de tai 07-08.2450 unicode ha van chinh

8/14/2019 de tai 07-08.2450 unicode ha van chinh

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S GIO DC O TO BNH NHTRNG TRUNG HC PH THNG S 3 PH CT

GIO VIN : H VN CHNHT: L HO

TRNG TRUNG HC PH THNG S 3 PH CT

NM HC : 2007-2008


2/12

p ng vi mc tiu , chng trnh, phng php , phng tin v cch kim tra nhgi bng hnh thc trc nghim khch quan , mi c th o c nhng mc kh nng khc nhau vgi tr ni dung kin thc . V vy s lng cu hi c tri di theo ton b chng trnh c bn

vt l ph thng . i hi hc sinh phi xt on v phn bit k cng khi chn la cu tr li ngnht hay hp l nht . lm c trn vn cc yu cu ra , hc sinh phi hiu , p dng cc nguyn l , suy

din v tng hp kin thc hc.Mi cu hi trc nghim c mt mc tiu r rng , nhm vo mt n v kin thc ca

chng trnh . V vy hc sinh rn luyn k nng vn dng kin thc, x l tnh hung t ra mkhng phm phi sai lm th hc sinh phi ch nhng vn no ?

1. .MC BIT1. Hy c nhanh qua mt ln ni dung thi v lm ngay nhng cu hi mc bit . V ch

cn nh li nhng ni dung , cng thc c trong SGK c th trn vn hoc mt phn , hoc didng thay i cht t .

V d : + Dao ng iu ho l g ?- Hc sinh ch cn nh dng sin hoc cosin (hm iu ho)+ Cng thc tnh bc sng - Hc sinh nh qung ng sng truyn c sau thi gian T

= V.T+ Trong phn ng ht nhn khng c nh lut bo ton no ?-V c ht khi nn khng c nh lut bo ton khi lng .+ Cng thc tnh nng lng in trng v nng lng t trng trong

mch dao ng LC .

- Nng lng in trng :WE =2

1Cu2

- Nng lng t trng :WL =2

1Li2

-V mt hnh thc cc biu thc nng lng trong mch dao ng v cc biu thc nng lng trongdao ng c hc lging nhau .

* V vy trong t in ,c in dung C t trng kh nng tch in ca t,khi c mt hiuin th hai u bn t c tch in th c mt nng lng in trng gi hai bn t: WE =

2

1Cu2

* Trong cun dy c h s t cm L khi c dng in in i chy qua th c nng lng t

trng trong cun dy l :WL =21 Li2

2. Hc sinh thng sai nhng trng hp sau :+ Nu cng thc tnh chu k dao ng iu ho con lc l xo

- Hc sinh khng nh r :T = 2K

m(a) hay T = 2

m

K.

- Vi mt l xo nht nh c cng K khng i . Nu treo mt vt ckh lng m cngnh v kch thch cho h dao ng th vt dao ng nhanh hn . Ngha l chu k dao ng nh . HayT t l thun vi m .V vy HS xc nh ngay biu thc (a) .

H thng n v : [ ]K :N/m , [ ]m :kg+ Cng thc tnh chu k dao ng con lc n .


3/12

- Chu k dao ng con lc n T = 2g

lhay T = 2

l

g.

- Hc sinh hnh dung cc din vin xic trn u . Ti sao si dy phi di ? thi gianhai ngi trao i qua li gia hai chic u . V vy si dy cng di , th thi gian chuyn ng cnglu hay chu k T cng ln .

Vy T t l thun vi l*Cch kim tra n gin nht l thay bng h thng n v chun : [ ]T :s , [ ]l :m , [ ]g :m/s2

+ Nu cng thc tnh chit sut tuyt i ca mt mi trng trong sut .

- Chit sut n =v

chay n =

c

v

- Hc sinh nn nh vn tc nh sng l ln nht v chit sut tuyt i cc mi trng philn hn 1 .

