cong thuc vat li

7/30/2019 cong thuc vat li

1/34

Cng thc gii nhanhcc bi ton trcnghim vt l


2/34

1

CHNG I: DAO NG C HCI. DAO NG IU HO1. Phng trnh dao ng: x = Asin(t + )2. Vn tc tc thi: v = Acos(t + )3. Gia tc tc thi: a = -2Asin(t + )4. Vt VTCB: x = 0; |v|Max = A; |a|Min = 0

Vt bin: x = A; |v|Min = 0; |a|Max = 2A

5. H thc c lp: 2 2 2( )vA x

= +

a = -2x6. Chiu di qu o: 2A

7. C nng: 2 21

2tE E E m A= + =

Vi 2 2 2 21

os ( ) os ( )2

E m A c t Ec t = + = +

2 2 2 21 sin ( ) sin ( )2t

E m A t E t = + = +

8. Dao ng iu ho c tn s gc l , tn s f, chu k T. Th ng nng v th nng bin thin vi tn s g2, tn s 2f, chu k T/2

9. ng nng v th nng trung bnh trong thi gian nT/2 ( nN*, T l chu k dao ng) l: 2 21

2 4

Em A=

10. Khong thi gian ngn nht vt i t v tr c to x1 n x2

2 1t

= = vi

11

22

sin

sin

A

A

= =

v ( 1 2,2 2

)

11. Qung ng i trong 1 chu k lun l 4A; trong 1/2 chu k lun l 2A

Qung ng i trong l/4 chu k l A khi vt xut pht t VTCB hoc v tr bin (tc l = 0; ; /2)12. Qung ng vt i c t thi im t1 n t2.

Xc nh: 1 1 2 2

1 1 2 2

Asin( ) A sin( )

os( ) os( )

x t x tv

v Ac t v Ac t

= + = +

= + = + (v1 v v2 ch cn xc nh du)

Phn tch: t2 t1 = nT + t (n N; 0 t < T)Qung ng i c trong thi gian nT l S1 = 4nA, trong thi gian t l S2.Qung ng tng cng l S = S1 + S2

* Nu v1v2 0 2 2 1

2 2 1

2

42

Tt S x x

Tt S A x x

< =

> =

* Nu v1v2 < 0 1 2 1 2

1 2 1 2

0 2

0 2

v S A x x

v S A x x

> = < = + +


3/34

13. Cc bc lp phng trnh dao ng dao ng iu ho:* Tnh * Tnh A (thng s dng h thc c lp)

* Tnh da vo iu kin u: lc t = t0 (thng t0 = 0)0

0

Asin( )

os( )

x t

v Ac t

= +

= +

Lu : + Vt chuyn ng theo chiu dng th v > 0, ngc li v < 0+ Trc khi tnh cn xc nh r thuc gc phn t th my ca ng trn lng gic

(thng ly - < )14. Cc bc gii bi ton tnh thi im vt i qua v tr bit x (hoc v, a, E, Et, E, F) ln th n

* Gii phng trnh lng gic ly cc nghim ca t (Vi t > 0 phm vi gi tr ca k )* Lit k n nghim u tin (thng n nh)* Thi im th n chnh l gi tr ln th nLu : ra thng cho gi tr n nh, cn nu n ln th tm quy lut suy ra nghim th n

15. Cc bc gii bi ton tm s ln vt i qua v tr bit x (hoc v, a, E, E t, E, F) t thi im t1 n t2.* Gii phng trnh lng gic c cc nghim* T t1 < t t2 Phm vi gi tr ca (Vi k Z)* Tng s gi tr ca k chnh l s ln vt i qua v tr .

16. Cc bc gii bi ton tm li dao ng sau thi im t mt khong thi gian t.Bit ti thi im t vt c li x = x0.

* T phng trnh dao ng iu ho: x = Asin(t + ) cho x = x0Ly nghim t + = (ng vi x ang tng, v cos(t + ) > 0)

hoc t + = - (ng vi x ang gim) vi2 2

* Li sau thi im t giy l: x = Asin(t + ) hoc x = Asin( - + t) = Asin(t - )17. Dao ng iu ho c phng trnh c bit:

* x = a Asin(t + ) vi a = constBin l A, tn s gc l , pha ban u x l to , x0 = Asin(t + ) l li .

To v tr cn bng x = a, to v tr bin x = a AVn tc v = x = x0, gia tc a = v = x = x0H thc c lp: a = -2x0

2 2 20 ( )

vA x

= +

* x = a Asin2(t + ) (ta h bc)Bin A/2; tn s gc 2, pha ban u 2.

II. CON LC L XO

1. Tn s gc:k

m

= ; chu k:2

2m

T

k

= = ; tn s:1 1

2 2

kf

T m

= = =

2. C nng: 2 2 21 1

2 2tE E E m A kA= + = =

Vi 2 2 2 21 1

os ( ) os ( )2 2

E mv kA c t Ec t = = + = +

2 2 2 21 1 sin ( ) sin ( )2 2t

E kx kA t E t = = + = +


4/34

3. * bin dng ca l xo thng ng:mg

lk

= 2l

T

=

* bin dng ca l xo nm trn mt phng nghing c gc nghing :

sinmgl

k

= 2

sin

lT

g

=

* Trng hp vt di:

+ Chiu di l xo ti VTCB: lCB = l0 + l(l0 l chiu di t nhin)+ Chiu di cc tiu (khi vt v tr cao nht): lMin = l0 + l A+ Chiu di cc i (khi vt v tr thp nht): lMax = l0 + l + A

lCB = (lMin + lMax)/2

+ Khi A > lth thi gian l xo nn l

tjD

D = , vi

cos =A

l

Thi gian l xo gin l T/2 - t, vi t l thi gian l xo nn (tnh nh trn)* Trng hp vt trn:lCB = l0 - l; lMin = l0 - l A; lMax = l0 - l + A lCB = (lMin + lMax)/2

4. Lc hi phc hay lc phc hi (l lc gy dao ng cho vt) l lc a vt v v tr cn bng (l hp lca cc lc tc dng ln vt xt phng dao ng), lun hng v VTCB, c ln Fhp = k|x| = m

2|x|.5. Lc n hi l lc a vt v v tr l xo khng bin dng.

C ln Fh = kx* (x* l bin dng ca l xo)

* Vi con lc l xo nm ngang th lc hi phc v lc n hi l mt (v ti VTCB l xo khng bin dng)* Vi con lc l xo thng ng hoc t trn mt phng nghing

+ ln lc n hi c biu thc:* Fh = k|l+ x| vi chiu dng hng xung* Fh = k|l- x| vi chiu dng hng ln

+ Lc n hi cc i (lc ko): FMax = k(l+ A) = FKMax+ Lc n hi cc tiu:

* Nu A < l FMin = k(l- A) = FKMin

* Nu A l FMin = 0 (lc vt i qua v tr l xo khng bin dng)Lc y (lc nn) n hi cc i: FNmax = k(A - l) (lc vt v tr cao nht)Lu : Khi vt trn: * FNmax = FMax = k(l+ A)

* Nu A < l FNmin = FMin = k(l- A)* Nu A l FKmax = k(A - l) cn FMin = 0

6. Mt l xo c cng k, chiu di lc ct thnh cc l xo c cng k1, k2, v chiu di tng ng ll1, l2, th ta c: kl = k1l1 = k2l2 = 7. Ghp l xo:

* Ni tip1 2

1 1 1...

k k k= + + cng treo mt vt khi lng nh nhau th: T2 = T1

2 + T22

* Song song: k = k1 + k2 + cng treo mt vt khi lng nh nhau th: 2 2 21 2

1 1 1...T T T= + +

8. Gn l xo k vo vt khi lng m1 c chu k T1, vo vt khi lng m2 c T2, vo vt khi lnm1+m2 c chu k T3, vo vt khi lng m1 m2 (m1 > m2)c chu k T4.Th ta c: 2 2 23 1 2T T T= + v

2 2 24 1 2T T T=

9. Vt m1 c t trn vt m2 dao ng iu ho theo phng thng ng. (Hnh 1) m1 lun nm yn trn m2 trong qu trnh dao ng th:

1 2ax 2

( )M

m m ggA

k

+= =

k

m

Vt di

m

k

Vt trn

k

m1m2

Hnh 1

m

m

k

Hnh 2


5/34

10. Vt m1 v m2 c gn vo hai u l xo t thng ng, m1 dao ng iu ho.(Hnh 2) m2 lun nm yn trn mt sn trong qu trnh m1 dao ng th:

1 2ax

( )M

m m gA

k

+=

11. Vt m1 t trn vt m2 dao ng iu ho theo phng ngang. H s ma st gia m1 v m2 l , b qua mst gia m2 v mt sn. (Hnh 3)

m1 khng trt trn m2 trong qu trnh dao ng th:

1 2ax 2 ( )M m m ggA k

+= =

III. CON LC N

1. Tn s gc:g

l= ; chu k:

22

lT

g

= = ; tn s:

1 1

2 2

gf

T l

= = =

2. Phng trnh dao ng:s = S0sin(t + ) hoc = 0sin(t + ) vi s = l, S0 = 0lv 10

0 v = s = S0cos(t + ) = l0cos(t + ) a = v = -2S0sin(t + ) = -

2l0sin(t + ) = -2s = -2l

Lu : S0 ng vai tr nh A cn s ng vai tr nh x3. H thc c lp:* a = -2s = -2l

* 2 2 20 ( )v

S s

= +

*2

2 20

v

gl = +

4. C nng: 2 2 2 2 2 2 0 0 0 01 1 1 1

2 2 2 2tmg

E E E m S S mgl m ll

= + = = = =

Vi 2 21

os ( )

2

E mv Ec t = = +

2(1 os ) sin ( )tE mgl c E t = = + 5. Ti cng mt ni con lc n chiu di l1 c chu k T1, con lc n chiu di l2c chu k T2, con lc chiu di l1 + l2 c chu k T2,con lc n chiu di l1 - l2(l1>l2) c chu k T4.Th ta c: 2 2 23 1 2T T T= + v

2 2 24 1 2T T T=

6. Vn tc v lc cng ca si dy con lc nv2 = 2gl(cos cos0) v TC = mg(3cos 2cos0)

7. Con lc n c chu k ng T cao h1, nhit t1. Khi a ti cao h2, nhit t2 th ta c:

2

T h t

T R

= +

Vi R = 6400km l bn knh Tri t, cn l h s n di ca thanh con lc.8. Con lc n c chu k ng T su d1, nhit t1. Khi a ti su d2, nhit t2 th ta c:

2 2

T d t

T R

= +

9. Con lc n c chu k ng T cao h, nhit t1. Khi a xung su d, nhit t2 th ta c:

2 2

T d h t

T R R

= +

10. Con lc n c chu k ng T su d, nhit t1. Khi a ln cao h, nhit t2 th ta c:

2 2

T h d t

T R R

= +

Hnh 3

m1km2


6/34

Lu : * Nu T > 0 th ng h chy chm (ng h m giy s dng con lc n)* Nu T < 0 th ng h chy nhanh* Nu T = 0 th ng h chy ng

* Thi gian chy sai mi ngy (24h = 86400s): 86400( )T

sT

=

11. Khi con lc n chu thm tc dng ca lc ph khng i:Lc ph khng i thng l:

* Lc qun tnh: F ma= ur r

, ln F = ma ( F aur r

)Lu : + Chuyn ng nhanh dn u a v

r r( v

rc hng chuyn ng)

+ Chuyn ng chm dn u a vr r

* Lc in trng: F qE=ur ur

, ln F = |q|E (Nu q > 0 F Eur ur

; cn nu q < 0 F Eur ur

)

* Lc y csimt: F = DgV (Fur

lung thng ng hng ln)Trong : D l khi lng ring ca cht lng hay cht kh.

g l gia tc ri t do.V l th tch ca phn vt chm trong cht lng hay cht kh .

Khi : 'P P F= +uur ur ur

gi l trng lc hiu dng hay trong lc biu kin (c vai tr nh trng lc Pur

)

' Fg g m= +

uruur ur

gi l gia tc trng trng hiu dng hay gia tc trng trng biu kin.

Chu k dao ng ca con lc n khi : ' 2'

lT =

Cc trng hp c bit:

* Fur

c phng ngang: + Ti VTCB dy treo lch vi phng thng ng mt gc c:F

tgP

=

+ 2 2' ( )F

g gm

= +

* F

ur

c phng thng ng th '

F

g g m=

+ Nu Fur

hng xung th 'F

g gm

= +

+ Nu Fur

hng ln th 'F

g gm

=

IV. TNG HP DAO NG1. Tng hp hai dao ng iu ho cng phng cng tn s x1 = A1sin(t + 1) v x2 = A2sin(t + 2) mt dao ng iu ho cng phng cng tn s x = Asin(t + ).

Trong : 2 2 21 2 1 2 2 12 os( )A A A A A c = + +

1 1 2 2

1 1 2 2

sin sinos os

A Atg

A c A c

+=+

vi 1 2 (nu 1 2 )

* Nu = 2k (x1, x2 cng pha) AMax = A1 + A2` * Nu = (2k+1) (x1, x2 ngc pha) AMin = |A1 - A2|2. Khi bit mt dao ng thnh phn x1 = A1sin(t + 1) v dao ng tng hp x = Asin(t + ) th dao nthnh phn cn li l x2 = A2sin(t + 2).

Trong : 2 2 22 1 1 12 os( )A A A AA c = +

1 12

1 1

sin sin

os os

A Atg

Ac A c

=

vi 1 2 ( nu 1 2 )


7/34

3. Nu mt vt tham gia ng thi nhiu dao ng iu ho cng phng cng tn s x1 = A1sin(t + 1;x2 = A2sin(t + 2) th dao ng tng hp cng l dao ng iu ho cng phng cng tn sx = Asin(t + ).Ta c: 1 1 2 2sin sin sin ...xA A A A = = + +

1 1 2 2os os os ...A Ac A c A c = = + +

2 2xA A A = + v

xAtgA

= vi [Min;Max]

V. DAO NG TT DN DAO NG CNG BC - CNG HNG1. Mt con lc l xo dao ng tt dn vi bin A, h s ma st . Qung ng vt i c n lc dng l

l:2 2 2

2 2

kA AS

mg g

= =

2. Mt vt dao ng tt dn th gim bin sau mi chu k l:2

4 4mg gA

k

= =

s dao ng thc hin c2

4 4

A Ak AN

A mg g

= = =

3. Hin tng cng hng xy ra khi: f = f0 hay = 0 hay T = T0Vi f, , T v f0, 0, T0 l tn s, tn s gc, chu k ca lc cng bc v ca h dao ng.


8/34

CHNG II: SNG C HCI. SNG C HC1. Bc sng: = vT = v/f

Trong : : Bc sng; T (s): Chu k ca sng; f (Hz): Tn s ca sngv: Vn tc truyn sng (c n v tng ng vi n v ca )

2. Phng trnh sngTi im O: uO = asin(t + )

Ti im M cch O mt on d trn phng truyn sng.* Sng truyn theo chiu dng ca trc Ox th uM = aMsin(t + -

d

v ) = aMsin(t + - 2

d

)

* Sng truyn theo chiu m ca trc Ox th uM = aMsin(t + +d

v ) = aMsin(t + + 2

d

)

3. lch pha gia hai im cch ngun mt khong d1, d21 2 1 22

d d d d

v

= =

Nu 2 im nm trn mt phng truyn sng v cch nhau mt khong d th:

2d d

v

= =

Lu :n v ca d, d1, d2, v v phi tng ng vi nhau4. Trong hin tng truyn sng trn si dy, dy c kch thch dao ng bi nam chm in vi tn s dnin l f th tn s dao ng ca dy l 2f.II. GIAO THOA SNGGiao thoa ca hai sng pht ra t hai ngun sng kt hp cch nhau mt khong l:Xt im M cch hai ngun ln lt d1, d2Gi x l s nguyn ln nht nh hn x (v d: 6 5; 4,05 4; 6,97 6= = = )

1. Hai ngun dao ng cng pha:

Bin dao ng ca im M: AM = 2aM|cos( 1 2d d

)|

* im dao ng cc i: d1 d2 = k (kZ)S im hoc s ng (khng tnh hai ngun):

l lk

< < hoc CN =2 1

l

+

* im dao ng cc tiu (khng dao ng): d1 d2 = (2k+1)2

(kZ)

S im hoc s ng (khng tnh hai ngun):

1 1

2 2

l lk

< < hoc CT

1N =2

2

l

+

2. Hai ngun dao ng ngc pha:

Bin dao ng ca im M: AM = 2aM|cos( 1 2 2d d + )|

* im dao ng cc i: d1 d2 = (2k+1)2

(kZ)

S im hoc s ng (khng tnh hai ngun):

1 1

2 2

l lk

< < hoc C

1N =2

2

l

+

* im dao ng cc tiu (khng dao ng): d1 d2 = k (kZ)S im hoc s ng (khng tnh hai ngun):

O

x

M

d


9/34

l lk

< < hoc CTN =2 1

l

+

3. Hai ngun dao ng vung pha:

Bin dao ng ca im M: AM = 2aM|cos( 1 24

d d

+ )|

S im (ng) dao ng cc i bng s im (ng) dao ng cc tiu (khng tnh hai ngun):1 1

4 4

l lk

< <

Ch : Vi bi ton tm s ng dao ng cc i v khng dao ng gia hai im M, N cch hai ngun llt l d1M, d2M, d1N, d2N.

t dM = d1M - d2M ; dN = d1N - d2N v gi s dM < dN.+ Hai ngun dao ng cng pha:

Cc i: dM < k < dN Cc tiu: dM < (k+0,5) < dN

+ Hai ngun dao ng ngc pha: Cc i:dM < (k+0,5) < dN Cc tiu: dM < k < dN

S gi tr nguyn ca k tho mn cc biu thc trn l s ng cn tm.III. SNG DNG1. * Gii hn c nh Nt sng

* Gii hn t do Bng sng* Ngun pht sng c coi gn ng l nt sng* B rng bng sng 4a (vi a l bin dao ng ca ngun)

2. iu kin c sng dng gia hai im cch nhau mt khong l:

* Hai im u l nt sng: *( )2

l k k N

=

S bng sng = s b sng = kS nt sng = k + 1

* Hai im u l bng sng: *( )2

l k k N =

S b sng nguyn = k 1S bng sng = k + 1S nt sng = k

* Mt im l nt sng cn mt im l bng sng: (2 1) ( )4

l k k N

= +

S b sng nguyn = kS bng sng = s nt sng = k + 1

3. Trong hin tng sng dng xy ra trn si dy AB vi u A l nt sng

Bin dao ng ca im M cch A mt on d l: 2 sin(2 )Md

A a = vi a l bin dao ng ca ngun.IV. SNG M

1. Cng m:E P

I= =tS S

Vi E (J), P (W) l nng lng, cng sut pht m ca ngunS (m2) l din tch mt vung gc vi phng truyn m (vi sng cu th S l din tch mt cu S=4R2

2. Mc cng m

0

( ) lgI

L BI

= Hoc0

( ) 10.lgI

L dBI

= (cng thc thng dng)

Vi I0 = 10-12 W/m2 f = 1000Hz: cng m chun.


10/34

CHNG III: IN XOAY CHIU1. Biu thc hiu in th tc thi v dng in tc thi:

u = U0sin(t + u) v i = I0sin(t + i)

Vi = u i l lch pha ca u so vi i, c2 2

2. Dng in xoay chiu i = I0sin(2ft + i)* Mi giy i chiu 2f ln

* Nu pha ban u i = 0 hoc i = th ch giy u tin i chiu 2f-1 ln.3. Cng thc tnh khong thi gian n hunh quang sng trong mt chu kKhi t hiu in th u = U0sin(t + u) vo hai u bng n, bit n ch sng ln khi u U1.

4t

= Vi 1

0

osU

cU

= , (0 < < /2)

4. Dng in xoay chiu trong on mch R,L,C* on mch ch c in tr thun R: uR cng pha vi i, ( = u i = 0)

UI

R= v 00

UI

R=

Lu : in tr R cho dng in khng i i qua v cU

I

R

=

* on mch ch c cun thun cm L: uL nhanh pha hn i/2, ( = u i = /2)

L

UI

Z= v 00

L

UI

Z= vi ZL = L l cm khng

Lu : Cun thun cm L cho dng in khng i i qua hon ton (khng cn tr).* on mch ch c t in C: uCchm pha hn i/2, ( = u i = -/2)

C

UI

Z= v 00

C

UI

Z= vi

1CZ C

= l dung khng

Lu : T in C khng cho dng in khng i i qua (cn tr hon ton).* on mch RLC khng phn nhnh

2 2 2 2 2 20 0 0 0( ) ( ) ( )L C R L C R L CZ R Z Z U U U U U U U U= + = + = +

;sin ; osL C L CZ Z Z Z R

tg cR Z Z

= = = vi2 2

+ Khi ZL > ZC hay1

LC> > 0 th u nhanh pha hn i

+ Khi ZL < ZC hay1

LC< < 0 th u chm pha hn i

+ Khi ZL = ZC hay1

LC= = 0 th u cng pha vi i.

Lc Max UI = Rgi l hin tng cng hng dng in

5. Cng sut to nhit trn on mch RLC: P = UIcos = I2R.6. Hiu in th u = U1 + U0sin(t + ) c coi gm mt hiu in th khng i U1 v mt hiu in thxoay chiu u = U0sin(t + ) ng thi t vo on mch.7. Tn s dng in do my pht in xoay chiu mt pha c P cp cc, rto quay vi vn tc n vng/pht ph

ra:60

pnf Hz=

T thng gi qua khung dy ca my pht in = NBScos(t +) = 0cos(t + )


11/34

Vi 0 = NBS l t thng cc i, N l s vng dy, B l cm ng t ca t trng, S l din tch ca vndy, = 2f

Sut in ng trong khung dy: e = NSBsin(t + ) = E0sin(t + )Vi E0 = NSB l sut in ng cc i.

8. Dng in xoay chiu ba pha

1 0

2 0

3 0

sin( )

2

sin( )32

sin( )3

i I t

i I t

i I t

=

=

= +

My pht mc hnh sao: Ud = 3 UpMy pht mc hnh tam gic: Ud = UpTi tiu th mc hnh sao: Id = Ip

Ti tiu th mc hnh tam gic: Id = 3 IpLu : my pht v ti tiu th thng chn cch mc tng ng vi nhau.

9. Cng thc my bin th: 1 1 2 1

2 2 1 2

U E I N

U E I N = = =

10. Cng sut hao ph trong qu trnh truyn ti in nng:2

2 2os

PP R

U c =

Thng xt: cos = 1 khi 2

2

PP R

U =

Trong : P l cng sut cn truyn ti ti ni tiu thU l hiu in th ni cung cpcos l h s cng sut ca dy ti in

lR

S= l in tr tng cng ca dy ti in (lu : dn in bng 2 dy)

gim th trn ng dy ti in: U = IRHiu sut ti in: .100%

P PH

P

=

11. on mch RLC c L thay i:

* Khi2

1L

C= th IMax URmax; PMax cn ULCMinLu : L v C mc lin tip nhau

* Khi2 2

CL

C

R ZZ

Z

+= th

2 2

axC

LM

U R ZU

R

+=

* Vi L = L1 hoc L = L2 th UL c cng gi tr th ULmax khi

1 2

1 2

1 2

21 1 1 1( )

2L L L

L LL

Z Z Z L L= + =

+

* Khi2 24

2C C

L

Z R ZZ

+ += th ax 2 2

2 R

4RLM

C C

UU

R Z Z=

+ Lu : R v L mc lin tip nhau

12. on mch RLC c C thay i:

* Khi2

1C

L= th IMax URmax; PMax cn ULCMinLu : L v C mc lin tip nhau

* Khi2 2

LC

L

R ZZ

Z

+= th

2 2

axL

CM

U R ZU

R

+=


12/34

* Khi C = C1 hoc C = C2 th UC c cng gi tr th UCmax khi1 2

1 21 1 1 1( )2 2C C C

C CC

Z Z Z

+= + =

* Khi2 24

2L L

C

Z R ZZ

+ += th ax 2 2

2 R

4RCM

L L

UU

R Z Z=

+ Lu : R v C mc lin tip nhau

13. Mch RLC c thay i:

* Khi 1LC

= th IMax URmax; PMax cn ULCMinLu : L v C mc lin tip nhau

* Khi2

1 1

2

C L RC

=

th ax 2 22 .

4LM

U LU

R LC R C=

* Khi21

2

L R

L C= th ax 2 2

2 .

4CM

U LU

R LC R C=

* Vi = 1 hoc = 2 th I hoc P hoc URc cng mt gi tr th IMax hoc PMax hoc URMax khi

1 2 = tn s 1 2f f f=

14. Hai on mch R1L1C1 v R2L2C2 cng u hoc cng i c pha lch nhau

Vi 1 111

L CZ ZtgR

= v 2 222

L CZ ZtgR

= (gi s 1 > 2)

C 1 2 = 1 21 21

tg tg tg

tg tg

=

+

Trng hp c bit = /2 (vung pha nhau) th tg1tg2 = -1.


13/34

CHNG IV: DAO NG IN T SNG IN T1. Dao ng in t* in tch tc thi q = Q0sin(t + )* Dng in tc thi i = q = Q0cos(t + ) = I0cos(t + )

* Hiu in th tc thi 0 0sin( ) sin( )Qq

u t U t C C

= = + = +

Trong :1

LC= l tn s gc ring,

2T LC= l chu k ring1

2f

LC= l tn s ring

00 0

QI Q

LC= =

0 00 0

Q I LU I

C C C= = =

* Nng lng in trng2

2

1 1

2 2 2

qE Cu qu

C= = =

220

sin ( )2

QE t

C = +

* Nng lng t trng2

2 201 os ( )2 2t

QE Li c t

C = = +

* Nng lng in t tE E E= + 2

2 20 0 0 0 0

1 1 1

2 2 2 2

QE CU Q U LI

C= = = =

Ch : Mch dao ng c tn s gc , tn s f v chu k T th nng lng in trng bin thin vi tn s

gc 2, tn s 2f v chu k T/22. Sng in tVn tc lan truyn trong khng gian v = c = 3.10-8m/sMy pht hoc my thu sng in t s dng mch dao ng LC th tn s sng in t pht hoc thu bng tns ring ca mch.

