mau bao cao thi nghiem thuc hanh vat li 9

MU BO CO CO TH NGHIM THC, THC HNH

BO CO TH NGHIMNgy thng nm 200

BI THC HNH S: 1.

TN TH NGHIM: KHO ST S PH THUC CA CNG DNG IN VO HIU IN TH GIA HAI U DY DNTIT 01 - BI 01: S PH THUC CA CNG DNG IN VO HIU IN TH GIA HAI U DY DNTng im (10)Chun b

(1)Trt t v sinh (1)Thao tc

(2)Cu hi (2)Kt qu (2)Nhn xt

(2)

I. MC CH YU CU: Thy c mi quan h gia cng dng in v hiu in th t vo hai u dy dn. Lm quen vi vic s dng th trong th nghim.II. CHUN B DNG C:

1. Gio vin: Mt b th nghim nh Hnh 1.1.2. Hc sinh: Mi nhm: - 1 bng lp in - 1 dy cnstantan ( c ung knh 0,3 mm, di 1800mm)

- 1 am pe k,1 vn k,1 - Ngun in 6 V

- 1 cng tc,6 on dy ni, 1 bin tr.III. NI DUNG THC HNH:

A. Cu hi chun b: S mch in: K A B Hnh 1 + - Cu 1: Quan st s mch in hnh 1, k tn, nu cng dng v cch mc ca tng b phn trong s .

Cu 2: Cht (+) ca dng c o in c trong S phi c mc v pha im A hay im B?

B. Cc bc tin hnh th nghim:- Lp mch in theo s hnh 1

- iu chnh bin tr hiu in th tng dn: 0V; 1,5V ; 3V; 4,5V; 6V. Ghi cng dng in tng ng ca mi trng hp vo bng 1. C. Kt qu th nghim:

Kt qu o

Ln oHiu in th (V)Cng dng in (A)

1

2

3

4

5

D. Nhn xt v rt ra kt lun: - Nhn xt: Bng ghi gi tr U, I ca cc ln o: ta thy khi U tng th I

- Kt lun: Khi tng (hoc gim) hiu in th gia hai u dy dn bao nhiu ln th cng dng in chy qua dy dn BO CO THC HNH

Ngy thng nm 200

BI THC HNH S: 2.

TIT 03 - BI 03: THC HNH: XC NH IN TR CA MT DY DN BNG AMPE K V VN KTng im (10)Chun b



(2)

I. MC CH YU CU: - Bit cch xc nh in tr ca mt dy dn bng ampe k v vn kII. CHUN B DNG C:

1. Gio vin: Mt b th nghim nh hc sinh. 2. Hc sinh: Mi nhm: 1 bng lp in, 1 dy cnstantan (c ung knh 0,3 mm, di 1800mm), 1 ampe k, 1 vn k, 1 ngun in 6 V, 1 cng tc, 6 on dy ni, 1 bin tr.III. NI DUNG THC HNH:

A. Cu hi chun b:Cu 1: Vit cng thc tnh in tr?

.

.

Cu 2: Mun o hiu in th gia hai u mt dy dn cn dng dng c g? Mc dng c nh th no vi vi dy dn cn o?

Tr li:

Cu 3: Mun o cng dng in chy qua mt dy dn cn dng dng c g? Mc dng c nh th no vi dy dn cn o?

Tr li:

B. Cc bc tin hnh thc hnh:1. S mch in: K

A B + - 2. Tin hnh thc hnh:

Bc 1: Lp mch in theo s hnh 1. Cht (+) ca am pe k v vn k ni vi cc dng ca ngun in.Bc 2: iu chnh bin tr hiu in th tng dn:U1 = 1,5V ; U2 = 2,5V; U3 = 3V; U4 = 4,5V; U5 = 6V. ng mch in, ghi gi tr hiu in th gia hai u in tr v cng dng in tng ng ca mi trng hp vo bng kt qu ca bo co. Bc 3: Tnh gi tr in tr ca dy dn ang xt trong mi ln o

R1= ; R2= ; R3 =; R4 = ; R5 =

Bc 4: Tnh gi tr trung bnh cng ca in tr : Rtb =

C. Kt qu thc hnh: Kt qu o Ln ohiu in th (V)Cng dng in (A)in tr()

1

2

3

4

5

Rtb = =

D. Nhn xt v rt ra kt lun:

E. Tr li cu hi:1. Nu cc nguyn nhn dn n sai s?

2. Cch khc phc?

BO CO TH NGHIM

Ngy thng nm 200

BI THC HNH S: 3.