V vy chn t s : n =v

c> 1 .

+ Nu cng thc tnh cng sut tiu th trn mt on mch in xoay chiu khng phnnhnh RLC .- Hc sinh nhm gia cng sut P v h s cng sut cos

- Cng sut :P = U.I cos hoc P = RI2 v H s cng sut: cos =Z

R

Nhn xt: Gi tr 1Cos

+ Cng thc v cng sut to nhit P v nhit lng to ra Q.- Hc sinh phn bit : Nhit lng l nng lng tnh bng Jun . Vy Q = RI2.t-Phn tiu th nng lng l in tr R . Nn sau thi gian t phi tiu th mt nng lng Q

= RI2.t v c cng sut to nhit P =t

Q= RI2 .

+ Cng thc v phng i nh v bi gic ca dng c quang hc .

- Nu phng i nh l t s cao nh so vi cao ca vt K =AB

BA ''

-Th bi gic l t s gc trng nh so vi gc trng trc tip vt 0 .

0

=G

2. .MC HIU


4/12

- Mc hiu i hi hc sinh khng nhng nh li v pht biu li nguyn dng vn hc , mcn c th thay i vn hc sang mt dng khc tng ng . i hi hc sinh phi c khnng din dch gii thch v ngoi suy .

* Nhng cu hi chn cu sai hay chn cu ng, th hc sinh phi c k tng ni dung phngn v suy lun loi tr .V d: + Tia + lch v pha bn dng hay lch v pha bn m ca t in.

- Hc sinh nn pht h chiu in trng . bit c lc in trng tc dng ln intch .Suy ra dng qu o chuyn ng .+ Sp xp cc bc sng ca cc bc x hng ngoi ,t ngoi ,Rn ghen ,tia

gamma. theo th t bc sng tng dn hay gim dn.- Hc sinh ch n bc sng hay tn s . Theo th t tng dn hay gim dn .- Hc sinh phi nm k thang sng in t v suy lun .+ S hnh thanh cc dy Lyman , Bamme , Passen . Khi elctrn dch chuyn mc

nng lng .- Hc sinh phi nh s dch chuyn elctrn t cc qu o dng bn ngoi ln lt v qu

o K , L , M hnh thnh dy Ly man , Bamme , Passen min t ngoi , nh sng nhn thy vhng ngoi .

t bit : Dy Bamme c mt phn nm min t ngoi ngoi 4 vch nhn thy c .+ Khi m thanh truyn t khng kh vo nc th bc sng v tn s c thay i khng ?- Hc sinh phi bit bc sng l qung ng sng truyn i c trong mt chu k dao

ng ca sng (ph thuc vo vn tc v thi gian ). M vn tc truyn sng ph thuc vo mitrng truyn sng .

+ Nhng cu hi v sng m .- Hc sinh cn hc k cc c tnh sinh l ca m c hnh thnh trn c s cc c tnh vt

l ca m .c trng sinh l m c trng vt l m

cao Tn s to Cng m , tn s m .m sc Bin m , tn s m .

- Hc sinh cha phn bit gia cng m I v mc cng m L

[I] : 2m

W; L = lg

0I

I(B) hoc L = 10lg

0I

I(dB)

+ Phn quang hnh hc :- Hc sinh cn c k dng c l gng hay knh xt ng i tia sng b phn x hay

khc x . Anh ca vt cng pha hay khc pha so vi dng c suy ra nh tht hay nh o .+ i vi phn mt v cc dng c quang hc .- Hc sinh cn hiu r v mt v cc tt ca mt .- Cn phn bit v tr t vt so vi mt hay so vi thu knh (xc nh gi tr d > 0)

- Mt ch thy c nh (o) ca vt . Vy d < 0 (so vi thu knh ) .- Ch : Mt trng thi khng iu tit l quau st vt im cc vin hay (mt bnh

thng) v mt iu tit ti a khi quan st vt im cc cn .- hiu c s iu tit ca mt hay s bin thin t ca mt . Ta xem nh ca vt qua

thu knh ng vai tr vt i vi thu tinh th (TKHT) ca mt . Mt iu tit nh (tht) hintrn vng mc .

