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第12章 波动光学


CHAPTER 12
Wave Optics
第十二章 波动光学

By Dr. G Y Chen

1

2

§12-1 Light Source and the Coherence of Light §12-2 Optical Distance and Optical P

ath Difference §12-3 Film Interference §12-4 Michelson’s Interferometer §12-5 Diffraction of Light, Huygens-Fresnel’s Principle

Introduction

§12-6 Single-Slit Fraunhofer Diffraction §12-7 Diffraction Grating §12-8 Polarized Light
§12-9 Polarization by Reflection or Refraction
3

§12-1 Optics Source and the Coherence of Light
1. electromagnetic wave and light:
(1) The source of electromagnetic wave:
All vibrating charges(振动电荷) or vibrating charge systems is the source that emit(发射) electromagnetic waves.

(2) The essence(本质) of electromagnetic wave:
electromagnetic waves is vector wave(矢量波) of electric field intensity (电场强度)and magnetic field intensity(磁场强度).

? ? r E (r , t ) ? E0 cos? (t ? ) u

? ? r H (r , t ) ? H 0 cos? (t ? ) u
4

? ? ①在任意给定点上,E 和 H 同时存在,具有相同的相位和
速度。

(3) Characteristics of electromagnetic wave:

? u H 、 三者满足右螺旋法则。
③对空间任一点,有

? ? ? ②电磁波是横波:E 和H 互相垂直,且与传播方向垂直, 、 E ?
? E ? ?H
1 1 1

④电磁波的传播速度取决于介质的介电常数ε和磁导率μ,且 满足:
u? 1

??

?

? r? 0 ?r ?0

?

? 0 ?0

?

? r ?r
5

(4) Visible light and electromagnetic wave:
Light is a kind of electromagnetic wave, which has a velocity of c=3×108m/s . Visible light(可见光) can cause the sense of sight(视觉), which is due to the electric field vector and is called light vector. The wavelength of visible light is 400—760nm, whose color corresponding different wavelength(frequency):

red orange yellow green wavelength turns shorter

cyan

blue

purple

frequency turns higher 6

2. light source

§12-1 Optics Source and the Coherence of Light

(1) Shiny objects is called light source. (2) Common mechanism of luminescence(发光).
能级跃迁辐射 波列

ν=E2-E1/h

L=t c

Luminescence of light source is due to molecules or atoms transit from the high energy level to low energy level, when the corresponding energy is released simultaneously.
7

(3) Properties:

①光源中原子的激发和辐射是独立的、随机的,间歇性的, 因此,同一时刻不同原子发的光其频率、振动方向相位差各 不相同。 ②每个原子每次发射的电磁波 为一段有限波长、振动方向和频率 均一定的正弦波列。

Question:多盏日光灯发 出的光不会发生干扰.
8

§12-1 Optics Source and the Coherence of Light
3. Interference of Light
The critical quantity in optical vibration is the electric vector E, not the magnetic vector H. Vector E is called light vector.

r1 E1 ? E01cos ?(t ? ) ? ?01] [ u r2 E2 ? E02cos ?(t ? ) ? ?02 ] [ u
For simplicity, we assume that two light vectors vibrating in the same directions, and have the same angular frequency, so the superposition is:

E ? E1 ? E2 ? E0 cos( t ? ? 0 ) ?
9

E ? E1 ? E2 ? E0 cos(?t ? ? 0 )
Where:

E0 ? E ? E ? 2 E01 E02 cos ??
r2 ? r1 ?? ? ? 02 ? ? 01 ? ? u E10 sin ?1 ? E20 sin ? 2 tg? 0 ? E10 cos?1 ? E20 cos? 2
2 0 2 10 2 20

2 01

2 02

Average light intensity:

1 I ? E ? E ? E ? 2 E10 E20 ? cos ??dt t 0
10

t

where E0 is determined by Δφ, while Δφ is given by

1 I ? E ? E ? E ? 2 E10 E20 ? cos ??dt t 0
2 0 2 10 2 20

t

r2 ? r1 ?? ? ?02 ? ?01 ? ? u

=0

Generally φ02-φ01 varies with t, so the integration will be 0.

