Final Exam and End Material Test
Wednesday, Dec 13, 3:00-5:00
Test rooms:
• Instructor Sections Room
• Dr. Musser B, D, G, R St. Pats Ballroom
• Dr. Hale P St. Pats Ballroom
• Dr. Wilemski C, F BCH 125
• Dr. Jentschura A, Q BCH 120
• Dr. Madison H, L EECH G-31
• Mr. Upshaw E, J, M, N G-3 Schrenk
• Dr. Hale K 104 Physics
• Special Accommodations Testing Center
(Contact me a.s.a.p. if you need
accommodations different than for exam 3)
Announcements
Final exam day events (Wednesday, Dec. 13, 3:00pm to
5:00pm)
• 50-point multiple choice end-material test (covering material
from chapters 33-36). (You get a free 8-point question!)
• 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.
No one admitted after 3:15pm!
You may spend your two hours however you see fit (all on
end-material, all on final exam, some mix).
Announcements
Your end material test points have been set to 8 already.
(should be visible after next spreadsheet update Tuesday afternoon)
You do not need to take the test to receive these points.
Posted grade spreadsheets are active – play with the scores
to see how many points you need for the next higher grade.
Zeroes for boardwork can still lower your total points.
Grade cutoffs will not be lowered under any circumstances.
If any of your scores need to be fixed contact your recitation
instructor NOW!
Announcements
PLC
PLC will run Monday afternoon and evening as usual
No PLC on Wednesday.
Teaching evaluations
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If you liked the class, please let us know! Constructive
criticism is highly appreciated as well!
The links are available until Sunday before Finals Week.
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Go to
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to fill out the
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It looks good on your
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and is fun!
Today’s agenda:
Introduction to diffraction
Single-slit diffraction
Diffraction grating
Diffraction
Light is an electromagnetic wave, and like all waves, “bends”
around obstacles.
<<d d >>d
most noticeable when the dimension of the obstacle is close
to the wavelength of the light
Diffraction pattern from a penny
positioned halfway between a
light source and a screen.
The shadow of the penny is the
circular dark spot.
Notice the circular bright and
dark fringes.
The central bright spot is a result of light “bending” around
the edges of the penny and interfering constructively in the
exact center of the shadow.
Good diffraction applets at http://ngsir.netfirms.com/englishhtm/Diffraction.htm
http://micro.magnet.fsu.edu/primer/java/diffraction/basicdiffraction/
http://www.physics.uq.edu.au/people/mcintyre/applets/grating/grating.html
Single Slit Diffraction
Recall: double-slit interference (lecture 26)
•slits were assumed infinitely thin (point sources)
Now: consider the effect of finite slit width
Single slit: a
• each point in slit acts as
source of light waves
• these different light
waves interfere.
Imagine dividing the slit in half.
Wave travels farther*
a/2
than wave by (a/2)sin.
a
Same for waves and . a/2
a
sin
If the path difference (a/2)sin equals 2
/2, these wave pairs cancel each
other destructive interference
a
Destructive interference: sin =
2 2
*All rays from the slit are converging at a point P very far to the right and out of the picture.
Destructive
interference:
a
sin = a/2
2 2 a
a/2
a sin =
a
sin
sin = 2
a
If you divide the slit into 4 equal parts, destructive
2
interference occurs when sin = .
a
If you divide the slit into 6 equal parts, destructive
3
interference occurs when sin = .
a
In general, destructive interference occurs when
a sin = m m =1, 2, 3, ...
• gives positions of dark
fringes
• no dark fringe for m=0 a/2
a
a/2
a
sin
2
The bright fringes are approximately halfway in between.
Applet.
a sin = m
http://www.walter-fendt.de/ph14e/singleslit.htm
Use this geometry for
tomorrow’s single-slit y
homework problems.
a
If is small,* then it is
O
valid to use the
approximation sin .
( must be expressed in
radians.)
x
*The approximation is quite good for angles of 10
or less, and not bad for even larger angles.
Single Slit Diffraction Intensity
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 twice
the width of the other (secondary) maxima.
Starting equations for single-slit intensity:
2
= a sin
2
sin /2
I = I0
/2
“Toy”
Example: 633 nm laser light is passed through a narrow slit
and a diffraction pattern is observed on a screen 6.0 m away.
The distance on the screen between the centers of the first
minima outside the central bright fringe is 32 mm. What is the
slit width?
y1 = (32 mm)/2 tan = y1/L tan sin for small
a sinθ mλ 1λ
L
sin = a =
a sin y1 /L y1
32 mm 6.0 m 633 10-9 m
a=
16 10 -3
m
a = 2.37 10-4 m
6m
Resolution of Single Slit (and Circular Aperture)
The ability of optical systems to distinguish closely spaced
objects is limited because of the wave nature of light.
If the sources are far enough apart so that their central
maxima do not overlap, their images can be distinguished and
they are said to be resolved.
