Mon. Feb. 18, 2008 1

Physics 208 Exam 1 Review

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)

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

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:

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

+

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

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

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

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

Mon. Feb. 18, 2008 10

Two-slit interference

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

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.

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

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.

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

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

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

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

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

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.

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

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

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

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

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

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.

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

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

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

Mon. Feb. 18, 2008 30

Pictorial representation of E: Electric Field Lines

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

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

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

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

Mon. Feb. 18, 2008 35

Force on a dipole

Dipole made of equal + and - charges

Force exerted on each charge

Uniform fieldcauses rotation

Dipole

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

+