Bc sng ca sng in t 2v

v LCf

= =

Lu : Mch dao ng c L bin i t LMin LMax v C bin i t CMin CMax th bc sng ca snin t pht (hoc thu)

Min tng ng vi LMin v CMinMax tng ng vi LMax v CMax


14/34

CHNG V: S PHN X V KHC X NH SNG1. Hin tng phn x nh snga) /n: L hin tng tia sng b i hng t ngt tr v mi trng c khi gp mt b mt nhn.b) nh lut phn x nh sng:* Tia phn x nm trong mt phng ti v bn kia php tuyn so vi tia ti* Gc phn x bng gc ti i = i2. Gng phnga) /n: L mt phn ca mt phng phn x tt nh sng chiu ti nb) Cng thc ca gng phng* V tr: d + d = 0

* phng i:' ' '

1A B d

kdAB

= = =

* Khong cch vt - nh: L = |d d| = 2|d| = 2|d|Quy c du: Vt tht d > 0, vt o d < 0, nh tht d > 0, nh o d , gng cu li 0

2

Rf = <

* V tr vt nh:1 1 1

'd d f+ =

dd ' '; ; '

' '

d f df f d d

d d d f d f = = =

+


15/34

* phng i:' ' ' 'A B d f f d

kd f d f AB

= = = =

1' ' ; (1 ) ; ' (1- )A B k AB d f d k f

k = = =

* Khong cch vt nh: L = |d d|Quy c du: ; ' 'd OA d OA= =

Vt tht d > 0; vt o d < 0nh tht d > 0; nh o d < 0Vt v nh cng chiu k > 0, vt v nh ngc chiu k < 0

Lu : T l din tch ca nh v vt bng bnh phng phng ie) S v tr vt nh* Gng cu lm:

* Gng cu li:

f) Tnh cht vt nh* Vt v nh cng tnh cht th ngc chiu v cng pha i vi gng.* Vt v nh tri tnh cht th cng chiu v khc pha i vi gng.

* Vt v nh l mt im nm ngoi trc chnh: Nu cng tnh cht th khc pha i vi trc chnh, cn ntri tnh cht th cng pha i vi trc chnh.* Xt chuyn ng theo phng trc chnh th vt v nh lun chuyn ng ngc chiu (Lu : khi vt chuyng qua tiu im th nh t ngt i chiu chuyn ng v i tnh cht).* Xt chuyn ng theo phng vung gc vi trc chnh: Nu vt v nh cng tnh cht th chuyn nngc chiu, cn nu tri tnh cht th chuyn ng cng chiu.* T l din tch ca nh v vt bng bnh phng phng i.* Vi gng cu lm: + Vt tht cho nh tht ln hoc nh hn vt

+ Vt tht cho nh o lun ln hn vt+ Vt o lun cho nh tht nh hn vt

* Vi gng cu li: + Vt tht lun cho nh o nh hn vt

+ Vt o cho nh tht lun ln hn vt+ Vt o cho nh o ln hoc nh hn vt

g) Th trng gng* Th trng ca gng ng vi mt v tr t mt l vng khng gian trc gng gii hn bi hnh nn (hnchp) ct c nh l nh ca mt qua gng.* Th trng ca gng ph thuc vo v tr t mt, loi gng v kch thc gng* Vi cc gng c cng kch thc v cng v tr t mt th th trng ca gng cu li > gng phng >gng cu lm.

+ -

Vt

nh

OC FI II III IV

12 34

+ -

Vt

nh

O CFI II III IV

12 34


16/34

h) Cc dng ton c bn v gng cu:Ni dung bi ton Phng php gii

Cho 2 trong 4 i lng d, d, f, k.Xc nh cc i lng cn li

S dng cc cng thc:dd ' '

; ; '' '

d f df f d d

d d d f d f = = =

+

' ' ' 'A B d f f dk

d f d f AB

= = = =

1' ' ; (1 ) ; ' (1- )A B k AB d f d k fk

= = =

Cho khong cch t vt v nh n tiu imchnh l a v b.Xc nh tiu c f

Ta c cng thc Niutnf2 = a.b

Lu : Trng hp vt tht v a b ch ng vi gngcu lm

Cho f v L (khong cch vt nh)Xc nh d, d

Gii h phng trnh:

'df

dd f

=

L = |d - d|

Cho k v LXc nh d, d, f

Gii h phng trnh:'d

kd

=

L = |d - d|dd '

'f

d d=

+

Cho phng i k1, k2 v dch chuyn cavt d = d2-d1 (hoc dch chuyn ca nhd = d2-d1).Xc nh f, d1...

Gii h phng trnh:

11 2 1

2 11 2

22

1(1 )

( )1

(1 )

d fk k kd d d f

k kd f

k

= = = =

'1 1

2 1 1 2'2 2

(1- )' ' ' ( )

(1- )

d k fd d d k k f

d k f

= = =

=

Lu :d, d c th m hoc dng

Cho dch chuyn ca vt d, dch chuynca nh d v t l cao ca 2 nh l n.Xc nh f, d1...

Thay k2 = nk1 hoc k1 = nk2 vo biu thc ca d v d

Ta c2 2( 1)

. 'n f

d dn

=

Lu : Khi 2 nhcng tnh cht th n > 0 d.d0

Cho dch chuyn ca vt d, dch chuynca nh d v tiu c f ca gng.Xc nh d1,d2 ...

Gii h phng trnh:2 1

2 11 2

' '2 1 1 2

( )

' ( )

k kd d d f

k k

d d d k k f

= = = =

Tnh c k1 v k2 ri thay vo cc phng trnh:


17/34

11

22

1(1 )

1(1 )

d fk

d fk

= =

Vt AB v mn M c nh cch nhau mtkhong L. C 2 v tr ca gng cu cch nhaumt khong l(l> L) c 2 nh A1B1, A2B2 rnt trn mn.Xc nh f, cao AB...

Gng v tr 1: Vt AB c v tr d1, nh A1B1 c v tr d1Gng v tr 2: Vt AB c v tr d2, nh A1B1 c v tr d2

Theo nguyn l thun nghch v chiu truyn nh sng:'' 2 21 12 1

' '2 1 1 1

4

L d dd d l Lf

ld d l d d

= = =

= = +

'1 1 1

11

1 2 1 1 2 2'2 2 2 1

2 '2 1

1 .

A B dk

dABk k AB A B A B

A B d dk

d dAB

= =

= =

= = =

4. Hin tng khc x nh snga) /n: L hin tng tia sng b i hng t ngt khi truyn qua mt phn cch ca hai mi trng tronsut.b) nh lut khc x nh sng* Tia khc x nm trong mt phng ti v bn kia php tuyn so vi tia ti

* 2211

sin

sinr

nin

n= =

Nu n2 > n1 r < i Mi trng 2 chit quang hn mi trng 1 (tia khc x lch gn php tuyn hn tia ti)Nu n2 < n1 r > i Mi trng 2 chit km hn mi trng 1 (tia khc x lch xa php tuyn hn tia ti)Nu i = 0 r = 0 nh sng chiu vung gc mt phn cch th truyn thng.

c) Chit sut tuyt ic

nv

= ; 2 11 2

n v

n v=

Trong c = 3.108m/s v v l vn tc nh sng truyn trong chn khng v trong mi trng trong suchit sut n.Lu : +/n khc v chit sut tuyt i: L t s gia vn tc nh sng trong chn khng v vn tc nh sntruyn trong mi trng trong sut .

+ ngha ca chit sut tuyt i: Cho bit vn tc nh snh truyn trong mi trng trong sut nhhn vn tc nh sng truyn trong chn khng bao nhiu ln.5. Lng cht phng* /n: L h thng gm hai mi trng trong sut ngn cch nhau bi mt phng.* c im nh: nh v vt c cng ln, cng chiu, cng pha nhng tri tnh cht* Cng thc ca lng cht phng:

/

1 2

OA OAn n

= Vt tht A t trong mi trng c chit sut n1

dch chuyn nh:1

' (1 )AA hn

=

Vi n = n21, h = OA l khong cch t vt ti mt phn cch.6. Bn mt song song* /n: L mt khi cht trong sut c gii hn bi hai mt phng song song* c im nh: nh v vt c cng ln, cng chiu nhng tri tnh cht


18/34

* dch chuyn nh: AA = e(1 -n

1).

Vi e l b dy bn mt song songn l chit sut t i ca bn i vi mi trng xung quanhNu n > 1 th nh dch gn bn, cn nu n < 1 th nh dch xa bn (ch xt vt tht)

7. Hin tng phn x ton phn

* /n: L hin tng khi chiu mt tia sng vo mt phn cch ca hai mi trng trong sut m ch c tia phnx khng c tia khc x.* iu kin c hin tng phn x ton phn:+ Tia sng c chiu t mi trng chit quang hn sang mi trng chit quang km.+ Gc ti ln hn hoc bng gc gii hn phn x ton phn: i igh.

