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Nhiu xa nh sng

1Hin tng nhiu x - Nguyn l Huygens-Fresnel1.1 Hin tng nhiu x nh sng1.2 Nguyn l Huygens-Fresnel1.3 Nhiu x Fresnel v nhiu x Fraunhofer

2 Nhiu x Fresnel2.1 Phng php i cu Fresnel.2.2 Nhiu x nh sng qua l trn v a trn nh

3 Nhiu x Fraunhofer3.1 Nhiu x qua mt khe hp3.2 Nhiu x qua mt l trn3.3 Nhiu x qua 2 khe3.4 Nhiu x qua nhiu khe3.5 Cch t nhiu x- my quang ph cch t3.6 Nhiu x tia X

TS. Hong Ch Hiu

1


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Bi ging Nhiu x nh sng TS. Hong Ch Hiu 2

1. Hin tng nhiu x-Nguyn l Huygens-Fresnel

Cac vi du v nhiu xa:

Nc chay qua mt khe p nho.

1.1 Hin tng nhiu xa anh sang.

Khi song anh sang truyn quamt l hep hoc mep man chn,anh sang khng truyn thng ma

bi lch i va c goi la hintng nhiu xa anh sang.


3/21


Mi im khng bi che khut cua mt mt song mt thi im cho trcc dung nhmt ngun phat song cu th cp vi cung tn s nh songscp. Bin cua trng quang hoc tai mt im bt ky bn ngoai la schng cht cua tt ca nhng song thcp nay khi tinh n bin vaphatng quan cua chung.

1.1 Nguyn ly Huygen-Fresnel.


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1.2 Nhiu xa Fresnel va nhiu xa Fraunhofer.

Nhiu x Fresnel (Nhiu x trng gn): Nu ngun sngv mn quan st t gn vt gy nhi x Fresnel. C thxem nhiu x Fresnel l nhiu x vi sng cu.

Nhiu x Fraunhofer (Nhiu x trng xa): Nu ngunsng v mn quan st kh xa vt gy nhiu x ta c nhiux Fraunhofer. Co th thy l chng no sng n v ikhi vt gy nhiu x c xem l sng phng th ta cnhiu x Fraunhofer.

Quy tc l: R> a2

/


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2. Nhiu xa Fresnel

2.1 Phng phap i cu FresnelLy B lm tm dng cc mt cubn knh ln lt b, b+/2, b+ ,b+ 3/2, b+j /2.

Cc mt cu ny ct mt sng thnh cc i cu. ng truynca cc sng th cp t cc ilin tip gi n B chnh nhau /2nn sng t hai i lin tip sngc pha nhau khi n B.

nB SSSSSSS +++= ...54321

y Sj vi j=1,2,3, l bin sng gi t i cu Fresnel th j n B.

j l din tch i th j, kj l tha s xin ca i th j.

jjj

jba

aKS

+

=

2

0

Nh vy, bin sng tng hp ti B l:

B

b

b+/2

b+b+3/2

A


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Din tich i thu j la hiu din tich 2 chom cu lin tip:

Nh vy khi j tng, h s xin Kj giam dn va do o bin Sj giam dn va tin n 0 khi BMj tip tuyn vi mt song .

Coi gn ung Sj bng trung binh cng cua hai s hang bn canh:

ba

abj

+=

nSSSS >>>> ...321

( )11

2

1+ += jjj SSS

2

2...

22222

1

5

4

33

2

11

nB

nB

SSS

SSSSSSSSS

+

++

++=

Du + khi n le va du khi n chn

Khi gia A va B khng co man chn: Sn 0,

vi I1 la cng sang do mt i

cu Fresnel u tin gi n4822

1

22

1 1ISS

IS

S bBB ====


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2.2 Nhiu x nh sng qua l trn v a trn

a. Nhiu xa qua l tron.

Gia s l tron m ra q i Fresnel u tin. Khi o bin dao ng song tng hp tai B la:

Nu kinh thc l nho, s i q khng ln lm (q


9/21

+ Nu s i q nguyn, chn th

Nu s i q khng nguyn th bng phng php th ta c th xc nhc bin sng tng hp OP v so snh n vi OI.


002

21

= BB ISS

Si roi tai B xp xi bng0 nn se thy ti tai B.