I ? I1 ? I 2

incoherent superposition(非相干叠加)

Only when φ02-φ01 does not vary with t, the total intensity

I ? I1 ? I 2 ? 2 I1 I 2 cos ?? coherent superposition

I1 ? I 2

I max

?? ? 4 I1 cos 2
2

(相干叠加)

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Three conditions that coherent lights must be met are
①Vibrating in the same direction(振动方向相同) ②have the same frequency(频率相同) ③have the invariable phase difference(相差恒定)

Results of interference
①stable interference pattern(稳定的干涉花样) Constructive 干涉相长 ②about the light intensity

?? ? 0, 2? , 4? ,... ? ?

I max ? 4I1

?? ? 0, ? , 3? ,... ? ?

I min ? 0

Destructive 干涉相消
12

§12-1 Optics Source and the Coherence of Light
4. Obtain of Coherent Light
The lack of lights from common sources is due to the fact that the emitting atoms in such sources act independently rather than cooperatively. So these lights have not the same frequency, vibration direction and phase difference, and they are not coherent light.

How to obtain Coherent Light?
13

4. Obtain of Coherent Light Way 1: Two beams interference by division of wavefront(从光波发出的同一波列的波面上取出 两个次波源,即分波阵面法)

——Young’s Interference(杨氏干涉)
Way 2: Two beams interference by division of amplitude(或把同一波列的波分为两束,即分振 幅法)

——Film Interference(薄膜干涉)
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1 Young’s Double-Slits Experiment 杨氏双缝实验
Experimental device

interference pattern
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1 Young’s Double-Slits Experiment 杨氏双缝实验

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1 Young’s Double-Slits Experiment 杨氏双缝实验
r1 P r2 O D x ·

x
x x0 ?x I

S1

?
S

d

? ?
S2

Wave path difference:

x ? ? r2 ? r1 ? d sin ? ? dtg? ? d D
17

Discussions:

bright fringe

(1)When δ=kλ,I=Imax

光强达最大,形成明条纹。

D? x ? ?k d
k=0:

k ? 0,1,2 ?
-3 -2 -1 0 1 2 3

central bright fringe;

k=±1: first-order bright fringes; k=±2: second-order bright fringes;
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Discussions:
(2)When δ=(2k-1)?λ/2, →I=Imin
干涉相消,形成暗纹。

dark fringe

D? x ? ?( 2 k ? 1) , k ? 1,2,3 ... 2d

-3 -2 -1 1 2 3 k=±1: first-order dark fringes;

k=±2: second-order dark fringes;
19

Features:
(1) The central distance between two adjacent bright (or dark) fringes is:

D ?x ? ? d

(2) Since λ is very small, only if D is large enough and d is small enough, can the fringes be easily distinguishable. 由于λ极小,故只有D足够大,d足够小,干涉条纹△x 才可能大到可以分辨。
20

Features:
(3) If two beams (λ1,λ2) pass the slits at the same time
Two kinds of interference pattern

?中央明条纹重叠; ?波长短的靠近中央; ?其它级次可能会重叠:

?x2

x ? k1?1 ? k2?2

?x1 0 1 2 3 -3 -2 -1

?如用白光,会出现彩虹。干涉条纹从内到外,颜色 由紫到红。
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Discussion:
Changes of the fringes in the following cases
杨氏实验在下列情况下条纹的变化

(1)使屏离双缝的间距D↑; (2)使光源波长λ↑; (3)两缝的间距d↑ ; (4)红色滤光片遮住一缝,蓝 色滤光片遮住另 一缝。
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2、Lloyd’s mirror experiment

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2、Lloyd’s mirror experiment
S

P O



S’
按杨氏双缝实验,显然O点应是中央明 纹,但事实上是暗纹。why?

