Download Physics 208 Exam 1 Review

Exam covers Ch. 21.5-7, 22-23,25-26,
Lecture, Discussion, HW, Lab
Physics 208 Exam 1 Review
Exam 1 is Wed. Feb. 20, 5:30-7 pm,
Sections , , , , , 2103 Chamberlin Hall
Sections , , , , , 180 Science Hall
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Chapter 21.5-7, 22
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Chapter 23
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Chapter 25
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Chapters 26
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Mon. Feb. 18, 2008
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Electric charges and forces
Electric fields
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Path length difference and phase
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Two slit interference
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Phase change on reflection
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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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different path length -> phase difference.
Alternating max and min due to path-length difference
π phase change when reflecting from medium with higher index of
refraction
Interference in thin films
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Different path lengths + reflection phase change
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Path length difference
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2
Chapter 22: Waves & interference
Wavelength, frequency, propagation speed
related as "f = v
Phase relation
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Reflection, refraction, and image formation
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Properties of waves
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Waves, interference, and diffraction
4
Question
You are listening to your favorite radio station, WOLX 94.9 FM (94.9x106 Hz)
while jogging toward 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)
Path length difference d
Phase difference = (d/λ)2π radians
Constructive for 2πn phase difference
Hint: wavelength = (3x108 m/s)/94.9x106 Hz
L
λ=3.16 m
Shorter path
A. 3 m
B. 1.6 m
Light
beam
Longer path
Foil with two
narrow slits
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C. 0.8 m
D. 0.5 m
Recording
plate
5
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d-x
x
d
path length diff = (d-x)-(d+x)
= 2x = λ/2
6
1
Two-slit interference
Two-slit interference: path length
L
y
θ
Constructive int:
Phase diff = 2"m, m = 0,±1,±2K
Path length diff = m", m = 0,±1,±2K
Destructive int.
Phase diff = 2" (m + 1/2), m = 0,±1,±2K
! diff = (m + 1/2) ", m = 0,±1,±2K
Path length
!
Path length difference " d sin # " d ( y /L )
!
d sin $ 2# d
Phase difference
" 2#
"
( y /L)
!
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%
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%
8
!
!
Thin film interference
Reflection phase shift
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Possible additional phase shift on reflection.
Start in medium with n1,
reflect from medium with n2
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λair
1/2 wavelength phase shift
from top surface reflection
Reflecting from n2
air: n1 =1
n2>n1, 1/2 wavelength phase shift
n2<n1, no phase shift
t
Difference in phase shift between different
paths is important.
n2>1 λair/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 + 1/2)( "air /n)
" 2t = ( m + 1/2)( #air /n )
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!
!
Overlapping diffraction patterns
Diffraction from a slit
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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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Two independent point
sources will produce two
diffraction patterns.
Angular
separation
If diffraction patterns overlap
too much, resolution is lost.
Image to right shows two
sources clearly resolved.
θ
Circular aperture diffraction limited: "min = 1.22
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#
D
12
!
2
Diffraction gratings
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Chap. 23: Refraction & Ray optics
Diffraction grating is pattern of multiple slits.
Very narrow, very closely spaced.
Same physics as two-slit interference
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Refraction
Ray tracing
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Types of images
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sin " bright = m
Lens equation
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Magnification
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#
d
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!
Real image: project onto screen
Virtual image: image with another lens
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d sin " bright = m#, m = 0,1,2K
Can locate image by following specific rays
Relates image distance, object distance, focal length
Ratio of images size to object size
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!
Lenses: focusing by refraction
Refraction
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Occurs when light moves into medium with different
index of refraction.
Light direction bends according to n1 sin "1 = n 2 sin " 2
F
Object
θi,1
θr
n1
n2
Special case:
Total internal reflection
!
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
θ2
Angle of refraction
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Different object positions
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Question
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Object
P.A.
Image (real, inverted)
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
Image (real, inverted)
B. Move screen closer, move object away
C. Move screen away, move object closer
Image
(virtual,
upright)
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Feb. 18, 2008
D. Move screen closer, move object closer
These rays seem to originate
from tip of a ‘virtual’ arrow.
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3
Question
Equations
Image and object
different sizes
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?
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Image (real, inverted)
p
Relation between
image distance
object distance
focal length
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Magnification = M =
!
A. f~0.1m
1 1 1
+ =
p q f
image height q image distance
= =
object height p object distance
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20
!
Charges conductors & insulators
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Add by superposition
Drops off with distance as 1/r2
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Charge free to move
Distributed over surface of conductor
Insulator
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Like charges repel
Unlike charges attract
Conductor:
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Insulators and conductors
Polarization of insulators, conductors
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Two types of charges, + and 
Total charge is conserved
Vector forces between charges
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Triboelectric effect: transfer charge
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Charges stuck in place where they are put
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Equal but opposite charges
are placed near a negative charge as shown.
What direction is the net force on the negative charge?
Electrical force between two stationary charged particles
A) Left
The SI unit of charge is the coulomb (C ), µC = 10-6 C
1 C corresponds to 6.24 x 1018 electrons or protons
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ke = Coulomb constant ≈ 9 x 109 N. m2/C2 = 1/(4πε o)
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Directed along line joining particles.
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Forces add by superposition
Electric force: magnitude & direction
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-
B) Right
εo = permittivity of free space = 8.854 x 10-12 C2 / N.m2
F=
C) Up
D) Down
E) Zero
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1
1
1
+
= 5.5 = " f = 0.18m
0.2m 2m
f
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Chapter 25: Electric Charges & Forces
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mag=10
-> q=10p->p=0.2m
!
D. f~1.0m
!
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q=2 m
B. f~0.2m
C. f~0.5m
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1 1 1
+ =
p q f
q
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+
kq1q2
r2
-
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Electric field
Chapter 26: The Electric Field
Defined as force per unit charge (N/C)
Calculated as superposition of contributions from
different charges
Examples
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Fe = qE
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If q is positive, F and E are in the same direction
Example: electric field from point charge
Single charge
Electric dipole
Line charge, sheet of charge
Qp=1.6x10-19 C
+
Electric field lines
Force on charged particles
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E
r = 1x10-10 m
E = (9×109 )(1.6×10-19)/(10-10) 2 N = 2.9×1011 N/C
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Pictorial representation of E: Electric Field Lines
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r
1 2p
3
4 "#o r
r
r
1 p
E ="
4 #$o r 3
-q
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
On x-axis far from dipole
!
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y
r
Electric dipole p = qs
moment
+q
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Electric field from a uniform ring of charge.
The magnitude of the electric field on the x-axis
On y-axis far from dipole
!
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
Quick quiz: continuous charge dist.
Electric dipole
r
E=
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Electric field lines
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(to the right)
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E. Has neither max nor min
Electric field magnitude drops off as 1/r3
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Force on charged particle
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Force on a dipole
Electric field produces force
qE on charged particle
Force produces an acceleration a = FE / m
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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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Dipole made of equal + and - charges
Force exerted
on each charge
Dipole
Uniform field
causes rotation
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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
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