When the central maximum of one image falls on the first
minimum of the other image the images are said to be just
resolved. This limiting condition of resolution is called
Rayleigh’s criterion.
From Rayleigh’s criterion:
minimum angular separation of sources for which the images
are resolved.
slit of width a:
=
a
circular aperture
of diameter D:
1.22
=
D
Resolution is wavelength limited!
These come from a = / sin
the small angle approximation,
and geometry. Photography:
closing the aperture too much
leads to unsharp pictures
If a single slit diffracts, what about a double slit?
Remember the double-slit interference pattern from the
chapter on interference?
2 d sin
I = Imax cos
If slit width (not spacing between
slits) is not infinitesimally small
but comparable to wavelength,
you must account for diffraction.
interference only
Double
Single Slit Diffraction with a
SlitDiffraction
r1
y
a
S1 r2
a
P
O
d
S2
L
x
Diffraction Gratings
diffraction grating: large number of equally spaced parallel slits
The path difference between
rays from any two adjacent
slits is = dsin .
If is equal to some integer
multiple of the wavelength
then waves from all slits will
d
arrive in phase at a point on
a distant screen.
= d sin
Interference maxima occur for d sin = m, m =1, 2, 3, ...
Diffraction is not the same as refraction!
Ok what’s with this equation monkey business?
double-slit interference
d sin = m, m =1, 2, 3, ... constructive
single-slit diffraction
a sin = m, m =1, 2, 3, ... destructive!
diffraction grating
d sin = m, m =1, 2, 3, ... constructive
d double slit and diffraction a a single slit but destructive
Diffraction Grating Intensity Distribution
Interference Maxima:
d sin = m
= d sin
The intensity maxima are
brighter and sharper than for
the two slit case. See here
and here.
Application: spectroscopy
visible light
hydrogen
helium
mercury
You can view the atomic spectra for each of the elements here.
http://h2physics.org/?cat=49
Example: the wavelengths of visible light are from
approximately 400 nm (violet) to 700 nm (red). Find the
angular width of the first-order visible spectrum produced by a
plane grating with 600 slits per millimeter when white light falls
normally on the grating.
700 nm* 400 nm
angle?
*Or 750 nm, or 800 nm, depending on who is observing.
Example: the wavelengths of visible light are from
approximately 400 nm (violet) to 700 nm (red). Find the
angular width of the first-order visible spectrum produced by a
plane grating with 600 slits per millimeter when white light falls
normally on the grating.
Interference Maxima: d sin = m
1
d= =1.67 10-6 m
600 slits/mm
V 1 400 10-9 m
First-order violet: sin V = m = -6
= 0.240
d 1.67 10 m
V =13.9
R 1 700 10-9 m
First-order red: sin R = m = -6
= 0.419
d 1.67 10 m
R = 24.8
R V = 24.8 -13.9 =10.9
10.9
Application: use of diffraction to probe materials.
La0.7Sr0.3Mn0.7Ni0.3O3
La, Sr
Mn, Cr
Application: use of diffraction to probe materials.
La0.7Sr0.3Mn0.7Ni0.3O3
La, Sr
Mn, Cr
Shoot a beam of x-rays or neutrons at an unknown material.
The x-rays or neutrons diffract.
Positions of peaks tell you what sets of planes exist in the
material. From this you can infer the crystal structure.
Intensities of peaks tell you atoms lie on the different planes,
and where they are located on the planes.
Application: use of diffraction to probe materials.
La0.7Sr0.3Mn0.7Cr0.3O3
(011) & (003)
Diffraction Grating Resolving Power
Diffraction gratings let us measure wavelengths by separating
the diffraction maxima associated with different wavelengths.
In order to distinguish two nearly equal wavelengths the
diffraction must have sufficient resolving power, R.
mercury
Consider two wavelengths λ1 and λ2 that are nearly equal.
1 + 2
The average wavelength is avg = and the difference is
2
= 2 - 1 .
avg definition of
The resolving power is defined as R = . resolving power
avg
R=
For a grating with N lines illuminated it can be shown that the
resolving power in the mth order diffraction is
resolving power
R = Nm. needed to resolve mth order
Dispersion
Spectroscopic instruments need to resolve spectral lines of
nearly the same wavelength.
mercury
The greater the angular dispersion,
angular dispersion = the better a spectrometer is at
resolving nearby lines.
Example: Light from mercury vapor lamps contain several
wavelengths in the visible region of the spectrum including two
yellow lines at 577 and 579 nm. What must be the resolving
power of a grating to distinguish these two lines?
mercury
577 nm + 579 nm
avg = = 578 nm
2
= 579 nm - 577 nm = 2 nm
avg 578 nm
R= = = 289
2 nm
Example: how many lines of the grating must be illuminated if
these two wavelengths are to be resolved in the first-order
spectrum?
mercury
R = 289
R 289
R = Nm N = = = 289
m 1