说明:光在平面镜表面时有半波损失(half-wave loss)。 Conclusion:从光疏介质到光密介质界在上反射时,反 射光有位相π的突变,这种现象称半波损失(half-wave loss)。
24

§12-2 Optical Path and Optical Path Difference
1 Optical Path
Light speed is depend on the medium, when the refractive index(折射率) of medium is n, the light speed in this medium can be given by u=c/n. The path of light in different mediums is not the same.

由于光在不同介质中的速度是不同的,在折射率为n的介质中, 光速u=c/n。因此,在相同时间内,光波在不同介质中传播的

路程是不同的。
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§12-2 Optical Path and Optical Path Difference
1 Optical Path
During the time t, if the path of the light in the medium is r, the path of the light in air in the same time should be

r c x ? ct ? c ? ? ? r ? nr u u
The product nr is called optical path.
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Conclusion:
1. In the same time, a path r in the medium vacuum. a path nr in

2. When the light pass in different mediums, the frequency keeps unchanged, so the phase change is the same. An equivalent optical path lead to a same phase change. 3. When light travel in different mediums, the optical path can be written

OpticalPat ? h

? nr

i i

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2. Optical Path Difference
The phase difference of two coherent light: r2 r1 ?? ? 2? ? 2? ?2 ?1
S1

r1

n1
n2

P

S2

r2

n2 r2 n1r1 n2 r2 n1r1 ? 2? ? 2? ? 2? ? 2? ?0 ?0 n2?2 n1?1 2? ? (n2 r2 ? n1r1 ) ?0

Optical Path Difference
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2. Optical Path Difference
suppose:
Optical Path Difference:

? ? n2 r2 ? n1r1
The relationship between the phase change and the optical path should be

? ? ?? ? 2? ?0

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§12-2 Optical Path and Optical Path Difference
3. Property of Thin Lens——aplanatism(等光程)
Theory and experiment show: A lens does’t cause any additional optical path difference or phase shift. bright point
In-phase 主光轴
Parallel light

focus
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§12-2 Optical Path and Optical Path Difference
3. Property of Thin Lens ——aplanatism
Parallel light Bright point

Focus
Main optical axis focal plane

In-phase

Does’t cause any additional optical path difference! Then ?? 31

§12-3 Film Interference
1. Equal-inclination interference, isoclinal interference (等倾干涉)

S

S

32

33

1. Equal-inclination interference, isoclinal interference (等倾干涉)

E

E′

A ray of monochromatic light is D incident on a thin transparent(透明 n 1 的)film at a small angle i. Some A ? C e n2 light is reflected(反射) at A and some is refracted (折射) toward B. B The refracted light is then reflected n1 at B, and refracted at C, then ? emerging from the film CE′ ? n2 ( AB ? BC ) ? n1 AD ? ? 2 paralleled to AE,The optical path difference is:

i

34

1. Equal-inclination interference, isoclinal interference (等倾干涉)
e AB ? BC ? cos? AD ? AC sin i

E

E′

n1
n2

i

D C B

AC ? 2etg?

A ?

n1 sin i ? n2 sin ?

e

? ? ? n2 ( AB ? BC ) ? n1 AD ? 2

n1

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1. Equal-inclination interference, isoclinal interference (等倾干涉)

? ? 2n2e cos? ?

?
2

? 2e

2 2 2 n2 ? n1 sin i

?

