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# 13 反射、折射、干涉、繞射

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(1)

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## Sections

### 3.

(3)

13-1 Reflection and Refraction (反射 與折射

1 1

2 2

1 1

(4)

(5)

(6)

### Rainbows

(7)

What produces the blue-green of a Morpho’s wing?

How do colorshifting inks shift colors?

airbrush

(8)

## Huygens’ principle

Fig. 35- 2

(9)

### principle

(10)

10

Fig. 35-3 35-

(11)

35- 11

Index of Refraction: c nv

1 2 1 1

1 2 2 2

ec hg

t t v

v v v

   

### 

1 1

2 2

sin (for triangle hce) sin (for triangle hcg)

hc hc

1 1 1

2 2 2

sin sin

v v

1 2

1 2

and

c c

n n

v v

 

1 1 2

2 2 1

sin sin

c n n c n n

### 

Law of Refraction: n1 sin

1n2 sin

2

(12)

(13)

13

35-

n

n n

n

n

(14)

14

35-

Fig. 35-4

### Since wavelengths in n1 and n2 are different, the two beams may no longer be in phase

1

1 1

1 1

Number of wavelengths in :

n

L L Ln

n N

n

### 

2

2 2

2 2

Number of wavelengths in :

n

L L Ln

n N

n

###  

2 2

2 1 2 1 2 1

Assuming : Ln Ln L

n n N N n n

###   

     

2 1 1/2 wavelength destructive interference

NN  

(15)

2

(16)

## Young’s Experiment

(17)

17

35-

0

(18)

18

Fig. 35-13

35-

0 0

1

0 0 2

sin sin ?

2 cos 2 cos

E t E t E t

E E E

w w f

 f

   

 

  f 

2 2 2 1

0 2

4 cos

EE f

2

2 1 2 1

2 0 2

2

0 0

4 cos 4 cos

I E

I I

IE  f   f

###  

phase path length difference difference

2

phase 2 path length difference difference

2 d sin

 

f  

   

   

   

   

    

   

Eq. 35-22

Eq. 35-23 Phasor diagram

(19)

E1 E2

19

35-

###  

1 0 sin and 2 0 sin

EE wt EE w ft

2 1

0 2

4 cos

II f 2

d sin

f  

 

###    

1 1 1

2 2 2

minima when: f m dsin m for m 0,1, 2, (minima)

1 2

maxima when: for 0,1, 2, 2 2 sin

sin for 0,1, 2, (maxima)

m m m d

d m m

f f

 

(20)

20

35-

Fig. 35-12

avg 2 0

I I

(21)

2

(22)

(23)

23

### Reflection Phase Shifts

35-

Fig. 35-16 n1 n1 > n2 n2

n1 n1 < n2 n2

Reflection Reflection Phase Shift Off lower index 0

Off higher index 0.5 wavelength

(24)

24

### Phase Difference in Thin-Film Interference

35-

Fig. 35-17

Three effects can contribute to the phase difference between r1 and r2.

1. Differences in reflection conditions 2. Difference in path length traveled.

3. Differences in the media in which the waves travel. One must use the wavelength in each medium (/ n), to calculate the phase.

2

0

(25)

25

### Equations for Thin-Film Interference

35- 2

odd number odd number

2 wavelength = (in-phase waves)

2 2 n

L   

½ wavelength phase difference to difference in reflection of r1 and r2

2L  integer wavelength = integer n2 (out-of-phase waves)

2

2

n n

  

12

### 

2

2L m for m 0,1, 2, (maxima-- bright film in air) n

   

2

2L m for m 0,1, 2, (minima-- dark film in air) n

  

(26)

26

### Color Shifting by Paper Currencies,paints and Morpho Butterflies

35-

weak mirror

looking directly down ： red or red-yellow tilting ：green

better mirror soap film

(27)

## 蝶

(28)

(29)

2

(30)

(31)

(32)

32

Fig. 35-23

### Michelson Interferometer

35-

1 2

2 2 (interferometer)

L d d

  

1

2 (slab of material of thickness placed in front of )

Lm L

L M

 

(33)

33

### Determining Material thickness L

35-

= 2 (number of wavelengths in same thickness of air)

a

N L

2 2

= = (number of wavelengths in slab of material)

m

m

L Ln

N  

2 2 2

###  

- = = n-1 (difference in wavelengths for paths with and without thin slab)

m a

Ln L L N N

 

(34)

34

### Problem 35-81

35-

In Fig. 35-49, an airtight chamber of

length d = 5.0 cm is placed in one of the arms of a Michelson interferometer. (The glass window on each end of the chamber has negligible thickness.) Light of

wavelength λ = 500 nm is used.

Evacuating the air from the chamber causes a shift of 60 bright fringes. From these data and to six significant figures, find the index of refraction of air at

atmospheric pressure.

(35)

35

35-

1 2

## b g 

4 1

 2

 

n L

N

n N

  L   

 

1

2 1 60 500 10

2 5 0 10 100030

9 2

 m

m

.

### h

. .

φ1 the phase difference with air ； 2 ：vacuum

N fringes

(36)

36

Diffraction Pattern from a single narrow slit.

