THÍ NGHIỆM CÁCH TỬ NHIỄU XẠ

TRNG I HC S PHM THNH PH H CH MINH KHOA VT L

TI:

TH NGHIM CCH T NHIU X

GING VIN HNG DN: TRN VN TN (L thuyt) NGUYN TH HO (Bi tp) SINH VIN THC HIN: TRN TH T ANH

NGUYN TH THY DNGLM TH M HNH NGUYN TH DIM HNG

TP.H CH MINH THNG 5-2012

Th nghim cch t nhiu x

MC LC. C S L THUYT .................................................................................................................. 2 .1.Nhiu x l g ? ...................................................................................................................... 2 .2.Cch t l g ? ........................................................................................................................ 2 .3.M t cch t: ........................................................................................................................ 2 .4. S phn b iu kin sng .................................................................................................... 2 .4.1. Biu thc bin ........................................................................................................... 2 .4.2. iu kin cc i v cc tiu ........................................................................................ 3 .5. Gc lch cc tiu .................................................................................................................. 3 . TH NGHIM CCH T NHIU X ..................................................................................... 4 .1. Mc ch th nghim ........................................................................................................... 4 .2. M t t nghim ................................................................................................................... 4 .3. Cch tin hnh ..................................................................................................................... 4 .2.1. iu chnh ng chun trc v knh ngm .................................................................... 4 .3.2. Xc nh n vch ca cch t......................................................................................... 4 .3.3. Dng nh nhiu x bc 1 v bc 2 xc nh bc sng ca mt vi n sc. ............. 6 . KT QU TH NGHIM ........................................................................................................ 7 .1 Xc nh s vch n trn mi mm ca cch t ..................................................................... 7 .1.1 Dng quang ph bc 1: ................................................................................................ 7 .1.2. Dng quang ph bc 2: .............................................................................................. 8 .2. Xc nh bc sng ca mt s vch (mu lc m v chm m) .................................. 9 .2.1. Vch mu chm m:................................................................................................. 9 .2.2. Vch mu xanh l cy: .............................................................................................. 9 V. KT LUN ............................................................................................................................ 10

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. C S L THUYT .1.Nhiu x l g ?Nhiu x l hin tng tia sng b lch khi phng truyn thng khi i gn cc chng ngi vt c kch thc nh (cc chng ngi vt c th l cc khe hp, l trn, c kch thc c bc song nh sng chiu ti.

.2.Cch t l g ?Cch t nhiu x l thit b quang hc c s dng tm hiu cc bc sng khc nhau hoc mu sc trong mt chm nh sng. Cch t truyn qua: trn tm thy tinh c nhng rnh khng trong su t, nh sng truyn qua phn trong sut v gy ra nhiu x (ch nghin cu cho nh sang nhn thy). Cch t phn x: to bi tm kim loi phng, nhn bng v c h s phn x cao, trn cc mt c vch cc rnh nh cch u nhau (dung nghin cu tia t ngoi).

.3.M t cch t:- L mt h thng gm n khe ging ht nhau c b rng l a cch u nhau vi khong cch gia 2 khe lin tip l l. l: chu k cch t ; = n: s cch t trn mt n v di.

Cu to ca cch t rt l tinh vi, trn b rng 1 mm c n hng trm, hng ngn khe, C nhiu loi cch t c cu to ring bit nhng u da trn nguyn tc: mt song ca chm tia sang ti c chia thnh nhng phn u n, ln lt truyn qua v b ngn li cch t.

.4. S phn b iu kin sng.4.1. Biu thc bin

= A0

2


.4.2. iu kin cc i v cc tiua.V tr cc tiu nhiu x:

b.V tr cc i chnh gjao thoa:

(k=c.V tr cc i ph giao thoa:

(k=1,2,)d.V tr cc cc tiu giao thoa:

(k=

.5. Gc lch cc tiuGc lch cc tiu v cch xc nh gc lch cc tiu Ti O vn sng trung tm mu trng hai bn vn trung tm l cc nh ca khe sng, l cc nh nhiu x bc 1, 2, . . Gi io l gc ti v i l gc nhiu x v l di sng ang dng (tnh bng mm) ta c cng thc xc nh cc i giao thoa th k (vn giao thoa bi n chm tia nhiu x) sin i - sin io = k n (1) Nu ta quay da theo mt chiu nht nh n thay i gc ti i, nh nhiu x di ch n mt lc no th dng li v chy tr lui. V tr dng li l v tr lch cc tiu. Gi Dm l gc lch cc tiu ta c : i = io=Dm D v 2.sin m = k . . n 2 2

(2)

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. TH NGHIM CCH T NHIU X .1. Mc ch th nghim o s vch n trn mi mm ca cch t bng cch dng nh sng bit trc bc sng nh sng. o mt vi bc sng nh sng cha bit.