+ Phn lng t nh sng :- Hc sinh nm c ni dung thuyt lng t nng lng Pln v quan im Anhxtanh

gii thch cc nh lut quang in .

- Khi vit cng thc : = hf =

hc= A +

2

1m 2maxoV


5/12

Hc sinh phi hiu rng : nng lng phtn truyn ht cho electrn v c s dng hai phn , mt phn thng lc lin kt thot ra ngoi v phn cn li tn ti di dng ng nng

ban u cc i ca quang electrn.+ Phn vt l ht nhn : Hc sinh thng nhm s ht ban u 0N , s ht cn li thi im t : tN v s ht b

phn r phng x N .

- V vy khng phn bit c :

- T s phn trm s ht cn li so vi s ht ban u .0N

N%

- Phn trm b phn r phng x .0N

N%

- Hc sinh d nhm s ht sinh ra bng s ht ban u theo phng trnh phng x. M thcra s phng x ht nhn sinh ra bng s ht b phn r phng x . Nhng khi lng htsinh ra ,khng bng khi lng ht nhn b phn r phng x .

*t bit trong cng thc tnh phng x . H = H0te

= Tt

H

2.0 .

- Hc sinh thng tnh sai phn n v thi gian .+ Phn dao ng cha phn bit r gia tn s dao ng f v tn s gc .

3. Mc p dng+ iu m hc sinh sai nhiu nht trong bi tp p dng l : Th s sai h thng n v . V

vy GV cn ni r h thng n v o lng chun h SI . Xut pht t 7 n v vt l c bn l : di : mt (m)Thi gian : giy (s)Khi lng : kilgam (kg)Cng dng in : Ampe (A)Cng sng : Cardopa (d)

Lng cht : mol (mol)Nhit : Kenvin (K)T suy ra s ph thuc ca n v mt i lng no vo cc n v c bn c gi l

th nguyn ca n v .V d : [vn tc] : m/s , [gia tc] : m/s2 , [cng] : (A = F.s , m F = m.a)

Kg. 2s

m.m = kg.m2.s-2

+ Hc sinh cha phn bit i lng c hng v i lng v hng . Nh vect ng

lng P v ln ng nng E = 2

1mv2

Hc sinh cha phn c gi tr ln v gi tr i s theo qui t v du (th hin tnhcht vt l)- Phn dao ng c hc .+ Hu ht hc sinh sai khi th s cc gi tr .- Khi lng m phi tnh bng kilgam .- cng l xo [k] : N/m- Chiu di con lc n [ l ] : m+ Phn dng in xoay chiu thng i h thng n v cc gi tr R , L , C , P , U , I theocc c s hoc bi s ca n .V vy GV cn ni r cc c s ca thng dng l :mili (m) , micr () , nan (n) , pic (p)

c s chun SI


6/12

Mili :1m 10-3

Micr : 1 10-6 Nan : 1n 10-9Pic : 1p 10-12

V d : 1F = 10-6F ; 1nH = 10-9 H ; 1p = 10-12Bi s chun

Kil 1K 103Mga 1M 1061 k = 1031 MeV = 106 e V

+ Phn quang hnh hc :- Hc sinh cha hiu k phn qui c v du (thun tu ton hc) n vic xc nh v tr ,tnh cht (tht , o) ca vt hay nh .- Hc sinh hiu c : t mt vt (l vt tht) thy nh trong gng l nh o , hng nhtrn mn l nh tht .c bit : hc sinh d nh c th ly v tr (mc) ca knh hoc gng . Hoc mc (trcchnh ) suy ra cc tnh cht vt v nh . Cng pha th tri tnh cht v tri pha th cngtnh cht ng theo quy lut trit hc v thng nht v u tranh cc mt i lp .+ Phn vt l ht nhn :- Hc sinh rt yu v k nng tnh ton s m . V vy GV cn trang b li cc php n ginv s m v cc hm siu vit . c bit l n v dng trong vt l ht nhn .V d : n v khi lng

me 9,1095.10-31kgme 0,000549 u

M : 1u =AN

12.

