13 反射、折射、干涉、繞射

13 反射、折射、干涉、繞射

Sections

1.

2.

3.

13-1 Reflection and Refraction (反射 與折射

)

n) (refractio

sin sin

n) (reflectio

1 1

2 2

1 1









n

n 

 

․The Index of Refraction

․The Stealth Aircraft F-117A

折射率

Chromatic Dispersion – prisms and

gratings

Rainbows

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

How do colorshifting inks shift colors?

airbrush

13-2 Interference – (干涉)

Huygens’ principle

All points on a wavefront serve as point sources of

spherical secondary wavelets.

After a time t, the new

position of the wavefront will be that of a surface tangent

to these secondary wavelets .

Law of Refraction from Huygens’

principle

10

Fig. 35-3 35-

35- 11

Index of Refraction: c n  v

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









Phase Difference, Wavelength and

Index of Refraction

13

Wavelength and Index of Refraction

35-

n

n n

v v

c c n

    

 ^{    }

n

n

v c n c

f f

  n 

   

The frequency of light in a medium

is the same as it is in vacuum

14

Phase Difference

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 ^







2

2 2

2 2

Number of wavelengths in :

n

L L Ln

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

N  N  

Ex.13-1 35-1

wavelength 550.0 nm n

=1.600 and

L = 2.600 m

Young’s Experiment

17

Coherence

35-

Two sources to produce an interference that is stable over time, if their light has a phase relationship that does not change with time: E(t)=E

cos( w t+ f )

Coherent sources: Phase f must be well defined and

constant. When waves from coherent sources meet, stable interference can occur － laser light (produced by

cooperative behavior of atoms)

Incoherent sources: f jitters randomly in time, no stable

interference occurs － sunlight

18

Fig. 35-13

Intensity and phase

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

E  E f

2

2 1 2 1

2 0 2

2

0 0

4 cos 4 cos

I E

I I

I  E  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

E₁ E₂

19

Intensity in Double-Slit Interference

35-

 

1 0 sin and 2 0 sin

E  E wt E  E w ft 

2 1

0 2

4 cos

I  I 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

Intensity in Double-Slit Interference

35-

Fig. 35-12

avg 2 0

I  I

Ex.13-2 35-2

wavelength 600 nm n

=1.5 and

m = 1 → m = 0

Interference from Thin Films

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

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

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

  





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

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

大

藍

魔

爾

蝴

蝶

雙狹縫干涉之強度

Ex.13-3 35-3 Brightest reflected light from a water film

thickness 320 nm n

=1.33

m = 0, 1700 nm, infrared

m = 1, 567 nm, yellow-green

m = 2, 340 nm, ultraviolet

Ex.13-4 35-4 anti-reflection

coating

Ex.13-5 35-5 thin air wedge

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

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

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

Solution to Problem 35-81

35-

f f

2 2 2 4 1

  L 

N M O

Q P ^{ }

L  n n L









b g 

4 1

 2

 

n L

 N

b g

n N

  L   

 



1 

2 1 60 500 10

2 5 0 10 100030

9 2

 m

m

c h

c

h

φ₁ the phase difference with air ； 2 ：vacuum

N fringes

36

Diffraction Pattern from a single narrow slit.

13-3 Diffraction

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)!

The Fresnel Bright Spot (1819)



Newton



corpuscle



Poisson



Fresnel



wave

Diffraction by a single slit

sin (1 minima)st

a   asin  2 (2 minima) ^nd

單 狹 縫 繞 射 之 強度

 Double-slit diffraction (with interference)

 Single-slit diffraction

雙狹縫與單狹縫

41 36-

Diffraction by a Single Slit:

Locating the first minimum

sin sin

2 2

a





 

(first minimum)

42 36-

Diffraction by a Single Slit:

Locating the Minima

(second minimum)

sin sin 2

4 2

a









(minima-dark fringes)

sin , for 1, 2,3 a





Ex.13-6 36-1 Slit width

Fig. 36-7 44

Intensity in Single-Slit Diffraction, Qualitatively

36-

phase 2 path length

difference difference





     

    

    ^{ }

^f

_ ^



^ 

 

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

45

Intensity and path length difference

36-

Fig. 36-9

1

sin 2

2 E

R

f

f

1 1 2

2

m sin

E_ E

f



f

 

 

m m

I E

I I

I E



 



 

    

 

2



f 



 

  

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

 

I







 

  

where 1 sin (36-6) 2



 f 

 



, 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:

Ex.13-7 36-2

1 , 1, 2,3,

m 2 m

  ^  ^ 

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

Rayleigh’s Criterion: two point sources are barely

resolvable if their angular separation θ

results in the central maximum of the diffraction pattern of one

source’s image is centered on the first minimum of the diffraction pattern of the other source’s image.

Resolvability

36-

Fig. 36- 11

small

sin 1 1.22 1.22 (Rayleigh's criterion)

R

R d d









13-3 Diffraction – (繞射)

Why do the colors in a pointillism

painting change with viewing distance?

Ex.13-8 36-3 pointillism

D = 2.0 mm

d = 1.5 mm

(diameter of

the pupil)

Ex.13-9 36-4

d = 32 mm f ^{= 24 cm}

λ

= 550 nm

(a) angular

separation (b) separation

in the focal

plane

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

Diffraction by a Double Slit

36-

Two vanishingly narrow slits a<<

Single slit a~

Two Single slits a~

 





I



 



 

  

d sin

  





a sin

  





Ex.13-10 36-5

d = 19.44

m a ^{= 4.050}

m

λ

= 405 nm

2

sin

sin for 0,1, 2, a

d m m

 

 



 

56

Diffraction Gratings

36-

Fig. 36-18 Fig. 36-19

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





Fig. 36-20

57

Width of Lines

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

Separates different wavelengths (colors) of light into distinct diffraction lines

Grating Spectroscope

36-

Fig. 36-23

Fig. 36-24

Compact Disc

60

Optically Variable Graphics

36-

Fig. 36-27

全像術

Viewing a holograph

A Holograph

64

Gratings: Dispersion

36-

sin

d





Differential of first equation (what change in angle

does a change in

wavelength produce?)

Angular position of maxima





d

 



For small angles





d

 



d













d



 

 



(dispersion defined)

D 



 

 (dispersion of a grating) (36-30)

cos D m

d 



65

Gratings: Resolving Power

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









avg (resolving power defined) R 

 



(resolving power of a grating) (36-32) R  Nm





d

 



66

Dispersion and Resolving Power Compared

36-

67

X-rays are electromagnetic radiation with wavelength ~1 Å

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

X-Ray Diffraction

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

d ~ 0.1 nm

→ three-dimensional diffraction grating

Diffraction of x-rays by crystal

36-

Fig. 36-30

2 sind





69 36-

Fig. 36-31

X-Ray Diffraction, cont’d

2 0

5

0 0

5 4 or 0.2236

20

d  a d  a  a

Structural Coloring by Diffraction