?
2

δ is only a function of the angle i—— Equalinclination interference: i

? ? k? (k ? 1, 2,3...)
?
2

bright fringe

? ? (2k ? 1) , (k ? 0,1, 2...) dark fringe
36

2. Equal-thickness interference(等厚干涉)
(1) Wedge Interference(劈尖干涉)
空气劈尖

很小

e、n L

37

(1) Wedge Interference Optical path difference:

棱边

? ? 2ne ?
Therefore:

?
2
劈尖角

e

n

? ? ?2 k 2 ? ? 2ne ? ? ? ? 2 ? (2k ? 1) ? 2 ?

k ? 1,2,3......bright fringes k ? 0,1,2,3...... dark fringes
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2. Equal-thickness interference(等厚干涉)
(1) Wedge Interference(劈尖干涉)
By all appearances,the same order bright or dark fringe is the position of points for which the film thickness is a constant. The interference is called Equal-thickness Interference(等厚 干涉) .

sinθ≈θ=e/L

The distance between two adjacent bright (or dark) fringes is:
Δe=Lsinθ=λ/(2n)→L≈λ/(2nθ)
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characteristics :
(1)The bright or dark fringes parallel to the seamed edge.(明暗条纹与棱边平行);

(2)Whether the seamed edge is bright or dark, is depend on the half-wave loss. 棱边是明还是暗,应看有无半波损失;
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characteristics :
(3)The distance between adjacent bright or dark fringes is L:

L sin? ? ek ?1 ? ek ?

?
2n

sin ? ? ?(rad )
? L? 2n?

or ? ?

?
2nL

? 越大,条纹越密, ? 越小,条纹越疏。
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2、Newton’s ring (牛顿环 )

wedge-shaped
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2、Newton’s ring (牛顿环 )
Have the half-wave loss? So, Optical path difference:

?

R ?

? ? ? 2ne ? 2
(k ? 1,2 ?) bright ?k? ? ? ?? ? ?(2 k ? 1) 2 (k ? 0,1?) dark ?
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2、Newton’s ring (牛顿环 )
r 2 ? R 2 ? ( R ? e)2 ? 2eR ? e 2 ? 2eR

?

R ?

i.e.:

r e? 2R

2

considering

? ? ? 2ne ? 2
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2、Newton’s ring (牛顿环 )

Hence:
R? rbright ? (2k ? 1) 2n k ? 1,2...

?

R ?

k R? rdark ? n

k ? 0,1,2....

45

Discussions:
1、如果没有半波损失,上面的公式如何? 2、平凸透镜曲率半径变小? 3、当平板玻璃下移,问牛顿 环是冒出还是缩进?

答:缩进。Why?

46

§12-4 Michelson’s Interferometer

47

§12-4 Michelson’s Interferometer
两相干光束 1 和 2 的干涉图样的 E 处观察 1) M1?和M2严格平行时,M2移动 表现为等倾干涉的圆环形条纹, 不断从中心冒出或向中心收缩。
M2 M?1 2 G1 G2 1

2) M1?和M2不严格平行时,则表 现为等厚干涉条纹,不断移过 S 视场中某一标记位置。 3) 平移距离 d 与条纹移动数 N 的关系满足: ?
半透半反膜 2?

M1

1?

d ?N?

2

E

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§12-5 Diffraction of Light

Huygens-Fresnel’s

Principle 一、Diffraction of Light

Wave can bend around obstacles, the

phenomena is called the diffraction phenomena of Wave(波的衍射现象).
49

Figure(a). When the slit is very large, a sharp shadow will be cast in the screen. When the width of the slit is not very largely compared to the wavelength, then the light intensity is not uniform and the incident light certainly ―flares out‖ into the geometric shadow of the slit., show in figure(b).并且中央光带的旁边还有些明 暗相间的条纹或彩色条纹。

(a)

The narrower the slit is, the more the light is flaring out.

(b) 50

二、Huygens-Fresnel’s Principle
(1)惠更斯提出:波在媒质中传播到的各点可看成 次级波源,它们将发射次级波。菲涅耳补充说:
(2)从同一波阵面上各点发出的 子波为相干波,经传播而在 空间某点相遇时,可以互相叠 加而产生干涉现象; (3)衍射区域各点的强度由 各子波在该点的相干叠加决定。 次级 波源 干涉

次级波

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§12-6 Single-Slit Fraunhofer(夫琅和费)diffraction
一、Single-Slit Fraunhofer diffraction
1、Concept: The Fraunhofer diffraction produced by a single slit. Incidence optics: s parallel light Screen: infinite distance

ideal :

E

l ??

