Vat Ly Dai Cuong A2 - HVBCVT

HC VIN CNG NGH BU CHNH VIN THNG

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SCH BI GING

VT L I CNG A2(Dng cho sinh vin h o to i hc t xa) Lu hnh ni b

H NI - 2005

HC VIN CNG NGH BU CHNH VIN THNG

BI GING

VT L I CNG A2

Bin son :

TS. V TH THANH H ThS. HONG TH LAN HNG TS. L TH MINH THANH

Hiu nh:

Li ni u

LI NI UTp VT L I CNG (A2) ny l tp hai ca b sch hng dn hc tp mn Vt l i cng cho sinh vin h o to i hc t xa ca Hc vin Cng ngh Bu chnh Vin thng, c bin son theo chng trnh ci cch gio dc do B Gio dc v o to thng qua (1990). B sch gm hai tp: Tp I: VT L I CNG (A1) bao gm cc phn C, NHIT, IN, T do Ts. V Vn Nhn, Ts. V inh Chu v Ks. Bi Xun Hi bin son. Tp II: VT L I CNG (A2) bao gm cc phn QUANG HC, THUYT TNG I HP, C HC LNG T V VT L NGUYN T do Ts. V Th Thanh H v ThS. Hong Th Lan Hng bin son. Tp sch Vt l i cng A2 gm 8 chng: - Chng I: Dao ng in t - Chng II: Giao thoa nh sng - Chng III: Nhiu x nh sng - Chng IV: Phn cc nh sng - Chng V: Thuyt tng i hp - Chng VI: Quang hc lng t - Chng VII: C hc lng t - Chng VIII: Vt l nguyn t. Trong mi chng u c: 1. Mc ch, yu cu gip sinh vin nm c trng tm ca chng. 2. Tm tt ni dung gip sinh vin nm bt c vn t ra, hng gii quyt v nhng kt qu chnh cn nm vng. 3. Cu hi l thuyt gip sinh vin t kim tra phn c v hiu ca mnh. 4. Bi tp gip sinh vin t kim tra kh nng vn dng kin thc l thuyt gii quyt nhng bi ton c th. Phn cng bin son tp VT L I CNG (A2) nh sau: V Th Thanh H bin son l thuyt cc chng II, III, IV, V, VI, VII, VIII. Hong Th Lan Hng bin son l thuyt chng I v bi tp ca tt c cc chng. 1

3

Li ni u Tp VT L I CNG (A2) ny mi in ln u, nn khng trnh khi nhng thiu st. Chng ti xin chn thnh cm n s ng gp qu bu ca bn c cho quyn sch ny. H Ni, ngy 1 thng 11 nm 2005 NHM TC GI

4

Chng 1: Dao ng in t

CHNG I: DAO NG IN TDao ng in t l s bin thin tun hon theo thi gian ca cc i lng in v t, c th nh in tch q trn cc bn t in, cng dng in i trong mt mch in xoay chiu, hiu in th gia hai u mt cun dy hay s bin thin tun hon ca in trng, t trng trong khng gian v.v... Tu theo cu to ca mch in, dao ng in t trong mch chia ra: dao ng in t iu ho, dao ng in t tt dn v dao ng in t cng bc. I. MC CH - YU CU 1. Nm c dao ng in t iu ho, dao dng in t tt dn, dao ng in t cng bc, hin tng cng hng. 2. Nm c phng php tng hp hai dao ng iu ho cng phng v cng tn s, hai dao ng iu ho cng tn s v c phng vung gc. II. NI DUNG:1. DAO NG IN T IU HO 1. Mch dao ng in t LC

Xt mt mch in gm mt t in c in dung C, mt cun dy c h s t cm L. B qua in tr trong mch. Trc ht, t in C c b ngun tch in n in tch Q0, hiu in th U0. Sau , ta b b ngun i v ng kho ca mch dao ng. Trong mch c bin thin tun hon theo thi gian ca cng dng in i, in tch q trn bn t in, hiu in th gia hai bn t, nng lng in trng ca t in, nng lng t trng ca ng dy ... Cc dao ng in t ny c dng hnh sin vi tn s 0 v bin dao ng khng i. Do , cc dao ng ny c gi l cc dao ng in t iu ho. Mt khc trong mch ch c mt cc yu t ring ca mch nh t in C v cun cm L, nn cc dao ng in t ny c gi l cc dao ng in t ring.

Hnh 1-1. Mch dao ng in t ring

5

Chng 1: Dao ng in t

Ta xt chi tit hn qu trnh dao ng ca mch trong mt chu k T. Ti thi im t = 0, in tch ca t l Q 0 , hiu in th gia hai bn l U 0 = Q 0 / C , nng lng in trng ca t in c gi tr cc i bng: E e(max ) =2 Q0 2C

(1-1)

Cho t phng in qua cun cm L. Dng in do t phng ra tng t ngt t khng, dng in bin i ny lm cho t thng gi qua cun cm L tng dn. Trong cun cm L c mt dng in t cm ngc chiu vi dng in do t C phng ra, nn dng in tng hp trong mch tng dn, in tch trn hai bn t gim dn. Lc ny nng lng in trng ca t in Ee= q 2 / 2C gim dn, cn nng lng t trng trong lng ng dy Em = Li 2 / 2 tng dn. Nh vy, c s chuyn ho dn t nng lng in trng sang nng lng t trng.

Hnh 1-2. Qu trnh to thnh dao ng in t ring Khi t C phng ht in tch, nng lng in trng Ee = 0, dng in trong mch t gi tr cc i I0, nng lng t trng trong ng dy t gi tr cc i2 E m(max ) = LI 0 / 2 , l thi im t = T/4. Sau dng in do t phng ra bt u

gim v trong cun dy li xut hin mt dng in t cm cng chiu vi dng in do t phng ra . V vy dng in trong mch gim dn t gi tr I0 v khng, qu trnh ny xy ra trong khong t t = T/4 n t = T/2. Trong qu trnh bin i ny cun L ng vai tr ca ngun np in cho t C nhng theo chiu ngc li, in tch ca t li tng dn t gi tr khng n gi tr cc i Q0. V mt nng lng th nng lng in trng tng dn, cn nng lng t trng gim dn. Nh vy c s chuyn ho t nng lng t trng thnh nng lng in trng, giai on ny kt thc ti thi im t = T/2, lc ny cun cm gii phng ht nng lng v in tch trn hai bn t li t gi tr cc i Q0 nhng i du hai bn, nng lng in trng li t gi tr cc2 i E e(max ) = Q 0 / 2C . Ti y, kt thc qu trnh dao ng trong mt na chu k u.

T C phng in vo cun cm theo chiu ngc vi na chu k u, cun cm li6

Chng 1: Dao ng in t

c tch nng lng ri li gii phng nng lng, t C li c tch in v n cui chu k (t = T) t C c tch in vi du in tch trn cc bn nh ti thi im ban u, mch dao ng in t tr li trng thi dao ng ban u. Mt dao ng in t ton phn c hon thnh. Di y ta thit lp phng trnh m t dao ng in t trn.2. Phng trnh dao ng in t iu ho

V khng c s mt mt nng lng trong mch, nn nng lng in t ca mch khng i: E e + E m = E = const Thay E e =q2 Li 2 v E m = vo (1-2), ta c: 2C 2 q 2 Li 2 + = const 2C 2

(1-2)

(1-3)

Ly o hm c hai v ca (1-3) theo thi gian ri thay dq / dt = i , ta thu c:q Ldi + =0 C dt

(1-4)

Ly o hm c hai v ca (1-4) theo thi gian ri thay dq/dt =i, ta c:d 2i dt 2 + 1 i=0 LC

(1-5)

t

1 2 = 0 , ta c: LCd 2i dt2 2 + 0 i = 0

(1-6)

l phng trnh vi phn cp hai thun nht c h s khng i. Nghim tng qut ca (1-6) c dng: i = I 0 cos(0 t + ) (1-7) trong I0 l bin ca cng dng in, l pha ban u ca dao ng, 0 l tn s gc ring ca dao ng:0 = 1 LC

(1-8)

7

Chng 1: Dao ng in t

T tm c chu k dao ng ring T0 ca dao ng in t iu ho:T0 = 2 = 2 LC 0 (1-9)

Cui cng ta nhn xt rng in tch ca t in, hiu in th gia hai bn t. cng bin thin vi thi gian theo nhng phng trnh c dng tng t nh (1-7).

Hnh 1-3. ng biu din dao ng iu ho

2. DAO NG IN T TT DN 1. Mch dao ng in t RLC

Trong mch dao ng by gi c thm mt in tr R tng trng cho in tr ca ton mch (hnh 1-4). Ta cng tin hnh np in cho t C, sau cho t in phng in qua in tr R v ng dy L. Tng t nh trnh by bi dao ng in t iu ho, y cng xut hin cc qu trnh chuyn ho gia nng lng in trng ca t in v nng lng t trng ca ng dy. Nhng do c s to nhit trn in tr R, nn cc dao ng ca cc i lng nh i, q, u,... khng Hnh 1-4. Mch dao ng in t tt dn cn dng hnh sin na, cc bin ca chng khng cn l cc i lng khng i nh trong trng hp dao ng in t iu ho, m gim dn theo thi gian. Do , loi dao ng ny c gi l dao ng in t tt dn. Mch dao ng RLC trn c gi l mch dao ng in t tt dn.2. Phng trnh dao ng in t tt dn

Do trong mch c in tr R, nn trong thi gian dt phn nng lng to nhit trn in tr Ri2dt bng gim nng lng in t -dE ca mch. Theo nh lut bo ton v chuyn ho nng lng, ta c:

dE = Ri 2 dtq 2 Li 2 + vo (1-10), ta c: Thay E = 2C 2

(1-10)

8

Chng 1: Dao ng in t

q 2 Li 2 = Ri 2 dt d + 2C 2

(1-11)

Chia c hai v ca phng trnh (1-11) cho dt, sau ly o hm theo thi gian v thay dq/dt = i, ta thu c:q di + L = Ri C dt

(1-12)

Ly o hm c hai v ca (1-12) theo thi gian v thay dq/dt = i, ta thu c:d 2i dt 2 + R di 1 + i=0 L dt LC

(1-13)

t

R 1 2 = 2 , = 0 , ta thu c phng trnh: L LCd 2i2

(1-14) dt l phng trnh vi phn cp hai thun nht c h s khng i. Vi iu kin h s tt nh sao cho 0 > hay (1-14) c dng:1 R > LC 2L 2

+ 2

di 2 + 0 i = 0 dt

th nghim tng qut ca phng trnh

i = I 0 e t cos(t + )

(1-15)

trong I0, l hng s tch phn ph thuc vo iu kin ban u, cn l tn s gc ca dao ng in t tt dn v c gi tr: = 1 R LC 2L 2 = 2

< 0

(1-16)

Chu k ca dao ng in t tt dn: T= 2 1 R LC 2L 2

=

22 0

2

(1-17)

Nh vy, chu k dao ng tt dn ln hn chu k dao ng ring trong mch. i lng I 0 e t l bin ca dao ng tt dn. N gim dn vi thi gian theo qui lut hm m. Tnh cht tt dn ca dao ng in t c c trng bng mt i lng gi l lng gim lga, k hiu bng ch : lng gim lga c gi tr bng lga t nhin ca t s gia hai tr s lin tip ca bin dao ng cch nhau mt khong thi gian bng mt chu k dao ng T. Theo nh ngha ta c:

9

Chng 1: Dao ng in t

= ln

I 0 e (t +T )

I 0 e t

= T

(1-18)

trong = R / 2L , r rng l nu R cng

ln th cng ln v dao ng tt cng nhanh. iu ph hp vi thc t. Ch : trong mch dao ng RLC ghp ni tip, ta ch c hin tng dao ng in t khi:1 L R > hay R < 2 LC 2L C2

Tr s R 0 = 2

L c gi l in tr ti C

Hnh 1-5. ng biu din dao ng in t tt dn

hn ca mch. Nu R R0 trong mch khng c dao ng.

3. DAO NG IN T CNG BC 1.Hin tng:

duy tr dao ng in t trong mch dao ng RLC, ngi ta phi cung cp nng lng cho mch in b li phn nng lng b tn hao trn in tr R. Mun vy, cn mc thm vo mch mt ngun in xoay chiu c sut in ng bin thin tun hon theo thi gian vi tn s gc v bin E0: E= E0sint

Lc u dao ng trong mch l chng cht ca hai dao ng: dao ng tt dn vi tn s gc v dao ng cng bc vi tn s gc . Giai on qu ny xy ra rt ngn, sau dao ng tt dn khng cn na v trong mch ch cn dao ng in t khng tt c tn s gc bng tn s gc ca ngun in. l dao ng in t cng bc.2. Phng trnh dao ng in t cng bc

Hnh 1-6. Mch dao ng in t cng bc

Trong thi gian dt, ngun in cung cp cho mch mt nng lng bng Eidt. Phn nng lng ny dng b p vo phn nng lng to nhit Joule - Lenx v

10

Chng 1: Dao ng in t

tng nng lng in t trong mch. Theo nh lut bo ton v chuyn ho nng lng, ta c :dE + Ri 2dt = Eidt

(1-19) (1-20)

q 2 Li2 + Ri 2dt = Eidt d + 2C 2

Thc hin php ly vi phn v thay E= E0sint ta c: q di + Ri + = E0 sin t dt C Ly o hm hai v theo thi gian ca (1-21), thay dq/dt = i, ta c: LL d 2i dt 2 +R di i + = E0 cos t dt C

(1-21)

(1-22)

t

1 R 2 = 2 , = 0 , ta thu c phng trnh: L LCd 2i dt 2+ 2

E di 2 + 0 i = 0 cos t dt L

(1-23)

Phng trnh vi phn (1-23) c nghim l tng ca hai nghim: - Nghim tng qut ca phng trnh thun nht. chnh l nghim ca phng trnh dao ng in t tt dn. - Nghim ring ca phng trnh khng thun nht. Nghim ny biu din mt dao ng in t khng tt do tc dng ca ngun in. Nghim ny c dng: i = I 0 cos(t + ) (1-24) trong l tn s gc ca ngun in kch thch, I0 l bin , l pha ban u ca dao ng, c xc nh bng:

I0 =

E01 R 2 + L C 2

, cot g =

L R

1 C

1 t Z = R 2 + L v gi l tng tr C ca mch dao ng, Z L = L v Z C = 1 ln C lt l cm khng v dung khng ca mch dao ng. Hnh 1-7. ng biu din dao ng in t cng bc

2

11

Chng 1: Dao ng in t 3. Hin tng cng hng

Cng thc trn chng t bin I0 ca dng in cng bc ph thuc vo gi tr tn s gc ca ngun xoay chiu kch thch. c bit vi mt in tr R nht nh, bin I0 t gi tr cc i khi tn s gc c gi tr sao cho tng tr Z ca mch dao ng cc tiu, gi tr ca phi tho mn iu kin:

L

1 = 0 hay = C

1 LC

(1-25)

ta thy gi tr ny ca ng bng tn s gc ca mch dao ng ring:

ch = 0Hin tng bin dng in ca mch dao ng in t cng bc t gi tr cc i c gi l hin tng cng hng in. Vy hin tng cng hng in xy ra khi tn s gc ca ngun xoay chiu kch thch c gi tr bng tn s gc ring ca mch dao ng. Gi tr ch ca ngun xoay chiu kch thch c gi l tn s cng hng. ng biu din (1-8) cho ta thy r s bin thin ca bin dng in I0 ca mch dao ng cng

(1-26)

