mon. feb. 18, 20081 physics 208 exam 1 review. mon. feb. 18, 20082 exam covers ch. 21.5-7,...
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Mon. Feb. 18, 2008 1
Physics 208 Exam 1 Review
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Mon. Feb. 18, 2008 2
Exam covers Ch. 21.5-7, 22-23,25-26,
Lecture, Discussion, HW, Lab
Chapter 21.5-7, 22 Waves, interference, and diffraction
Chapter 23 Reflection, refraction, and image formation
Chapter 25 Electric charges and forces
Chapters 26 Electric fields
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)
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Mon. Feb. 18, 2008 3
Quick Quiz
An electric dipole is in a uniform electric field as shown. The dipole
DipoleA. accelerates left
B. accelerates right
C. stays fixed
D. accelerates up
E. none of the above
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Mon. Feb. 18, 2008 4
Electric torque on dipoles
Total torque is sum of these
€
rτ =
rr ×
r F ,
magnitude rF sinθ
€
rF + = qE,
r r = s /2
⇒ r τ + = qE s /2( )sinθ
r F − = −qE,
r r = s /2
⇒ r τ + = qE s /2( )sinθ
€
rr +
€
rr −
€
rτ =qsE sinθ
€
rp = qs⇒
€
rτ =
rp ×
r E Torque on dipole
in uniform field
Remember torque?
Here there are two torques, both into page:
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Mon. Feb. 18, 2008 5
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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Mon. Feb. 18, 2008 6
Properties of waves
Wavelength, frequency, propagation speed related as
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
€
λf = v
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Mon. Feb. 18, 2008 7
Chapter 22: Waves & interference
Path length difference and phase different path length -> phase difference.
Two slit interference Alternating max and min due to path-length difference
Phase change on reflection 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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Mon. Feb. 18, 2008 8
Path length difference
Path length difference d Phase difference = (d/λ)2π radians Constructive for 2πn phase difference
L
Shorter path
Longer path
Light beam
Foil with two narrow slits
Recording plate
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Mon. Feb. 18, 2008 9
QuestionYou 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)
A. 3 m
B. 1.6 m
C. 0.8 m
D. 0.5 m
d
xd-x
path length diff = (d+x)-(d-x)= 2x
λ=3.16 mHint: wavelength = (3x108 m/s)/94.9x106 Hz
Destructive 2x=λ/2x=λ/4Constructive make 2x=λx=λ/2
x increases by λ/4 = 3.16m/4=0.79m
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Mon. Feb. 18, 2008 10
Two-slit interference
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Mon. Feb. 18, 2008 11
Two-slit interference: path length
y
L
€
≈d sinθ ≈ d y /L( )Path length difference
€
≈2πd sinθ
λ≈
2π d
λy /L( )Phase difference
Constructive int:
Destructive int.
Phase diff = Path length diff =
Phase diff = Path length diff =
€
mλ , m = 0,±1,±2K
€
2πm, m = 0,±1,±2K
€
2π (m +1/2), m = 0,±1,±2K
€
(m +1/2)λ , m = 0,±1,±2K
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Mon. Feb. 18, 2008 12
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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Mon. Feb. 18, 2008 13
Thin film interference
air: n1=1
n2>1
1/2 wavelength phase shiftfrom top surface reflection
t
λair
λair/n
Extra path lengthExtra phase shift needed for constructive interference is
No phase shift frombottom interface
€
m +1/2( ) λ air /n( )
€
⇒ 2t = m +1/2( ) λ air /n( )
air: n1=1
Reflecting from n2
Reflecting from n1
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Mon. Feb. 18, 2008 14
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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Mon. Feb. 18, 2008 15
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.
