Exam Review I - UW-Madison Department of Physics
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Transcript Exam Review I - UW-Madison Department of Physics
Physics 208 Exam 1 Review
Mon. Feb. 18, 2008
1
Exam covers Ch. 21.5-7, 22-23,25-26,
Lecture, Discussion, HW, Lab
Exam 1 is Wed. Feb. 20, 5:30-7 pm,
2103 Ch: Adam(301,310), Eli(302,311), Stephen(303,306),
180 Science Hall: Amanda(305,307), Mike(304,309), Ye(308)
Chapter 21.5-7, 22
Chapter 23
Reflection, refraction, and image formation
Chapter 25
Waves, interference, and diffraction
Electric charges and forces
Chapters 26
Mon. Feb. 18, 2008
Electric fields
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Quick Quiz
An electric dipole is in a uniform electric field as shown.
The dipole
Dipole
A. accelerates left
B. accelerates right
C. stays fixed
D. accelerates up
E. none of the above
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Electric torque on dipoles
Remember torque? r F ,
m agnitude rF sin
r
Here there are two torques, both into page:
F qE, r s /2
qEs /2sin
r
F qE, r s /2
qEs /2sin
Total torque is sum of these
qsEsin
p qs Torque on dipole
in uniform field
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p E
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Dipole in non-uniform field
A permanent dipole is near a positive point charge in
a viscous fluid. The dipole will
A. rotate CW & move toward charge
B. rotate CW & move away
C. rotate CCW & move toward
+
D. rotate CCW & move away
E. none of the above
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Properties of waves
Wavelength, frequency, propagation speed
related as f v
Phase relation
In-phase: crests line up
180˚ Out-of-phase: crests line up with trough
Time-delay leads to phase difference
Path-length difference leads to phase difference
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Chapter 22: Waves & interference
Path length difference and phase
Two slit interference
Alternating max and min due to path-length difference
Phase change on reflection
different path length -> phase difference.
π phase change when reflecting from medium with higher index of
refraction
Interference in thin films
Different path lengths + reflection phase change
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Path length difference
Path length difference d
Phase difference = (d/)2 radians
Constructive for 2n phase difference
L
Shorter path
Light
beam
Longer path
Foil with two
narrow slits
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Recording
plate
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Question
You are listening to your favorite radio station, WOLX 94.9 FM (94.9x106 Hz)
while jogging away from a reflecting wall, when the signal fades out. About
how far must you jog to have the signal full strength again? (assume no
phase change when the signal reflects from the wall)
Hint: wavelength = (3x108 m/s)/94.9x106 Hz
A. 3 m
B. 1.6 m
C. 0.8 m
D. 0.5 m
Mon. Feb. 18, 2008
=3.16 m
d-x
x
d
path length diff = (d+x)-(d-x)= 2x
Destructive 2x=/2x=/4
Constructive make 2x=x=/2
x increases by /4 = 3.16m/4=0.79m
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Two-slit interference
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Two-slit interference: path length
L
y
Constructive int:
Destructive int.
Phase diff = 2m, m 0,1,2
Path length diff = m, m 0,1,2
Phase diff = 2 (m 1/2), m 0,1,2
diff = (m 1/2) , m 0,1,2
Path length
Path length difference d sin d y /L
dsin 2 d
Phase difference
2
y /L
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Reflection phase shift
Possible additional phase shift on reflection.
Start in medium with n1,
reflect from medium with n2
n2>n1, 1/2 wavelength phase shift
n2<n1, no phase shift
Difference in phase shift between different
paths is important.
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Thin film interference
air
1/2 wavelength phase shift
from top surface reflection
Reflecting from n2
air: n1=1
t
n2>1 air/n
air: n1=1
Extra path length
No phase shift from
bottom interface
Reflecting from n1
Extra phase shift needed for
constructive interference is
m 1/2air /n
2t m 1/2air /n
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Diffraction from a slit
Each point inside slit
acts as a source
Net result is series of
minima and maxima
Similar to two-slit
interference.
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Overlapping diffraction patterns
Two independent point
sources will produce two
diffraction patterns.
If diffraction patterns overlap
too much, resolution is lost.
Image to right shows two
sources clearly resolved.
Angular
separation
Circular aperture diffraction limited: min 1.22
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D
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Diffraction gratings
Diffraction grating is pattern of multiple slits.
Very narrow, very closely spaced.
Same physics as two-slit interference
d sinbright m, m 0,1,2
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sinbright m
d
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Chap. 23: Refraction & Ray optics
Refraction
Ray tracing
Types of images
Real image: project onto screen
Virtual image: image with another lens
Lens equation
Can locate image by following specific rays
Relates image distance, object distance, focal length
Magnification
Ratio of images size to object size
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Refraction
Occurs when light moves into medium with different
index of refraction.
