This presentation is the property of its rightful owner.
Sponsored Links
1 / 33

衍射光栅 PowerPoint PPT Presentation


  • 245 Views
  • Uploaded on
  • Presentation posted in: General

衍射光栅. 制作者: 赣南师范学院物理与电子信息学院 王形华. 第四章 衍射光栅. §O 序 一、光栅的定义 广义地说,具有周期性的空间结构或光学性能(如透射率、折射率)的衍射屏,统称为光栅。. 二、常见的光栅种类. 1 、在一块不透明的 障板上刻出一系列 等宽等间隔的平行狭 缝 —— 一维 透射 光栅。 (见图 0-1a ) 2 、在一块很平的铝面 上刻一系列等间隔的平行 槽纹 —— 反射光栅。

Download Presentation

衍射光栅

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


5742177


5742177

O


5742177

1

0-1a

2

0-1b


5742177

3

4


5742177

1

a

b

d=a+b


5742177

1-2


5742177

1

2NN

3N-1N-2

4


5 1 3

5 1-3


5742177

N

1

a=a0sin/

I=a2=a02(sin/)2

=asin/

2N


5742177

3NN

4N,N


5742177

5

L=dsin =2L / =2dsin/ =2 OCB1

2OCsin= OB1 =a

OC= a /2sin

OCBN

N =2N

A=OBN

=2OCsin(N)

=asin(N)/sin


5742177

I=a2 (sinN/sin)2

: A=a0(sin/) (sinN/sin)

I=a02(sin/)2 (sinN/sin)2

=asin/ =dsin/

(sin/)2 :

(sinN/sin)2 :


5742177


5742177

1

=k(k=0123)

sinN=sinNk=0 sin=0

sinN/sin=N

1 sin=k/ddsin=k


5742177

2I=N2I10 N2

3 /2 sin 1

k d /

d , k


5742177

2.

1N

sinN=0 sin0 sinN/ sin=0

=k+m/N

k=0123

m=123N-1


5742177

sin=(k+m/N) /d

(=dsin/)

N-1

N-2


5742177

2

sin

k=dsink/

sink=/d =k

sink=k/d k k/d

k+ (k+1/N) /d

/(Nd)(Note: m=1)


5742177

sink=k/d

sin(k+)=(k+1/N) /d

sin(k+)- sink(dsin/d)=k

=cosk

=/Nd cosk

k0 cosk1

/Nd


5742177

1

1-6


5742177

2d


5742177

1

UP=CU0Qeikrd

(0)

2d

U0xdN

12 3N P

P L1L2L3LN

L2=L1+L, L3=L+2 L, LN=L1+(N-1) L

L=dsin ( 1-7)


5742177

3. P

U= CU0xeikrdx

N ()

= C U0xjexp(ikrj)dxj

j=1()

rj=Lj-xjsin xj

U0xexp(ikrj)dx =eikLjU0xexp(-ikxjsin)dxj

(j) (j)

d/2

= eikLj U0xexp(-ikxjsin)dxj

- d/2


5742177

U()j

N d/2

U() = C ( eikLj) U0xexp(-ikrsin)dx

j=1- d/2

=N() u()

d/2

u() =C U0xexp(-ikrsin)dx

- d/2


5742177

N

N()= eikLj= eikL1 [1+eikL+ e2ikL++e(N-1)ikL]

j=1

N = dsin/ k L=2


5742177

U0x

3


5742177


5742177

[1+cos(2x/d)]

=kdsin/2=dsin/


5742177

=0

N()0

u()N()u()01, 101/4(1-9)


5742177

P17 45


  • Login