12

1(g) = 1,66055.10-27 kg = 0,511. MeV/C2

Nn : 1. )/(10.3

)(10.6022,128

13

2 sm

J

C

MeV

= 1,7827.10-3kg

Vy : 1kg 0,561.1030 MeV/C2+ Cc k hiu i lng vt l trong nhng chuyn xc sut khc nhau c th ging nhau .V d : Ch f : l tn s phn dao ng n v Hc (Hz)

f : l tiu c TK phn quang hc n v chiu diCh D : l t TK , n v ip

[D ] =f

1

Gc lch D tnh theo n v gc : hoc radian

Ch K phn quang hnh l phng iK phn my bin th l h s my bin th .

Ch phn sng l Bc sngCh phn vt l ht nhn l hng s phn r phng x

V d : Ch f : l tn s phn dao ng n v Hc (Hz)f : l tiu c TK phn quang hc n v chiu di

Ch D : l t TK , n v ip

[D ] =f

1

Gc lch D tnh theo n v gc : hoc radian


7/12

Ch K phn quang hnh l phng iK phn my bin th l h s my bin th .

4. Mc phn tch v tng hp - y l dng bi bi tp i hi hc sinh phi bit vn dng linh hot nhng iu hc v vic giiquyt vn t ra .Bi ton 1 : Mt ngi cn th eo knh cn s 4 mi nhn thy r nhng vt xa v cng . Khi eo

knh st mt , ngi ch c c trang sch t cch mt t nht l 25cm . Tm gii hn nhn rngi ny khi khng eo knh .+ Hc sinh cn phi hiu c sa tt cn th phi dng thu knh g ?- sa tt cn th phi dng thu knh phn k+ Nu l thu knh hi t hay phn k th tiu c phi gi tr dng hay m ?- V thu knh phn k nn tiu c c gi tr m .+ Qui c v knh smy l ch cho gi tr ln ca i lng vt l no ?- Knh s my l ch v gi tr ln ca t .

Vy TKPK c t D = -4 ip => f =4

11

=

D= 0,25m = -25cm

+ eo knh st mt v cch mt c tc dng g ?- eo knh st mt . Nu v tr nh ca vt qua TK cch TK chnh l khong cch n mt (0m = 0K) .+ Mt nhn qua knh s thy nh ca vt l tht hay o ?- Anh ca vt qua TK l nh o . Nhng li ng vai tr vt tht i vi mt . Vy mt mun nhnthy vt th nh ca n phi hin trn vng mc ca mt .+ Vt xa v cng th nh qua thu knh hin u ?- Vt xa v cng (d = ) th nh hin ti tiu im nh ca TK . Hay im cc vin ca mt d1 = th d1 = f = -25cm .+ Anh ca vt qua thu knh ng vai tr g i vi mt ?-Vt cch thu knh l 25cm th d2 = 25cm . Anh qua thu knh hin ti im cc cn ca mt Khi d2

= 25cm , f = -25cm th d2 = )25(25

)25.(25.

2

2

= fd

fd

= 12,5cmVy : Gii hn nhn r ngi ny khi khng eo knh l t 12,5cm n 50cm .+ V tr vt t gn nht trc knh th nh (o) phi hin ti v tr no ca mt ?+ Phn bit c gii hn nhn r ngn nht ca mt ? V gii hn nhn r ca mt .

Nh vy gii bi tp trn hc sinh phi hiu v vn dng chnh xc tng vn da vo s phntch v tng hp tt c cc kin thc c lin quan v quang hc .

