Lecture28 Diffraction

8/3/2019 Lecture28 Diffraction

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Dont forget to put your used Physics 24 textbook to gooduse!

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needs fixed, fix it now (next week is too late)!

See your recitation instructor! (Not me!)


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Know the exam time!

Find your room ahead of time!

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Final Exam is 1:30 pm, Tuesday December 13. No one admitted after 1:45 pm!

y Physics 24 Final Room Assignments, Fall 2011:

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Two lectures ago I showed you these two plots of the intensitydistribution in the double-slit experiment:

Which is correct?

Peak intensity varies with angle. Peak intensity independent of angle.


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Diffraction

Light is an electro

magnetic wave, and like all waves, bendsaround obstacles.

P

d

Pd

This bending, which is most noticeable when the dimensionof the obstacle is close to the wavelength of the light, iscalled diffraction. Only waves diffract.


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Diffraction pattern from a pennypositioned halfway between a

light source and a screen.

The shadow of the penny is thecircular dark spot.

Notice the circular bright anddark fringes.

The central bright spot is not a defect in the picture. It is aresult of light bending around the edges of the penny andinterfering constructively in the exact center of the shadow.

Good diffraction applets athttp://ngsir.netfirms.com/englishhtm/Diffraction.htmhttp://micro.magnet.fsu.edu/primer/java/diffraction/basicdiffraction/

http://www.physics.uq.edu.au/people/mcintyre/applets/grating/grating.html


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Single Slit Diffraction

In the previous chapter we calculated the interference pattern

from a pair of slits.

One of the assumptions inthe calculation was thatthe slit width was very

small compared with thewavelength of the light.

Now we consider theeffect of finite slitwidth. We start with asingle slit.

Ua XYZ

[

\

Each part of the slit acts as a source of light rays, and these

different light rays interfere. Fall 2011 skip to slide 18.


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UY

Z

[

\

a/2

a/2

a

sin2a U

X

Divide the slit in half.

Ray X travels farther*than ray Z by (a/2)sinU.Likewise for rays Y and[.

If this path difference is exactly half awavelength (corresponding to aphase difference of 180) then thetwo waves will cancel each other and

destructive interference results.

Destructive interference:P

Ua

sin =2 2

*All rays from the slit are converging at a pointP very far to the right and out of the picture.


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UY

Z

[

\

a/2

a/2

a

sin2a U

X

Destructiveinterference:

PUa sin =2 2

U Pa sin =

If you divide the slit into 4 equal parts, destructive

interference occurs when.

P

U2

sin = a

PUsin =a

.

PU

3sin =

a

If you divide the slit into 6 equal parts, destructive

interference occurs when


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UY

Z

[

\

a/2

a/2

a

sin2a U

X

In general, destructive interference occurs when

, , , ,U Pa sin = m m =1 2 3 ...

The above equation gives the positions of the dark fringes.The bright fringes are approximately halfway in between.


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http://www.walter-fendt.de/ph14e/singleslit.htm

U Pa sin = m


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a U

O

x

y

Use this geometry for

tomorrows single-slithomework problems.

IfU is small,* then it is

valid to use theapproximation sin U } U.(U must be expressed inradians.)

*The approximation is quite good for angles of 10ror less, and not bad for even larger angles.


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Your text gives the intensity distribution for the single slit.

The general features of that distribution are shown below.

Most of the intensity is in the central maximum. It is twicethe width of the other (secondary) maxima.

Single Slit Diffraction Intensity


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New starting equations for single-slit intensity:

TF U

P

2= a sin

F

F-

2

0

sin /2I = I

/2

Toy


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Example: 633 nm laser light is passed through a narrow slitand a diffraction pattern is observed on a screen 6.0 m away.The distance on the screen between the centers of the firstminima outside the central bright fringe is 32 mm. What is theslit width?

y1 = (32 mm)/2 tanU = y1/L tanU } sinU } U for small U

P P P PU } !

U 1 1

Lsin = a=

a sin y /L y

vv

-9

-36.0 m 633 10 ma=

16 10 m

v -4a=2.37 10 m


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Quiz time (for points)


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Seasons Greetings

from a couple of old friends.


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The ability of optical systems to distinguish closely spaced

objects is limited because of the wave nature of light. Fall2011 skip to slide 29.

If the sources are far enough apart so that their centralmaxima do not overlap, their images can be distinguished and

they are said to be resolved.

Resolution of Single Slit (and Circular Aperture)


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When the central maximum of one image falls on the firstminimum of the other image the images are said to bejust

resolved. This limiting condition of resolution is calledRayleighs criterion.


