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

    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

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

1

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 2

Question

Two 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?

n=1

λ

n=1.25

A.

λ/4 B.

λ/2 C.

λ D. 2 λ 3

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 4

Phase difference & interference

   Path length difference

d

Phase difference =

d(

2  /  ) radians Constructive for 2 

n

phase difference

L

Shorter path

Light beam Foil with two narrow slits

Longer path

Recording plate

5

Question

You are listening to your favorite radio station, WOLX 94.9 FM (94.9x10

6 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 = (3x10 8 m/s)/94.9x10

6 Hz  =3.16 m A. 3 m B. 1.6 m C. 0.8 m D. 0.5 m x d-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

6

Two-slit interference

7

Two-slit interference: path length

L y  Constructive int: Phase diff = Path length diff = 2 

m m

 , ,

m m

  0,  1,  0,  1,  2 2 Destructive int.

2  (

m

(

m

 1/2),

m

 1/2)  ,

m

  0,  1,  2 0,  1,  2 Path length difference 

d

sin     Phase difference /

L

  2 

d

sin    2 

d

 /

L

8  

Reflection phase shift

 Possible additional phase shift on reflection.

 Start in medium with n 1 , reflect from medium with n 2  n 2 >n 1 , 1/2 wavelength phase shift  n 2

different

paths is important.

9

Thin film interference

 air air: n 1 =1 1/2 wavelength phase shift from top surface reflection

Reflecting from n 2

t n 2 >1  air /n No phase shift from bottom interface

Reflecting from n 1

air: n 1 =1 Extra path length  2

t

 

m

Extra path length needed for constructive interference is 

m

 1/2   

air

 1/2   

air

/

n

 /

n

  10 

 

Thin-film interference example

eye

      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

Incident light

n film =1.2

n air =1 n glass =1.5

t=275nm

2

t



m



air

/

n film m

 1    660

nm

11

 

Thin Film Interference II

  Same coated glass underwater Now only one path has 180˚ phase shift 2

t

 

m

 1/ 2  

air

/

n film



air

  2

tn film

/ 

m

 1/2  

nm

 1.2

/ 

m

 1/2 

Incident light eye

n air =1 n water =1.33

n film =1.2

n glass =1.5

m=0 gives 1320 nm, too long. m=1 gives 440 nm Color changes underwater!

12

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)

13

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



D

 14

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  sin 

bright



m



d

15

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 16

Refraction

  Occurs when light moves into medium with different index of refraction.

Light direction bends according to

n

1 sin  1 

n

2 sin  2  i,1  r n 1 n 2  2

Angle of refraction



Special case: Total internal reflection

17

Total internal reflection

Total 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 18

Lenses: focusing by refraction

F P.A.

Image Object 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

19

Object

Different object positions

Image (real, inverted) Image (virtual, upright) Image (real, inverted) These rays seem to originate from tip of a ‘virtual’ arrow.

20

Question

You 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 21

Equations

Image and object different sizes Image (real, inverted) s  Relation between image distance object distance focal length s’ 1 

s

1

s

  1

f



image object height height

 

s



s

 

image distance object distance

22 

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?

A. f~0.1m

B. f~0.2m

C. f~0.5m

D. f~1.0m

 1 

s

1

s

  1

f

s’=2 m mag=10 -> s’=10s ->s=0.2m

1 0.2

m

 1 2

m

 5.5

 1

f



f

 0.18

m

23 

Diverging lens

Object Image Optical Axis Focal length defined to be

negative

Then thin-lens equation can be used: 1 

s

1

s

  1

f

 Thurs. Sep. 17, 2009 24 Physics 208, Lecture 5

p q Objective

Compound Microscope

Real, inverted, image Object Objective lateral mag.

=-q/p~L/f

objective

Virtual image ‘Object’ Eyepiece: simple magnifier.

Angular Mag.=25cm/p ~25cm/f eyepiece Mon. Feb. 4, 2008 Physics 208, Lecture 4

Eyepiece

25

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/r 2 Insulators and conductors  Polarization of insulators, conductors 26

Electric force: magnitude & direction

 Electrical force between two stationary charged particles     The SI unit of charge is the

coulomb

(C ),

µ

C = 10 -6 C 1 C corresponds to 6.24 x 10 18

k e

 electrons or protons =

Coulomb constant

≈ 9 x 10 9 e o 

permittivity of free space =

N .

m 2 /C 2 = 1/(4

π

8.854 x 10 -12 C 2 e o ) / N .

m 2 Directed along line joining particles.

27

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 B) Right C) Up D) Down E) Zero + -

F



kq

1

q

2

r

2 28

The electric dipole

 Can all be

approximated

by electric dipole.

 Two opposite charges magnitude

q

separated by distance

s

Dipole moment

p

Vector  Points from - charge to + charge Has magnitude

qs

Force on an electric dipole

• What is the direction of the force on the electric dipole from the positive point charge?

A. Up B. Down C. Left D. Right E. Force is zero

+



p