Hnh1-8. ng biu din cng hng in

bc theo tn s gc ca ngun xoay chiu kch thch. Trong thc t, mun xy ra cng hng in, ta dng hai phng php sau: - Hoc thay i tn s gc ca ngun kch thch sao cho n bng tn s gc ring 0 ca mch dao ng. - Hoc thay i h s t cm L v in dung C ca mch dao ng sao cho tn s gc ring 0 ng bng tn s gc ca ngun kch thch. Hin tng cng hng in c ng dng rt rng ri trong k thut v tuyn in, th d trong vic thu sng in t ( mch chn sng).4. S TNG HP DAO NG 1.Tng hp hai dao ng iu ho cng phng v cng tn s

Gi s c mt cht im tham gia ng thi hai dao ng iu ho cng phng v cng tn s: x1 = A1 cos(0 t + 1 ) x 2 = A 2 cos(0 t + 2 )12

(1-27) (1-28)

Chng 1: Dao ng in t

Hai dao ng ny cng phng Ox v cng tn s gc 0, nhng khc bin v pha ban u. Dao ng tng hp ca cht im bng tng ca hai dao ng thnh phnx = x1 + x2 = A cos(0t + )

(1-29)

C th tm dng ca x bng phng php cng lng gic. Nhng thun tin, ta dng phng php gin Fresnel. r r V hai vc t OM1, OM 2 cng t ti im O, c ln bng bin A1, A2 ca hai dao ng . thi im t = 0, chng hp vi trc Ox cc gc 1 v 2 l pha ban u. r r Khi tng hp ca OM1, OM 2 l mt vc t r r r OM = OM1 + OM 2 (1-30) r vc t OM trng vi ng cho ca hnh bnh hnh OM1MM2, c ln bng A v hp vi trc Ox mt gc v c xc nh bi h thc:2 A = A1 + A 2 + 2A1A 2 cos(2 1 ) , tg = 2

A1 sin 1 + A 2 sin 2 A1 cos 1 + A 2 cos 2

(1.31)

Hnh 1-9. Tng hp hai dao ng iu ho cng phng, cng tn s. r r Hai vc t OM1 v OM 2 quay xung quanh im O theo chiu dng vi cng vn tc gc khng i bng tn s gc 0 . thi im t, hai vc t ny s hp vi trc Ox cc gc (0t + 1) v (0t + 2) ng bng pha dao ng x1 v x2. Hnh chiu trn r r phng Ox ca hai vc t OM1 v OM 2 c gi tr bng: r hc ox OM1 = A1 cos(0 t + 1 ) = x1 (1-32) r hc ox OM 2 = A 2 cos(0 t + 2 ) = x 2 (1-33)

13

Chng 1: Dao ng in t

r r V hai vc t OM1 v OM 2 quay theo chiu dng vi cng vn tc gc 0 , nn hnh

bnh hnh OM1MM2 gi nguyn dng khi n quay quanh im O. Do , thi im t, r vc t tng hp OM vn c ln bng A v hp vi trc Ox mt gc (0t + ). Hnh r chiu trn phng Ox ca vc t tng hp OM c tr s bng: r hc ox OM = A cos(0 t + ) = x (1-34) Mt khc theo nh l v hnh chiu, ta c: r r r hc ox OM = hc ox OM1 + hc ox OM 2 (1-35)

Nh vy, tng hp hai dao ng iu ho x1 v x2 cng phng, cng tn s gc cng l mt dao ng iu ho x c cng phng v cng tn s gc 0 vi cc dao ng thnh phn, cn bin A v pha ban u ca n c xc nh bi (1-31) . H thc (1-31) cho thy bin A ca dao ng tng hp x ph thuc vo hiu pha (1 2 ) ca hai dao ng thnh phn x1 v x2:

- Nu (2 1) = 2k , vi k = 0, 1, 2, 3,... , th cos(2 1 ) = 1 v bin A t cc i: A = A1 + A 2 = A max (1-36) Trong trng hp ny, hai dao ng x1 v x2 cng phng, cng chiu v c gi l hai dao ng cng pha. - Nu (2 1) = (2k + 1) , vi k = 0, 1, 2, 3,... , th cos(2 1 ) = 1 v bin A t cc tiu:A = A1 A 2 = A min

(1-37)

Trong trng hp ny, hai dao ng x1v x2 cng phng ngc chiu v gi l hai dao ng ngc pha.2. Tng hp hai dao ng iu ho c phng vung gc v cng tn s gc

Gi s mt cht im tham gia ng thi hai dao ng iu ho x v y c phng vung gc v cng tn s gc 0 : x = A1 cos(0 t + 1 ) x = cos 0 t cos 1 sin 0 t sin 1 A1 y = cos 0 t cos 2 sin 0 t sin 2 A2 (1.38) (1-39)

y = A 2 cos(0 t + 2 )

14

Chng 1: Dao ng in t Ln lt nhn (1-38) v (1-39) vi cos 2 v cos 1 , ri cng v vi v:

y x cos 2 cos 1 = sin 0 t sin ( 2 1 ) A1 A2

(1-40)

Tng t, ln lt nhn (1-38) v (1-39) vi sin 2 v sin 1 , ri cng v vi v: y x sin 2 sin 1 = cos 0 t sin ( 2 1 ) A1 A2 (1-41) Hnh 1-10. Hai dao ng iu ho vung gc

Bnh phng hai v (1-40) , (1-41) ri cng v vi v:

x22 A1

+

y2 A2 2

2 xy cos( 2 1 ) = sin 2 ( 2 1 ) A1A 2

(1-42)

Phng trnh (1-42) chng t qu o chuyn ng tng hp ca hai dao ng iu ho c phng vung gc v c cng tn s gc l mt ng elip. Dng ca elip ny ph thuc vo gi tr ca hiu pha ( 2 1 ) ca hai dao ng thnh phn x v y. - Nu ( 2 1 ) = 2k , vi k = 0, 1, 2, 3,... , th (1-42) tr thnh: x22 A1

+

y2 A2 2

y 2xy x = 0 hay =0 A1 A 2 A1 A 2

(1-43)

Phng trnh (1-43) chng t cht im dao ng theo ng thng nm trong cung phn t I v III, i qua v tr cn bng bn ca cht im ti gc O v trng vi ng cho ca hnh ch nht c hai cnh bng 2A1 v Hnh 1-11. Qu o ca cht im khi 2 1 =2k2A 2 .

- Nu ( 2 1 ) = (2k + 1) , vi k = 0, 1, 2, 3,... , th (1-42) tr thnh: x22 A1

+

y2 A2 2

+

2xy y x = 0 hay + =0 A1 A 2 A1 A 2

(1-44)

15

Chng 1: Dao ng in t

Phng trnh (1-44) chng t cht im dao ng theo ng thng nm trong cung phn t II v IV, i qua v tr cn bng bn ca cht im ti gc O v trng vi ng cho ca hnh ch nht c hai cnh bng 2A1 v Hnh 1-12. Qu o ca cht im khi 2 1 =(2k+1) - Nu ( 2 1 ) = (2k + 1) x22 A1

2A 2 .

, vi k = 0, 1, 2, 3,... , th (1-42) tr thnh: 2 =1 (1-45)

+

y2 A2 2

Hnh 1-13: Qu o ca cht im khi 2-1=(2k+1)/2

Qu o ca cht im khi 2-1=(2k+1)/2 v A1=A2

Phng trnh (1-45) chng t cht im dao ng trn mt qu o lip dng chnh tc c hai bn trc l A1 v A 2 . c bit nu A1 = A 2 = A th (1-45) tr thnh: x 2 + y2 = A2 (1-46)

Trong trng hp ny, qu o ca cht im l ng trn c tm ti gc to O v bn knh bng A. Nu ( 2 1 ) c cc gi tr khc vi cc gi tr nu trn th cht im s chuyn ng trn nhng qu o lip xin.

16

Chng 1: Dao ng in t

2 1 = 0

0 < 2 - 1 < /2

2 1=/2

/2 < 2 1 <

2 1 =

< 2 - 1 nkk. Ngun sng S rng v cch xa. Mn E c t vung gc vi gng. Mt im M trn mn E s nhn c hai tia sng t S gi n. Tia truyn trc tip SM v tia SIM phn x trn gng, sau n M. Hai tia ny giao thoa vi nhau.

Hnh 2-8. Th nghim ca Lloyd Theo l thuyt: nu

r1 r2 = L1 L 2 = k

th

im

M

sng,

nu

th im M s ti. Tuy nhin thc nghim li thy rng: 2 nhng im l thuyt d on l sng th kt qu li l ti v ngc li, nhng im l thuyt d on l ti th li l sng. Vy hiu pha dao ng ca hai tia sng trong trng hp ny 2 2 (L1 L 2 ) m phi l = (L1 L 2 ) + . thm mt khng phi l = lng th pha dao ng ca mt trong hai tia phi thay i mt lng . V tia SM truyn trc tip t ngun n im M, nn ch c tia phn x trn gng mi thay i, c th l pha dao ng ca n sau khi phn x s thay i mt lng . Tng ng vi vic pha thay i mt lng l th quang l ca n s thay i mt lng l: r1 r2 = L1 L 2 = (2k + 1) 1 = 2 L1 2 '1 = 2 2 L1 + = L1 (2-14)

L1 = L1 +

' Trong 1 v L1 l pha v quang l khi cha tnh n s thay i pha do phn x, cn 1

v L'1 l pha v quang l ca tia sng khi c tnh n s phn x trn thy tinh l mi 31

Chng 2: Giao thoa nh sng trng chit quang hn mi trng nh sng ti. Trong trng hp phn x trn mi trng c chit sut nh hn chit sut mi trng nh sng ti, v d ta cho nh sng truyn trong mi trng thy tinh n mt phn cch gia thy tinh v khng kh ri phn x li, khi pha dao ng v quang l ca tia phn x khng c g thay i. Kt lun: Khi phn x trn mi trng chit quang hn mi trng nh sng ti, pha dao ng ca nh sng thay i mt lng , iu cng tng ng vi vic coi tia phn x di thm mt on . 2 2. Giao thoa gy bi nm khng kh Nm khng kh l mt lp khng kh hnh nm gii hn bi hai bn thu tinh phng G1, G2 c dy khng ng k, t nghing vi nhau mt gc nh . Chiu chm tia sng n sc song song, vung gc vi mt G2 . Tia sng t ngun S i vo bn thu tinh G1 ti M chia lm hai: Mt tia phn x i ra ngoi (tia R1), mt tia i tip vo nm khng kh, n K trn G2 v phn x ti ri i ra ngoi Hnh 2-9. Nm khng kh (tia R2). Ti M c s gp nhau ca hai tia phn x ni trn v chng giao thoa vi nhau. Trn mt G1 ta nhn c vn giao thoa. Tia R2 (l tia phn x trn mt G2) phi i thm mt on 2d so vi tia R1 (l tia phn x trn mt G1) v v tia R2 phn x trn mt trn ca G2 (thy tinh) chit quang hn mi trng nh sng n (khng kh) nn quang l ca tia ny di thm mt on l / 2 . Cn tia R1 phn x trn mt di ca G1 th khng thay i pha v y l phn x trn mi trng khng kh, km chit quang hn mi trng nh sng ti (mi trng thy tinh). Hiu quang l ca hai tia l:

L 2 L1 = 2d +

2 = (2k + 1) 2 2

(2-15)

d l b dy ca lp khng kh ti M. Cc im ti tho mn iu kin:

L 2 L1 = 2d +Do :

dt = k

2

vi

k = 0,1,2...

(2-16)

Tp hp cc im c cng b dy d ca lp khng kh l mt on thng song song vi cnh nm. Ti cnh nm d = 0, ta c mt vn ti. Cc im sng tho mn iu kin:

32

Chng 2: Giao thoa nh sng

L 2 L1 = 2d +Do :

= k 2

d s = ( 2 k 1)

4

vi

k =1,2,3...

(2-17)

Vn sng cng l nhng on thng song song vi cnh nm v nm xen k vi vn ti. 3. Vn trn Newton H cho vn trn Newton gm mt thu knh phng - li t tip xc vi mt bn thy tinh phng (hnh 2-10). Lp khng kh gia thu knh v bn thy tinh l bn mng c b dy thay i. Chiu mt chm tia sng n sc song song vung gc vi bn thy tinh. Cc tia sng phn x mt trn v mt di ca bn mng ny s giao thoa vi nhau, to thnh cc vn giao thoa c cng dy, nh x mt cong ca thu knh phng- li. Ging nh nm khng kh, cc tiu vn giao thoa (vn ti) nm ti v tr ng vi b dy ca lp khng kh:

dt = k

2 4

vi k = 0,1,2...

(2-18)

v cc i vn giao thoa (vn sng) nm ti v tr ng vi b dy lp khng kh:

d s = ( 2 k 1)

vi k = 1,2,3...

(2-19)

Hnh 2-10. Vn trn Newton Do tnh cht i xng ca bn mng nn cc vn giao thoa l nhng vng trn ng tm gi l vn trn Newton. Ta tnh bn knh ca vn th k:

33

Chng 2: Giao thoa nh sng2 rk = R 2 (R d k ) 2

trong R l bn knh cong ca thu knh, dk l b dy ca lp khng kh ti vn th k. V d k a2 > a3 ... Khi k kh ln th a k 0 . V khong cch t i cu n im M v gc nghing tng rt chm nn ak gim chm, ta c th coi ak do i cu th k gy ra l trung bnh cng ca ak-1 v ak+1:

1 a k = (a k 1 + a k +1 ) 2

(3-3)

Khong cch ca hai i cu k tip ti im M khc nhau / 2 . Cc i cu u nm trn mt sng , ngha l pha dao ng ca tt c cc im trn mi i cu u nh nhau. Kt qu, hiu pha ca hai dao ng sng do hai i cu k tip gy ra ti M l:

=

2 2 (L1 L 2 ) = . = 2

(3-4)

Nh vy hai dao ng sng ngc pha nhau nn chng s kh ln nhau. V M kh xa mt , ta coi cc dao ng sng do cc i cu gy ra ti M cng phng, do dao ng sng tng hp do cc i gy ra ti M s l: a = a1 - a2 + a3 - a4+... (3-5) Sau y chng ta s s dng phng php i cu Fresnel kho st hin tng nhiu x ca nh sng qua l trn, a trn v qua khe hp. 3. Nhiu x qua l trn Xt ngun sng im S, pht nh sng n sc qua l trn AB trn mn chn P n im M, S v M nm trn trc ca l trn. Ly S lm tm dng mt cu ta vo l trn AB. Ly M lm tm v cc i cu Fresnel trn mt . Gi s l cha n i cu. Bin dao ng sng tng hp ti M l:

47

Chng 3: Nhiu x nh sng

a = a1 a 2 + a 3 a 4 + ... a n

Hnh 3-4. Nhiu x qua l trn Ta c th vit:

an a3 a3 a5 a1 a1 2 + a2 + a4 + a= + + ... + a n 1 a 2 2 2 2 2 an n 2 2V cc biu thc trong du ngoc bng khng, nn:

a a a= 1 n 2 2 Ly du + nu i n l l v du - nu i n l chn. Ta xt cc trng hp sau:

(3-6)

* Khi khng c mn chn P hoc kch thc l trn rt ln: n , a n 0 nn cng sng ti M:

a2 I0 = a 2 = 1 4* Nu l cha s l i cu

(3-7)

a a a= 1+ n 2 2a a I= 1 + n 2 22

(3-8)

I > I0, im M sng hn khi khng c mn P. c bit nu l cha mt i cu

a a a = 1 + 1 = a1 2 2* Nu l cha s chn i cu

v

2 I = a1 = 4 I 0

(3-9)

Cng sng gp 4 ln so vi khi khng c l trn, nh vy im M rt sng.

a a a= 1 n 2 2

(3-10)

48


a a I= 1 n 2 2

2

(3-11)

I < I0, im M ti hn khi khng c l trn. Nu l trn cha hai i cu th

a a a = 1 2 0 , do I = 0, im M ti nht. 2 2Tm li im M c th sng hn hoc ti hn so vi khi khng c l trn ty theo kch thc ca l v v tr ca mn quan st. 4. Nhiu x qua mt a trn Gia ngun sng S v im M c mt a trn chn sng bn knh ro. Gi s a che khut m i cu Fresnel u tin. Bin dao ng ti M l:

a = a m +1 a m + 2 + a m + 3 .... a a a a = m +1 + m +1 a m + 2 + m + 3 + ... 2 2 2Hnh 3-5. Nhiu x qua mt a trn

V cc biu thc trong ngoc c th coi bng khng, do :

a a = m+1 2

(3-12)

Nu a ch che t i cu th am+1 khng khc a1 l my, do cng sng ti M cng ging nh trng hp khng c chng ngi vt gia S v M. Trong trng hp a che nhiu i cu th am+1 0 do cng sng ti M bng khng. 2. NHIU X NH SNG CA SNG PHNG 1. Nhiu x ca sng phng qua mt khe hp to ra chm sng song song, ngi ta t ngun sng S ti tiu im ca thu knh hi t Lo. Chiu chm sng n sc song song bc sng vo khe hp c b rng b (hnh 3-6). Sau khi i qua khe hp, tia sng s b nhiu x theo nhiu phng. Tch cc tia nhiu x theo mt phng no chng s gp nhau v cng. Mun quan st nh nhiu x chng ta s dng thu knh hi t L, chm tia nhiu x s hi t ti im M trn mt phng tiu ca thu knh hi t L. Vi cc gi tr khc nhau chm nhiu x s hi t ti cc im khc nhau. Ty theo gi tr ca im M c th sng hoc ti. Nhng im sng ti ny nm dc trn ng thng vung gc vi chiu di khe hp v c gi l cc cc i v cc tiu nhiu x.

