# 第12章 波动光学

CHAPTER 12
Wave Optics

By Dr. G Y Chen

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§12-1 Light Source and the Coherence of Light §12-2 Optical Distance and Optical Path 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
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§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
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? ? ①在任意给定点上，E 和 H 同时存在，具有相同的相位和

(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
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(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).

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.
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(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.
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§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 ) ?
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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
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§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?
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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
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Discussions:

bright fringe

(1)When δ=kλ,I=Imax

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

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;
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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。
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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

(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’

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§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.

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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

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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

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

e、n L

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(1) Wedge Interference Optical path difference:

? ? 2ne ?
Therefore:

?
2

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

? 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....

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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？

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§12-4 Michelson’s Interferometer

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§12-4 Michelson’s Interferometer

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： ?

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).
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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

（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

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§12-6 Single-Slit Fraunhofer(夫琅和费)diffraction

1、Concept: The Fraunhofer diffraction produced by a single slit. Incidence optics: s parallel light Screen: infinite distance

ideal ：

E

l ??

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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.

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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
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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。

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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。

k ? 1,2? dark fringes

a sin ? ? ?(2k ? 1)

?
2

k ? 1,2?bright fringes

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（1）中央明纹宽度为相邻两暗纹的间距

（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。

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58

L

?

I/I0
1.22 0

?
d sin?

D

f

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Ariy disk(爱里斑)：

? 0 ? sin ? 0 ? 1.22

?
D

? ? ? 1.22

?

D
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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?
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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

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(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
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d S1

?R

?1
0 I

S2

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§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.

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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,?
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1、主极大条纹

k?

(a ? b) sin ?

?

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（2）可见条纹的最大级次
?k ? (a ? b) sin ?

?
k? sin ? ? ? d

? kmax ?

( a ? b)

?

（3）Since

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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.
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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:

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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

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

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

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3、暗纹条件

(a ? b) sin ?

?? ?

2? (a ? b) sin ?

?
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E ? ? Ei ? 0
i ?1

N ? ?? ? ?m ? 2? N ? (a ? b) sin ? ? ?m ? 2? m为不为N 的整数

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?角越小，故紫色在内侧，红色在外侧。

d

76

I
I

0
0

1 2 sin ( sin ??? ) a

I
I I 0 I 0

4
4

8
8

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

?8 ?4

0
0

4
4

8
8

sin?

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( a ? b ) ( s in ? ? s in ? ) ? ? k ?

k max ?

a?b

?

(sin ? ? 1)

k max ?

a?b

?

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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?

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

§12-8 Polarized Light

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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

E

z

x 振

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diagrammatize：图示法

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2、 Partially（部分） Polarized Light ：

84

Partially Polarized Light

decompose

diagrammatize：

85

86

The polarized direction

Nature Light

The polarized direction

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The polarized direction

I0

I

?

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

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§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

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93

n1 sin i ? n2 sin r

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2.Brewster’s Law(布儒斯特定律)

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
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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） 。

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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?

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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

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a b b a

?h
ek-1 ek

?h

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2010秋第12章波动光学1

12第十二章 波动光学