+ Nu khng co man chn:

+ Nu l cha ung 1 i:422

2

1 ISIS

S oo

B

BoB===

1

2

12

IS

ISS BBB ===


10/21


b. Nhiu xa qua mt l tron chn sang

BA

Gi s a mt p i u tin, bin

dao ng tng hp s l:

V n , nn Sn 0. Do

Nu l c knh thc nh, p nh th:

2

1 np

B

SSS

=

+

2

1+

=p

B

S

S

2

111

SSSSBp +

roi tai B khng khac nhiu so vi khi khng co man chn D va cho ta mt im sang,goi la im sang Poison.Khi kich thc ia chn ln, thi B se ti, vi: 001 == + BpB ISS


11/21


3. Nhiu xa Fraunhofer

3.1 Nhiu xa qua mt khe hep

Chia mt song gii han bi khe thanh Nphn bng nhau rng x=a/N.Xet song phng th cp theo goc gpnhau tai P. hiu quang trinh gi t hai i

lin tip la:

Song tai P la chng cht cua N song thcp gi t khe n P. H s xin la nhnhau, nn bin cac song th cp la nhnhau.

x

sin22

sinsin.

N

aL

N

axL

==

==

N

oN

EEEEE

EEEEErrrrr

++++=

=====

...

..

321

321


12/21


+ Ti O trn mn. Khi N sngth cp ti O l ng pha =0, sin=0 =0. Cng sng ti Ol cc i.

+ Ti phng 0, cc sng thcp lch pha nhau mt lngkhng i. Tng ca N vector lmt vector di Er


13/21


M

H

C

R

Khi N tin ti v hn, chui vector cc sng th cp s tin ti m

t cung trn tm C v bin sng tng hp l di dy trngcung :

sin22 ROHOME ===

2/=

sin

2. aN ==

Mt khac theo inh nghia:

22 maxmax

ER

R

E

R

OM===

2

2222

sin

sin

sinsinsin

22

=

=

==

a

a

IIEEI mmm

+ Tai O, =0; =0 ta co vn sang trung tmmax01

sinII == =

O


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+ Cc tiu cng sang xut hin khi sin=0 va 0 ng vi:

+ Khi m=1 ta c cc tiu u tin vi gc 1 tha mn

ma

mma

=

===

sin

,...3,2,1,sin

a

=1sin

Nu khe rt hp; a sin1=11=/2. iu ny cos ngha l cci chnh gia m v hn, nh

sng chiu khp mn. Khi a tngdn, cc i chnh hp li, xuthin cc cc i th xen k.

Cc ai th cp

+=2

1sin ma


15/21


3.2. Nhiu xa qua mt l tron

Cc tiu u tin:

phn ly theo tiu chun Rayleigh:

Cc ai cua anh thnht trungung vicc tiu au tin cuaanh th2 la gii han phn ly.


16/21


3.3. Nhiu xa qua 2 khe hp

2

sin

sinsin

=

a

a

II mP

=

2cos4

2 oP II

sin

22b==

Cng sng phn b theo hin tng giao thoa:

Do hin tng nhiu x trn khe c rng a nncng sng qua mi khe l:

Cc tiu nhiu x:,...3,2,1sin == m

am

Cc i giao thoa:

,...3,2,1;sin == mb

m


17/21


3.4. Nhiu xa qua nhiu khe hp

( )

[ ]

=

=

=

)1(

2

1

.

................

.

Nti

N

ti

ti

eES

eES

eES

( )[ ]

=

=+++== i

iNtiNiiti

N

i

iPe

eeEeeeESS

1

1...1

1

1

Bin ca sng pht ra t mi khe theo phng l:

lch pha ca 2 khe lin tip l:

sin2

b=

sin,

sin aEE m ==

Cc khe c th xem l mt ngun sng gi ti P nh sau


18/21


==

2sin

2sin

2 2

2

2*

N

ESS

I PPP

( )22

sin

sinsin

=

NII oP

Gia 2 cc i chnh lin tip s c (N-1)cc tiu, c (N-2) cc i ph

Vi2

20

0

EI =


19/21


3.5 Cach t nhiu xa

Cc ai nhiu xa

d: chu ky cua cach t

md =sin


20/21


tn sc gc ca cch t:

bn rng gc ca mt vch nhiu x:

Khong cch gc gia hai vn sng cng bc:

Nng sut phn gii ca cch t :

Vng ph t do ti bc m:

m

m

b

m

d

dD

cos

==

m

mNb

cos=

m

mdmddDd

cos==

mNR =

=

min

)(

( )m

td

=


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3.6 Nhiu xa tia X

Cng thc Bragg:

2dsin = m

d = Khong cch gia cc mt phng nguyn t = Gc gia chm tia X vi mt phng nguyn tm = Bc giao thoa= bc sng

Mt phng trn

Mt phng di

Chm tia ti Chm phn x

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