49


Hnh 3-6. Nhiu x qua mt khe hp V nh sng gi n khe l sng phng nn mt phng khe l mt sng, cc sng th cp trn mt phng khe dao ng cng pha. Xt cc tia nhiu x theo phng =0, chng hi t ti im F. Mt phng khe v mt quan st l hai mt trc giao do theo nh l Malus, cc tia sng gi t mt phng khe ti im F c quang l bng nhau v dao ng cng pha nn chng tng cng nhau. im F rt sng v c gi l cc i gia. Xt trng hp 0 . p dng tng ca phng php i cu Fresnel ta v cc mt phng 0 , 1 , 2 ,... vung gc vi chm tia nhiu x v cch u nhau mt khong

/2, chng s chia mt khe thnh cc di sng nm song song vi b rng ca khe hp. B rng ca mi di l l = v s di trn khe s l: 2 sin

N=

b 2b sin = l

(3-13)

Theo nguyn l Huygens, nhng di ny l ngun sng th cp dao ng cng pha (v nm trn cng mt mt sng) v pht nh sng n im M. V quang l ca hai tia sng t hai di k tip n im M khc nhau /2 nn dao ng sng do hai di k tip gi ti M ngc pha nhau v chng s kh nhau. Kt qu l nu khe cha s chn di (N = 2k) th dao ng sng do tng cp di k tip gy ra ti M s kh ln nhau v im M s ti v l cc tiu nhiu x. iu kin im M ti l:

N=hay

2b sin = 2k bvi k = 1, 2, 3...

sin = k

(3-14)

Nu khe cha mt s l di (N = 2k+1) th dao ng sng do tng cp di k tip gi ti im M s kh ln nhau, cn dao ng sng do di cui cng gi ti th khng b kh. Kt qu im M s sng v c gi l cc i nhiu x bc k. Cng sng ca cc cc i ny nh hn rt nhiu so vi cc i gia. iu kin im M sng l:

50


N=hay

2b sin = 2k + 1 2b sin = 0 sin = , 2 , 3 ,... b b b sin = 3 , 5 , ... 2b 2b

sin = (2k + 1)

vi

k = 1, 2, 3...

(3-15)

Tm li ta c cc iu kin cc i, cc tiu nhiu x qua mt khe hp nh sau: - Cc i gia (k=0) : - Cc tiu nhiu x : - Cc i nhiu x :

th phn b cng sng trn mn quan st cho bi hnh 3-7.

Hnh 3-7. Hnh nhiu x ca sng phng qua mt khe hp Nhn xt thy cc cc i nhiu x bc k = 1,2,3...nm xen gia cc cc tiu nhiu x v phn b i xng hai bn cc i gia. Cc i gia c b rng gp i cc cc i khc. Theo tnh ton l thuyt, cng sng ca cc cc i nhiu x tun theo h thc sau I0 : I1 : I2 : I3 : ....= 1 : 0,045 : 0,016 : 0,008 : ... 2. Nhiu x ca sng phng truyn qua cch t phng Cch t phng l mt h nhiu khe hp ging nhau c rng b, nm song song cch u trn cng mt mt phng. Khong cch d gia hai khe k tip c gi l chu k ca cch t. S khe hp trn mt n v chiu di: n =

1 d51

Chng 3: Nhiu x nh sng Xt mt cch t phng c N khe hp. B rng ca mt khe l b, chu k ca cch t l d. Chiu chm sng n sc song song bc sng vung gc vi mt cch t. V cc khe c th coi l ngun kt hp, do ngoi hin tng nhiu x gy bi mt khe cn c hin tng giao thoa gy bi cc khe. Do nh nhiu x qua cch t s phc tp hn nhiu so vi nh nhiu x qua mt khe hp. Ta s kho st nh nhiu x qua cch t:

Hnh 3-8. Nhiu x qua cch t

- Tt c N khe hp u cho cc tiu nhiu x ti nhng im trn mn nh tha mn iu kin:

sin = k

b

vi k = 1,2,3...

(3-16)

Nhng cc tiu ny c gi l cc tiu chnh. - Xt phn b cng sng gia hai cc tiu chnh: Hiu quang l ca hai tia sng xut pht t hai khe k tip n im M l L1 L 2 = d sin . Nu hiu quang l bng s nguyn ln bc sng

L1 L 2 = d sin = m th dao ng sng do hai tia gy ra ti M cng pha v tng cng ln nhau. Kt qu im M sng. Cc im c gi l cc i chnh. V tr cc cc i chnh l:

sin = m

d

vi m = 0, 1, 2, 3....

(3-17)

S nguyn m l bc ca cc i chnh. Cc i chnh gia (m = 0) nm ti tiu im F ca thu knh. V d > b nn gia hai cc tiu chnh c th c nhiu cc i chnh. V d: k = 1 d nn m < k = 3 , ngha l m = 0, 1, 2. Nh vy gia hai v d/b = 3 . Do m < k b b d cc tiu chnh c 5 cc i chnh.

Hnh 3-9 nh nhiu x qua ba khe hp

52


- Xt phn b cng sng gia hai cc i chnh: Ti im chnh gia hai cc i chnh k tip, gc nhiu x tha mn iu kin:

sin = (2m + 1)

2d

vi m = 0,1,2...

Ti cc im ny, hiu quang l ca hai tia gi t hai khe k tip c gi tr l:

. y l iu kin cc tiu giao thoa, hai tia s kh ln nhau. Tuy 2 nhin im chnh gia cha chc ti. minh ha c th ta xt hai trng hp n gin sau: d sin = (2m + 1)+ Nu s khe hp N = 2 (s chn) th cc dao ng sng do hai khe hp gi ti s kh nhau hon ton v im chnh gia s ti. im ti c gi l cc tiu ph. + Nu s khe hp N = 3 (s l) th cc dao ng sng do hai khe hp gi ti s kh nhau, cn dao ng sng do khe th ba gy ra khng b kh. Kt qu l gia hai cc i chnh l mt cc i. Cc i ny c cng kh nh, nn c gi l cc i ph. R rng gia cc i ph ny v hai cc i chnh hai bn phi c hai cc tiu ph. Ngi ta chng minh c rng, nu cch t c N khe hp th gia hai cc i chnh s c N-1 cc tiu ph v N-2 cc i ph. Hnh 3-9 biu din nh nhiu x qua ba khe hp. Cch t phng c th dng o bc sng nh sng, ng dng trong my n sc... T cng thc (3-17) nu ta bit c chu k ca cch t, bng cch o gc ng vi cc i chnh bc m ta c th xc nh c bc sng nh sng. 3. Nhiu x trn tinh th Cc nguyn t (phn t hay ion) cu to nn vt rn tinh th c sp xp theo mt cu trc tun hon gi l mng tinh th, trong v tr ca cc nguyn t (phn t hay ion) gi l nt mng. Khong cch gia cc nt mng, c trng cho tnh tun hon, c gi l chu k ca mng tinh th. Chiu ln tinh th mt chm tia Rnghen, mi nt mng tr thnh tm nhiu x v mng tinh th ng vai tr nh mt cch t vi chu k l chu k ca mng tinh th. Chm tia

Hnh 3-10. Nhiu x trn tinh th

Rnghen s nhiu x theo nhiu phng, tuy nhin ch theo phng phn x gng (phng m gc phn x bng gc ti), cng ca tia nhiu x ln ta c th quan st c nh nhiu x. Nhng tia nhiu x ny s giao thoa vi nhau v cho cc i nhiu x nu hai tia nhiu x k tip c hiu quang l bng s nguyn ln bc sng L = 2d sin = k hay

53


sin = k

2d

(3-18)

d l khong cch gia hai mt phng nguyn t ca vt rn tinh th (chu k mng tinh th). Cng thc (3-18) gi l cng thc Vulf-Bragg. y l cng thc c bn phn tch cu trc ca vt rn tinh th bng tia Rnghen. Nu bit bc sng ca tia Rnghen v o gc ta c th xc nh c chu k d ca mng tinh th.

III. TM TT NI DUNG1. Hin tng nhiu x nh sng * nh ngha: Hin tng nhiu x nh sng l hin tng tia sng b lch khi phng truyn thng khi i qua cc chng ngi vt c kch thc nh nh l trn, khe hp, a trn... * Nguyn l Huygens - Fresnel: - Mi im trong khng gian c sng nh sng t ngun thc gi n u tr thnh ngun sng th cp pht sng nh sng v pha trc. - Bin v pha ca ngun th cp l bin v pha do ngun thc gy ra ti v tr ca ngun th cp. 2. Phng php i cu Fresnel Din tch cc i cu bng nhau v bng:

S =

Rb R+b

Bn knh rk ca i cu th k bng:

rk =

Rb k R+b

k=1,2,3...

trong R l bn knh ca mt sng bao quanh ngun sng im S b l khong cch t im c chiu sng M ti i cu th nht. l bc sng do ngun S pht ra. 3. Nhiu x sng cu qua l trn p dng phng php i cu Fresnel, ta tnh c bin ca nh sng tng hp ti M, cch ngun S mt khong R+b:

a a a= 1 n 2 2Ly du + nu n l l v du - nu n l chn. Ta xt cc trng hp sau: * Khi khng c mn chn P hoc l trn rt ln: n , a n 0 sng ti M: nn cng

54


a2 I0 = a 2 = 1 4* Nu l cha s l i cu:

a a a= 1+ n 2 2a a I= 1 + n 2 22

I > I0, im M sng hn khi khng c mn P. c bit nu l cha mt i cu

a a a = 1 + 1 = a1 2 2* Nu l cha s chn i cu

v

2 I = a1 = 4 I 0

Cng sng gp 4 ln so vi khi khng c l trn, nh vy im M rt sng.

a a a= 1 n 2 2a a I= 1 n 2 22

I < I0, im M ti hn khi khng c l trn. Nu l trn cha hai i cu th

a a a = 1 2 0 , do I = 0, im M ti nht. 2 2Tm li im M c th sng hn hoc ti hn so vi khi khng c l trn tu theo kch thc ca l v v tr ca mn quan st. 4. Nhiu x ca sng phng qua mt khe hp p dng phng php i cu Fresnel ta tnh ton c bin dao ng sng tng hp ti mt im M trn mn quan st. Kt qu ta c cc iu kin cc i, cc tiu nhiu x qua mt khe hp nh sau: - Cc i gia (k=0) : - Cc tiu nhiu x : - Cc i nhiu x :

sin = 0 sin = , 2 , 3 ,... b b b sin = 3 , 5 , ... 2b 2b

Trn th phn b cng sng ta thy cc i gia rt sng, cc cc i nhiu x bc k=1,2,3...nm xen gia cc cc tiu nhiu x v phn b i xng hai bn cc i gia. Cc i gia c b rng gp i cc cc i khc. Theo tnh ton l thuyt, cng sng ca cc cc i nhiu x tun theo h thc sau: I0 : I1 : I2 : I3 : ....= 1 : 0,045 : 0,016 : 0,008 : ... 5. Nhiu x qua nhiu khe Cch t 55

Chng 3: Nhiu x nh sng Cch t phng l mt h nhiu khe hp ging nhau c rng b, nm song song cch u trn cng mt mt phng. Khong cch d gia hai khe k tip c gi l chu k ca cch t. Ngi ta c th ch to c cc cch t di 10cm, trn mi mm c t 500 1200 vch. Cc cch t ny c th s dng xc nh bc sng nh sng n sc, xc nh thnh phn cu to ca cc cht v dng trong my quang ph... i vi vt rn tinh th, mng tinh th ng vai tr mt cch t khng gian ba chiu. S nhiu x ca cc tia X trn cc nt mng cho ta kt qu:

2d sin = kd l khong cch gia hai nt mng, gi l hng s mng. y l cng thc Vulf-Bragg, c dng xc nh cu trc ca vt rn tinh th.

IV. CU HI L THUYT1. Nu nh ngha hin tng nhiu x nh sng. Dng nguyn l Huygens gii thch nh tnh hin tng nhiu x. 2. Pht biu nguyn l Huygens-Fresnel. 3. Trnh by phng php i cu Fresnel. 4. Gii thch hin tng nhiu x nh sng qua l trn nh. Xt cc trng hp l trn cha mt s l i cu, mt s chn i cu, c bit cha mt i cu v hai i cu. 5. M t hin tng nhiu x nh sng qua mt khe hp. Tm iu kin cc i, cc tiu nhiu x. V nh nhiu x ca sng phng qua mt khe hp. 6. nh ngha cch t phng v nu ng dng ca cch t. 7. Trnh by nhiu x ca tia X trn tinh th. Cng thc Vulf- Bragg. Nu ng dng ca hin tng nhiu x tia X.

V. BI TPTh d 1: Mt ngun sng im chiu nh sng n sc bc sng = 0,5m vo mt l trn c bn knh r = 0,5mm. Khong cch t ngun sng n l trn R = 1m.Tm khong cch t l trn n mn quan st tm nhiu x l ti nht. p s: tm ca hnh nhiu x l ti nht th l trn ch cha 2 i cu Fresnel, bn knh ca l trn bng bn knh ca i cu th 22 Rr2 2Rb 0,25.106 1 r2 = =rb= = = m 2 6 6 3 R+b 2R r2 2.0,5.10 0,25.10

Th d 2: Mt chm tia sng n sc c bc sng = 0,5m c chiu vung gc vi mt khe hp ch nht c b rng b = 0,1mm, ngay sau khe hp t mt thu knh hi t. Tm b rng ca vn cc i gia trn mn quan st t ti mt phng tiu ca thu knh v cch thu knh D = 1m.