52

A beam of parallel light is incident from the left of an opaque plate having a narrow vertical slit, a screen is placed at the focus plane of the lens on the right of the slit.

53

2、菲涅耳半波带 作一组平行于AC的平面, 相邻平面间距为λ/2,这一 组平面称半波带。这样,对 于应P点,狭缝边缘的两条 光线的光程差δ=asin ? ,所 包含的半波带的个数:

A a

B

φ

C

a sin ? k? ?/2
54

Discuss k:
(2)k为整数时,相邻波带的间距为λ/2,即位 相相差π。故:

(1)k可正可负,可为整数,也可能不是整数。

55

k为偶数时,则两个半波带相消,p为暗点。 k为奇数时,则两个半波带相消,结果留下 一个半波带未被抵消,故p为明纹。
即 a sin ? ? ?k?

k ? 1,2? dark fringes

a sin ? ? ?(2k ? 1)

?
2

k ? 1,2?bright fringes

但是 k=0时,a sin ? =0,即所有光线都同相 ,called center bright fringe。
56

(1)中央明纹宽度为相邻两暗纹的间距
即 -λ< asinφ<λ (2)若k不为整数,则p点介于明暗之间。

(3)用白光作光源,除中央明纹仍为白光外, 其余明纹为从内到呈从紫到红分布。

57

二、Diffraction from a Circle

58

衍射屏

L

中央亮斑 ( Ariy disk )
?

I/I0
1.22 0

?
d sin?

爱里斑

圆孔孔径

D

f

59

Ariy disk(爱里斑):

? 0 ? sin ? 0 ? 1.22

?
D

刚可分辨

当一个物点的爱里斑恰好在另 不可分辨 一个物点的爱里斑边缘时恰能 ? 分辨, ? is called the minimum resolving angle(最小分辨角).

? ? ? 1.22

?

D
60

Example:
A slit of width a=2.5μm is illuminated by white light. (1) What is the diffraction angle ?1 that the first minimum for the red light λ=6500A falls at? (2) What is the wavelength λ of the light whose first diffraction maximum falls at ?1 ,thus coinciding with the first minimum for the red light?
61

Solution:
(1) At the first minimum, from equation asin?=kλ,k=1,we have:
k? ? sin ?1 ? ? a a

?1 ? sin

?1 ?

a

? 15

o

62

(2)At the first maximum, for the equation asin?=(2k+1)λ/2, where k=1, ?’= ?1 =15o, we have 3 a sin ?1 ? ? ? 2 a sin ?1 ? ?
Therefore:

2 2 ? ? ? ? ? ? 6500 ? 4330 ( A) 3 3
63

d S1

?R

?1
0 I

S2

64

§12-7 Diffraction Grating (衍射光栅)
? A diffraction grating are made by ruling( 划 线 ) equally spaced parallel grooves on a polished glass plate. ? The width of a groove is b.(刻痕是b,不透 光), the width of a slit is a, d=a+b is called grating constant.

65

66

二、grating equation
The optical path difference between rays from adjacent slits is (a+b)sinφ。 The φ is called diffraction angle. The grating equation is:
φ

P

(a ? b) sin ? ? ?k? k ? 0,1,2,?
67

1、主极大条纹
满足光栅方程的明条纹称主极大条纹,也称 光谱线。 (1) 当?=0时,k=0为中央明纹

k?

(a ? b) sin ?

?

68

(2)可见条纹的最大级次
?k ? (a ? b) sin ?

?
k? sin ? ? ? d

? kmax ?

( a ? b)

?