.2. M t t nghim-Ngun sng: ngun halogen - Gic k +Knh ngm +ng chun trc +Bn phng +a chia v du xch -Cch t (cha bit n)

.3. Cch tin hnh.2.1. iu chnh ng chun trc v knh ngm- t ng chun trc quay ra ngoi, qua knh ngm ta nhn ra xa v cc xc nh v tr ngm sao cho nh ngm c r nht. - Sau , t ng chun trc thng hng vi ng truyn tia sng ca n Neon thy ngn.

.3.2. Xc nh n vch ca cch tDng nh nhiu x bc 1 v nh nhiu x bc 2 xc nh n + Dng nh nhiu x bc 1(k=1) Dng knh ngm tm quang ph bc 1 bn phi - t cch t trn a xoay trn gic k (hi xoay v bn tri). Mt nhn qua ng ngm, ta quan st thy mt vch sng trng, xoay ng ngm t t sang bn phi, quan st thy quang ph bc mt, khi thy vch mu vng kp, ta iu chnh cho tm ch X ca knh ngm trng vi vch vng 1 (vch gn vch mu ). Sau xoay tht nh cch t theo ngc chiu kim ng h sao cho vch mu vng di chuyn sang bn tri. Ch quan st khi vch mu vng dng li v i chiu di chuyn ngc li. Ti thi im dng li ca vch vng, ta t t quay ng ngm tm ng ngm trung vi vch mu vng lc ny. - Tip theo, ta c ch s gc lch ca knh ngm trn vng trn ca gic k. Dng knh ngm tm quang ph bc 1 bn tri - Thc hin tng t nh tm quang ph bc 1 bn phi, nhng thc hin cc bc theo chiu ngc li. Ta tm c gc lch ca vch vng 1.

4

Th nghim cch t nhiu x Ta c:

= 2D i = io = T (1) v (2) => 2 sin ( Vi k = 1 => v 2.sin

(1) = k..n (2)

' ) = k. . n (3) 4

=

(4)

+ Dng nh nhiu x bc 2 (k=2) Dng knh ngm tm quang ph bc 2 bn phi - t cch t trn a xoay trn gic k (hi xoay v bn tri). Mt nhn qua ng ngm, ta quan st thy mt vch sng trng, xoay ng ngm t t sang bn phi, quan st thy quang ph bc mt, xoay tip ng ngm ta s quan st thy quang ph bc 2, khi thy vch mu vng kp, ta iu chnh cho tm ca knh ngm trng vi vch vng 1 (vch gn vch mu ). Sau xoay tht nh cch t theo ngc chiu kim ng h sao cho vch mu vng di chuyn sang bn tri. Ch quan st khi vch mu vng dng li v i chiu di chuyn ngc li. Ti thi im dng li ca vch vng, ta t t quay ng ngm tm ng ngm trng vi vch mu vng lc ny. - Tip theo, ta c ch s gc lch ca knh ngm trn vng trn ca gic k. Dng knh ngm tm quang ph bc 2 bn tri - Thc hin tng t nh tm quang ph bc 2 bn phi, nhng thc hin cc bc theo chiu ngc li. Ta tm c gc lch ca vch vng 1. T cng thc (3): vi k = 2 ta c:

=

Sai s trn n 2 ' T cng thc n sin( ) 4 dn d - - Suy ra: = + cotan ( ) d( ) n 4 4 n - ( ') Suy ra: = + cotan ( ). n 4 4 C th ly = 3 A0 Sai s ( - ) gm sai s lc ngm v sai s lc c. i vi nh bc 1 ta ly ( - ) = 2 pht5


i vi nh bc 2 ta ly ( - ) = 4 pht (v nh m hn)

.3.3. Dng nh nhiu x bc 1 v bc 2 xc nh bc sng ca mt vi n sc.Trong TN ny ta o bc sng ca vch lc m v chm m

2 ' sin( ) n.k 4 Dng nh nhiu x bc 1 xc nh bc sng ca vch chm m: (k = 1) 2 ' sin( ) n1 4 Dng nh nhiu x bc 2 xc nh bc sng ca vch lc m: (k = 2) 1 ' sin( ) n2 4 Mun xc nh bc sng nh sng cn tm, ta o gi tr v ng vi ng sng . Sai s trn T (3) ta c cng thc tnh bc sng:

2 ' T cng thc sin( ) n 4 - - d dn Suy ra: = + cotan ( ) d( ) 4 4 n - n ( ') Suy ra: = + cotan ( ). 4 n 4Sai s ( - ) gm sai s lc ngm v sai s lc c. i vi nh bc 1 ta ly ( - ) = 2 pht i vi nh bc 2 ta ly ( - ) = 4 pht (v nh m hn)