Nu bi tp m rng :Mt quan st nh ca vt trng thi iu tit ti a v trng thi khng iu titTh hc sinh phi bit :+ Quan st vt trng thi khng iu tit ti a . Th t ca thu tinh th l ln nht v tiu c

(mt) l nh nht . Mt ang quan st vt cc cn .Bi ton 2 :Cho on mch xoay chiu gm 2 phn t X v Y mc ni tip . Khi t vo hai u on mch mthiu in th xoay chiu c gi tr hiu dng l U th hiu in th hiu dng gia hai u phn t Xv Y tng ng l g ?

Hc sinh cn nm r cc phn t trong mch in xoay chiu l g .+ Gm : in tr thun R , cun dy (thun cm hoc khng thun cm ) , t in c in dung ( C).-Phng php biu din dao ng iu ho bng gin vect quay .+ Hiu in th gia hai u in tr thun bin thin iu ho v cng pha vi dng in .


8/12

+ Hiu in th gia hai u cun dy thun cm bin thin iu ho v nhanh pha hn dng in l

:Z

H

c bit : Nu cun dy c in tr th hiu in th hai u cun dy s nhanh pha hn dng in

mt lng l xc nh bi tg =OR

ZL

+ Hiu in th gia hai u t in bin thin iu ho v chm pha hn dng in lZ

H

- Nhng iu nu trn ch l l thuyt c bn SGK . V vy hc sinh phi vn dng vo tnh hungc th nh th no ?+ u tin gi s cun dy thun cm v a ra cc cp tnh hun c th c .(Ch : Cn biu din di vct bng ln gi rt cho)+ Trng hp 1 : Mch R,C

ln ; UAB = 22 CR UU +

Nn : UAB = 22 )2()( VVz + = U 7 (V) U (v l)

+ Trng hp 2 : Mch R , Llm tng t :UAB = U 7 (V) U (v l)

+ Trng hp 3 : Mch L , C ln : UAB = {VL VC { = { 3 U 2U{ U (v l)

Vy cun dy phi c in tr thun : xt mch LC

Theo gin vct :222

0 ABRLRUUU +=

Hay : (2U)2 = ( 3 U)2 + U2Vy trong Y l R0L v trong X l C . Tho mn

Nhn xt : hc sinh lm c nhanh cc vn t ra . Khng yu cu hc sinh phi lm y cc tnh hung t ra . M c th dng phng php loi tr .* Mc nh gi :- bi ton 1 : c v nhn xt 4 phng n a ra+ Phng n A : V ngi ny b tt cn th nn gii hn nhn r ngi ny khi khng eo knhkhng th ln hn 12cm . Nh vt chng l ngi cn th thy c vt cch mt t 12cm n vcng .+ Phng n D : Ngi cn th phi c im cc cn gn hn mt bnh thng nn khng th l35cm . Mt khc khng th thy vt xa v cng .+ Phng n C : V khi eo knh mi c c trang sch cch mt 25cm . Nu khng eo knh thv tr vt t gn nht khng phi cch mt l 25cm . Mt khc nu hc sinh c tiu c TKPK l f

= -25cm th nh ca vt phi l nh o v ch nm trong khong t tiu im nh n TK ({d{ 25cm) . Vy chn ngay phng n B .- bi ton 2 : c v nhn xt 4 phng n a ra sau khi loi ln lt cc phng n A,B,C bngcch dng gin vct quay . Th chn ngay phng n D m khng cn phi tnh ton c th n

phng n D .Mc nh gi , nhn nh v gi tr ca mt t tng , mt phng php , mt ni dung kin thc .y l mt bc pht trin tr tu sng to ca hc sinh trong vic chn lc v lnh hi kin htc .Hc sinh phi nm bt c bn cht ca i tng , s vt , hin tng lin quan n kh nng

phn on gi tr ca cc ti liu , cc phng php , i vi nhng mc ch nht nh no , xemth ti liu y hoc phng php y c hi nhng tiu ch ra khng .Cc tiu ch nh gi cth do hc sinh t ra hoc cho sn .Kt lun :