(3)Since

可见光栅常数越小,各级明纹衍射角越大, 即各级明纹分得愈开。
69

Example:
A white light incident normally on a grating. The grating constant (a+b)= 2.4×10-4cm, the focal length f=0.25 m. Find the distances from the 3th order line of violet light and the 2th order line of red light to the center Po on the screen, respectively.
70

3?1 0 (1)?13 ? arcsin( ) ? 30 a?b x13 ? f ? tan?13 ? 0.25 ? tan 30 ? 0.1443 m 2?2 o ? (2)? 22 ? arcsin( ) ? 39 18 a?b o ? x22 ? f ? tan? 22 ? 0.25 ? tan 39 18 ? 0.2046 m

Solution:

71

2、Missing Order(谱线的缺级 )

? 以上讨论的是光栅各缝发出的光因干涉在屏 上形成主极大。但是每条缝本身又有衍射现 象,它将对明纹有影响。考虑其中一条缝, 这时应出现衍射条纹。 ? 这样一来,满足光栅方程的?角,本应出现 明纹时,但又由于满足单缝衍射的暗纹条件, 有

72

(a ? b) sin ? ? ?k? a sin ? ? ?k ' ?

主极大
暗纹

There is no principle maximum on the diffraction screen, or such principle maximum disappars. This phenomenon is called missing order(谱线的缺级).

a?b Missing Order : k ? k' a

73

3、暗纹条件
在光栅衍射中,相邻主极大间存在一些暗条 纹,也称极小,它是由各缝射出的光聚焦于 一点,因干涉相消形成的。

相邻两缝的光程差为:
对应的位相差为:

(a ? b) sin ?

?? ?

2? (a ? b) sin ?

?
74

用旋转矢量法,△?为各缝射出的光矢量的夹 角,若 ? N ?

E ? ? Ei ? 0
i ?1

N ? ?? ? ?m ? 2? N ? (a ? b) sin ? ? ?m ? 2? m为不为N 的整数
显然,相邻主极大之间有N-1个极小, 有N-2个次极大。
75

则:

三、光栅光谱
入射光为白光,除0级外,不同波长的光 各级主极大位置不同,我们把不同波长的同 级谱线集合称光栅光谱。 k? 由于 sin ? ? ? ,同级谱线,λ越小,

?角越小,故紫色在内侧,红色在外侧。

d

76

单缝衍射光强曲线
单缝衍射光强曲线 ?2 ?1 多光束干涉光强曲线 ?2 ?1 多光束干涉光强曲线 ?8 ?4

I
I

0
0

根据 N=4, d=a+b=4a, 1 画出的光强分布sin? 2
1 2 sin ( sin ??? ) a

I
I I 0 I 0

4
4

8
8

sin?
sin?
sin ? (? ) sin? d

光束衍射光强曲线 ?8 ?4 ?8 ?4 光束衍射光强曲线
?8 ?4

0
0

4
4

8
8

sin?

77

四、斜入射时的光栅方程

( a ? b ) ( s in ? ? s in ? ) ? ? k ?
规定:衍射光与入射光在光 栅平面法线同侧时,?为正, 斜入射可看到更 反之为负。
问题:衍射花样将发生什么变化?

高级次的明纹!

最大级次:k max

k max ?

a?b

?

(sin ? ? 1)
取整

正入射

k max ?

a?b

?

取整
78

Example:
A parallel beam of wavelength λ=5500A is incident obliquely on a diffraction grating whose constant is d=2.10μm and the of a slit is a=0.7μm. The angle of incidence is θ=30o. How many orders of fringes can be observed?

79

Solution:
K max ?

d

?

(sin ? ? sin ? )
2.1?10 ?6 5 ?10 2.1?10 ?6 5 ?10
?7 ?7

? K max ? ? K max ?

(sin 30 ? 1) ? 6.3 (sin 30 ? 1) ? ?2.1

The +kmax and –kmax are integer.