TN TH NGHIM: TH NGHIM KIM TRA IN TR TNG NG

CA ON MCH MC NI TIP

TIT 04 - BI 04: ON MCH NI TIPTng im (10)Chun b



(2)

I. MC CH YU CU:

- Qua th nghim xy dng c cng thc tnh in tr tng ng ca on mch mc ni tip: Rt = R1 + R2II. CHUN B DNG C:

1. Gio vin: Dng c lm th nghim nh mi nhm hc sinh.

2. Hc sinh: Mi nhm: 1 bng lp in, 3 in tr mu: R1= 6 ; R2 = 10; R3 = 16; 1 am pe k, 1 vn k, 1 ngun in 6V, 1 cng tc, 7 on dy ni, III. NI DUNG TH NGHIM:

A. Cu hi chun b: Cu 1: Vit cng thc tnh in tr ca on mch gm 2 in tr mc ni tip?

Cu 2: Nu phng n lm th nghim kim tra cng thc: Rt = R1 + R2?

B. Cc bc tin hnh th nghim : ... A B C K + -

B1: Lp mch in nh hnh v: R1 =6(), R2 =10().

B2: ng cng tc, xc nh s ch ca ampe k I1 ? B3: Gi nguyn hiu in th, thay hai in tr trn bng in tr 16. Xc nh s ch ca ampe k I2 ? B4: So snh I1 v I2 ? C. Kt qu th nghim: Loi in trCng dng in v hiu in thSo snh cng dng in qua cc ln o

R1 =6() , R2 =10().

I1= I1 . I2

R3 = 16

I2 =

D. Nhn xt v rt ra kt lun: - Nhn xt:

T kt qu th nghim trn ta thy:

U1 . U2 => R3 . R1 + R2 . I1 . I2 - Kt lun: on mch gm hai in tr mc ni tip c in tr tng ng bng: BO CO TH NGHIM

Ngy thng nm 200

BI THC HNH S: 4.

TN TH NGHIM: TH NGHIM KIM TRA IN TR TNG NG

CA ON MCH MC SONG SONGTIT 05 - BI 05: ON MCH SONG SONGTng im (10)Chun b



(2)

I. MC CH YU CU:

- Qua th nghim kim tra cng thc tnh in tr tng ng ca on mch gm hai in tr mc song song.

II.CHUN B DNG C:1. Gio vin: 1 b th nghim nh mi nhm hc sinh.

2. Hc sinh: : mi nhm: 1 bng lp in, 3 in tr mu: R1=10 ; R2 =15; R3 = 6 , 1 am pe k, 1 vn k, 1 ngun in 6 V, 1 cng tc, 7 on dy ni.

III. NI DUNG TH NGHIM:

A.Cu hi chun b:Cu 1: Vit cng thc tnh in tr ca on mch gm hai in tr mc song song?......................................................................................................................................................................................................................................................................................Cu 2: Hai in tr R1=10 ;R2 =15 mc song song. Tnh in tr tng ng ca on mch.

B. Cc bc tin hnh th nghim: B1: Lp mch in nh hnh v: R1 =10() ; R2 =15().

B2: ng cng tc, c v ghi s ch ca am pe k I1.

B3: Gi nguyn hiu in th, thay 2 in tr trn bng in tr R3 = 6 . c v ghi s ch ca am pe k I2.

B4: So snh I1 v I2. C. Kt qu th nghim:

Loi in trCng dng inSo snh I1 v I1.

R1 =10() ; R2 =15()

I1 =.....................

R3 = 6 ()

I2 =.....................