56

Chng 3: Nhiu x nh sng Bi gii: B rng ca vn cc i gia l khong cch gia hai cc tiu nhiu x u tin hai bn cc i gia. ln ca gc nhiu x ng vi cc cc tiu nhiu x l: sin = . b T hnh v ta thyl = 2Dtg 2D sin l= 2D 2.1.0,5.10 6 = = 1cm b 0,1.10 3

Th d 3: Cho mt chm tia sng n sc song song c bc sng = 0,5m, chiu vng gc vi mt ca mt cch t phng truyn qua. st pha sau ca cch t ngi ta t mt thu knh hi t c tiu c f = 50cm. Khi trn mn quan st t ti mt phng tiu ca thu knh, hai vch quang ph bc nht cch nhau mt khong a = 10,1cm. Xc nh: 1. Chu k cch t v s khe trn 1cm chiu di ca cch t. 2. S vch cc i chnh trong quang ph nhiu x. Bi gii 1.V tr cc cc i chnh trong quang ph nhiu x xc nh bi cng thc:

sin =

m , m = 0, 1, 2, 3... d

Do vy v tr hai vch cc i chnh ca quang ph bc nht ng vi gc lch 1 bng: sin 1 = , v 1 rt nh nn d tg1 sin 1 . T hnh v, ta c tg1 =

M1F L = OF 2f

So snh tg 1 vi sin 1 ta c chu k cch t:

d=

2f 2.50.10 2.0,5.10 6 = = 4,95m L 10,1.10 21 = 2020 khe / cm d

S khe trn 1cm chiu di ca cch t: n = 2. T cng thc: sin =

m d 4,95.10 6 , m sin 1 m = = 9,9 0,5.10 6 d

57

Chng 3: Nhiu x nh sng V m nguyn nn c th ly cc gi tr: 0, 1,2 ,3 ,4, 5, 6, 7, 8, 9. Do cc vch cc i chnh ti a trong quang ph nhiu x ca cch t bng: Nmax = 2.9 + 1 = 19 vch. Bi tp t gii 1. Chiu nh sng n sc bc sng = 0,5m vo mt l trn bn knh cha bit. Ngun sng im t cch l trn 2m, sau l trn 2m t mn quan st. Hi bn knh ca l trn bng bao nhiu tm ca hnh nhiu x l ti nht.

p s: tm ca hnh nhiu x l ti nht th l trn ch cha 2 i cu Fresnel:

r=

kRb = R+b

2.2.2.0,5.10 6 = 10 3 m 4

2. Mt mn nh c t cch ngun sng im n sc bc sng = 0,5m mt khong 2m. Chnh gia mn nh v ngun sng t mt l trn ng knh 0,2cm. Tm s i cu Fresnel m l trn cha c. p s: Bn knh ca l trn bng bn knh ca i cu th k

r=

r 2 ( R + b) kRb =4 k= Rb R+b

3. Mt ngun sng im chiu nh sng n sc bc sng = 0,5m vo mt l trn c bn knh r = 1mm. Khong cch t ngun sng n l trn R= 1m. Tm khong cch t l trn n mn quan st l trn cha ba i Fresnel. p s: l trn ch cha ba i cu Fresnel c ngha l bn knh ca l trn bng bn2 Rr3 10 6 3Rb knh ca i cu th ba: r3 = b= = = 2m 2 R+b 3R r3 3.0,5.10 6 10 6

4. t mt mn quan st cch mt ngun sng im pht ra nh sng n sc bc sng = 0,6m mt khong x. Chnh gia khong x t mt a trn nh chn sng ng knh 1mm. Hi x bng bao nhiu im M0 trn mn quan st c sng gn ging nh cha t a trn, bit im M0 v ngun sng u nm trn trc ca a trn. p s: Mun cng sng ti M0 gn ging nh khi cha c a trn th a trn ch chn mt i cu Fresnel:

4.(0,5.103 ) 2 2r 2 Rb = 1,67m ; R = b R = ; x = 2R = r1 = R+b 0,6.1065. Mt chm tia sng n sc song song bc sng = 0,589m chiu thng gc vi mt khe hp c b rng b = 2m. Hi nhng cc tiu nhiu x c quan st di nhng gc nhiu x bng bao nhiu? (so vi phng ban u) p s: p dng cng thc: 58

Chng 3: Nhiu x nh sngsin = k , sin 1 1 = 17 08, 2 = 36 0 5, 3 = 62 0 b

6. Chiu mt chm tia sng n sc song song vung gc vi mt khe hp. Bc sng nh 1 b rng ca khe hp. Hi cc tiu nhiu x th ba c quan st di gc sng bng 6 lch bng bao nhiu? p s: = 300 7. Mt chm tia sng c ri vung gc vi mt cch t. Bit rng gc nhiu x i vi vch quang ph 1 = 0,65m trong quang ph bc hai bng 1 = 450. Xc nh gc nhiu x ng vi vch quang ph 2 = 0,5m trong quang ph bc ba. p s:

3 2 sin 1 2 = 54 0 40 2 1 8. Cho mt chm tia sng n sc song song c bc sng = 0,7m chiu vung gc vi mt ca mt cch t truyn qua. Trn mt phng tiu ca thu knh hi t t st pha sau sin 1 = m1n1 = 2n1 , sin 2 = m 2 n 2 = 3n 2 sin 2 =cch t, ngi ta quan st thy vch quang ph bc ba lch = 48 0 36 . Xc nh: 1. Chu k cch t v s khe trn 1cm chiu di ca cch t. 2. S cc i chnh nm trong khong gia hai cc tiu chnh bc nht trong nh nhiu x. Cho bit mi khe ca cch t c rng b = 0,7m, sin 48 0 36 = 0,75 p s: 1.Gc nhiu x ng vi cc cc i chnh c xc nh bi cng thc:

sin =

m m d= = 2,8.10 4 cm sin d 1 3571 khe / cm d

S khe trn 1cm chiu di ca cch t: n =

2. Gc nhiu x ng vi cc tiu chnh trong nh nhiu x c xc nh bi cng thc: k , s cc i chnh nm gia hai cc tiu chnh bc nht phi tho mn iu kin: sin = b m k kd m = 4 . Vy gia hai cc tiu chnh bc nht c 7 cc i chnh. d b b 9. Cho mt cch t phng c chu k cch t d = 2m. Sau cch t t mt thu knh hi t, trn mn quan st t ti mt phng tiu ca thu knh ngi ta quan st thy khong cch gia hai quang ph bc nht ng vi bc sng 1 = 0,4044m v 2 = 0,4047m bng 0,1mm. Xc nh tiu c ca thu knh. p s:

59

Chng 3: Nhiu x nh sng Gc nhiu x ng vi cc i:

sin =

m d

V tr cc i ng vi gc nhiu x:

y = MF = f .tg y 2 y1 , y 2 y1 = 0,1mm, tg 2 tg1 sin 2 = 2 , sin 1 = 1 d d f =

f = 0,65m10. Mt chm nh sng trng song song chiu vung gc vo mt mt cch t phng. Cho bit trn mi milimet chiu di ca cch t c n = 50 khe. Pha sau cch t t mt thu knh hi t. Xc nh hiu s cc gc nhiu x ng vi vch c bc sng 1 = 0,76m nm cui quang ph bc nht v vch tm c bc sng 2 = 0,4m nm u quang ph bc hai. p s: d =

1 = 0,02mm n

Gc nhiu x cui quang ph bc nht ng vi nh sng :

0,76.10 6 sin 1 = 1 = = 0,038 1 = 2 011 3 d 0,02.10Gc nhiu x u quang ph bc hai ng vi nh sng tm:

sin 2 =

2 2 2.0,4.10 6 = = 0,04 2 = 2 018 3 d 0,02.10

Hiu s ca cc gc nhiu x: = 2 1 = 7 11. Cho mt chm tia sng n sc song song chiu vung gc vo mt ca mt cch t phng c chu k d = 2m. Xc nh bc ln nht ca cc vch cc i trong quang ph nhiu x cho bi cch t i vi nh sng c bc sng 1 = 0,7m v i vi nh sng tm c bc sng 2 = 0,42m. p s: sin = m

d. sin d , m sin 1 , nn m m= d d = 2,86 m1(max ) = 2 1 d = 4,76 m 2(max ) = 4 2

i vi nh sng : m1 i vi nh sng tm: m 2

60

Chng 4: Phn cc nh sng

CHNG IV: PHN CC NH SNGTrong hai chng trc chng ta nghin cu hin tng giao thoa v hin tng nhiu x nh sng ch da vo bn cht sng ca nh sng m khng cn phn bit sng nh sng l sng ngang hay sng dc. Trong chng ny chng ta s chng minh nh sng l sng ngang qua hin tng phn cc nh sng. Ta bit sng in t l sng ngang, l sng c cc vect cng in trng v vect cng t trng dao ng vung gc vi phng truyn sng. Ch c sng ngang mi c th th hin tnh phn cc cho nn nghin cu s phn cc ca nh sng chng ta mt ln na khng nh bn cht sng in t ca nh sng.

I. MC CH - YU CU1. Nm c s phn cc nh sng th hin nh sng l sng ngang. Phn bit nh sng t nhin v nh sng phn cc (mt phn, ton phn). Thit lp nh lut Malus. 2. Nm c s phn cc nh sng do phn x, khc x, phn cc do lng chit t nhin. 3. Nm c ng dng ca hin tng quay mt phng phn cc xc nh nng ca cc cht hot quang trong phn cc k (ng k).

II. NI DUNG1. NH SNG PHN CC 1. nh sng t nhin v nh sng phn cc

1(a) E Tia sng Tia sng (b) E E1

Hnh 4-1. nh sng t nhin (a) v nh sng phn cc thng (b) 61

Chng 4: Phn cc nh sng nh sng do mt ngun sng pht ra l tp hp ca v s cc on sng ni tip nhau. Trong mi on sng, vect cng in trng E lun dao ng theo mt phng xc nh vung gc vi tia sng. Nhng do tnh hn lon ca chuyn ng bn trong mi nguyn t nn vect E trong cc on sng do mt nguyn t pht ra c th dao ng theo cc phng khc nhau vung gc vi tia sng. Mt khc ngun sng bao gm nhiu nguyn t, do phng dao ng ca vect E trong cc on sng do cc nguyn t pht ra cng thay i hn lon v phn b u xung quanh tia sng. nh sng c vect cng in trng dao ng u n theo mi phng vung gc tia sng c gi l nh sng t nhin. Hnh 4-1a biu din nh sng t nhin, trong mt phng vung gc vi tia sng cc vect E c tr s bng nhau v phn b u n xung quanh tia sng. nh sng t nhin khi i qua mi trng bt ng hng v mt quang hc (v d bn tinh th Tuamalin), trong nhng iu kin nht nh no do tc dng ca mi trng nn vect E ch dao ng theo mt phng xc nh. nh sng c vect E ch dao ng theo mt phng xc nh c gi l nh sng phn cc thng hay nh sng phn cc ton phn. Hnh 4-1b biu din nh sng phn cc ton phn E1 . Hin tng nh sng t nhin bin thnh nh sng phn cc gi l hin tng phn cc nh sng. Mt phng cha tia sng v phng dao ng ca E c gi l mt phng dao ng, cn mt phng cha tia sng v vung gc vi mt phng dao ng gi l mt phng phn cc. Vi nh ngha nh sng phn cc ton phn th mi on sng do nguyn t pht ra l mt nh sng phn cc ton phn. Nh vy nh sng t nhin do cc nguyn t ca mt ngun sng pht ra l tp hp ca v s nh sng phn cc ton phn, dao ng u n theo tt c mi phng vung gc vi tia sng. Trong mt s trng hp do tc dng ca mi trng ln nh sng truyn qua n, vect cng in trng vn dao ng theo tt c cc phng vung gc vi tia sng nhng c phng dao ng yu, c phng dao ng mnh. nh sng ny c gi l nh sng phn cc mt phn. 2. nh lut Malus v phn cc nh sng Thc nghim chng t rng, bn tinh th Tuamalin (hp cht silicbrat aluminium) vi chiu dy 1mm c th bin nh sng t nhin thnh nh sng phn cc thng. Nguyn nhn ca hin tng ny l do tnh hp th nh sng khng u theo cc phng khc nhau trong tinh th (gi l tnh hp th d hng). Trong bn Tuamalin c mt phng c bit gi l quang trc ca tinh th (k hiu l ) . Theo phng quang trc, nh sng khng b hp th v truyn t do qua bn tinh th, cn theo phng vung gc vi quang trc, nh sng b hp th hon ton. Khi ta chiu mt chm tia sng t nhin vung gc vi mt ABCD ca bn tinh th tuamalin c quang trc song song cnh AB, v nh sng l sng ngang nn tia sng sau bn tuamalin c vect sng E song song vi quang trc ca bn (hnh 4-1b). Di y ta s xt k hn v s truyn nh sng qua bn tuamalin.

62

Chng 4: Phn cc nh sng Xt nh sng t nhin truyn ti bn tuamalin T1, bt k vect sng E no ca nh sng t nhin cng u c th phn tch thnh hai thnh phn: E1x vung gc vi quang trc 1 v E1y song song vi quang trc2 2 E 2 = E1x + E1y

1 . Khi (4-1)

Do nh sng t nhin c E phn b u n xung quanh tia sng nn ta c th ly trung bnh:2 2 E1x = E1y =

1 2 E 2

(4-2)

Do tnh hp th d hng ca bn tinh th tuamalin, thnh phn E1x vung gc vi quang trc b hp th hon ton, cn thnh phn E1y song song vi quang trc c truyn hon ton qua bn tuamalin T1, nh sng t nhin bin thnh nh sng phn cc ton phn c vect sng E1 = E1y song song vi quang trc 1 (hnh 4-2) v cng sng I1 sau bn T1 bng:2 2 I1 = E1 = E1y =

1 2 1 E = I0 2 2

(4-3)

trong I 0 = E 2 l cng ca nh sng t nhin truyn ti bn T1. Ly mt bn tuamalin T2 c quang trc 2 t sau T1. Gi l gc gia quang trc

1 v 2 . Vect sng E1 sau bn tuamalin T1 s c phn tch thnh hai thnh phn:E 2 = E1 cos ,, ,

song

song

vi

quang

trc,

2 v

E 2 = E1 sin vung gc vi 2 . Thnh phn E 2 s truynqua bn T2, cn thnh phn E 2 s b hp th hon ton. Nh vy sau bn T2 ta cng nhn c nh sng phn cc ton phn c vect sng E 2 = E 2 v cng sng I2 bng2 I 2 = E 2 = E1 cos 2 = I1 cos 2 (4-4) 2 I1 l cng sng sau bn tuamalin T1. Nh vy nu gi c nh bn T1 v quay bn T2 xung quanh tia sng th I2 s thay i. Khi hai quang trc song song vi nhau, = 0 th I2 s , ,,

Hnh 4-2

t gi tr cc i v bng I1. Cn lc hai quang trc vung gc vi nhau, = bng 0. T1 c gi l knh phn cc, T2 c gi l knh phn tch (hnh 4-3) Cng thc (4-4) biu din mt nh lut gi l nh lut Malus.

th I2 s 2

63

Chng 4: Phn cc nh sng nh lut Malus: Khi cho mt chm tia sng t nhin truyn qua hai bn tuamalin c quang trc hp vi nhau mt gc th cng sng nhn c t l vi cos2. Do tnh i xng ca nh sng t nhin xung quanh phng truyn nn nu ta quay bn tuamalin xung quanh tia sng th v tr no cng c nh sng truyn qua. Cn khi tia sng chiu n bn tuamalin l nh sng phn cc th khi quay bn tuamalin cng sng sau bn s thay i. Nh vy bn tuamalin c th gip ta phn bit c chm sng t nhin v chm sng phn cc.

