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
2
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
Mon. Feb. 18, 2008
3
Electric torque on dipoles
Remember torque?   r  F ,
magnitude rF sin 
r
Here there are two torques, both into page:
F  qE, r  s /2

   qE s /2sin 
r
F  qE, r  s /2

   qE s /2sin 

Total torque is sum of these
  qsE sin 

p  qs 

Mon. Feb. 18, 2008


Torque on dipole
in uniform field
  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
Mon. Feb. 18, 2008
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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

Mon. Feb. 18, 2008
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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
Mon. Feb. 18, 2008
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Path length difference



Path length difference d
Phase difference = (d/)2 radians
Constructive for 2n phase difference
L
Shorter path
Light
beam
Longer path
Foil with two
narrow slits
Mon. Feb. 18, 2008
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=/2x=/4
Constructive make 2x=x=/2
x increases by /4 = 3.16m/4=0.79m
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Two-slit interference
Mon. Feb. 18, 2008
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Two-slit interference: path length
L
y

Constructive int:
Phase diff = 2m, 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

d sin  2 d
Phase difference
 2


y /L
Destructive int.
Mon. Feb. 18, 2008


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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.
Mon. Feb. 18, 2008
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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  1/2air /n
 2t  m  1/2air /n 
Mon. Feb. 18, 2008
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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
Mon. Feb. 18, 2008

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 sin  bright  m, m  0,1,2
Mon. Feb. 18, 2008
sin bright  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
Mon. Feb. 18, 2008
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Refraction


Occurs when light moves into medium with different
index of refraction.
Light direction bends according to n1 sin 1  n2 sin 2
i,1
r
n1
n2

Special case:
Total internal reflection
2
Angle of refraction
Mon. Feb. 18, 2008
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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
Mon. Feb. 18, 2008
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Different object positions
Object
Image (real, inverted)
Image (real, inverted)
Image
(virtual, upright)
Mon. Feb. 18, 2008
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
Mon. Feb. 18, 2008
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Equations
Image and object
different sizes
Image (real, inverted)
p
Relation between
image distance
object distance
focal length

Magnification
= M =
Mon. Feb. 18, 2008
q
1 1 1
 
p q f
image height q image distance
 
object height p object distance
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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
Mon. Feb. 18, 2008
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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
Mon. Feb. 18, 2008
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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.
Mon. Feb. 18, 2008
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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
Mon. Feb. 18, 2008
+
-
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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
Mon. Feb. 18, 2008
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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 = (9109)(1.610-19)/(10-10)2 N = 2.91011 N/C
Mon. Feb. 18, 2008
(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
Mon. Feb. 18, 2008
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Electric dipole
E
1 2p
4eo 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
Mon. Feb. 18, 2008
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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
Mon. Feb. 18, 2008
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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
Mon. Feb. 18, 2008
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Force on a dipole

Dipole made of equal + and - charges

Force exerted
on each charge

Uniform field
causes rotation
Mon. Feb. 18, 2008
Dipole
35
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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