D. Nhn xt v rt ra kt lun:

Hiu in th khng thay i, m I = I1....... I2. Nu R1 =10() mc song song vi R2 = 15() th hai in tr ny .........................vi in tr R3 = 6 ().Vy cng thc tnh in tr tng ng ca on mch gm 2 in tr mc ni tip l: ................................

...............................................................................................................................................BO CO TH NGHIM

Ngy thng nm 200

BI THC HNH S: 5.

TN TH NGHIM: TH NGHIM KIM TRA S PH THUC CA IN TR VO CHIU DI DY DNTIT 07 - BI 07: S PH THUC CA IN TR

VO CHIU DI DY DNTng im (10)Chun b



(2)

I. MC CH YU CU:

- Qua th nghim thy c s ph thuc ca in tr vo chiu di dy dn. II. CHUN B DNG C:

1. Gio vin: mt b th nghim nh ca nhm hc sinh.

2. Hc sinh: Mi nhm: 1 bng lp in, 1 dy dn constantan ng knh 0,3 mm, di l1=900mm; 1 dy dn constantan ng knh 0,3 mm, di l2=1800mm; 1 dy dn constantan ng knh 0,3 mm, di l3=2700mm, 1 ampe k, 1 vn k, 1 ngun in 6 V, 1 cng tc, 7 on dy ni.III. NI DUNG TH NGHIM:

A. Cu hi chun b: Cu 1: Lm th no xc nh c in tr ca dy dn?

Cu 2: Ba dy dn c chiu di l, 2l, 3l tit din nh nhau v cng lm t mt loi vt liu. Bit dy l c in tr R. Vy dy 2l c in tr bao nhiu? Dy 3l c in tr bao nhiu?

B. Cc bc tin hnh th nghim: B1: Lp mch in nh s hnh v: vi l1= 900mm. B2: o cng dng in I1, hiu in th U1

v tnh R1 = ri ghi vo bng . B3: Thay ln lt l1 bng l2= 1800mm; l3= 2700mm, ri ghi gi tr dng in I2; I3 v hiu in th U2, U3. Tnh R2=; R3 = ri ghi vo bng 1.

B4: So snh R1, R2, R3 v l1, l2, l3 .

C. Kt qu th nghim:

Ln th nghimHiu in th(v)Cng dng in (A)in tr()

Dy dn l1Dy dn l2Dy dn l3

U1 =

U2 =

U3 =I1 =

I2 =

I3 =R1 =

R2 =

R3 =

D. Nhn xt v rt ra kt lun: - Nhn xt : l3 = 3l1 => R3 = .. R1. l2 = 2l1 => R2 = .. R1. - Kt lun: in tr ca dy dn t l thun vi ..ca dy.

BO CO THC HNH

Ngy thng nm 200

BI THC HNH S: 6.

TIT 15 - BI 15:

THC HNH: XC NH CNG SUT CA CC DNG C INTng im (10)Chun b



(2)

I. MC CH YU CU:

Bit cch dng vn k v am pe k xc nh cng sut ca cc vt liu tiu th in. II. CHUN B DNG C: 1. Gio vin: Mt b th nghim nh ca nhm hc sinh.

2. Hc sinh: Mi nhm: 1 bng lp in, 1 bin tr 20 - 2A, 1 bng n 2,5V, 1 ngun in 3V, 1 cng tc, 9 on dy ni, 1 qut in nh, 1 ampe k, 1 vn k. III. NI DUNG THC HNH:

A. Cu hi chun b:Cu 1: Cng sut P ca mt dng c in hoc ca mt on mch lin h vi hiu in th U v cng dng in I bng h thc no?....

Cu 2: Nu quy tc s dng ampe k o cng dng in v quy tc s dng vn k o hiu in th?

Cu 3: Trn mt bng n c ghi 220V-25W. Cc con s cho bit iu g?....................................................................................................................

B. Cc bc tin hnh thc hnh: B1: Lp mch in nh s hnh v:

B2: ng mch in, di chuyn con chy

vn k ch hiu in th hai u bng n l: U1 =1V, ghi hiu in th U1 v dng in I1 vo bng bo co. B3: Tng dn hiu in th hai u bng n ln 1,5V; 2V; 2,5V ri ghi cc gi tr hiu in th v dng in tng ng vo bng.

