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Exam 1 Tue. Sep. 29, 5:30-7 pm, 145 Birge

Covers 21.5-7, 22, 23.1-4, 23.7, 24.1-5, 26 + lecture, lab, discussion, HW

8 1/2 x 11 handwritten note sheet (both sides) allowed

Chap 21.5-7, 22 Waves, interference, and diffraction

Chap 23 Reflection, refraction, and image formation

Chap 24 Optical instruments

Chap 26 Electric charges and forces

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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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QuestionTwo waves of wavelength λ are traveling through a medium with index

of refraction n=1. One of the waves passes a medium of thickness d=λ and index n=1.25..

What is the phase difference between the waves far to the right?

A. λ/4

B. λ/2

C. λ

D. 2λ

n=1 λ

n=1.25

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Ch 22, 21.5-7: 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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Phase difference & interference

Path length difference d Phase difference = d(2π /λ) radians Constructive for 2πn phase difference

L

Shorter path

Longer path

Light beam

Foil with two narrow slits

Recording plate

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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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Two-slit interference

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Two-slit interference: path length

y

L

€

≈d sinθ ≈ d y /L( )Path length difference

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≈2πd sinθ

λ≈

2π d

λy /L( )Phase difference

Constructive int:

Destructive int.

Phase diff = Path length diff =

Phase diff = Path length diff =

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mλ , m = 0,±1,±2K

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2πm, m = 0,±1,±2K

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2π (m +1/2), m = 0,±1,±2K

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(m +1/2)λ , m = 0,±1,±2K

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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: n1=1

n2>1

1/2 wavelength phase shiftfrom top surface reflection

t

λair

λair/n

Extra path lengthExtra path length needed for constructive interference is

No phase shift frombottom interface

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m +1/2( ) λ air /n( )

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⇒ 2t = m +1/2( ) λ air /n( )

air: n1=1

Reflecting from n2

Reflecting from n1

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Thin-film interference example

Coated glass in air, coating thickness = 275nm

Incident white light 400-700nm Glass infinitely thick What color reflected light do you see? Both paths have 180˚ phase shifts So only path length difference is

important

nfilm=1.2

nglass=1.5

nair=1Incident light

eye

t=275nm

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2t = mλ air /n film

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m =1⇒ λ = 660nm

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Thin Film Interference II Same coated glass

underwater Now only one path has

180˚ phase shift

nfilm=1.2

nair=1

Incident light

eye

nglass=1.5

nwater=1.33

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2t = m+1/ 2( )λair / nfilm

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λair = 2tn film / m +1/2( )

= 2 275nm( ) 1.2( ) / m +1/2( )

m=0 gives 1320 nm, too long.

m=1 gives 440 nm

Color changes underwater!

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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.

Angular locations of

minima (destructive 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.

Angularseparation

Circular aperture diffraction limited:

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min =1.22λ

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

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d sinθbright = mλ , m = 0,1,2K

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sinθbright = mλ

d

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Chap. 23-24: 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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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

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n1 sinθ1 = n2 sinθ2

Special case:Total internal reflection

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Total internal reflectionTotal internal reflection occurs

A) at angles of incidence greater than that for which the angle of refraction = 90˚

B) at angles of incidence less than that for which the angle of refraction = 90˚

C) at angles of incidence equal to 90˚

D) when the refractive indices of the two media are matched

D) none of the above

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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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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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QuestionYou have a near point of 25cm. You hold a 5 cm focal length

converging lens of focal length a negligible distance from your eye to view a penny more closely. If you hold the penny so that it appears sharp when you focus your eye at infinity (relaxed eye) how many times larger does the penny appear than the best you can do without the converging lens?

A. 2

B. 3

C. 4

D. 5

E. 10

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Equations

Magnification = M =

Image and object different sizes

Image (real, inverted)

s s’

€

−image height

object height= −

′ s

s= −

image distance

object distance€

1

s+

1′ s =

1

fRelation between

image distanceobject distancefocal length

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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

s+

1′ s =

1

fA. f~0.1m

B. f~0.2m

C. f~0.5m

D. f~1.0m

s’=2 m

mag=10 -> s’=10s ->s=0.2m

€

1

0.2m+

1

2m= 5.5 =

1

f⇒ f = 0.18m

Thurs. Sep. 17, 2009 Physics 208, Lecture 524

Diverging lens

OpticalAxis

Then thin-lens equation can be used:

€

1

s+

1′ s

=1

f

Object Image

Focal length defined to be negative

Mon. Feb. 4, 2008 Physics 208, Lecture 4 25

Virtual image ‘Object’

Eyepiece

CompoundMicroscope

qp

Object Real, inverted, image

Eyepiece: simple magnifier.

Angular Mag.=25cm/p

~25cm/feyepiece

Objective lateral mag.

=-q/p~L/fobjective

Objective

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Chapter 26: 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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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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+ -

-

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

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rqkq

F =

Forces add by superposition

The electric dipole

Dipole moment

Vector

Points from - charge to + charge

Has magnitude qs

€

rp

Can all be approximated by electric dipole.

Two opposite charges magnitude q separated by distance s

Force on an electric dipole• What is the direction of the force on the electric

dipole from the positive point charge?

€

rp

+

A. Up

B. Down

C. Left

D. Right

E. Force is zero

How does the magnitude of the force depend on ?

€

rp