9/12

hc c thi ng n khi chun b thi theo phng php trc nghim l phi hc k ,c suy lun , phn tch v tng hp cc kin thc khoa hc , ch khng hc vt (hc thuc lng) . Scu hi trong bi thi trc nghim tri rng ton b ni dung kin thc c bn trong gio trnh hc .V phn chia thnh cc mc khc nhau , vi cu trc ngu nhin theo m khng theo mt tht nht nh d hay kh . Vn hc trc hay hc sau , v vy hc sinh phi c nhanh v k phndn ca mi cu hi nhng cu l thuyt hoc bi tp p dng n gin , hnh dung cu tr li

ng phi c nhng yu t no , tnh cht no hoc tnh ton n gin tm ra p s v chn ngayphng n tr li . i khi c mt s cu hi hc sinh khng cn n tnh chnh xc , m ch cn tnhgn ng ri tm phng n tr li gn vi kt qu tnh ton . V vy nu hc sinh thy khnghon ton chc chn nhng c kin thc tin chc n 75% hy mnh dn chn v nh du cutr li .Trn y ch l nhng rt nh gip c hc sinh phn no trong phng php lm bi trc nghim.Rt mong s chn thnh gp ca tt c bn c . Xin chn thnh cm n .

C :* Tnh chu k dao ng con lc l xo . Bit khi lng qu cu m = 0,1kg . L xo c cng K =

0,25N/cm ; T = 2.k

m = 225

100 = 4.(A) = 12,50(s)

Hc sinh s sai khi th s : T = 225,0

1,0(sai)

Nu c h t trn mt phng nghing = 300 . B qua ma st th chu k dao ng l bao nhiu ?Hc sinh lm tng phi tnh gi tr lin quan l gc * Tnh c nng ca h bit K = 0,25N/cm , A = 2cm

E -2

1KA2 =

2

1.25.(2.10-2) 2 = 50.10-4(J)

Hc sinh s sai l : E =21 .0,25 (2)a

* Mt cht im dao ng iu ho theo qui lut hm sin . Xc nh bin , tn s , pha ban u ?x = 5.cos.10t (cm)+ Bin A = 5cm+ Tn s gc : = 10 = 2f f = 5Hz

+ V : x = 5.cos10t = 5.sin(10t +z

) cm . nn pha ban u : =

z

rad

- Hc sinh s nhm tn s f v tn s gc . Pha ban u l = 0 . Vi qui lut hm cosin , nhng

theo qui lut hm sin th = z

* Sng c hc :- Cng thc quan h gia bc sng , vn tc truyn sng , chu k sng (T) hay tn s sng f .

+ Hc sinh hc vt khng nh : = V.T hay : =T

V.

hc sinh nh chnh xc : GV cn nhc li rng bc sng chnh l qung ng m sng truyni c trng mt chu k dao ng ca sng . M : S = V.T- i vi bi tp v qu trnh truyn sng gy ra t 1 ngun v 2 ngun kt hp .Hc sinh cn ch .+ Khong cch gia 2 gn sng trong qu trnh truyn sng do mt ngun sng gy ra l mt bc

sng .


10/12

+ Khong cch gia 2 gn li hay im ng yn trong qu trnh dao thoa sng l na bc sng .+ Kt qu chng minh v hiu ng i do 2 ngun sng kt hp cng bin , cng tn s , cng

pha . Ti nhng im c bin dao ng cc i ld2 d1 = K

Nhng nu 2 ngun sng kt hp cng bin , cng tn s , ngc pha th ti nhng im daong vi bin cc i l

d2 d1 = (2K + 1)2

* Phn in xoay chiu .Hc sinh thng sai cng thc xc nh gc lch pha gia hiu in th u v cng dng in i

l : tg =K

ZZCL

Khng c dng cng thc : cos =Z

R tnh

V d 1 : on mch ch c R , C

Hc sinh ghi : tg =K

C2(sai)

GV cn phn tch . on mch c C v R s c tnh dung khng . Nn u phi chm pha hn i .

