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8/3/2019 Lecture28 Diffraction
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Announcements
Important notes on grades spreadsheets:
You can enter projected scores into the spreadsheet tosee how many points you need for the next higher grade(or to see if your grade is already certain).
Your scores for the end material test have been set to 8(your free points) so that the spreadsheet can correctlydiscard low scores.
Zeroes for boardwork can still lower your total points.
You have already received your 8 free end materialpoints. Do not take the test if that is all you need!
Important: sample.
Course grade cutoffs will not be lowered under any circumstances.
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Announcements
PLC
Faculty will check PLC periodically Wednesdayafternoon and provide help to any students who show up.
PLC will run Monday afternoon and evening as usual.
Wednesday evening PLC will be cancelled unless I gete-mails from students saying they will attend.
Wednesday PLC status will be announced in Wednesdays
lecture. If there is PLC, most likely...
I don't know half of you half as well as I should like, and I like less than half of
you half as well as you deserve.Bilbo Baggins
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Announcements
Fill out electronic course evaluations
Use the following link to complete the evaluation(s):https://itweb.mst.edu/auth-cgi-
bin/cgiwrap/distanceed/evals/survey.pl
The link will be available from Sunday, November 28,through Sunday, December 11 2011.
For feedback or problems with the link, please contact Dr.Scott Miller [email protected]
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Announcements
Dont forget to put your used Physics 24 textbook to gooduse!
If anything about your grade
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!
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.
Final Exam is 1:30 pm, Tuesday December 13. No one admitted after 1:45 pm!
y Physics 24 Final Room Assignments, Fall 2011:
Instructor Sections RoomDr. Hagen K, L G31 Electrical EngineeringDr. Hale E, G 104 Physics (same as exams)Dr. Hor J, M St. Pats Ballroom CDr. Parris B, H 125 Butler-Carlton (same)
Dr. Peacher D, F 204 McNutt Dr. Schmitt A, C G-3 Schrenk
Special Needs Testing Center
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Todays agenda: no purple boxes.
No purple boxes!Youve seen your last purple box agenda!
FinalExam
FinalExam
Final
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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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LEAD Tutors/Peer Instructors Needed!
You can tutor or be aPL
C peer instructor if you have at least a3.6 GPA and get an A in the course you want to tutor.
Contactme or go to
http://lead.mst.edu/to fill out theapplication form.
It looks good on yourresume, pays well,and is fun!