We can observe seven order fringes corresponding with k=+5,+4,+2,+1,0,-1,-2.
80

d 2.1? 10 ?6 ? ? ?3 ?7 a 0.7 ? 10

§12-8 Polarized Light
一、 Nature Light An unpolarized wave, which can be considered to be a random superposition of many polarized waves.
普通光源的发光机理是由大量的原子或分子的自发辐射。它们之间无论是发 光先后次序(相位),振动方向(偏振)、振动的大小(振幅)及发光持续 时间都相互独立,故从垂直于传播方向的平面上看,各个方向的振动均匀相 等。 自然光中任一光矢量可分解为互相垂直的振动。

81

二、 Polarized Light
1、Linearly Polarized Light : 在垂直于传播方向的平面内,光矢量只沿一 个固定方向振动。线偏振光又称平面偏振光。
? c
播 方 向 传

E

z

x 振




82

diagrammatize:图示法
光振动垂直板面

光振动平行板面 面对光的传播方向看

83

2、 Partially(部分) Polarized Light :
光波包括一切方向的振动,但不同方向上的 振幅不等,分解在两个垂直的方向上,一个 方向上的振幅大于另一个方向上的振幅。

84

Partially Polarized Light

decompose

diagrammatize:

平行板面的光振动较强

垂直板面的光振动较强
85

三、起偏和检偏
把自然光变为偏振光的过程称为起偏; 检验(观察)偏振光的过程称为检偏。

86

The polarized direction

Nature Light

起偏器

检偏器

The polarized direction
当一束自然光投射到偏振片时,垂直于某一

特殊方向光振动分量全被吸收,只让平行于偏振
化方向的光振动分量通过从而获得线偏振光。
87

四、 Law of Malus 马吕斯定律
The polarized direction

I0

I

起偏片

检偏片

?

I ? I 0 cos ?
2

Called Law of Malus

88

Discuss:
? 如何判断是自然光还是偏振光?

89

Example:
If the angle between two polarizes is 3 0 0 ,the intensity of the transmitted light is I1, Maintain the incident un-polarized light unchanged and align(tilt) the two polarizes so that the angle between them is 4 5 , then the intensity of the transmitted light is 2 I 1 /3
0

90

Solution:
I
0

1I 2

0

I
1

1 I co s 3 0 I ? 2
2 1 0
2 x 0

0

1 I co s 4 5 ? I ? 2 I I ? 2 3
0 x

1

91

§12-9 Polarization by Reflection Or refraction

? 一束自然光入射到两种介质的分界上,产生 反射和折射。 ? 将自然光分解成两个互相垂直的振动方向其 中一个垂直于入射面,称s光;另一分量的 振动在入射面内,称之为p光。 ? 通常:反射光中,s光成分比p光多 ? 折射光中,p光成分比s光成分多

92

93

二、Snell’s law(refraction law)

n1 sin i ? n2 sin r

94

2.Brewster’s Law(布儒斯特定律)

起偏振角 i0 n1 i0

By experiment :
2

S 线偏振光

For the polarizing n angle it is found experimentally that the reflected and the refracted beams are at right angles, or i+r=90o
95

From Snell’s Law

n1 sin i ? n2 sin r
? ib ? r ? 90
o

n2 tg ib ? n21 ? n1

The reflected (反射) beam is fully polarized , ib is called 全偏振角 or 布儒斯特角。
96

here:
1 ) The reflected ( 反 射 ) beam is fully polarized(S光),其振动方向垂直于入 射面; 2)折射光和反射光的传播方向相互垂直. 3)The refracted (折射) beam is partially polarized(S光和P光) 。

97

Example:
We wish to use a plate of glass (n=1.57) as a polarizer. (1)What is the Brewater angle? That is , at what angle of incidence will the reflected beam be fully polarized? (2) What angle of refraction corresponds to this angle of incidence?

98

Solution:

(1)ib ? tan

?1

n ? tan 1.57 ? 57.5
o o o

?1

o

( 2)ib ? r ? 90
o

r ? 90 ? 57 .5 ? 32 .5

99

a b b a

?h
ek-1 ek

?h

100


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