Vi 2211

sin ghn

i nn

= = (khi chiu nh sng t mi trng trong sut chit sut n ra khng kh th1

sin ghi n= )

8. Lng knha) /n: L khi cht trong sut hnh lng tr ng c tit din thng l mt tam gicHoc: L khi cht trong sut c gii hn bi hai mt phng khng song songb) iu kin ca lng knh v tia sng qua lng knh

* Chit sut lng knh n > 1* nh sng n sc* Tia sng nm trong tit din thng* Tia sng t y i lnKhi m bo 4 iu kin trn th tia l ra khi lng knh lch v pha yc) Cng thc ca lng knh

sini1 = nsinr1sini2 = nsinr2A = r1 + r2D = i1 + i2 A

Khi tia ti v tia l i xng vi nhau qua mt phng phn gic ca gc chit quang i1 = i2 r1 = r2 th DMi

sin( ) sin2 2

MinD A An+ =

Ch : Khi i, A 100 th i1 = nr1i2 = nr2A = r1 + r2D = (n-1)A

9) Thu knh mnga) /n: L mt khi cht trong sut c gii hn bi hai mt cong thng l hai mt cu, mt trong hai mt cth l mt phng.b) Cc tia c bit* Tia ti song song vi trc chnh cho tia l c phng i qua tiu im nh chnh F.* Tia ti c phng i qua tiu im vt chnh F cho tia l song song vi trc chnh* Tia ti qua quang tm O th cho tia l truyn thngc) Tia bt k* Tia ti song song vi trc ph cho tia l c phng i qua tiu im nh ph 'nF thuc trc ph

* Tia ti c phng i qua tiu im vt ph Fn cho tia l song song vi trc ph cha tiu im ph d) Cng thc ca thu knh

* t:1

Df

= (ip - mt)


19/34

1 2

1 1 1( 1)( )D n

f R R= = +

Trong : n l chit sut ca thu knhR1, R2 l bn knh cc mt cu (Mt li: R1, R2 > 0; mt lm R1, R2 < 0; mt phng R1, R2=)

* V tr vt nh:1 1 1

'd d f+ =

dd ' '; ; '' '

d f df f d dd d d f d f = = =+

* phng i:' ' ' 'A B d f f d

kd f d f AB

= = = =

1' ' ; (1 ) ; ' (1- )A B k AB d f d k f

k = = =

* Khong cch vt nh: L = |d +d|Quy c du: ; ' 'd OA d OA= =

Vt tht d > 0; vt o d < 0nh tht d > 0; nh o d < 0

Vt v nh cng chiu k > 0, vt v nh ngc chiu k < 0Lu : T l din tch ca nh v vt bng bnh phng phng ie) S v tr vt nh* Thu knh hi t:

* Thu knh phn k:

f) Tnh cht vt nh* Vt v nh cng tnh cht th ngc chiu v khc pha i vi thu knh.* Vt v nh tri tnh cht th cng chiu v cng pha i vi thu knh.* Vt v nh l mt im nm ngoi trc chnh: Nu cng tnh cht th khc pha i vi trc chnh, cn ntri tnh cht th cng pha i vi trc chnh.* Xt chuyn ng theo phng trc chnh th vt v nh lun chuyn ng cng chiu (Lu : khi vt chuyng qua tiu im vt th nh t ngt i chiu chuyn ng v i tnh cht).* Xt chuyn ng theo phng vung gc vi trc chnh: Nu vt v nh cng tnh cht th chuyn nngc chiu, cn nu tri tnh cht th chuyn ng cng chiu.* T l din tch ca nh v vt bng bnh phng ca phng i.* Vi thu knh hi t: + Vt tht cho nh tht ln hoc nh hn vt

+ Vt tht cho nh o lun ln hn vt

+ -Vt

nh

O2F F

I II III IV

1 23 4

F 2F

+-

+ -Vt

nh

O2FF

I II III IV

1 23 4

F2F

+-


20/34

+ Vt o lun cho nh tht nh hn vt* Vi thu knh phn k: + Vt tht lun cho nh o nh hn vt

+ Vt o cho nh tht lun ln hn vt+ Vt o cho nh o ln hoc nh hn vt

h) Cc dng ton c bn v thu knh:Ni dung bi ton Phng php gii

Cho 3 trong 4 i lng f, D, n, R1, R2Xc nh cc i lng cn li

S dng cng thc

1 2

1 1 1( 1)( )D n

f R R= = +

Lu : n l chit sut t i ca cht lm thu knh i vimi trng xung quanh.

Cho 2 trong 4 i lng d, d, f, k.

Xc nh cc i lng cn li

S dng cc cng thc:dd ' '

; ; '' '

d f df f d d

d d d f d f = = =

+

' ' ' 'A B d f f dk d f d f AB

= = = =

1' ' ; (1 ) ; ' (1- )A B k AB d f d k f

k= = =

Cho f v L (khong cch vt nh)Xc nh d, d

Gii h phng trnh:

'df

dd f

=

v L = |d + d|

Cho khong cch t vt n tiu im vt chnhF v khong cch t nh n tiu im nhchnh F l a v b.Xc nh tiu c f

Ta c cng thc Niutnf2 = a.b

Lu : Trng hp vt tht v a b ch ng vi TKHT

Cho k v LXc nh d, d, f

Gii h phng trnh:'d

kd

=

L = |d + d|dd '

'f

d d=

+

Cho phng i k1, k2 v dch chuyn ca

vt d = d2-d1 (hoc dch chuyn ca nhd = d2 - d1).Xc nh f, d1...

Gii h phng trnh:

11 2 1

2 1

1 22

2

1(1 )

( )

1(1 )

d fk k k

d d d f

k kd fk

= = =

= '

1 12 1 1 2'

2 2

(1- )' ' ' ( )

(1- )

d k fd d d k k f

d k f

= = =

=

Lu :d, d c th m hoc dng

Cho dch chuyn ca vt d, dch chuynca nh d v t l cao ca 2 nh l n.Xc nh f, d1...

Thay k2 = nk1 hoc k1 = nk2 vo biu thc ca d v d

Ta c2 2( 1)

. 'n f

d dn

=

Lu : Khi 2 nhcng tnh cht th n > 0 d.d


21/34

Khi 2 nh tri tnh cht th n < 0 d.d>0

Cho dch chuyn ca vt d, dch chuynca nh d v tiu c f ca thu knh.Xc nh d1,d2 ...

Gii h phng trnh:2 1

2 11 2

' '2 1 1 2

( )

' ( )

k kd d d f

k k

d d d k k f

= = = =

Tnh c k1 v k2 ri thay vo cc phng trnh:

11

22

1(1 )

1(1 )

d f

k

d fk

=

=

Vt AB v mn M c nh cch nhau mtkhong L. C 2 v tr ca thu knh cch nhaumt khong l(l< L) c 2 nh A1B1, A2B2 rnt trn mn.

Xc nh f, cao AB...

TK v tr 1: Vt AB c v tr d1, nh A1B1 c v tr d1TK v tr 2: Vt AB c v tr d2, nh A1B1 c v tr d2Theo nguyn l thun nghch v chiu truyn nh sng:

'' 2 21 12 1

''1 12 1

4

L d dd d L lf

Ll d dd d

= + = =

= =

'1 1 1

11

1 2 1 1 2 2'2 2 2 1

2 '2 1

1 .

A B d

k dABk k AB A B A B

A B d dk

d dAB

= = = =

= = =

10. Quang h ng trca) S to nh qua quang h ng trc* nh ca phn t trc s tr thnh vt i vi phn t sau

S to nh: 1 1' '1 1 2 2

1 1 2 2....O O

d d d d AB A B A B

* Dng cng thc ca tng phn t cho mi ln to nh v cng thc chuyn tip

'1 1 1

n n nd d f+ = (Lu : Vi gng phng 1 0

f= )

dn + dn+1 = ln(n+1) , Vi ln(n+1) l khong cch gia 2 quang c th n v n1. VD: d1 + d2 = l12 = O1O2

* phng i' ' '

1 21 1 2 21 2

1 21 1 11

...... ... ( 1)

...nn n n n n

nnn n

A B A B d d dA B A Bk k k k

d d dAB AB A B A B = = = =

Vi n l s ln to nh (s nh)Ch :Nu k > 0: nh cui cng cng chiu vi vt

Nu k < 0: nh cui cng ngc chiu vi vtNu dn > 0: nh cui cng l nh thtNu dn < 0: nh cui cng l nh o

b) Mt s lu * Nu quang h c quang c phn x th vt phi t trc quang c ny v s ln to nh ln hn s quang c.* Nu vt t ngoi quang h th cho mt nh cui cng. Nu vt t gia h th cho 2 nh cui cng.* Vi h gm 2 gng th phi ch s ln to nh trn mi gng v to nh trn gng no trc.* Vi quang h ghp st: (khong cch gia cc quang c l= 0)

+ H thu knh ghp st: Tng ng 1 TK c tD = D1 + D2 + ...