Hnh 4-3 3. S phn cc nh sng do phn x v khc x Thc nghim chng t rng khi cho mt tia sng t nhin chiu ti mt phn cch gia hai mi trng di gc ti i 0 th tia phn x v tia khc x u l nh sng phn cc mt phn. Vect cng in trng ca tia phn x c bin dao ng ln nht theo phng vung gc vi mt phng ti, cn vect cng in trng ca tia khc x c bin dao ng ln nht theo phng nm trong mt phng ti (hnh 4-4) . Khi thay i gc ti i th mc phn cc ca tia phn x v tia khc x cng thay i. Khi gc ti i tha mn iu kin: tg iB = n21 (4-5) th tia phn x s phn cc ton phn,

n n 21 = 2 l chit sut t i ca mi n1 trng hai i vi mi trng mt, iB c gi l gc ti Brewster hay gc phn cc ton phn. V d khi phn x t khng kh trn thy tinh th iB = 57o. Tia khc x khng bao gi l nh sng phn cc ton phn, nhng khi i = iB th tia khc x cng b phn cc mnh nht.B B

Hnh 4-4: Phn cc do phn x v khc x

64

Chng 4: Phn cc nh sng 2. PHN CC DO LNG CHIT

Thc nghim chng t rng mt s tinh th nh bng lan, thch anh... c tnh cht c bit l nu chiu mt tia sng n tinh th th ni chung ta s c hai tia. Hin tng ny gi l hin tng lng chit. Nguyn nhn l do tnh bt ng hng ca tinh th v mt quang hc (tc l tnh cht quang ca tinh th cc hng khc nhau th s khc nhau). nghin cu hin tng lng chit chng ta xt tinh th bng lan.Tinh th bng lan l dng kt tinh ca canxi cacbnat (CaCO3). Mi ht tinh th bng lan c dng mt khi su mt hnh thoi (hnh 4-5), trong ng thng ni hai nh A v A1 gi l quang trc ca tinh th. Mt tia sng truyn vo tinh th bng lan theo phng song song vi quang trc s khng b tch thnh hai tia khc x. Chiu mt tia sng t nhin vung gc vi mt

Hnh 4-5 Tinh th bng lan

ABCD ca tinh th. Thc nghim chng t rng tia ny s b tch thnh hai tia khc x (hnh 4-6): - Tia truyn thng khng b lch khi phng truyn gi l tia thng (k hiu l tia o). Tia ny tun theo nh lut khc x nh sng. Tia thng phn cc ton phn, c vect sng E vung gc vi mt mt phng c bit gi l mt phng chnh ca tia (mt phng cha tia thng v quang trc). - Tia lch khi phng truyn gi l tia bt thng (k hiu l tia e). Tia ny khng tun theo nh lut khc x nh sng. Tia bt thng phn cc ton phn, c vect sng E nm trong mt phng chnh ca n (mt phng cha quang trc v tia bt thng). Khi l ra khi tinh th, hai tia thng v tia bt thng ch khc nhau v phng phn cc. Chit sut ca tinh th bng lan i vi tia thng lun khng i v bng no=1,659. Chit sut ne ca tinh th bng lan i vi tia bt thng ph thuc vo phng truyn ca n trong tinh th v thay i t 1,659 (theo phng quang trc) n 1,486 (theo phng vung gc vi quang trc). Nh vy i vi tinh th bng lan ta c: ne no (4-6) V chit sut n = c/v, vi c l vn tc nh sng trong chn khng v v l vn tc nh sng trong mi trng, do : ve vo (4-7) ngha l trong tinh th bng lan, vn tc ca tia bt thng ni chung ln hn vn tc ca tia thng. Tinh th bng lan, thch anh, tuamalin... l nhng tinh th n trc. Trong t nhin cn c tinh th lng trc, l nhng tinh th c hai quang trc theo hai hng khc nhau. Mt tia sng t nhin truyn qua tinh th lng trc cng b tch thnh hai tia khc x nhng c hai tia ny u l nhng tia bt thng.

65


Hnh 4-6. Tnh lng chit ca tinh th 3. KNH PHN CC Ngi ta s dng cc tinh th lng chit ch to knh phn cc. Knh phn cc l nhng dng c c th bin nh sng t nhin thnh nh sng phn cc, v d nh bn tuamalin, bn plarit, lng knh nicn... 1. Bn plarit Mt s tinh th lng chit c tnh hp th d hng mnh i vi mt trong hai tia thng v bt thng. V d bn tinh th tuamalin dy hn 1mm hu nh hp th hon ton tia thng v ch cho tia bt thng truyn qua n. V vy bn tuamalin c th dng lm knh phn cc. Trong nhng nm gn y ngi ta ch to nhng knh phn cc lm bng xenluylit, trn c ph mt lp tinh th nh hng sunfat-it-kinin c tnh hp th d hng mnh. Nhng bn ny gi l bn plarit. Bn plarit dy khong 0,1 mm c th hp th hon ton tia thng v to ra nh sng phn cc ton phn sau khi i ra khi bn. Bn plarit tng i r nn c s dng nhiu trong ngnh vn ti. khc phc hin tng ngi li xe t b lo mt do nh sng t cc n pha ca cc t khc chy ngc chiu gy ra, ngi ta dn cc bn plarit ln mt knh n pha t v knh chn gi pha trc ngi li t sao cho quang trc ca cc bn song song v cng nghing 45o so vi phng ngang. Khi hai t chy ngc chiu ti gp nhau th cc bn plarit trn hai t ny c quang trc bt cho nhau. Nh vy nh sng phn cc pht ra t n pha ca t th nht chy ti khng th truyn qua knh chn gi ca t th hai chy ngc chiu chiu vo mt ngi li xe. Trong khi ngi li xe th hai vn c th nhn thy nh sng phn cc pht ra t n pha ca xe mnh chiu sang cc vt pha trc, v nh sng phn cc ny sau khi phn x trn cc vt vn gi nguyn phng dao ng song song vi quang trc ca knh chn gi trc mt ngi li xe. 2. Lng knh Nicol (Nicn) Lng knh Nicol (gi tt l nicn) l mt khi tinh th bng lan c ct theo mt cho thnh hai na v dn li vi nhau bng mt lp nha canaa trong sut c chit sut n= 1,550.

66

Chng 4: Phn cc nh sng Tia sng t nhin SI chiu vo mt AC ca nicn theo phng song song vi mt y CA' b tch thnh hai: tia thng v tia bt thng. Chit sut ca tinh th i vi tia thng no=1,659, cn chit sut ca tinh th i vi tia bt thng ne ph thuc vo hng, n thay i t 1,486 n 1,659. V no > ne nn tia thng b khc x mnh hn tia bt thng. Chit sut ca tinh th i vi tia thng ln hn chit sut ca lp nha v hnh dng, kch thc ca nicn c chn sao cho tia thng khi n lp nha canaa b phn x ton phn v sau b hp th trn lp sn en ca mt y CA'. Cn tia bt thng (ne < n) truyn qua lp nha canaa v l ra khi nicn theo phng song song vi tia ti SI.

Hnh 4-7. Lng knh Nicol Nh vy, nicn bin nh sng t nhin (hoc phn cc mt phn) truyn qua n thnh nh sng phn cc ton phn c mt phng dao ng trng vi mt phng chnh ca nicn. Nu cho mt chm sng t nhin qua h hai nicn N1 v N2 th cng sng I2 pha sau bn nicn N2 cng c xc nh theo nh lut Malus (cng thc 4-4), vi l gc gia hai mt phng chnh ca nicn N1 v N2. Khi hai nicn N1 v N2 t v tr song song, ng vi = 0, cng sng sau nicn N2 t cc i I2 = Imax (sng nht). Khi hai nicn t v tr bt cho, ng vi =/2, cng sng sau nicn N2 t cc tiu I2 = Imin (ti nht).

Hnh 4-8. a) Hai nicn song song

b) Hai nicn bt cho

67

Chng 4: Phn cc nh sng 4. NH SNG PHN CC ELIP Trong cc tit trc chng ta nghin cu nh sng phn cc thng, l nh sng c vect sng E dao ng theo mt phng xc nh, tc l E dao ng trn ng thng. Thc nghim ch ra rng ta c th to ra nh sng phn cc trong u mt vect sng E chuyn ng trn mt ng elip (hay ng trn), nh sng phn cc ny c gi l nh sng phn cc elip hay phn cc trn. Xt bn tinh th T c quang trc v dy d. Chiu vung gc vi mt trc ca bn tinh th mt tia sng phn cc ton phn c vect sng E hp vi quang trc mt gc . Khi vo bn tinh th, tia sng ny b tch thnh hai: tia thng v tia bt thng. Tia thng c vect sng E o vung gc vi quang trc, cn tia bt thng c vect sng E e song song vi quang trc v c hai vect sng u nm trong mt phng vung gc vi tia sng (hnh 4-9).

Hnh 4-9. nh sng phn cc elip Vect sng tng hp ca tia thng v tia bt thng ti im M sau bn tinh th bng:E = Eo + Ee

(4-8)

trong bn tinh th, hai tia ny truyn i vi vn tc khc nhau (do chit sut ca tinh th i vi hai tia khc nhau, ne no ) v khi l ra khi bn chng li truyn i vi cng vn tc. Do , hiu quang l ca tia thng v tia bt thng ti mt im M sau bn bng:

L = L o - L e = (n o - n e )dtng ng vi hiu pha l

(4-9)

=

2 2 (L o - L e ) = ( n o - n e )d

(4-10)

trong l bc sng nh sng trong chn khng.

68

Chng 4: Phn cc nh sng Cc vect sng E o v E e dao ng theo hai phng vung gc vi nhau, do u mt vect sng tng hp s chuyn ng trn mt ng elip xc nh bi phng trnh:

x22 A1

+

y2 A2 2

-

2 xy cos = sin 2 A 1A 2

(4-11)

vi A1 v A2 ln lt l bin v = o - e l hiu pha dao ng ca hai vect sng

E o v E e . Nu trc khi vo bn tinh th, nh sng phn cc ton phn c bin l A th A1=A.sin v A2=A.cos .Nh vy, nh sng phn cc thng sau khi truyn qua bn tinh th s bin thnh nh sng phn cc elip. Chng ta s xt mt vi trng hp ring ph thuc vo dy d ca bn tinh th. 1. Bn phn t bc sng Bn phn t bc sng l bn tinh th c dy d sao cho hiu quang l ca tia thng v tia bt thng truyn qua bn bng mt s l ln ca phn t bc sng:

L = (n o - n e )d = (2k + 1)Khi hiu pha ca hai tia bng:

4

(4-12)

= (2k + 1)v phng trnh (4-11) s thnh:

2

(4-13)

x22 A1

+

y2 A2 2

=1

(4-14)

Hnh 4-10a: Phn cc elip dng chnh tc

Hnh 4-10b: Phn cc trn

Trong trng hp ny, u mt ca vect sng tng hp E pha sau bn tinh th chuyn ng trn mt elip dng chnh tc c hai bn trc l A1 v A2 c xc nh bi phng trnh (4-14) (hnh 4-10a). c bit, nu = 45o th A1 = A2 = A0 v phng trnh (4-14) s thnh:2 x 2 + y2 = A0

(4-15) 69

Chng 4: Phn cc nh sng Khi u mt ca vect sng tng hp E pha sau bn tinh th chuyn ng trn ng trn tm O, bn knh A0 c xc nh bi phng trnh (4-15) (hnh 4-10b). Nh vy, sau khi truyn qua bn phn t bc sng, nh sng phn cc thng b bin i thnh nh sng phn cc elip dng chnh tc hoc phn cc trn. 2. Bn na bc sng Bn na bc sng l bn tinh th c dy d sao cho hiu quang l ca tia thng v tia bt thng truyn qua bn bng mt s l ln na bc sng:

L = (n o - n e )d = (2k + 1)Khi hiu pha ca hai tia bng:

2(4-17)

(4-16)

= (2k + 1)v phng trnh (4-11) s thnh:

x y + =0 A1 A 2

(4-18) Hnh 4-11

y l phng trnh ca ng thng, mt vect sng tng hp E pha sau bn s chuyn ng trn ng thng nm

trong gc phn t th hai v th t ca h ta Oxy (hnh 4-11), ng thng hp vi quang trc mt gc . Trc khi vo bn tinh th, mt vect sng ca nh sng phn cc thng dao ng trn ng thng. Nh vy sau khi truyn qua bn na bc sng nh sng phn cc thng vn l nh sng phn cc thng, nhng phng dao ng quay i mt gc 2 so vi trc khi i vo bn. 3. Bn mt bc sng Bn mt bc sng l bn tinh th c dy d sao cho hiu quang l ca tia thng v tia bt thng truyn qua bn bng mt s nguyn ln bc sng:

L = (n o - n e )d = kkhi hiu pha ca hai tia bng:

(4-19) (4-20)

= 2kv phng trnh (4-11) s thnh:

Hnh 4-12

x y =0 A1 A 2

(4-21)

y l phng trnh ca ng thng, nm trong gc phn t th nht v th ba ca h ta Oxy (hnh 4-12), ng thng hp vi quang trc mt gc . Nh vy sau khi truyn qua bn mt bc sng nh sng phn cc thng gi nguyn khng i.

70

Chng 4: Phn cc nh sng 5. LNG CHIT DO IN TRNG Mt s cht lng nh sulfua ccbon, benzn... khi chu tc dng ca in trng th tr nn bt ng hng v mt quang hc. Hin tng ny c Kerr tm ra nm 1875 v gi l hiu ng Kerr. S th nghim v hiu ng Kerr c trnh by trn hnh 4-13.

Hnh 4-13. Th nghim v hiu ng Kerr Khi cha c in trng, cc phn t cht lng chuyn ng nhit hn lon nn cht lng l ng hng v khng lm thay i phng ca nh sng phn cc ton phn sau nicn N1 truyn ti n. Do nh sng phn cc ton phn ny khng th truyn tip qua nicn N2 (bt cho vi N1) v sau nicn N2 s hon ton ti. Khi cht lng chu tc dng ca in trng gia hai bn cc ca t in, cc phn t ca n tr thnh cc lng cc in nm dc theo phng ca in trng. Cht lng tr thnh mi trng bt ng hng vi quang trc l phng ca in trng. Trong trng hp ny, chm nh sng phn cc ton phn sau nicn N1 truyn ti cht lng b tch thnh tia thng v tia bt thng. Tng hp ca hai tia ny s l nh sng phn cc elip, c th truyn tip qua nicn N2 (bt cho vi N1), nn sau nicn N2 s sng. Thc nghim chng t vi mi nh sng n sc, hiu s chit sut no - ne ca cht lng (chu tc dng ca in trng) i vi tia thng v tia bt thng truyn trong n c ln t l vi bnh phng cng in trng E tc dng ln cht lng:n o n e = kE 2

(4-22)

vi k l h s t l ph thuc vo bn cht ca cht lng. Hiu pha gia hai dao ng ca tia thng v tia bt thng sau khi i qua lp cht lng c b dy d s l: = 2 2 2 ( n o n e )d = kE d = 2BE 2 d

(4-23)

trong B = k/ gi l hng s Kerr. Gi tr ca B ph thuc nhit ca cht lng v bc sng nh sng. Thi gian cc phn t nh hng theo phng ca in trng v thi gian cc phn t tr v trng thi chuyn ng hn lon ch vo c 10-10s. Tnh cht ny ca hiu ng Kerr c ng dng ch to van quang hc dng ng ngt nh sng rt nhanh khng c qun tnh.