V vy xt v mt thun tu ton hc : tg =K

ZC < 0

Do : < 0 nn u chm pha hn i .V d 2 : on mch ch c L , C

Hc sinh ghi : tg = =

0CL ZZ =

Z

rad.

iu m hc sinh qun l : phi so snh gi tr ZL v ZC , xem on mch c tnh cht cm khng(ZL > ZC) . Hoc kt qu trn t s dng hay m dn n tg = + (ZL > ZC) hay tg = - (ZC >

ZL ) .V vy hc sinh cn c k gi thuyt v s lch pha ca u so vi i hay ca i so vi u . Hoc cau1 so vi u2 . t bit l gi tr hiu dng trn dng c o v gi tr cc i trong biu thc : i = Iosint(A) ; hoc : u = Vo sint (V)- Hc sinh cn phn bit gi tr hiu dng X ca dng in l s ch trn Ampe k v gi tr ny cth t ln nht . Khi xy ra cng hng . Ngha l cng hiu dng t gi tr cc i (ln nht)

l : Imax . Lc ny gi tr Io trong biu thc tc thi s l : Imax =2

max0I

I0max = Imax 2 .Ni n gin : Imax I0 .

Lp lun tng t cho gi tr Umax v U0 (s ch cc i trn vn k v gi tr U0 trong biu thc tcthi u = v0sint (v) .* Phn quang hnh hc .- vn dng cc cng thc quan h v v tr ca vt v nh , phng i , quan h tiu c v t .

+ Vt o hoc nh o(d < 0 ) (d < 0 )

Mt chm tia sng hi t b chn bi gng cu li bn knh R = 50cm , cch tm gng 25cm . Xcnh v tr , tnh cht , phng i nh v chiu vt so vi nh .- Hc sinh s sai : ngh rng im sng cch gng l vt tht (d > 0) . Nh vy chm hi t t xa vcng gp nhau ti S v t S truyn i xa th S l im sng ng vai tr vt tht . Nhng chm hi t

b chn bi dng c quang hc th S l vt o (d < 0 ) . Nhng hc sinh phi bit rng d l khong


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cch t vt n TK hoc gng . Vy khi S cch tm gng v phi gn nh gng th d1 = -25cm. Nhng khi S v pha xa nh gng hn th d2 = -75cm

- Khi hc sinh tnh : d =fd

fd

.=

)25(25

)25).(25(

=

V vy hc sinh cn lp lun . Nu vt (o) tiu im ca F th nh . Ngha l chm tia phnx song song vi trc chnh .

c bit khng c dng cng thc : d =fd

fd

. . Khi d = th d = ?

M hc sinh ch c s dng cng thc :fdd

1

'

11=+ . Khi d = ; th d = f

Hoc lp lun : Vt v cng th nh hin ti tiu im chnh (tht hoc o)

- Xt trng hp d = -75cm , th d =fd

fd

.=

2575

)25).(75(

+

d = -37,5cm < 0Vy hc sinh phi kt lun v v tr v tnh cht ca nh (nh o, cch nh gng l 37,5cm)

- phng i nh : K = - dd' = - 2175 )5,37(

= > 0

Vy nh cng chiu vi vy v bng2

1ln vt . Nhng bi tp v TK hoc gng t ch xt trng

hp vt t vung gc vi trc chnh , k c bi ton gng phng th mt phng t vt phi vunggc vi mt phng ca gng .

Trong qu trnh phn tch khng trnh khi thiu st .Mong cc thy c v cc bn gp

GIO VIN : H VN CHNH


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de tai 07-08.2450 unicode ha van chinh

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