+ H gm 1 thu knh v gng ghp st: Tng ng mt gng cu c t


22/34

D = 2DTK+ Dg (Lu : Gng phng Dg = 0)c) H v tiuL h khng c tiu im.Chm tia ti song song th cho chm tia l khi h cng l chm song songnh to bi h v tiu c cao khng ph thuc vo v tr t vtKhong cch gia cc quang c v phng i ca h v tiu:

* H gm 2 thu knh: l= f1 + f2 v 2

1

fk

f

=

* H gm thu knh v gng phng: l= f v k = -1* H gm thu knh v gng cu: l= fTK+ 2fg v k = 1

Hoc l= fTKv k = -1


23/34

CHNG VI: MT V CC DNG C QUANG HC1. Mt* im cc cn CC: + Mt iu tit ti a

+ Tiu c ca mt fMin+ OCC = : khong nhn r ngn nht

* im cc vin CV: + Mt khng iu tit+ Tiu c ca mt fMax+ OCV: khong nhn r di nht

* Mt khng c tt l mt khi khng iu tit c tiu im nm trn vng mc: OCC = 25cm, OCV = * Gii hn nhn r ca mt [CC;CV]* Khi chuyn t trng thi quan st vt v tr cch mt d1 sang trng thi quan st vt v tr cch mt d2 th bin thin t ca mt l:

2 1

1 1D

d dD = - Lu : d1 v d2 tnh bng n v mt (m)

p dng: Khi chuyn t trng thi khng iu tit sang trng thi iu tit ti a th:

1 1

C V

DOC OC

D = - Lu : OCC v OCV tnh bng n v mt (m)

* mt khng nhn thy vt khi vt c t bt k v tr no trc knh th knh eo cch mt mt khong

c t:1

C

DOC l

< --

* Mt cn th l mt khi khng iu tit c tiu im nm trc vng mc.+ fMax < OV vi OV l khong cch t quang tm thu tinh th ti vng mc+ OCC = < 25cm+ OCV c gi tr hu hn+ Cch sa (c 2 cch, cch 1 c li nht thng c s dng)

C1) eo thu knh phn k nhn xa nh ngi bnh thng, tc l vt v cc cho nh o qua knh nm im cc vin.

d = , d = - OKCV = - (OCV l) vi l= OOKl khong cch t knh ti mt.Tiu c ca knh fk= d = - (OCV l)Knh eo st mt l= 0: fk= - OCV

C2) eo thu knh phn k nhn gn nh ngi bnh thng, tc l vt t cch mt 25cm cho nh o quaknh nm im cc cn.

d = (25- l)cm, d = - OKCC = -(OCC - l)

Tiu c ca knh:dd '

0'K

fd d

= OV+ OCC = > 25cm

+ Khng c im CV (o nm sau mt)+ Cch sa

eo thu knh hi t nhn gn nh ngi bnh thng, tc l vt t cch mt 25cm cho nh o qua knhnm im cc cn.

d = (25-l)cm, d = - OKCC = -(OCC - l) vi l= OOKl khong cch t knh ti mt.

Tiu c ca knh:dd '

0'K

fd d

= >+

* Mt lo (mt bnh thng khi v gi) l mt khng c tt+ fMax = OV+ OCC = > 25cm (ging mt vin th)


24/34

+ OCV = + Cch sa nh sa tt vin th.

* Gc trng vt :L gc hp bi hai tia sng i qua mp ca vt v quang tm ca thu tinh th

Vi AB l on thng t vung gc vi trc chnh ca mt c gc trng th ;AB AB

tg l OAOA l

a = = =

* Nng sut phn li ca mt Min

L gc trng nh nht gia hai im m mt cn c th phn bit c hai im .Lu : mt phn bit c 2 im A, B th A, B [CC; CV] v Min* bi gic G ca mt dng c quang hc:

L t s gia gc trng nh qua quang c v gc trng vt khi vt t im cc cn.

0 0

' ' .

' '

tg A BG k

tg AB OA d l

a a

a a= = = =

+

Vi = OCC khong nhn r ngn nht ca mt ngi quan st.ll khong cch t quang c ti mt.k l phng i nh ca quang c .OA = |d| + ll khong cch t nh cui cng qua quang c ti mt.

Lu :nh ngha v cng thc tnh bi gic trn khng ng vi knh thin vn.

Knh thin vn th gc trng vt 0 l trc tip 0 0

tgG

tga aa a

= =

2. Knh lp* L dng c quang hc b tr cho mt lm tng gc trng nh ca cc vt nh.* Cch ngm chng:

Thay i khong cch t vt AB n knh lp nh AB l nh o nm trong gii hn nhn r ca mt.Vt AB nm trong tiu im vt F ca knh lp.+ Ngm chng im CC (mt iu tit ti a): nh qua quang c nm im CC+ Ngm chng im CV (mt khng iu tit): nh qua quang c nm im CV

Vi mt khng c tt CV nn ngm chng CV l ngm chng v cc

mi mt th ngi quan st chn cch ngm chng im CV* bi gic

+ Cng thc tng qut:

'G k

d l=

+

+ Ngm chng CC: GC = k

+ Ngm chng CV:

.VV

G kOC

=

+ Ngm chng v cc:

Gf

= , thng ly = OCC = 25cm. (khng ph thuc vo v tr t mt)

+ Khi mt t ti tiu im nh ca knh lp th bi gic khng ph thuc vo cch ngm chng.

Gf

= vi = OCC ca mt ngi quan st.

Lu : - Vi ll khong cch t mt ti knh lp th khi: 0 l< f GC > GVl= f GC = GVl> f GC < GV

- Trn vnh knh thng ghi gi tr25

( )G

f cm=

V d: Ghi X10 th25

10 2,5( )

G f cmf cm

= = =


25/34

3. Knh hin vi* L dng c quang hc b tr cho mt lm tng gc trng nh ca cc vt rt nh.

(c bi gic ln hn nhiu so vi s bi gic ca knh lp)* Cu to:

+ Vt knh O1 l TKHT c tiu c rt ngn.+ Th knh O2 l TKHT c tiu c ngn (c tc dng nh knh lp).+ Vt knh v th knh c t ng trc v c khong cch khng i.

* S to nh:1 1

' '1 1 2 2

1 1 2 2O O

d d d d AB A B A B

* Cch ngm chng:Thay i khong cch t vt AB n vt knh O1 nh cui cng A2B2 l nh o ngc chiu vi AB nm

trong gii hn nhn r ca mt.AB nm ngoi v rt gn tiu im vt F1 ca vt knh O1A1B1 l nh tht ngc chiu vi AB nm trong tiu im vt F2 ca th knh O2

* bi gic :

+ Cng thc tng qut:'2

G k

d l=

+

Vi ll khong cch t th knh ti mt+ Ngm chng CC:

' '1 2

1 21 2

d d

d dCG k k k = = =

+ Ngm chng CV:

VV

G kOC

=

+ Ngm chng v cc:1 2

G

f f

d = c p dng cho mt c bt k v OCV = .

Hoc 1 2.G k G = , ch tnh cho mt c = 25cm v OCV = .

Vi k1 l s phng i nh A1B1 qua vt knh (thng ghi trn vnh vt knh)

22 2

25( )

Gf f cm

= = l bi gic ca th knh khi ngm chng v cc (thng ghi trn vnh th knh)

= F1F2 = O1O2 f1 f2 l di quang hc ca knh hin vi.VD: Trn vnh vt knh v th knh ca knh hin vi ghi X100 v X5

th vi ngi mt bnh thng ( = 25cm) c G = 500.

Cn ngi mt c = 20cm v OCV = th500.20

40025

cmG

cm= =

Lu : Mt s bi ton v knh lp v knh hin vi yu cu

- Xc nh gc trng khi bit AB th t0

. .

AB GG

AB

a aa

a= = =

- Xc nh ABMin khi bit nng sut phn li Min:0

.. MinMinG ABAB G

aa aa

= = =

4. Knh thin vn* L dng c quang hc b tr cho mt lm tng gc trng nh ca cc vt rt xa.* Cu to:

+ Vt knh O1 l TKHT c tiu c di.+ Th knh O2 l TKHT c tiu c ngn (c tc dng nh knh lp).+ Vt knh v th knh c t ng trc v c khong cch thay i c.

* S to nh:


26/34

1 1' '

1 1 2 21 1 2 2

O O

d d d d AB A B A B

AB d1 = d1 = f1 v c O1O2 = d1 + d2 = f1 + d2* Cch ngm chng:

Thay i khong cch gia vt knh O1 v th knh O2 nh o cui cng A2B2 nm trong gii hn nhn rca mt.