71

Chng 4: Phn cc nh sng 6. S QUAY MT PHNG PHN CC Mt s tinh th hoc dung dch c tc dng lm quay mt phng phn cc ca chm nh sng phn cc ton phn truyn qua chng. Hin tng ny gi l hin tng quay mt phng phn cc. Cc cht lm quay mt phng phn cc ca nh sng phn cc gi l cht hot quang, th d nh thch anh, dung dch ng... Hin tng quay mt phng phn cc c th hin nh sau: Cho nh sng t nhin i qua knh phn cc T1 v knh phn tch T2 t vung gc vi nhau. Kt qu l nh sng khng i qua c knh phn tch T2, sau bn T2 s ti. By gi nu t gia knh phn cc T1 v knh phn tch T2 mt bn tinh th thch anh c quang trc nm dc theo phng truyn ca tia sng th thy nh sng i qua c knh phn tch T2, sau bn T2 s sng. Mun cho nh sng khng i qua c ta phi quay knh phn tch mt gc . iu chng t di tc dng ca bn tinh th nh sng phn cc thng sau bn T1 b quay i mt gc (hnh 4-14), hay ta ni bn tinh th lm quay mt phng phn cc mt gc . l hin Hnh 4-14. Hin tng quay mt phng phn cc tng quay mt phng phn cc. Thc nghim cho thy gc quay ca mt phng phn cc t l thun vi dy d ca bn tinh th:

= d

(4-24)

l h s quay, n c gi tr ph thuc bn cht, nhit ca cht rn quang hot v bc sng ca nh sng. V d i vi bn thch anh 200C: = 21,7 /mm ng vi = 0,589 m; = 48,9 /mm ng vi = 0,4047 m. i vi cc dung dch, gc quay ca mt phng phn cc t l vi dy d ca lp dung dch c nh sng phn cc truyn qua v t l vi nng c ca dung dch:

= [ ] cd

(4-25)

trong [] c gi l h s quay ring, n c gi tr ph thuc bn cht v nhit ca dung dch hot quang, ng thi ph thuc bc sng ca nh sng. V d i vi nh sng vng Na ( = 0,589m) 200C, [] ca dung dch ng l 66,50cm2/g. Hin tng quay mt phng phn cc c ng dng trong mt dng c gi l ng k xc nh nng ng trong dung dch. nh sng t bng n S truyn qua knh lc sc F v knh phn cc P bin i thnh nh sng n sc phn cc ton phn. Quan st trong ng ngm O, ng thi quay knh phn tch A cho ti khi th trng trong ng ngm tr nn ti hon ton. Khi knh phn tch A nm v tr bt cho vi knh phn cc P v mt phng chnh ca chng vung gc vi nhau. Gc 1 xc nh v tr ca knh phn tch A c c trn thc o gc K. t 72

Chng 4: Phn cc nh sng ng thu tinh H cha y dung dch hot quang cn nghin cu vo khong gia hai knh A v P, th trng trong ng ngm O li sng. Nguyn nhn l do dung dch hot quang lm mt phng dao ng ca nh sng phn cc ton phn truyn qua n quay i mt gc ti v tr khng vung gc vi mt phng chnh ca knh phn tch A na. By gi ta quay knh phn tch A cho n khi th trng trong ng ngm O ti hon ton. c gc 2, xc nh v tr ny ca knh phn tch A. T tm ra c gc quay ca mt phng phn cc = 2 - 1.

Hnh 4-15. M hnh ca ng k Theo cng thc (4-25), nu bit dy d v hng s quay ring [ ] ca dung dch hot quang, ta d dng xc nh c nng c ca dung dch :

c=

1 = 2 [].d [].d

(4-26)

III. TM TT NI DUNG1. S phn cc nh sng * nh sng c vect cng in trng dao ng u n theo mi phng vung gc tia sng c gi l nh sng t nhin. * nh sng c vect cng in trng ch dao ng theo mt phng xc nh c gi l nh sng phn cc thng hay nh sng phn cc ton phn. * nh sng c vect cng in trng dao ng theo tt c cc phng vung gc vi tia sng nhng c phng dao ng yu, c phng dao ng mnh c gi l nh sng phn cc mt phn. * Mt phng cha tia sng v phng dao ng ca E c gi l mt phng dao ng, cn mt phng cha tia sng v vung gc vi mt phng dao ng gi l mt phng phn cc. * Trong bn Tuamalin c mt phng c bit gi l quang trc ca tinh th (k hiu l ) . Theo phng quang trc, nh sng khng b hp th, m truyn qua hon ton cn theo phng vung gc vi quang trc, nh sng b hp th hon ton. * nh lut Malus: Khi cho mt chm tia sng t nhin truyn qua hai bn tuamalin c quang trc hp vi nhau mt gc th cng sng nhn c t l vi cos2. 73


I 2 = I1 cos 2 2. S phn cc do phn x, khc x:Thc nghim chng t rng khi cho mt tia sng t nhin chiu ti mt phn cch gia hai mi trng di gc ti i 0 th tia phn x v tia khc x u l nh sng phn cc mt phn. Vect cng in trng ca tia phn x c bin dao ng ln nht theo phng vung gc vi mt phng ti, cn vect cng in trng ca tia khc x c bin dao ng ln nht theo phng nm trong mt phng ti. Khi thay i gc ti i th mc phn cc ca tia phn x v tia khc x cng thay i. Khi gc ti i tha mn iu kin: tg iB = n21 th tia phn x s phn cc ton phn, n21 l chit sut t i ca mi trng hai i vi mi trng mt, iB c gi l gc ti Brewster. Tia khc x khng bao gi l nh sng phn cc ton phn, nhng khi i = iB th tia khc x cng b phn cc mnh nht.B B

3. S phn cc do lng chit Thc nghim chng t rng mt s tinh th nh bng lan, thch anh... c tnh cht c bit l nu chiu mt tia sng n tinh th th ni chung ta s thu c hai tia. Hin tng ny gi l hin tng lng chit. Nguyn nhn l do tnh bt ng hng ca tinh th v mt quang hc (tc l tnh cht quang ca tinh th cc hng khc nhau th s khc nhau). Tia sng khi chiu vo tinh th lng chit s b tch thnh hai tia khc x: - Tia tun theo nh lut khc x gi l tia thng. Tia thng phn cc ton phn, c vect sng E vung gc vi mt phng chnh ca tia thng. - Tia khng theo nh lut khc x gi l tia bt thng. Tia bt thng phn cc ton phn, c vect sng E nm trong mt phng chnh ca n. Khi l ra khi tinh th, hai tia thng v tia bt thng ch khc nhau v phng phn cc. Chit sut ca tinh th bng lan i vi tia thng lun khng i v bng no=1,659. Chit sut ne ca tinh th bng lan i vi tia bt thng ph thuc vo phng truyn ca n trong tinh th v thay i t 1,659 (theo phng quang trc) n 1,486 (theo phng vung gc vi quang trc). Nh vy i vi tinh th bng lan ta c: ne no V chit sut n = c/v, vi c l vn tc nh sng trong chn khng v v l vn tc nh sng trong mi trng, do : ve vo ngha l trong tinh th bng lan, vn tc ca tia bt thng ni chung ln hn vn tc ca tia thng. Tinh th bng lan, thch anh, tuamalin... l nhng tinh th n trc. Trong t nhin cn c tinh th lng trc, l nhng tinh th c hai quang trc theo hai hng khc nhau. Mt tia sng t nhin truyn qua tinh th lng trc cng b tch thnh hai tia khc x nhng c hai tia ny u l nhng tia bt thng.

74

Chng 4: Phn cc nh sng Ngi ta s dng cc tinh th lng chit ch to knh phn cc. Knh phn cc l nhng dng c c th bin nh sng t nhin thnh nh sng phn cc, v d nh bn tuamalin, bn plarit, lng knh nicol... Mt s cht lng nh sulfua ccbon, benzn... khi chu tc dng ca in trng th tr nn bt ng hng v mt quang hc (c tnh lng chit). Hiu ng ny gi l hiu ng Kerr v c ng dng ch to van quang hc 4. nh sng phn cc elip nh sng phn cc trong u mt vect sng E chuyn ng trn mt ng elip (hay ng trn) c gi l nh sng phn cc elip (hay phn cc trn) . Chiu vung gc vi mt trc ca bn tinh th mt tia sng phn cc ton phn c vect sng E hp vi quang trc mt gc . Khi vo bn tinh th, tia sng ny b tch thnh hai: tia thng v tia bt thng. Tia thng v tia bt thng l hai tia sng kt hp, chng giao thoa vi nhau. Cc vect sng E o ca tia thng v E e dao ng theo hai phng vung gc vi nhau, do u mt vect sng tng hp s chuyn ng trn mt ng elip xc nh bi phng trnh:

x22 A1

+

2 xy cos = sin 2 2 A A A2 1 2 -

y2

(1)

x, y l di dao ng, A1, A2 l bin dao ng ca E o v E e . Hiu pha ca cc tia thng v tia bt thng l

=* Bn phn t bc sng

2 2 (L o - L e ) = ( n o - n e )d

(2)

Bn phn t bc sng l bn tinh th c dy d sao cho hiu quang l ca tia thng v tia bt thng truyn qua bn bng mt s l ln ca phn t bc sng:

L = (n o - n e )d = (2k + 1)

4

(3a)

Thay (3a) vo (2), sau vo (1) ta thu c phng trnh ca ng elip dng chnh tc. Do sau khi truyn qua bn phn t bc sng, nh sng phn cc thng b bin i thnh nh sng phn cc elip dng chnh tc hoc phn cc trn. * Bn na bc sng Bn na bc sng l bn tinh th c dy d sao cho hiu quang l ca tia thng v tia bt thng truyn qua bn bng mt s l ln na bc sng:

L = (n o - n e )d = (2k + 1)

2

(3b)

Thay (3b) vo (2), sau vo (1) ta thu c phng trnh ca ng thng, quay mt gc 2. Do khi truyn qua bn na bc sng nh sng phn cc thng vn l nh sng phn cc thng, nhng phng dao ng quay i mt gc 2 so vi trc khi i vo bn. 75

Chng 4: Phn cc nh sng * Bn mt bc sng Bn mt bc sng l bn tinh th c dy d sao cho hiu quang l ca tia thng v tia bt thng truyn qua bn bng mt s nguyn ln bc sng:

L = (n o - n e )d = k

(3c)

Thay (3c) vo (2), sau vo (1) ta thu c phng trnh ca ng thng. Vy sau khi truyn qua bn mt bc sng nh sng phn cc thng gi nguyn khng i. 5. S quay mt phng phn cc Mt s tinh th hoc dung dch c tc dng lm quay mt phng phn cc ca chm nh sng phn cc ton phn truyn qua chng. Hin tng ny gi l s quay mt phng phn cc. Cc cht lm quay mt phng phn cc ca nh sng phn cc gi l cht hot quang, th d nh thch anh, dung dch ng... Thc nghim cho thy gc quay ca mt phng phn cc t l thun vi dy d ca bn tinh th:

= d

l h s quay, n c gi tr ph thuc bn cht v nhit ca cht rn quang hot v bc sng ca nh sng. i vi cc dung dch, gc quay ca mt phng phn cc t l vi dy d ca lp dung dch c nh sng phn cc truyn qua v t l vi nng c ca dung dch:

= [ ] cdtrong [] c gi l h s quay ring, n c gi tr ph thuc bn cht v nhit ca dung dch hot quang, ng thi ph thuc bc sng ca nh sng. Hin tng quay mt phng phn cc c ng dng trong mt dng c gi l ng k xc nh nng ng trong dung dch.

IV. CU HI L THUYT1. Hin tng phn cc chng t bn cht g ca nh sng ? nh sng l sng ngang hay sng dc ? Gii thch ti sao ? 2. Phn bit nh sng t nhin v nh sng phn cc ton phn, nh sng phn cc mt phn. 3. Pht biu v vit biu thc ca nh lut Malus. 4. Trnh by s phn cc do phn x, khc x. 5. Trnh by tnh lng chit ca tinh th. 6. Nu s ging nhau v khc nhau ca hai tia thng v bt thng khi i qua tinh th bng lan. 7. Trnh by hiu ng Kerr. 8. nh ngha nh sng phn cc elip, phn cc trn. Trnh by cch to ra nh sng phn cc elip. Xt cc trng hp b dy bn mt phn t bc sng, bn na bc sng v bn mt bc sng. 76

Chng 4: Phn cc nh sng 8. Nu ng dng ca hin tng quay mt phng phn cc.

V. BI TPTh d 1: Hi gc nghing ca mt tri so vi chn tri phi bng bao nhiu nhng tia sng mt tri phn chiu trn mt h b phn cc ton phn. Bit rng chit sut ca nc h n = 1,33. Bi gii: Theo nh lut Brewster, mun tia sng phn chiu b phn cc ton phn th gc ti ca n phi bng gc ti Brewster, xc nh bi cng thc:

tgi B = n = 1,33 i B = 530 5Do gc nghing ca mt tri so vi ng chn tri: = 900 i B = 36055 Th d 2: Cho mt chm tia sng phn cc thng c bc sng trong chn khng l 0 = 0,589m chiu vung gc vi quang trc ca mt bn tinh th bng lan. Chit sut ca tinh th bng lan i vi tia thng v tia bt thng ln lt bng n0 = 1,658 v ne = 1,488. Xc nh bc sng ca tia thng v tia bt thng. Bi gii: Bc sng ca nh sng truyn trong mi trng c chit sut n lin h vi bc sng 0 ca nh sng trong chn khng: = 0 n Bc sng ca tia thng trong tinh th bng lan: t =

0 0,589 = = 0,355m n 0 1,658 0 = 0,396m ne

Bc sng ca tia bt thng trong tinh th bng lan: bt = Bi tp t gii

1. Cho bit khi nh sng truyn t mt cht c chit sut n ra ngoi khng kh th xy ra hin tng phn x ton phn ca nh sng ng vi gc gii hn igh = 450. Xc nh gc ti Brewster ca cht ny, mi trng cha tia ti l khng kh. p s: Gc gii hn:

n 1 n = 2 = 1,414 sin i gh = kk = n 2 tgi B = n n kk = 1,414 i B = 54 0 43

77

Chng 4: Phn cc nh sng 2. nh sng t nhin truyn t khng kh ti chiu vo mt bn thu tinh. Cho bit nh sng phn x b phn cc ton phn khi gc khc x r = 330. Xc nh chit sut ca bn thu tinh. p s: Khi chm phn x b phn cc ton phn th gc ti i tho mn:

tgi = tgi B = n n =Theo nh lut khc x nh sng:

sin i B sin i B = cos i B 1 sin 2 i B

sin i = n , m sin i B = sin i = n. sin 330 sin r0 2

n=

n. sin 330 1 n. sin 33

(

)

n 1,56

3. Xc nh gc ti Brewster ca mt mt thu tinh c chit sut n1 = 1,57 khi mi trng nh sng ti l: 1. Khng kh. 2. Nc c chit sut n2 = 4/3. p s: i B = 57 0 30, i B = 49 0 43 4. Mt chm tia sng sau khi truyn qua mt cht lng ng trong mt bnh thu tinh, phn x trn y bnh. Tia phn x b phn cc ton phn khi gc ti trn y bnh bng 42 0 37 , chit sut ca bnh thu tinh n = 1,5. Tnh: 1. Chit sut ca cht lng. 2. Gc ti trn y bnh chm tia phn x trn phn x ton phn. p s: n/ = 1,63, i = 66056/ 5. Cho mt chm tia sng t nhin chiu vo mt ca mt bn thu tinh nhng trong cht lng. Chit sut ca thu tinh l n1 = 1,5. Cho bit chm tia phn x trn mt thu tinh b phn cc ton phn khi cc tia phn x hp vi cc tia ti mt gc = 97 0 . Xc nh chit sut n2 ca cht lng. p s: tgi = tgi B =

n1 n2

Theo iu kin u bi: i = i B =

n tg = 1 n 2 = 2 2 n2

n1 97 0 tg 2

= 1,33 .

