom the slits isshown to the right. What is N? (L>> d,y, a)

(a)N= 2 (b)N = 3 (c)N =40

I

Imax

0 +6-6

Y (cm)

2. Now the slit spacing dis halved, but the slit width ais kept constant.Which of the graphs best represents the new intensity distribution?

(a)

0 +6-6

cm0

I

Imax (c)

0 +6-6

Y cm 0

I

Imax(b)

0 +6-6

Y cm 0

I

Imax

Home Exercise - Solution


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(a)N= 2 (b)N = 3 (c)N =40

I

Imax

0 +6-6

Y (cm)


(a)

0 +6-6

cm0

I

Imax (c)

0 +6-6

Y cm 0

I

Imax(b)

0 +6-6

Y cm 0

I

Imax

N is determined from the number of minima between two principalmaxima. Here, there are two minima between principal maxima.Therefore, N = 3 .



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0

I

Imax

0 +6-6

Y (cm)


(a)

0 +6-6

Y (cm)

0

I

Imax (c)

0 +6-6

Y (cm)

0

I

Imax(b)

0 +6-6

Y (cm)

0

I

Imax

Decreasing dwill increasespacing between maxima.

The spacingbetween maximais increased, but diffraction

profile shouldnt expand, asseen here.

This one does it all.Increased spacing

between maxima andconstant diffraction.



(a)N= 2 (b)N = 3 (c)N =4

Appendix


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Appendix General method for treating interferenceof many sources with the same or different

intensities

Phasors

Application to multiple-slit problems

Diffraction from a single slit, with simple derivation ofkey features from geometrical reasoning

Appendix: Multi-Slit Interference P


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Appendix: Multi Slit InterferenceWhat changes if we increase thenumber of slits (e.g., N = 3, 4, 1000?)

First look at the principle maxima:

If slit 1 and 2 are in phase witheach other, than slit 3 will also bein phase.

Conclusion: Position of principleinterference maxima are the samefor any number of slits!

(i.e., d sinq= m )

S3

S2

P

Incident wave(wavelength ) y

L

d

S1

A1

A1

A1

A

For other directions, simple geometrycan tell us the resultant amplitude Ain terms of A1and .

Use this to find interference minima

Note: phasorangle is withrespect toadjacent slit!

Used in homeworks, etc.

Atot= 3 A1

Itot= 9 I1

For N slitsI

tot

= N2I1

A1 A

1

A1

What about amplitude of principlemaxima? Draw phasor diagram:

Appendix: Multi-Slit Interference


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Appendix: Multi Slit Interference Intensity for N equally spaced slits is easily found from phasor analysis:

Draw normal lines bisecting the phasors. They intersect, defining R as shown:

221 si nR

A

22

NsinR

AN

)/sin(

)/Nsin(AAN

2

21

Solve for R

and plug inhere:

Result:

2

122 )/sin()/Nsin(

IIN

N AN

R

A1

R

N slits:

A1/2

RR

2

2

2

2

2

N

For more on phasors, seetext, secs 35.3, 36.3, 36.4and slides in lecture 2

Appendix Exercise Multiple Slits


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Appendix Exercise Multiple Slits

1. In 2-slit interference, the first minimum

corresponds to p. For 3-slits, we havea secondary maximum at p (seediagram). What is the intensity of thissecondary maximum? (Hint: Use phasors.)

(a)=p/2 (b)=2p/3 (c)=3p/4

2. What value of corresponds to the first zeroof the 3-slit interference pattern?

(a)I1 (b)1.5 I1 (c)3 I1 0 2p2p

I

0

9I1

=?

I =?

(a)=p/2 (b)=2p/3 (c)=3p/4


Appendix Exercise Multiple Slits Solution


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Appendix Exercise Multiple Slits - Solution

1. In 2-slit interference, the first minimum

corresponds to p. For 3-slits, we havea secondary maximum at p (seediagram). What is the intensity of thissecondary maximum? (Hint: Use phasors.)

(a)I1 (b)1.5 I1 (c)3 I1 0 2p2p

I

0

9I1

=?

I =?

What does the phasor diagram look like?

Two of the three phasors cancel, leaving only one I1

A1=

(a)=p/2 (b)=2p/3 (c)=3p/4


Appendix Exercise Multiple Slits - Solution?


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Appendix Exercise Multiple Slits Solution1. For 3-slits, the intensity of this

secondary maximum is:

(a)=p/2 (b)=2

p/3

(c)=3p/42. What value of corresponds to the first zero

of the 3-slit interference pattern?

(a)I1 (b)1.5 I1 (c)3 I1

0 2p2p

I

0

9I1

=?

I =?

A p/2

No. A is not zero.

=2p/3

Yes! Equilateraltriangle gives A = 0.

=3p/4

No, triangle does not close.

A


(a)=p/2 (b)=2p/3 (c)=3p/4

Appendix Exercise Multiple Slits - Solution


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Appendix Exercise Multiple Slits Solution

(a)=p/2 (b)=2p/3 (c)=3p/42. What value of corresponds to the first zero

of the 3-slit interference pattern?

0 2p2p

I

0

9I1

=?

I =?

(a)=p/2 (b)=2p/3 (c)=3p/43. What value of

corresponds to the first zeroof the 4-slit interference pattern?

p/2

Yes. Squaregives A = 0.

=2p/3No. With 4 slits westart to wrap around(note: A = A1)

=3p/4

No. A0

A

For N slits the first zero is at 2p/N.

1. For 3-slits, the intensity of thissecondary maximum is:

(a)I1 (b)1.5 I1 (c)3 I1

Appendix: Single-slit Diffraction


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Appendix: Single slit Diffraction To analyze diffraction, we treat it as

interferenceof light from many sources (i.e.the Huygens wavelets that originate from eachpoint in the slit opening).

Model the single slit as M point sources withspacing between the sources of a/M. We willlet M go to infinity on the next slide.

The phase difference bbetween first and lastsource is given by b/2p a/= a sinq / aq/ .

Destructive interference occurs when thepolygon is closed (b= 2p):

Slits Demo

10-slits

a

A1(1 slit)

Aa(1 source)

aM b

P

Screen(far away)

a= a sinq aq

q

a

L >> a implies rays areparallel.

L

a

,smallFor

sinameanshi sT

q

q

q

1

DestructiveInterference

Appendix: Single-slit Diffraction


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Appendix: Single slit Diffraction

We have turned the single-slit problem into the M-slitproblem that we already solved in this lecture.

However, as we let M , the problem becomes muchsimpler because the polygon becomes the arc of a circle.

The radius of the circle is determined by the relationbetween angle and arc length: b= Ao/R.

2

012/

)2/sin(II

b

b Graph this function

Intensity is related to amplitude: I = A2. So, heres the final answer:

Ao

q

pb a

2

A1b/2

R

Trigonometry: A1/2 = R sin(b/2)

With R = Ao/b, A1= (2Ao/b) sin(b/2)

2

201

/

)/sin(AA

b

b

summer12 lectures lect03

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