A1B1 l nh tht nm ti tiu im vt F2 ca th knh O2* bi gic :

+ Cng thc tng qut: 12 '2

fG kd l

=+

Vi'2

22

dk

d= - l phng i nh A2B2 qua th knh O2

ll khong cch t th knh ti mt

Trng hp c bit, mt st th knh l= 0 th 1

2

fG

d= v O1O2 = f1 + d2

+ Ngm chng v cc: 1

2

fG

f= v O1O2 = f1 + f2


27/34

CHNG VII: TNH CHT SNG CA NH SNG1. Hin tng tn sc nh sng.* /n: L hin tng nh sng b tch thnh nhiu mu khc nhau khi i qua mt phn cch ca hai mi trngtrong sut.* nh sng n sc l nh sng khng b tn sc

nh sng n sc c tn s xc nh, ch c mt mu.

Bc sng ca nh sng n scv

f

l = , truyn trong chn khng 0c

f

l = 0 0c

v n

l ll

l

= =

* Chit sut ca mi trng trong sut ph thuc vo mu sc nh sng. i vi nh sng mu l nh nht,mu tm l ln nht.* nh sng trng l tp hp ca v s nh sng n sc c mu bin thin lin tc t n tm.

Bc sng ca nh sng trng: 0,4 m 0,76 m.2. Hin tng giao thoa nh sng (ch xt giao thoa nh sng trong th nghim Ing).* /n: L s tng hp ca hai hay nhiu sng nh sng kt hp trong khng gian trong xut hin nhng vchsng v nhng vch ti xen k nhau.

Cc vch sng (vn sng) v cc vch ti (vn ti) gi l vn giao thoa.* Hiu ng i ca nh sng (hiu quang trnh)

2 1

axd d d

DD = - =

Trong : a = S1S2 l khong cch gia hai khe sng

D = OI l khong cch t hai khe sng S1, S2 n mn quan st

S1M = d1; S2M = d2

x = OM l (to ) khong cch t vn trung tm n im M ta xt

* V tr (to ) vn sng: d = k ,D

k k Za

l=

k = 0: Vn sng trung tm

k = 1: Vn sng bc (th) 1k = 2: Vn sng bc (th) 2

* V tr (to ) vn ti: d = (k + 0,5) ( 0,5) ,D

k k Za

l= +

k = 0, k = -1: Vn ti th (bc) nhtk = 1, k = -2: Vn ti th (bc) haik = 2, k = -3: Vn ti th (bc) ba

* Khong vn i: L khong cch gia hai vn sng hoc hai vn ti lin tip:D

ia

l=

* Nu th nghim c tin hnh trong mi trng trong sut c chit sut n th bc sng v khong vn:

nn n D iin a nlll = = =

* Khi ngun sng S di chuyn theo phng song song vi S1S2 th h vn di chuyn ngc chiu v khong vni vn khng i.

di ca h vn l: 01

Dx d

D=

Trong : D l khong cch t 2 khe ti mnD1 l khong cch t ngun sng ti 2 khed l dch chuyn ca ngun sng

S1

D

S2

d1

d2I O

xM

a


28/34

* Khi trn ng truyn ca nh sng t khe S1 (hoc S2) c t mt bn mng dy e, chit sut n th h vn

s dch chuyn v pha S1 (hoc S2) mt on: 0( 1)n eD

xa

-=

* Xc nh s vn sng, vn ti trong vng giao thoa (trng giao thoa) c b rng L (i xng qua vn trungtm)

+ S vn sng (l s l): 2 12SL

Ni

= +

+ S vn ti (l s chn): 2 0,52tLN

i= +

Trong [x] l phn nguyn ca x. V d: [6] = 6; [5,05] = 5; [7,99] = 7* Xc nh s vn sng, vn ti gia hai im M, N c to x1, x2 (gi s x1 < x2)

+ Vn sng: x1 < ki < x2+ Vn ti: x1 < (k+0,5)i < x2

S gi tr k Z l s vn sng (vn ti) cn tmLu : M v N cng pha vi vn trung tm th x1 v x2 cng du.

M v N khc pha vi vn trung tm th x1 v x2 khc du.* Xc nh khong vn itrong khong c b rng L. Bit trong khong L c n vn sng.

+ Nu 2 u l hai vn sng th: 1Li n= -

+ Nu 2 u l hai vn ti th:L

in

=

+ Nu mt u l vn sng cn mt u l vn ti th:0,5

Li

n=

-

* S trng nhau ca cc bc x 1, 2 ... (khong vn tng ng l i1, i2 ...)+ Trng nhau ca vn sng: xs = k1i1 = k2i2 = ... k11 = k22 = ...+ Trng nhau ca vn ti: xt = (k1 + 0,5)i1 = (k2 + 0,5)i2 = ... (k1 + 0,5)1 = (k2 + 0,5)2 = ...

Lu : V tr c mu cng mu vi vn sng trung tm l v tr trng nhau ca tt c cc vn sng ca cc bc

x.* Trong hin tng giao thoa nh sng trng (0,4 m 0,76 m)

- B rng quang ph bc k: ( )tD

x ka

l lD = - vi v t l bc sng nh sng v tm

- Xc nh s vn sng, s vn ti v cc bc x tng ng ti mt v tr xc nh ( bit x)

+ Vn sng:ax

, k ZD

x ka kD

ll= =

Vi 0,4 m 0,76 m cc gi tr ca k

+ Vn ti:ax

( 0,5) , k Z( 0,5)

Dx k

a k D

ll= + =

+

Vi 0,4 m 0,76 m cc gi tr ca k- Khong cch di nht v ngn nht gia vn sng v vn ti cng bc k:

[k ( 0,5) ]Min tD

x ka

=

ax [k ( 0,5) ]M tD

x ka

= + Khi vn sng v vn ti nm khc pha i vi vn trung tm.

ax [k ( 0,5) ]M tD

x ka

= Khi vn sng v vn ti nm cng pha i vi vn trung tm.


29/34

CHNG VIII: LNG T NH SNG1. Nng lng mt lng t nh sng (ht phtn)

2hchf mcel

= = =

Trong h = 6,625.10-34 Js l hng s Plng.c = 3.108m/s l vn tc nh sng trong chn khng.f, l tn s, bc sng ca nh sng (ca bc x).

m l khi lng ca phtn2. Tia Rnghen (tia X)Bc sng nh nht ca tia Rnghen

Min

hc

El =

Trong 220

2 2

mvmvE e U= = + l ng nng ca electron khi p vo i catt (i m cc)

U l hiu in th gia ant v cattv l vn tc electron khi p vo i cattv0 l vn tc ca electron khi ri catt (thng v0 = 0)m = 9,1.10-31 kg l khi lng electron

3. Hin tng quang in*Cng thc Anhxtanh

20 ax

2Mmvhchf Ae

l= = = +

Trong 0

hcA

l= l cng thot ca kim loi dng lm catt

0 l gii hn quang in ca kim loi dng lm cattv0Max l vn tc ban u ca electron quang in khi thot khi cattf, l tn s, bc sng ca nh sng kch thch

* dng quang in trit tiu th UAK Uh (Uh < 0), Uh gi l hiu in th hm20 ax

2M

h

mveU =

Lu : Trong mt s bi ton ngi ta ly Uh > 0 th l ln.* Xt vt c lp v in, c in th cc i VMax v khong cch cc i dMax m electron chuyn ng trongin trng cn c cng E c tnh theo cng thc:

2ax 0 ax ax

1

2M M Me V mv e Ed = =

* Vi U l hiu in th gia ant v catt, vA l vn tc cc i ca electron khi p vo ant, vK= v0Max l vtc ban u cc i ca electron khi ri catt th:

2 21 1

2 2A Ke U mv mv= -* Hiu sut lng t (hiu sut quang in)

0

nH

n=

Vi n v n0 l s electron quang in bt khi catt v s phtn p vo catt trong cng mt khong thigian t.

Cng sut ca ngun bc x: 0 0 0n n hf n hc

pt t t

e

l= = =


30/34

Cng dng quang in bo ho: bhn eq

It t

= =

bh bh bhI I hf I hcHp e p e p e

e

l = = =

* Bn knh qu o ca electron khi chuyn ng vi vn tc v trong t trng u B

, = ( ,B)

sin

mvR v

e B

a

a

=r ur

Xt electron va ri khi catt th v = v0Max

Khi sin 1mv

v B Re B

a^ = =r ur

Lu : Hin tng quang in xy ra khi c chiu ng thi nhiu bc x th khi tnh cc i lng: Vn tcban u cc i v0Max, hiu in th hm Uh, in th cc i VMax, u c tnh ng vi bc x c Min(hoc fMax)4. Tin Bo - Quang ph nguyn t Hir* Tin Bo

mn m n

mn

hchf E E e

l

= = = -

* Bn knh qu o dng th n ca electron trong nguyn t hir:rn = n

2r0Vi r0 =5,3.10

-11m l bn knh Bo ( qu o K)* Nng lng electron trong nguyn t hir:

2

13,6( )nE eVn

= - Vi n N*.

* S mc nng lng- Dy Laiman: Nm trong vng t ngoi

ng vi e chuyn t qu o bn ngoi v qu o K

Lu : Vch di nht LKkhi e chuyn t L KVch ngn nht Kkhi e chuyn t K.