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From Rayleighs criterion we can determine the minimumangular separation of the sources at the slit for which theimages are resolved.

For a slit of width a:

For a circular aperture of diameter D:

Resolution is wavelength lim

ited!

P

U= aP

U1.22

=D

P

Ua=

sinThese come from

the small angle approximation,

and geometry.


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If a single slit diffracts, what about a double slit?

Remember the double-slit interference pattern from thechapter on interference?

T U

P

2max

d sinI = I cos

If the slit width (not the spacingbetween slits) is small (i.e.,

comparable to the wavelength ofthe light), you must account fordiffraction.

interference only


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S2

S1

d

a

UP

L

y

r1

r2

Double Slit Diffraction


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A diffraction grating consists of a large number of equally

spaced parallel slits.

U

d

H =dsinU

The path difference betweenrays fromany two adjacentslits is H = dsin U.

Interference maxima occur for

Diffraction Gratings

IfH is equal to some integermultiple of the wavelengththen waves fromall slits will

arrive in phase at a point ona distant screen.

, , , ,U Pd sin = m m =1 2 3 ...


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Ok whats with this equation monkey business?

, , , ,

U Pa sin = m m =1 2 3 ...

, , , ,U Pd sin = m m =1 2 3 ...

, , , ,U Pd sin = m m =1 2 3 ... double-slit interferenceconstructive

single-slit diffractiondestructive!

diffraction gratingconstructive


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U

d

H =dsinUThe intensity maxima arebrighter and sharper than forthe two slit case.

Interference Maxi

ma:

Diffraction Grating Intensity Distribution

U Pd sin = m


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Application: spectroscopy

You can view the atomic spectra for each of the elements here.

visible light

hydrogen

helium

mercury


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Interference Maxima:

First-order violet:

Example: the wavelengths of visible light are fromapproximately 400 nm (violet) to 700 nm (red). Find theangular width of the first-order visible spectrum produced by a

plane grating with 600 slits per millimeter when white light fallsnormally on the grating.

U Pd sin = m

v -61d = =1.67 10 m600 slits/mm

v

PU v

-9

VV -6

1 400 10 msin = m = =0.240

d 1.67 10 m

rUV =13.9


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First-order red: vP

Uv

-9

RR -6

1 700 10 msin = m = =0.419

d 1.67 10m

rUR =24.8

r r r(U ! U UR V =24.8 -13.9 =10.9

visible light

10.9r


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No matter what the grating spacing, d, the largest angle forthe 2nd order spectrum (for the red end) is always greaterthan the smallest angle for the 3rd order spectrum (for theviolet end), so 2nd and 3rd orders always overlap.

Example: for this diffraction grating show that the violet end ofthe third-order spectrum overlaps the red end of the second-order spectrum.

Third-order violet: vP v

U-9 -6

VV

3 400 10 m 1.20 10 msin = m = =

d d d

Second-order red: vP v

U

-9 -6

VR

2 700 10 m 1.40 10 msin = m = =d d d

U UR2 V3sin > sin


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Diffraction gratings let us measure wavelengths by separating

the diffraction maxima associated with different wavelengths.In order to distinguish two nearly equal wavelengths thediffraction must have sufficient resolving power, R.

Consider two wavelengths 1 and 2 that are nearly equal.

The average wavelength is and the difference is

The resolving power is defined as

Diffraction Grating Resolving Power

P PP 1 2avg

+=

2

(P P P2 1= - .

P

(Pavg

R = .definition ofresolving power

mercury


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For a grating with N lines illuminated it can be shown that theresolving power in the mth order diffraction is

P

(Pavg

R =

R =Nm.

Dispersion

mercury

Spectroscopic instruments need to resolve spectral lines ofnearly the same wavelength.

The greater the angular dispersion,the better a spectrometer is atresolving nearby lines.

(U(P

angular dispersion=

resolving powerneeded to resolve mth order


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Example: Light frommercury vapor lamps contain severalwavelengths in the visible region of the spectrum including twoyellow lines at 577 and 579 nm. Whatmust be the resolving

power of a grating to distinguish these two lines?

mercury

P

(Pavg 578 nmR = = =289

2 nm

Pavg577 nm + 579 nm

= = 578 nm2

(P =579 nm - 577 nm = 2 nm


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Example: how many lines of the grating must be illuminated ifthese two wavelengths are to be resolved in the first-order

spectrum

?

mercury

R=289

R 289

R =Nm N= = =289m 1


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Lecture28 Diffraction

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