Angularseparation
Circular aperture diffraction limited:
€
min =1.22λ
D
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Mon. Feb. 18, 2008 16
Diffraction gratings Diffraction grating is pattern of multiple slits. Very narrow, very closely spaced. Same physics as two-slit interference
€
d sinθbright = mλ , m = 0,1,2K
€
sinθbright = mλ
d
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Mon. Feb. 18, 2008 17
Chap. 23: Refraction & Ray optics
Refraction Ray tracing
Can locate image by following specific rays Types of images
Real image: project onto screen Virtual image: image with another lens
Lens equation Relates image distance, object distance, focal length
Magnification Ratio of images size to object size
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Mon. Feb. 18, 2008 18
Refraction Occurs when light moves into medium with different
index of refraction. Light direction bends according to
i,1 r
2
Angle of refraction
n1
n2
€
n1 sinθ1 = n2 sinθ2
Special case:Total internal reflection
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Mon. Feb. 18, 2008 19
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.
Lenses: focusing by refraction
F
F
Object
ImageP.A.
Here image is real, inverted, enlarged
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Mon. Feb. 18, 2008 20
Different object positions
Image (real, inverted)
Image (real, inverted)Object
Image (virtual, upright)
These rays seem to originatefrom tip of a ‘virtual’ arrow.
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Mon. Feb. 18, 2008 21
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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Mon. Feb. 18, 2008 22
Equations
Magnification = M =
Image and object different sizes
Image (real, inverted)
p q
€
image height
object height=
q
p=
image distance
object distance€
1
p+
1
q=
1
fRelation between
image distanceobject distancefocal length
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Mon. Feb. 18, 2008 23
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
p+
1
q=
1
fA. f~0.1m
B. f~0.2m
C. f~0.5m
D. f~1.0m
q=2 m
mag=10 -> q=10p->p=0.2m
€
1
0.2m+
1
2m= 5.5 =
1
f⇒ f = 0.18m
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Mon. Feb. 18, 2008 24
Chapter 25: Electric Charges & Forces
Triboelectric effect: transfer charge Total charge is conserved
Vector forces between charges Add by superposition Drops off with distance as 1/r2
Insulators and conductors Polarization of insulators, conductors
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Mon. Feb. 18, 2008 25
Charges conductors & insulators
Two types of charges, + and - Like charges repel Unlike charges attract
Conductor: Charge free to move Distributed over surface of conductor
Insulator Charges stuck in place where they are put
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Mon. Feb. 18, 2008 26
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/(4πo) o permittivity of free space = 8.854 x 10-12 C2 / N.m2
Directed along line joining particles.
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Mon. Feb. 18, 2008 27
+ -
-
Equal but opposite charges are placed near a negative charge as shown.
What direction is the net force on the negative charge?
A) Left
B) Right
C) Up
D) Down
E) Zero
221
rqkq
F =
Forces add by superposition
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Mon. Feb. 18, 2008 28
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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Mon. Feb. 18, 2008 29
Electric field
Fe = qE
If q is positive, F and E are in the same direction
+
r = 1x10-10 m
Qp=1.6x10-19 C
E
E = (9109)(1.610-19)/(10-10)2 N = 2.91011 N/C
(to the right)
Example: electric field from point charge
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Mon. Feb. 18, 2008 30
Pictorial representation of E: Electric Field Lines
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Mon. Feb. 18, 2008 31
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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Mon. Feb. 18, 2008 32
Electric dipole
€
rE =
1
4πεo
2r p
r3
On y-axis far from dipole
€
rE = −
1
4πεo
r p
r3
On x-axis far from dipole
Electric field magnitude drops off as 1/r3
€
rp = qsElectric dipole
moment+q
-q
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Mon. Feb. 18, 2008 33
Quick quiz: continuous charge dist.
Electric field from a uniform ring of charge. The magnitude of the electric field on the x-axis
x
yA. Has a maximum at x=0
B. Has a maximum at x=
C. Has a maximum at finite nonzero x
D. Has a minimum at finite nonzero x
E. Has neither max nor min
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Mon. Feb. 18, 2008 34
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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Mon. Feb. 18, 2008 35
Force on a dipole
Dipole made of equal + and - charges
Force exerted on each charge
Uniform fieldcauses rotation
Dipole
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Mon. Feb. 18, 2008 36
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
+