- Dy Banme: Mt phn nm trong vng t ngoi, mtphn nm trong vng nh sng nhn thy

ng vi e chuyn t qu o bn ngoi v qu o LVng nh sng nhn thy c 4 vch:Vch H ng vi e: M LVch lam H ng vi e: N LVch chm H ng vi e: O LVch tm H ng vi e: P L

Lu : Vch di nht ML (Vch H)Vch ngn nht L khi e chuyn t L.

- Dy Pasen: Nm trong vng hng ngoing vi e chuyn t qu o bn ngoi v qu o M

Lu : Vch di nht NM khi e chuyn t N M.Vch ngn nht M khi e chuyn t M.

Mi lin h gia cc bc sng v tn s ca cc vch quang ph ca nguyn t hir:

13 12 23

1 1 1

= + v f13 = f12 +f23 (nh cng vct)

hfmn hfmn

nhn phtn pht phtnEm

En

Em > En

Laiman

K

M

N

O

L

P

Banme

Pasen

HHHH

n=1

n=2

n=3

n=4

n=5n=6


31/34


32/34

CHNG IX. VT L HT NHN1. Hin tng phng x* S nguyn t cht phng x cn li sau thi gian t

0 0.2 .t

tTN N e l- -= =

* S ht nguyn t b phn r bng s ht nhn con c to thnh v bng s ht ( hoc e- hoc e+) c tothnh:

0 0 (1 )

t

N N N N e

l-

D = - = -* Khi lng cht phng x cn li sau thi gian t

0 0.2 .t

tTm m m e l- -= =

Trong : N0, m0 l s nguyn t, khi lng cht phng x ban uT l chu k bn r

2 0,693ln

T Tl = = l hng s phng x

v T khng ph thuc vo cc tc ng bn ngoi m ch ph thuc bn cht bn trong ca chtphng x.

* Khi lng cht b phng x sau thi gian t

0 0 (1 )tm m m m e l-D = - = -

* Phn trm cht phng x b phn r:0

1 tm

em

l-D = -

Phn trm cht phng x cn li:0

2t

tTm

em

l- -= =

* Khi lng cht mi c to thnh sau thi gian t

1 0 11 1 0(1 ) (1 )

t t

A A

A N ANm A e m e

N N Al l- -D= = - = -

Trong : A, A1 l s khi ca cht phng x ban u v ca cht mi c to thnh

NA = 6,022.10-23

mol-1

l s Avgar.Lu : Trng hp phng x +, - th A = A1 m1 = m* phng x H

L i lng c trng cho tnh phng x mnh hay yu ca mt lng cht phng x, o bng s phn rtrong 1 giy.

0 0.2 .t

tTH H H e Nl l- -= = =

H0 = N0 l phng x ban u.n v: Becren (Bq); 1Bq = 1 phn r/giy

Curi (Ci); 1 Ci = 3,7.1010 BqLu : Khi tnh phng x H, H0 (Bq) th chu k phng x T phi i ra n v giy(s).

2. H thc Anhxtanh, ht khi, nng lng lin kt* H thc Anhxtanh gia khi lng v nng lngVt c khi lng m th c nng lng ngh E = m.c2

Vi c = 3.108 m/s l vn tc nh sng trong chn khng.* ht khi ca ht nhn AZX

m = m0 mTrong m0 = Zmp + Nmn = Zmp + (A-Z)mn l khi lng cc nucln.

m l khi lng ht nhn X.* Nng lng lin kt E = m.c2 = (m0-m)c

2


33/34

* Nng lng lin kt ring (l nng lng lin kt tnh cho 1 nucln):E

AD

Lu : Nng lng lin kt ring cng ln th ht nhn cng bn vng.3. Phn ng ht nhn* Phng trnh phn ng: 31 2 4

1 2 3 41 2 3 4AA A A

Z Z Z ZX X X X+ +

Trong s cc ht ny c th l ht s cp nh nucln, eletrn, phtn ...Trng hp c bit l s phng x: X1 X2 + X3

X1 l ht nhn m, X2 l ht nhn con, X3 l ht hoc * Cc nh lut bo ton

+ Bo ton s nucln (s khi): A1 + A2 = A3 + A4+ Bo ton in tch (nguyn t s): Z1 + Z2 = Z3 + Z4+ Bo ton ng lng: 1 2 3 4 1 1 2 2 4 3 4 4m m m mp p p p hay v v v v+ = + + = +

+ Bo ton nng lng:1 2 3 4X X X X

K K E K K+ + D = +

Trong : E l nng lng phn ng ht nhn21

2X x xK m v= l ng nng chuyn ng ca ht X

Lu : - Khng c nh lut bo ton khi lng.- Mi quan h gia ng lng pX v ng nng KX ca ht X l: 2 2X X Xp m K=

- Khi tnh vn tc v hay ng nng K thng p dng quy tc hnh bnh hnh

V d: 1 2p p p= + bit

1 2,p p=uur uur

2 2 2

1 2 1 22p p p p p cosj= + +

hay 2 2 21 1 2 2 1 2 1 2( ) ( ) ( ) 2mv m v m v m m v v cosj= + +

hay 1 1 2 2 1 2 1 22mK m K m K m m K K cosj= + +

Tng t khi bit

1 1 ,p p=uur ur

hoc

2 2 ,p p=uur ur

Trng hp c bit: 1 2p p^ 2 2 2

1 2p p p= +Tng t khi 1p p^ hoc 2p p^

v = 0 (p = 0) p1 = p2 1 1 2 22 2 1 1

K v m A

K v m A= =

Tng t v1 = 0 hoc v2 = 0.* Nng lng phn ng ht nhn

E = (M0 - M)c2

Trong :1 20 X X

m m= + l tng khi lng cc ht nhn trc phn ng.

3 4X Xm m= + l tng khi lng cc ht nhn sau phn ng.

Lu : - Nu M0 > M th phn ng to nng lng E di dng ng nng ca cc ht X3, X4 hoc phtn .Cc ht sinh ra c ht khi ln hn nn bn vng hn.- Nu M0 < M th phn ng thu nng lng |E| di dng ng nng ca cc ht X1, X2 hoc phtn .

Cc ht sinh ra c ht khi nh hn nn km bn vng.* Trong phn ng ht nhn 31 2 4

1 2 3 41 2 3 4AA A A

Z Z Z ZX X X X+ +

Cc ht nhn X1, X2, X3, X4 c:Nng lng lin kt ring tng ng l 1, 2, 3, 4.Nng lng lin kt tng ng l E1, E2, E3, E4 ht khi tng ng l m1, m2, m3, m4Nng lng ca phn ng ht nhn

p

1p

2p


34/34

E = A33 +A44 - A11 - A22E = E3 + E4 E1 E2E = (m3 + m4 - m1 - m2)c

2* Quy tc dch chuyn ca s phng x

+ Phng x ( 42He ):4 42 2

A AZ ZX He Y

-- +

So vi ht nhn m, ht nhn con li 2 trong bng tun hon v c s khi gim 4 n v.+ Phng x - ( 10 e

- ): 01 1A AZ ZX e Y- + +

So vi ht nhn m, ht nhn con tin 1 trong bng tun hon v c cng s khi.Thc cht ca phng x - l mt ht ntrn bin thnh mt ht prtn, mt ht electrn v mt ht ntrin:

n p e v- + +

Lu : - Bn cht (thc cht) ca tia phng x - l ht electrn (e-)- Ht ntrin (v) khng mang in, khng khi lng (hoc rt nh) chuyn ng vi vn tc ca nh

sng v hu nh khng tng tc vi vt cht.+ Phng x + ( 10 e

+ ): 01 1A AZ ZX e Y+ - +

So vi ht nhn m, ht nhn con li 1 trong bng tun hon v c cng s khi.Thc cht ca phng x + l mt ht prtn bin thnh mt ht ntrn, mt ht pzitrn v mt ht ntrin

p n e v+ + +

Lu : Bn cht (thc cht) ca tia phng x + l ht pzitrn (e+)+ Phng x (ht phtn)

Ht nhn con sinh ra trng thi kch thch c mc nng lng E1 chuyn xung mc nng lng E2 ngthi phng ra mt phtn c nng lng

1 2

hchf E E e

l= = = -

Lu : Trong phng x khng c s bin i ht nhn phng x thng i km theo phng x v 4. Cc hng s v n v thng s dng* S Avgar: NA = 6,022.10

23 mol-1* n v nng lng: 1eV = 1,6.10-19 J; 1MeV = 1,6.10-13 J

* n v khi lng nguyn t (n v Cacbon): 1u = 1,66055.10-27

kg = 931 MeV/c2

* in tch nguyn t: |e| = 1,6.10-19 C* Khi lng prtn: mp = 1,0073u* Khi lng ntrn: mn = 1,0087u* Khi lng electrn: me = 9,1.10

-31kg = 0,0005u

cong thuc vat li

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