Dng nh nhiu x bc 1 xc nh bc sng ca vch chm m: (k=1)+ Dng knh ngm tm quang ph bc 1 bn phi - t cch t trn a xoay trn gic k (hi xoay v bn tri). Mt nhn qua ng ngm, ta quan st thy mt vch sng trng, xoay ng ngm t t sang bn phi, quan st thy quang ph bc mt, khi thy vch mu chm m, ta iu chnh cho tm ca knh ngm trng vi vch chm m. Sau xoay tht nh cch t theo ngc chiu kim ng h sao cho vch mu chm m di chuyn sang bn tri. Ch quan st khi vch mu chm m dng li v i chiu di chuyn ngc li. Ti thi im dng li ca vch chm m, ta t t quay ng ngm tm ng ngm trung vi vch mu chm m lc ny. - Tip theo, ta c ch s gc lch ca knh ngm trn vng trn ca gic k. + Dng knh ngm tm quang ph bc 1 bn tri 6

Th nghim cch t nhiu x - Thc hin tng t nh tm quang ph bc 1 bn phi, nhng thc hin cc bc theo chiu ngc li. Ta tm c gc lch ca vch chm m.

=Dng nh nhiu x bc 2 xc nh bc sng ca vch lc m: (k=2)+ Dng knh ngm tm quang ph bc 2 bn phi - t cch t trn a xoay trn gic k (hi xoay v bn tri). Mt nhn qua ng ngm, ta quan st thy mt vch sng trng, xoay ng ngm t t sang bn phi, quan st thy quang ph bc mt, xoay tip cho n khi thy quan ph bc 2, khi thy vch mu lc m, ta iu chnh cho tm ca knh ngm trng vi vch lc m. Sau xoay tht nh cch t theo ngc chiu kim ng h sao cho vch mu lc m di chuyn sang bn tri. Ch quan st khi vch mu lc m dng li v i chiu di chuyn ngc li. Ti thi im dng li ca vch lc m, ta t t quay ng ngm tm ng ngm trung vi vch mu lc m lc ny. - Tip theo, ta c ch s gc lch ca knh ngm trn vng trn ca gic k. + Dng knh ngm tm quang ph bc 2 bn tri - Thc hin tng t nh tm quang ph bc 2 bn phi, nhng thc hin cc bc theo chiu ngc li. Ta tm c gc lch ca vch lc m.

= . KT QU TH NGHIM .1 Xc nh s vch n trn mi mm ca cch t.1.1 Dng quang ph bc 1: Chn vng = 5790( A0)Ln o 1 2 3 Trung bnh Tnh n1 T cng thc n1

255370 255380 255360

'275330 275310 275300

n1

2

sin(

'4

)

= 299 (vch/mm)

= 255370

' =275313

2

sin(

'4

)7

Th nghim cch t nhiu x dn d - - = + cotan ( 4 ) d( 4 ) n n - ( ') n = + cotan ( 4 ) . 4

Suy ra: Suy ra:

C th ly = 3 A0 Sai s ( - ) gm sai s lc ngm v sai s lc c. i vi nh bc 1 ta ly ( - ) = 2 pht i vi nh bc 2 ta ly ( - ) = 4 pht (v nh m hn) - ( ') T ta c: n1 = . ( ( ) ) 4 Vy:

=

= 299

0.71 vch/mm

.1.2. Dng quang ph bc 2: Chn vng = 5790 A0

Ln o 1 2 3 Trung bnh Tnh

245330 245350 245310

'285330 285310 285320

n2

1

sin(

'4

)

= 300 ( vch/mm )

= 245330

' =285320

(

( ') ) 4

Vy:

=

300 0.7 vch/mm

8


.2. Xc nh bc sng ca mt s vch (mu lc m v chm m).2.1. Vch mu chm m: Dng quang ph bc 1: Ln o 1 25850 2 25830 3 25840 Trung = 25840 bnh

'27330 27320 27350

2 ' sin( ) 4 n1

= 4370,2 ( A0)

' = 27333

Tnh (lu : i vi nh bc 1 ta ly ( - ) = 2 pht Vy: 4370.2 69 A0

.2.2. Vch mu xanh l cy: Dng quang ph bc 2: Ln o 1 24840 2 24860 3 24850 Trung bnh = 24850

'28340 28350 28370

1 ' sin( ) 4 n2

= 5203,1 ( A0)

' =28353

Tnh (lu : i vi nh bc 2 ta ly ( - ) = 4 pht Vy: 5203,1 55,78 A0

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V. KT LUN Trong giao thoa c nhiu x. Thng qua th nghim ta c th xc nh c s vch n trn tng mm ca mt cch t, ng thi cng xc nh c bc sng ca mt s vch sng mu. Cch t dng trong th nghim c n=300 vch/ mm Xc nh c bc sng ca mt s nh sng nhn thy : Vch mu chm tm c bc sng l: 4370.2 69 A0 Vch mu xanh l cy c bc sng l: 5203,1 55,78 A

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