### and the Wave Theory of Light

36- Central

maximum

Side or secondary maxima

Light

Fresnel Bright Spot.

Bright spot

Light These patterns

cannot be explained using geometrical optics (Ch. 34)!

(37)

The Fresnel Bright Spot (1819)

### wave

(38)

Diffraction by a single slit

sin (1 minima)st

a   asin 2 (2 minima) nd

(39)

(40)

(41)

41 36-

sin sin

2 2

a

 

a

(42)

42 36-

(second minimum)

sin sin 2

4 2

a

 

a

### 

(minima-dark fringes)

sin , for 1, 2,3 a

m

m

(43)

(44)

Fig. 36-7 44

### Intensity in Single-Slit Diffraction, Qualitatively

36-

phase 2 path length

difference difference

     

    

     

2

xsin

###  

 

N=18 = 0 small 1st min. 1st side max.

(45)

45

36-

Fig. 36-9

1

sin 2

2 E

R

Em

R

1 1 2

2

m sin

E E

22

m sin 2

m m

I E

I I

I E

 

    

2

asin

### 

 

  

(46)

46

Here we will show that the intensity at the screen due to a single slit is:

Fig. 36-8 36-

Intensity in Single-Slit Diffraction, Quantitatively

m sin 2 (36-5)

I

I

### 

 

  

where 1 sin (36-6) 2

a

 

, for 1, 2,3

m m

### 

In Eq. 36-5, minima occur when:

sin , for 1, 2, 3

or sin , for 1, 2, 3 (minima-dark fringes)

m a m

a m m

###  

 

 

If we put this into Eq. 36-6 we find:

(47)

1 , 1, 2,3,

m 2 m

     

(48)

48

### Diffraction by a Circular Aperture

36-

Distant point source, e,g., star

lens

Image is not a point, as expected from geometrical optics! Diffraction is

responsible for this image pattern

d

Light

a

Light

a

sin 1.22 (1st min.- circ. aperture) d

### 

sin (1st min.- single slit) a

(49)

49

R

36-

### Fig. 36-11

small

sin 1 1.22 1.22 (Rayleigh's criterion)

R

R d d

 

### 

(50)

13-3 Diffraction – (繞射)

(51)

(52)

λ

(53)

53

### The telescopes on some commercial and military surveillance satellites

36-

D

L   R  122. d

 = 550 × 10–9 m.

(a) L = 400 × 103 m ， D = 0.85 m → d = 0.32 m.

(b) D = 0.10 m → d = 2.7 m.

Resolution of 85 cm and 10 cm respectively

(54)

54

### Diffraction by a Double Slit

36-

Two vanishingly narrow slits a<<

Single slit a~

Two Single slits a~

m

cos2

### 

sin 2 (double slit)

I

I

 

  

d sin

a sin

(55)

μ

μ

λ

### = 405 nm

2

sin

sin for 0,1, 2, a

d m m

 

(56)

56

### Diffraction Gratings

36-

Fig. 36-18 Fig. 36-19

sin for 0,1, 2 (maxima-lines) d

m

m

Fig. 36-20

(57)

57

36-

### Fig. 36-22

sin hw , sin hw hw Nd

 

### 

hw (half width of central line) Nd

###  

 

hw (half width of line at ) cos

Nd

 

Fig. 36-21

(58)

58

36-

Fig. 36-23

Fig. 36-24

(59)

Compact Disc

(60)

60

36-

### Fig. 36-27

(61)

(62)

Viewing a holograph

(63)

A Holograph

(64)

64

36-

sin

d

m

### 

Differential of first equation (what change in angle

does a change in

wavelength produce?)

Angular position of maxima

cos

d

dmd

For small angles

cos

d

  m

and

d

 

d

 

cosm

d

###  

 

(dispersion defined)

D

(dispersion of a grating) (36-30)

cos D m

d

(65)

65

36-

hw Nd cos

 

### 

Substituting for  in calculation on previous slide

Rayleigh's criterion for half- width to resolve two lines

hw

N m

  

  

R

Nm

### 

avg (resolving power defined) R

(resolving power of a grating) (36-32) RNm

cos

d

  m

(66)

66

### Dispersion and Resolving Power Compared

36-

(67)

67

X-rays are electromagnetic radiation with wavelength ~1 Å

= 10-10 m (visible light ~5.5x10-7 m)

36-

Fig. 36-29

### X-ray generation

X-ray wavelengths too short to be

resolved by a standard optical grating

1 1 1 0.1 nm

sin sin 0.0019

3000 nm m

d

### 

 

(68)

68

d ~ 0.1 nm

→ three-dimensional diffraction grating

36-

Fig. 36-30

2 sind

m

### 

for m  0,1, 2 (Bragg's law)

(69)

69 36-

Fig. 36-31

2 0

5

0 0

5 4 or 0.2236

20

da daa

(70)

### Structural Coloring by Diffraction

(71)

Objects in mirror are closer than they appear.  The mirror is ：

Objects in mirror are closer than they appear.  The mirror is ：

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