6. Mt bn thch anh c ct song song vi quang trc v c dy d = 1mm. Chiu nh sng n sc c bc sng = 0,6m vung gc vi mt bn. Tnh hiu quang l v hiu

78

Chng 4: Phn cc nh sng pha ca tia thng v tia bt thng truyn qua bn thch anh, bit rng chit sut ca bn i vi tia thng v tia bt thng ln lt bng n0 = 1,544, ne = 1,535. p s: Hiu quang l ca tia thng v tia bt thng truyn qua bn thch anh c gi tr bng: L = (n 0 n e ).d = 0,009mm Hiu pha ca tia thng v tia bt thng truyn qua bn thch anh c gi tr bng: 2 = .L = 30 (rad) 7. Cho bit i vi nh sng n sc c bc sng = 0,545m th chit sut ca bn phn t bc sng i vi tia thng v tia bt thng truyn trong bn c gi tr ln lt bng n0 = 1,658 v ne = 1,488. Hi bn phn t bc sng c dy nh nht bng bao nhiu?

p s: L = (n 0 n e )d = (2k + 1) , k = 0,1, 2, 3,... 4Bn phn t bc sng c dy nh nht khi k = 0. Vy dmin = 800nm. 8. Mt bn na bc sng c dy nh nht bng dmin = 1,732m. Cho bit chit sut ca bn i vi tia thng v tia bt thng ln lt bng n0 = 1,658 v ne = 1,488. Xc nh bc sng ca nh sng truyn ti bn ny.

p s: L = (n 0 n e ).d = (2k + 1) , k = 0,1, 2, 3,... 2Bn na bc sng c dy nh nht khi k = 0. Vy d min =

= 2.d min (n 0 n e ) = 0,589m 2(n 0 n e )

9. Mt bn tinh th c ct song song vi quang trc v c dy d = 0,25mm. Ngi ta dng bn tinh th ny lm bn phn t bc sng i vi nh sng c bc sng = 0,53m. Xc nh i vi nhng bc sng no trong vng quang ph thy c, bn tinh th ny cng l bn phn t bc sng. Coi rng hiu s chit sut ca bn i vi tia bt thng v tia thng khng i v bng ne n0 = 0,009 ng vi mi bc sng trong vng quang ph thy c c gi tr t 0,4m n 0,76m. p s: L = (n e n 0 )d = (2k + 1)

4

Bc sng ca nh sng truyn ti bn: =

4d(n e n o ) 9 , m 0,4 0,76, suy = 2k + 1 2k + 1 ra 5,42 k 10,75, m k nguyn nn k = 6, 7, 8,9, 10.

79

Chng 4: Phn cc nh sng9 9 = 0,69m, k = 7 = = 0,6m 2.6 + 1 2.7 + 1 9 9 k =8 = = 0,53m, k = 9 = = 0,47m 2 .8 + 1 2.9 + 1 9 k = 10 = = 0,43m 2.10 + 1 k =6=

Vy bn tinh th cn l bn phn t bc sng i vi cc nh sng c bc sng trn. 10. Mt bn thch anh c ct song song vi quang trc ca n vi dy khng vt qu 0,5mm. Xc nh dy ln nht ca bn thch anh ny chm nh sng phn cc phn cc thng c bc sng = 0,589m sau khi truyn qua bn tho mn iu kin sau: 1. Mt phng phn cc b quay i mt gc no . 2. Tr thnh nh sng phn cc trn. Cho bit hiu s chit sut ca tia thng v tia bt thng i vi bn thch anh ne n0 = 0,009. p s: 1. dmax = 0,49mm. 2.dmax = 0,47mm 11. Gia hai knh nicn song song ngi ta t mt bn thch anh c cc mt vung gc vi quang trc. Khi bn thch anh c dy d1 = 2mm th mt phng phn cc ca nh sng n sc truyn qua n b quay i mt gc 1 = 530. Xc nh dy d2 ca bn thch anh ny nh sng n sc khng truyn qua c knh nicn phn tch. p s: Khi truyn theo quang trc ca thch anh mt phng phn cc ca nh sng b quay mt gc 1: 1 = .d1 nh sng n sc khng truyn qua c knh phn tch th bn thch anh phi c dy d2 sao cho mt phng phn cc quay i mt gc 2 = 900, m 2 =.d2d 2 = 2 d 2 = 3,4mm d1 1

80

Chng 5: Thuyt tng i hp Einstein

CHNG V: THUYT TNG I HP EINSTEINTheo c hc c in (c hc Newton) th khng gian, thi gian v vt cht khng ph thuc vo chuyn ng; khng gian v thi gian l tuyt i, kch thc v khi lng ca vt l bt bin. Nhng n cui th k 19 v u th k 20, khoa hc k thut pht trin mnh, ngi ta gp nhng vt chuyn ng nhanh vi vn tc c vn tc nh sng trong chn khng (3.108 m/s), khi xut hin s mu thun vi cc quan im ca c hc Newton: Khng gian, thi gian v khi lng ca vt khi chuyn ng vi vn tc gn bng vn tc nh sng th ph thuc vo chuyn ng. Nm 1905, Einstein mi 25 tui xut l thuyt tng i ca mnh. L thuyt tng i c xem l mt l thuyt tuyt p v khng gian v thi gian. L thuyt ng vng qua nhiu th thch thc nghim trong sut 100 nm qua. L thuyt tng i da trn hai nguyn l: nguyn l tng i v nguyn l v s bt bin ca vn tc nh sng.

I. MC CH - YU CU1. Hiu c ngha ca nguyn l tng i Einstein, nguyn l v tnh bt bin ca vn tc nh sng. 2. Hiu v vn dng c php bin i Lorentz. Tnh tng i ca khng gian, thi gian. 3. Nm c khi lng, ng lng tng i tnh, h thc Einstein v ng dng.

II. NI DUNG1. CC TIN EINSTEIN 1. Nguyn l tng i: Mi nh lut vt l u nh nhau trong cc h qui chiu qun tnh. Galileo tha nhn rng nhng nh lut ca c hc hon ton ging nhau trong mi h qui chiu qun tnh. Einstein m rng tng ny cho ton b cc nh lut vt l trong cc lnh vc in t, quang hc... 2. Nguyn l v s bt bin ca vn tc nh sng: Vn tc nh sng trong chn khng u bng nhau i vi mi h qun tnh. N c gi tr bng c = 3.108 m/s v l gi tr vn tc cc i trong t nhin.

81

Chng 5: Thuyt tng i hp Einstein 2. NG HC TNG I TNH PHP BIN I LORENTZ 1. S mu thun ca php bin i Galileo vi thuyt tng i Einstein Xt hai h qui chiu qun tnh K v K'. H K' chuyn ng thng u vi vn tc V so vi h K, dc theo phng x. Theo php bin i Galileo, thi gian din bin mt qu trnh vt l trong cc h qui chiu qun tnh K v K u nh nhau: t = t. Khong cch gia hai im 1 v 2 no o c trong hai h K v K u bng nhau:

l = x 2 x1 = l = x 2 x1trong h K trong h K/ Vn tc v ca cht im chuyn ng trong h K bng tng cc vn tc v' ca cht im trong h K v vn tc V ca h K' i vi h K:

v = v'+ VTt c cc kt qu trn y u ng i vi v c, cn nh sng truyn n C vi vn tc c -V< c. iu ny mu thun vi nguyn l th 2 trong thuyt tng i Einstein.

82

Chng 5: Thuyt tng i hp Einstein 2. Php bin i Lorentz Lorentz tm ra php bin i cc ta khng gian v thi gian khi chuyn t h qun tnh ny sang h qun tnh khc, tha mn cc yu cu ca thuyt tng i Einstein. Php bin i ny c gi l php bin i Lorentz. Php bin i Lorentz da trn hai tin ca Einstein.Xt hai h qui chiu qun tnh K v K. Ti t = 0, hai gc O, O trng nhau, K chuyn ng thng u so vi K vi vn tc V theo phng x. Theo thuyt tng i thi gian khng c tnh cht tuyt i m ph thuc vo h qui chiu, ngha l t t. Gi s ta x l hm ca x v t theo phng trnh: x = f(x,t) (5-1) tm dng ca phng trnh trn ta hy vit phng trnh chuyn ng ca hai gc ta O v O. i vi h K, gc O chuyn ng vi vn tc V. Ta c: x = Vt hay x Vt = 0 (5-2) x l ta ca gc O trong h K. i vi h K, gc O ng yn, do ta x ca n s l: x = 0 (5-3)

Phng trnh (5-1) cng phi ng i vi im O, iu c ngha l khi ta thay x = 0 vo phng trnh (5-1) th phi thu c phng trnh (5-2), mun vy th:x ' = ( x Vt )

(5-4)

trong l hng s. i vi h K, gc O chuyn ng vi vn tc V. Nhng i vi h K, gc O l ng yn. Lp lun tng t nh trn ta cx = ( x '+ Vt ' )

(5-5)

trong l hng s. Theo tin th nht ca Einstein th mi h qui chiu qun tnh u tng ng nhau, ngha l t (5-4) c th suy ra (5-5) v ngc li bng cch thay V-V, x x, t t. Suy ra: = . Theo tin hai: x = ct t = x/c x = ct t = x/c Thay t v t vo (5-4) v (5-5) ta c:xV x' = x , c x' V x = x '+ c

Nhn v vi v ca hai h thc trn, sau rt gn ta nhn c:= 1 1 V2 c2

Thay vo cc cng thc trn ta nhn c cc cng thc ca php bin i Lorentz.

Php bin i Lorentz:

x' =

x Vt 1 V c2 2

,

x=

x '+ Vt ' 1 V2

(5-6)

c2

83

Chng 5: Thuyt tng i hp Einstein

tv

V c2 V c

x2

t '+,

V c2

x'2

t' =

t=

(5-7)

1

2

1

V

c2

V h K chuyn ng dc theo trc x nn y = y v z = z. T kt qu trn ta nhn thy nu c (tng tc tc thi) hay khi V c 0 (s gn ng c in khi V c, ta x, t tr nn o, do khng th c cc chuyn ng vi vn tc ln hn vn tc nh sng. 3. CC H QU CA PHP BIN I LORENTZ 1. Khi nim v tnh ng thi v quan h nhn qu Gi s trong h qun tnh K c hai bin c A1(x1, y1, z1, t1) v bin c A2(x2, y2, z2, t2) vi x1 x 2 . Chng ta hy tm khong thi gian t t1 gia hai bin c trong h K' 2 chuyn ng u i vi h K vi vn tc V dc theo trc x. T cc cng thc bin i Lorentz ta c

t '2 t '1 =

t 2 t1

V c2

( x 2 x1 ) V2

(5-8)

1

c2

T (5-8) ta suy ra rng nhng bin c xy ra ng thi trong h K (t1 = t2) s khng ng thi trong h K v t '2 t '1 0 , ch c mt trng hp ngoi l l khi hai bin c xy ra ng thi ti nhng im c cng gi tr ca x (y c th khc nhau). Nh vy khi nim ng thi l mt khi nim tng i, hai bin c xy ra ng thi trong mt h qui chiu qun tnh ny ni chung c th khng ng thi trong mt h qui chiu qun tnh khc. Nhn vo cng thc (5-8) ta thy gi s trong h K: t2 - t1>0 (tc l bin c A1 xy ra trc bin c A2), nhng trong h K: t2 - t1 cha chc ln hn 0, n ph thuc vo du V v ln ca ( x 2 x1 ) . Nh vy trong h K th t ca cc bin c c th bt k. c2 Tuy nhin iu ny khng c xt cho cc bin c c quan h nhn qu vi nhau. Mi quan h nhn qu l mi quan h c nguyn nhn v kt qu. Nguyn nhn bao gi cng xy ra trc, kt qu xy ra sau. Nh vy: Th t ca cc bin c c quan h nhn qu bao gi cng c m bo trong mi h qui chiu qun tnh. Th d: vin n c 84

Chng 5: Thuyt tng i hp Einstein bn ra (nguyn nhn), vin n trng ch (kt qu). Gi A1(x1, t1) l bin c vin n bn ra v A2(x2, t2) l bin c vin n trng ch. Trong h K: t2 > t1. Gi u l vn tc vin n v gi s x2 > x1, ta c x2 - x1 = u(t2-t1). Thay vo (5-8) ta c:

t '2 t '1 =

t 2 t1

V c2

.u ( t 2 t1 ) V c2

V.u ( t 2 t1 ) 1 c2 = 1 V2

(5-9)

1

2

c2

' Ta lun c u t1 th ta cng c t '2 > t1 . Trong c hai h K v K bao gi bin c vin n trng ch cng xy ra sau bin c vin n c bn ra.

2. S co ca di (s co ngn Lorentz) Xt hai h qui chiu qun tnh K v K'. H K' chuyn ng thng u vi vn tc V so vi h K dc theo trc x. Gi s c mt thanh ng yn trong h K t dc theo trc x, di ca n trong h K bng: l o = x ' 2 x '1 . Gi l l di ca thanh trong h K. T php bin i Lorentz ta c:

x Vt 2 , x '2 = 2 2 V 1 c2

x Vt1 x '1 = 1 V2 1 c2

Ta phi xc nh v tr cc u ca thanh trong h K ti cng mt thi im: t2 = t1, do :

x '2 x '1 =

x 2 x1 1 V2 c2

l = lo 1

V2 c2

< lo

(5-10)

H K' chuyn ng so vi h K, nu ta ng h K quan st th thy thanh chuyn ng cng h K'. Chiu di ca thanh h K nh hn chiu di ca n trong h K'. Vy: di (dc theo phng chuyn ng) ca thanh trong h qui chiu m thanh chuyn ng ngn hn di ca thanh trong h m thanh ng yn. Ni mt cch khc khi vt chuyn ng, kch thc ca n b co ngn theo phng chuyn ng. V d: mt vt c vn tc gn bng vn tc nh sng V=260000 km/s th

0,5 khi l = 0,5 l o , kch thc ca vt s b co ngn i mt na. Nu quan c2 st mt vt hnh hp vung chuyn ng vi vn tc ln nh vy ta s thy n c dng mt hnh hp ch nht, cn mt khi cu s c dng hnh elipxoit trn xoay.Nh vy kch thc ca mt vt s khc nhau tu thuc vo ch ta quan st n trong h ng yn hay chuyn ng. iu ni ln rng khng gian c tnh tng i, n

1

V2

85

Chng 5: Thuyt tng i hp Einstein ph thuc vo chuyn ng. Khi vt chuyn ng vi vn tc nh (V tng ng vi s tn x tia X ln

Hnh 6-8

cc electrn lin kt yu vi ht nhn. Nng lng lin kt ca cc electrn ny rt nh so vi nng lng ca chm tia X chiu ti, do cc electrn c th coi nh t do. V y l va chm n hi gia phtn v electrn t do nn ta s p dng hai nh lut bo ton nng lng v bo ton ng lng cho h kn tia X - e-". Gi thit trc va chm electrn (e-) ng yn. Tia X c nng lng ln, khi tn x trn electrn t do tia X s truyn nng lng cho electrn nn sau va chm vn tc ca electrn rt ln, do ta phi p dng hiu ng tng i tnh trong trng hp ny. Chng ta xt ng lng, nng lng ca ht phtn v electrn trc v sau va chm: Trc va chm: e- ng yn : Phtn : Nng lng : m o c 2 ng lng : 0 Nng lng : E = h ng lng : p = mc = Sau va chm: Phtn tn x: Nng lng : E ' = h ng lng : p =h h = c

h h = c

105

Chng 6: Quang hc lng t e- : Nng lng :

mo 1v2 c2

c 2 = mc 2

ng lng : p e =

mo 1v2 c2

v = mv

(mo l khi lng ngh ca e- ) Theo nh lut bo ton nng lng v ng lng:

h + m o c 2 = h + mc 2 p = p + p e

(6-20) (6-21)

Gi l gc gia p v p' (hnh 6-8). Sau khi bin i cc biu thc (6-20) v (6-21) v s dng cng thc lin h gia nng lng v ng lng trong c hc tng i tnh (5-22), cui cng ta c:

m o c 2 ( - ' ) = h ' (1 - cos ) = 2h ' sin 2Thay =

2

(6-22)

c vo biu thc trn ta c: '- = 2 h sin 2 = 2 c sin 2 moc 2 2(6-23)

trong c =

h = 2,426.10 12 m l hng s chung cho mi cht, gi l bc sng moc

Compton. i lng = '- l bin thin ca bc sng trong tn x, n ch ph thuc vo gc tn x m khng ph thuc vo vt liu lm bia. Khi phtn vo su trong nguyn t v va chm vi cc electrn lin kt mnh vi ht nhn, ta phi coi va chm ny l va chm ca phtn vi nguyn t (ch khng phi vi electrn), cng thc (6-23) vn ng nhng phi thay khi lng ca electrn bng khi lng ca nguyn t, n ln hn nhiu ln so vi khi lng ca electrn. Do hu nh khng c s thay i bc sng. Nh vy trong bc x tn x c mt nhng phtn vi bc sng khng i. Qua hiu ng Compton ngi ta chng minh c ht phtn c ng lng p= h / . ng lng l mt c trng ca ht. Nh vy tnh cht ht ca nh sng c xc nhn trn vn khi da vo thuyt phtn gii thch thnh cng hiu ng Compton.