Light direction bends according to n1 sin 1 n 2 sin 2
i,1
r
n1
Special case:
Total internal reflection
n2
2
Angle of refraction
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Lenses: focusing by refraction
F
Object
P.A.
Image
F
1) Rays parallel to principal axis pass through focal point.
2) Rays through center of lens are not refracted.
3) Rays through F emerge parallel to principal axis.
Here image is real, inverted, enlarged
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Different object positions
Object
Image (real, inverted)
Image (real, inverted)
Image
(virtual, upright)
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These rays seem to originate
from tip of a ‘virtual’ arrow.
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Question
You have a focused image on the screen, but you want
the image to be bigger. Relative to the lens, you should
A. Move screen away, move object away
B. Move screen closer, move object away
C. Move screen away, move object closer
D. Move screen closer, move object closer
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Equations
Image and object
different sizes
Image (real, inverted)
p
Relation between
image distance
object distance
focal length
Magnification
= M =
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q
1 1 1
p q f
image height q image distance
object height p object distance
22
Question
You want an image on a screen to be ten times larger than
your object, and the screen is 2 m away. About what focal
length lens do you need?
1 1 1
p q f
A. f~0.1m
q=2 m
B. f~0.2m
C. f~0.5m
D. f~1.0m
Mon. Feb. 18, 2008
mag=10
-> q=10p->p=0.2m
1
1
1
5.5 f 0.18m
0.2m 2m
f
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Chapter 25: Electric Charges & Forces
Triboelectric effect: transfer charge
Vector forces between charges
Total charge is conserved
Add by superposition
Drops off with distance as 1/r2
Insulators and conductors
Polarization of insulators, conductors
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Charges conductors & insulators
Two types of charges, + and
Conductor:
Like charges repel
Unlike charges attract
Charge free to move
Distributed over surface of conductor
Insulator
Charges stuck in place where they are put
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Electric force: magnitude & direction
Electrical force between two stationary charged particles
The SI unit of charge is the coulomb (C ), µC = 10-6 C
1 C corresponds to 6.24 x 1018 electrons or protons
ke = Coulomb constant ≈ 9 x 109 N.m2/C2 = 1/(4πeo)
-12 C2 / N.m2
eo permittivity of free space = 8.854 x 10
Directed along line joining particles.
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Forces add by superposition
Equal but opposite charges
are placed near a negative charge as shown.
What direction is the net force on the negative charge?
A) Left
kq1q2
F 2
r
-
B) Right
C) Up
D) Down
E) Zero
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+
-
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Chapter 26: The Electric Field
Defined as force per unit charge (N/C)
Calculated as superposition of contributions from
different charges
Examples
Single charge
Electric dipole
Line charge, sheet of charge
Electric field lines
Force on charged particles
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Electric field
Fe = qE
If q is positive, F and E are in the same direction
Example: electric field from point charge
Qp=1.6x10-19 C
+
E
r = 1x10-10 m
E = (9109)(1.610-19)/(10-10)2 N = 2.91011 N/C
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(to the right)
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Pictorial representation of E: Electric Field Lines
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Electric field lines
Local electric field tangent to field line
Density of lines proportional to electric field
strength
Fields lines can only start on + charge
Can only end on - charge (but some don’t end!).
Electric field lines can never cross
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Electric dipole
E
1 2p
4eo r 3
On y-axis far from dipole
Electric dipole p qs
moment
+q
E
-q
1
p
4 eo r 3
On x-axis far from dipole
Electric field magnitude drops off as 1/r3
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32
Quick quiz: continuous charge dist.
Electric field from a uniform ring of charge.
The magnitude of the electric field on the x-axis
y
A. Has a maximum at x=0
B. Has a maximum at x=
C. Has a maximum at finite
nonzero x
x
D. Has a minimum at finite
nonzero x
E. Has neither max nor min
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Force on charged particle
Electric field produces force
qE on charged particle
Force produces an acceleration a = FE / m
Uniform E-field (direction&magnitude)
produces constant acceleration if no other forces
Positive charge accelerates in same direction as field
Negative charge accelerates in direction opposite to
electric field
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Force on a dipole
Dipole made of equal + and - charges
Force exerted
on each charge
Uniform field
causes rotation
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Dipole
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Dipole in non-uniform field
A dipole is near a positive point charge in a viscous
fluid. The dipole will
A. rotate CW & move toward charge
B. rotate CW & move away
C. rotate CCW & move toward
+
D. rotate CCW & move away
E. none of the above
Mon. Feb. 18, 2008
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