Transcript Slide 1

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 Physics 2135 Final Room Assignments, Spring 2015:
Instructor
Dr. Kurter
Dr. Madison
Dr. Parris
Mr. Upshaw
Mr. Upshaw
Dr. Waddill
Sections
F, H
E, G
K, M
A, C
J, L
B, D
Exam is from
Room
7:30-9:30 am
120 BCH
Friday
May 15!
G-5 H/SS
103 Eng. Mgt.
No one
admitted after
G-3 Schrenk
7:45 am!
St. Pat’s C
103 & 104 Centennial
See notes on next slide for information about room locations.
Special Accommodations
Testing Center
Know the exam time!
Find your room ahead of time!
If at 7:30 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.
 Physics 2135 Final Room Assignments, 7:30 AM, May 15
Instructor
Dr. Kurter
Sections
F, H
Room
120 BCH
BCH is Butler-Carlton Civil Engineering Building
Dr. Madison
E, G
G-5 H/SS
H/SS is the Humanities/Social Science Building
Dr. Parris
K, M
103 Eng. Mgt.
Eng. Mgt. Is the Engineering Management Building
Mr. Upshaw
A, C
G-3 Schrenk
G-3 Schrenk is your ”normal” exam room
Mr. Upshaw
J, L
St. Pat’s C
St. Pat’s C is St. Pat’s Ballroom C in Havener Center
Dr. Waddill
B, D
103 & 104 Centennial
this is two classrooms in Centennial Hall with a sliding partition between them
Special Accommodations
Testing Center
Today’s agenda: no purple boxes.
No purple boxes!
I am taking a break from purple boxes. You’ve seen your last purple box agenda!
Final
Exam
Final
Final
Exam
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Two lectures ago I showed you these two plots of the intensity
distribution in the double-slit experiment:
Peak intensity varies with angle.
Which is correct?
Peak intensity independent of angle.
Diffraction
Light is an electromagnetic wave, and like all waves, “bends”
around obstacles.

d
<<d
d
>>d
This bending, which is most noticeable when the dimension
of the obstacle is close to the wavelength of the light, is
called “diffraction.” Only waves diffract.
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 not a defect in the picture. It 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
In the previous chapter we calculated the interference pattern
from a pair of slits.
One of the assumptions in
the calculation was that
the slit width was very
small compared with the
wavelength of the light.
a

Now we consider the
effect of finite slit
width. We start with a
single slit.
Each part of the slit acts as a source of light rays, and these
different light rays interfere.





Divide the slit in half.
Ray  travels farther*
than ray  by (a/2)sin.
Likewise for rays  and
.
a/2
a/2
If this path difference is exactly half a
wavelength (corresponding to a
phase difference of 180°) then the
two waves will cancel each other and
destructive interference results.
Destructive interference:

a





a
sin 
2
a

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 =
2
2
a sin = 

sin =
a
a/2

a
a/2
a
sin 
2
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





a/2

a
a/2
a
sin 
2
In general, destructive interference occurs when
a sin = m, m=1, 2, 3, ...
The above equation gives the positions of the dark fringes.
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
homework problems.
y
a
If  is small,* then it is
valid to use the
approximation sin   .
( must be expressed in
radians.)

O
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

 sin  /2  
I = I0 


/2
 
 
“Toy”
2
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  sin   for small 
tan = y1/L
a sinθ  mλ  1λ
32 mm



L
sin =  a =


a
sin y1 /L y1
a=
6m
 6.0 m  633 10-9 m
16 10
-3
a = 2.37 10-4 m
m
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 we can determine the minimum
angular separation of the sources at the slit for which the

images are resolved.
These come from a =
For a slit of width a:  =

a
sin
the small angle approximation,
and geometry.
For a circular aperture of diameter D:  =
1.22 
D
Resolution is wavelength limited!
If a single slit diffracts, what about a double slit?
Remember the double-slit interference pattern from the
chapter on interference?
I = Imax
  d sin 
cos 




2
If the slit width (not the spacing
between slits) is small (i.e.,
comparable to the wavelength of
the light), you must account for
diffraction.
interference only
Double
SlitDiffraction
Diffraction with a  
Single Slit
r1
S1
a
y
a
r2

P
O
d
S2
L
x
Diffraction Gratings
A diffraction grating consists of a large number of equally
spaced parallel slits.
The path difference between
rays from any two adjacent
slits is  = dsin .

d
 = d sin 
If  is equal to some integer
multiple of the wavelength
then waves from all slits will
arrive in phase at a point on
a distant screen.
Interference maxima occur for d sin =m, m=1, 2, 3, ...
Ok what’s with this equation monkey business?
d sin =m, m=1, 2, 3, ...
double-slit interference
constructive
a sin = m, m=1, 2, 3, ...
single-slit diffraction
destructive!
d sin =m, m=1, 2, 3, ...
diffraction grating
constructive
d  double slit and diffraction
a  a single slit but destructive
Diffraction Grating Intensity Distribution
Interference Maxima:
d sin = m

d
 = 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.
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.
400 nm
700 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
First-order violet:
V
sin V = m =
d
V =13.9
1  400 10-9 m
-6
1.67 10 m
= 0.240
First-order red:
R
sin R =m =
d
1  700 10-9 m
1.67 10 m
R = 24.8
  R  V =24.8 -13.9 =10.9
10.9
-6
= 0.419
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
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 .
The resolving power is defined as R =
 avg

.
definition of
resolving power
R=
 avg

For a grating with N lines illuminated it can be shown that the
resolving power in the mth order diffraction is
R = Nm.
resolving power
needed to resolve mth order
Dispersion
Spectroscopic instruments need to resolve spectral lines of
nearly the same wavelength.
mercury

angular dispersion =

The greater the 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
 avg =
577 nm + 579 nm
= 578 nm
2
 = 579 nm - 577 nm = 2 nm
R=
 avg

=
578 nm
= 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