III. TM TT NI DUNG1. Hin tng bc x nhit 106

Chng 6: Quang hc lng t * Sng in t do cc vt pht ra gi chung l bc x. Dng bc x do cc nguyn t v phn t b kch thch bi tc dng nhit c gi l bc x nhit. Nu phn nng lng ca vt b mt i do pht x bng phn nng lng vt thu c do hp th th bc x nhit khng i v c gi l bc x nhit cn bng. * Cc i lng c trng cho bc x nhit : - Nng sut pht x ton phn ca vt nhit T:

RT =

dT dS dR T d d ', T d , T

dT l nng lng do din tch dS ca vt pht x trong mt n v thi gian.- H s pht x n sc nhit T, ng vi bc sng : r, T = - H s hp th n sc nhit T, ng vi bc sng : a , T =

d , T l nng lng ca bc x ti, d', T l nng lng vt hp th.Thc t vt khng hp th hon ton bc x ti nn a,T < 1. Vt c a,T =1 vi mi nhit T v mi bc sng gi l vt en tuyt i. * nh lut Kirchhoff: T s ca h s pht x n sc v h s hp th n sc ca mt vt trng thi cn bng nhit khng ph thuc vo bn cht ca vt m ch ph thuc vo r, T = f , T , trong f,T l hm s nhit v bc sng ca chm bc x, ngha l a , T chung cho mi vt, nn c gi l hm ph bin. i vi vt en tuyt i: r,T = f,T Nng sut pht x ton phn ca vt en tuyt i bng R T = dR T = f , T d0

* Cc nh lut pht x ca vt en tuyt i Stephan-Boltzmann thit lp c nh lut lin h gia RT v nhit T ca vt:

R T = T 4 . Hng s c gi l hng s Stephan-Boltzmann.Wien tm c nh lut lin h gia bc sng m ca chm bc x mang nhiu b nng lng nht (f,T ln nht) vi nhit tuyt i T ca vt : m = , trong T b c gi l hng s Wien.

* Da vo quan nim c in coi cc nguyn t v phn t pht x v hp th nng lng mt cch lin tc, Rayleigh-Jeans tm c mt cng thc xc nh h s pht x n sc ca vt en tuyt i: f , T =

2 2 c2

kT

Nhng cng thc ny gp hai kh khn ch yu:

107

Chng 6: Quang hc lng t - Cng thc ny ch ph hp vi thc nghim vng tn s nh (bc sng di), cn vng tn s ln (bc sng ngn), tc l vng sng t ngoi, n sai lch rt nhiu. B tc ny c gi l s khng hong vng t ngoi. - T cng thc ny ta c th tnh c nng sut pht x ton phn ca mt vt en tuyt i nhit T:0

R T = f , T d =

2kT 2 d = c2 0

Nng lng pht x ton phn ca vt mt nhit T nht nh li bng v cng. S d c kt qu v l l do quan nim vt l c in v s pht x v hp th nng lng bc x mt cch lin tc. gii quyt nhng b tc trn Planck ph nh l thuyt c in v bc x v ra mt l thuyt mi gi l thuyt lng t nng lng. * Thuyt lng t ca Planck: cc nguyn t v phn t pht x hay hp th nng = h = hc / lng mt cch gin on . Xut pht t thuyt lng t, Planck tm ra cng thc ca hm ph bin, tc l h s pht x n sc ca vt en tuyt i:

c e 1 Cng thc ca Planck khc phc c kh khn vng t ngoi, ng c trng ph pht x ca vt en tuyt i tnh t cng thc ny ph hp vi kt qu thc nghim mi vng nhit , mi vng tn s khc nhau. T cng thc Planck ta c th tm li c cc cng thc Stephan-Boltzmann v cng thc Wien.2. Hiu ng quang in l hiu ng bn ra cc electrn t mt tm kim loi khi ri vo tm kim loi mt bc x in t thch hp. Ngi ta tm c ba nh lut quang in: * nh lut v gii hn quang in: Hin tng quang in ch xy ra khi bc sng ca nh sng ti phi tha mn: < o hoc > o o, o ty thuc vo tng kim loi v c gi l gii hn quang in ca kim loi . * nh lut v dng quang in bo ha: Cng dng quang in bo ha t l vi cng nh sng chiu ti kim loi. * nh lut v ng nng ban u cc i: ng nng ban u cc i ca cc quang electron khng ph thuc vo cng nh sng chiu ti m ch ph thuc bc sng ca nh sng chiu ti v bn cht kim loi. gii thch ba nh lut trn, Einstein a ra thuyt phtn. Thuyt ny cho rng nh sng bao gm nhng ht phtn. Mi phtn mang nng lng = h = hc / , chuyn ng vi vn tc c=3.108 m/s. Cng ca chm sng t l vi s phtn do ngun sng pht ra trong mt n v thi gian. 108

f , T =

2 22

hh / kT

Chng 6: Quang hc lng t Nh vy nh sng va c tnh cht sng va c tnh cht ht. 3. Hiu ng Compton Chm nh sng (chm ht phtn) sau khi tn x ln cc ht electrn t do th bc sng ca n tng ln

= 2 c sin 2

2

Thc nghim xc nh c tng bc sng ny. tng bc sng khng ph thuc vt liu lm bia m ch ph thuc vo gc tn x. gii thch hiu ng Compton, ngi ta da trn hai nh lut bo ton: bo ton nng lng (v va chm n hi) v bo ton ng lng (v l h kn gm ht phtn v ht electrn). Qua hiu ng ny ngi ta chng minh c ht phtn c ng lng p = mc = h / c = h / . ng lng l mt c trng ca ht. Nh vy tnh cht ht ca nh sng c xc nhn trn vn khi da vo thuyt phtn gii thch thnh cng hiu ng Compton.

IV. CU HI L THUYT1. nh ngha bc x nhit cn bng. 2. Vit biu thc v nu ngha ca cc i lng: nng sut pht x ton phn, h s pht x n sc, h s hp th n sc ca bc x nhit cn bng nhit T. 3. nh ngha vt en tuyt i. 4. Pht biu nh lut Kirchhoff. Nu ngha ca hm ph bin. V th ng c trng ph pht x ca vt en tuyt i. 5. Pht biu cc nh lut pht x ca vt en tuyt i . 6. Nu quan nim c in v bn cht ca bc x. Vit cng thc ca Rayleigh-Jeans. Nu nhng kh khn m cng thc gp phi i vi hin tng bc x nhit. 7. Pht biu thuyt lng t ca Planck. Vit cng thc Planck. Nu nhng thnh cng ca thuyt lng t. 8. nh ngha hin tng quang in. Pht biu ba nh lut quang in. 9. Pht biu thuyt phtn ca Einstein. Vn dng thuyt phtn gii thch ba nh lut quang in. 10. Trnh by ni dung hiu ng Compton. Trong hiu ng ny, chm tia X tn x ln electrn t do hay lin kt ? 11. Gii thch hiu ng Compton. 12. Ti sao coi hiu ng Compton l mt bng chng thc nghim xc nhn trn vn tnh ht ca nh sng.

109

Chng 6: Quang hc lng t

IV. BI TPTh d 1: Hi nhit ca l nung bng bao nhiu cho bit mi giy l pht ra mt nng lng bng 8,28 calo qua mt l nh c kch thc bng 6,1cm2. Coi bc x c pht ra t mt vt en tuyt i. Bi gii: Nng sut pht x ton phn ca vt en tuyt i: R = T 4 , R l nng sut do mt n v din tch pht ra trong mt n v thi gian, nn R lin h vi cng sut pht x l: P = R.S

T=4

P 8,28.4,18 =4 = 1004 (K ) .S 5,67.10 8.6,1.10 4

Th d 2: Cng thot ca kim loi dng lm catt ca t bo quang in A = 5eV. Tm: 1. Gii hn quang in ca tm kim loi . 2. Vn tc ban u cc i ca cc quang electrn khi catt c chiu bng nh sng n sc bc sng = 0,2m. 3. Hiu in th hm khng c mt electrn no n c ant. Bi gii 1. Gii hn quang in ca catt: 0 =

hc 6,625.10 34.3.108 = = 2,48.10 7 m 19 A 5.1,6.10

2. Vn tc ban u cc i ca cc electrn:

hc 1 2 hc 2 = A + m e v 0 max v 0 max = A 2 me v 0 max = 6,625.1034.3.108 5.1,6.1019 = 0,65.106 m / s 9,1.1031 0,2.106 2

3. Hiu in th hm:

1 hc hc 1 6,625.1034.3.108 = A + eUh U h = ( A) = 5.1,6.1019 = 1,2 V 1,6.1019 e 0,2.106 Th d 3: Phtn mang nng lng 0,15MeV n tn x trn electrn t do. Sau khi tn x bc sng ca chm phtn tn x tng thm = 0,015A0. Xc nh bc sng ca phtn v gc tn x ca phtn. Bi gii: =

hc hc 6,625.10 34.3.108 = = = 8,28.10 12 m 13 0,15.1,6.10 sin 2 = = 0,31 sin = 0,556 = 67 0 33 2 2 2 c 2110

= 2 c sin 2

Chng 6: Quang hc lng t Bi tp t gii 1. Tm cng sut bc x ca mt l nung, cho bit nhit ca l bng t = 7270C, din tch ca ca l bng 250cm2. Coi l l vt en tuyt i. p s: P = T 4S = 1417,5 ( W ) 2.Vt en tuyt i c dng mt qu cu ng knh d = 10cm nhit T khng i. Tm nhit T, cho bit cng sut bc x nhit cho bng 12kcalo/pht. p s: P =

12.10 3.4,18 = 836 ( W ) , T = 60 4

P d .4 22

= 828 (K )

3. Nhit ca si dy tc vonfram ca bng n in lun bin i v c t nng bng dng in xoay chiu. Hiu s gia nhit cao nht v thp nht bng 800, nhit trung bnh bng 2300K. Hi cng sut bc x bin i bao nhiu ln, coi dy tc bng n l vt en tuyt i. p s: Tmax Tmin = 80 K,

Tmax + Tmin = 2300 K Tmax = 2340 K, Tmin = 2260 K 2 = 1,15 4

Pmax Tmax = Pmin Tmin

4. Nhit ca vt en tuyt i tng t 1000 K n 3000 K. Hi: 1. Nng sut pht x ton phn ca n tng bao nhiu ln? 2. Bc sng ng vi nng sut pht x cc i thay i bao nhiu ln?4

T R p s: 1. 2 = 2 = 81 ln , R 1 T1 T 2. m1 = 2 = 3 ln m 2 T1

5. Mt vt en tuyt i nhit T1 = 2900 K. Do vt b ngui i nn bc sng ng vi nng sut pht x cc i thay i = 9m. Hi vt lnh n nhit bng bao nhiu? p s: m1 =

1 bT1 1 b b , m2 = = b T T T2 = T + b = 290 (K ) T1 T2 1 1 2

6. Tm gii hn quang in i vi cc kim loi c cng thot 2,4eV, 2,3eV, 2eV. p s: 01 =

hc hc = 5,18.10 7 m , 02 = = 5,4.10 7 m , A1 A2 hc = 6,21.10 7 m A3111

03 =

Chng 6: Quang hc lng t 7. Gii hn quang in ca kim loi dng lm catt ca t bo quang in 0 = 0,5m. Tm: 1. Cng thot ca electrn khi tm kim loi . 2. Vn tc ban u cc i ca cc quang electrn khi catt c chiu bng nh sng n sc bc sng = 0,25m. p s: 1. 0 = 2.

hc hc 6,625.10 34.3.108 A= = = 39,75.10 20 J 6 A 0 0,5.10

hc 1 2 hc 2 6 = A + me v0 max v0 max = A = 0,93.10 m / s 2 me

8. Chiu mt bc x in t n sc bc sng = 0,41m ln mt kim loi dng lm catt ca t bo quang in th c hin tng quang in xy ra. Nu dng mt hiu in th hm 0,76V th cc quang electrn bn ra u b gi li.Tm: 1. Cng thot ca electrn i vi kim loi . 2. Vn tc ban u cc i ca cc quang electrn khi bn ra khi catt. p s: 1. 2.2 m e v 0 max = eU h v 0 max = 2

hc hc = A + eU h A = eU h = 36,32.10 20 J 2eU h = me 2.1,6.10 19.0,76 9,1.10 31

= 0,52.10 6 m / s

9. Cng thot ca kim loi dng lm catt ca t bo quang in A= 2,48eV. Tm: 1. Gii hn quan in ca tm kim loi . 2.Vn tc ban u cc i ca cc quang electrn khi catt c chiu bng nh sng n sc bc sng = 0,36m. 3. Hiu in th hm khng c mt electrn no n c ant. p s: 1. 0 =

hc 6,625.10 34.3.108 = = 0,5.10 6 m 19 A 2,48.1,6.10 2 me hc 6 A = 0,584.10 m / s

2. 3.

hc 1 2 = A + m e v 0 max v 0 max = 2 hc = A + eU h U h == 0,97 V

10. Khi chiu mt chm nh sng c bc sng = 0,234m vo mt kim loi dng lm catt ca t bo quang in th c hin tng quang in xy ra. Bit tn s gii hn ca catt 0= 6.1014Hz. Tm: 1. Cng thot ca electrn i vi kim loi . 2. Hiu in th hm khng c mt electrn no n c ant. 112

Chng 6: Quang hc lng t 3. Vn tc

Vat Ly Dai Cuong A2 - HVBCVT

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