- T cc gi tr o c, tnh cng sut bng n hiu in th U1; U2; U3; U4. - Nhn xt v cng sut bng n ng vi hiu in th t vo hai u bng.

* Thay bng n bng qut in nh: - Di chuyn con chy ca bin tr hiu in th t vo qut l 3V. Ghi gi tr dng in v cng sut ca qut. - Lm th nghim nh trn ba ln. Tnh gi tr cng sut trung bnh ca qut. C. Kt qu th nghim: 1. Xc nh cng sut ca bng n pin Bng 1Ln oHiu in th (V)Dng in(A)Cng sut(W)

11I1 =P1 =

21,5I2 =P2 =

32I3 =P3 =

42,5I4 =P4 =

2. Xc nh cng xut ca qut in:

Bng 2Ln oHiu in th (V)Dng in(A)Cng sut(W)

13I1 =P1 =

23I2 =P2 =

33I3 =P3 =

- Tnh Pq = ..............................................................................D. Nhn xt v kt lun:

E. Tr li cu hi:

Nu cc nguyn nhn lm nh hng n sai s trong kt qu thc hnh trn? Cch khc phc?

BO CO THC HNH

Ngy thng nm 200

BI THC HNH S: 7.

TIT 20 - BI 18: THC HNH:

KIM NGHIM MI QUAN H Q ~ I2 TRONG NH LUT JUN - LEN - XTng im (10)Chun b



(2)

I. MC CH YU CU:

- Qua th nghim thy c nhit lng to ra dy dn t l vi bnh phng cng dng in.II. CHUN B DNG C: 1. Gio vin: Mt b th nghim nh ca nhm hc sinh.

2. Hc sinh: Mi nhm: 1 bnh nhit lng k 250ml, dy t nng c in tr 6 bng nicm, que khuy, nhit k c phm vi o t 150C ti 1000C v c chia nh nht l 10C, ngun in khng i 12V - 2A, 1 bin tr 20 - 2A, 1 cng tc, 5 on dy ni, 1 am pe k, 1 ng h bm giy, 170ml nc sch.III. NI DUNG THC HNH:

A. Cu hi chun b:Cu 1: Nhit lng to ra dy dn khi c dng in chy qua ph thuc vo nhng yu t no v s ph thuc c biu th bng biu thc no?Cu 2: Nhit lng Q c dng un nng nc c khi lng m1 v lm nng cc ng nc c khi lng m2, khi nhit ca nc v cc tng t t10C ti t20C. Nhit dung ring ca nc l C1 v nhit dung ring ca cht lm cc l C2. H thc no biu th mi lin h gia Q v cc i lng m1 , m2, C1, C2, t1, t2?Cu 3: Nu ton b nhit lng to ra bi dy dn in tr R c dng in cng I chy qua trong thi gian t c dng un nng nc v cc trn y th tng nhit t0 = t20 - t10 lin h vi cng dng in I bi h thc no?......

B. Cc bc tin hnh thc hnh:

B1. nc vo cc un, sao cho khi y np cc th ton b dy t ngp hon ton trong nc. B2. Lp nhit k qua l np cc un, iu chnh bu ca nhit k ngp trong nc v khng chm vo dy t cng nh khng chm y cc .

B3. t nh nhng cc un vo trong v ngoi cch nhit ca nhit lng k, kim tra m bo v tr ng ca nhit lng k.

B4. Mc dy t vo mch in nh s hnh: 18.1(sgk) B5. ng cng tc, iu chnh bin tr am pe k c s ch I1 = 0,6A. Dng que khuy nc nh nhng trong khong 1 pht. Sau , bm ng h, o thi gian un th ngay khi c v ghi nhit ban u t10 vo bng 1. Trong khi un, thng xuyn khuy nc c nhit ng u. un nc trong 7 pht, ngay cui thi gian ny c v ghi nhit t20 ca nc vo bng 1 B6. Trong ln th nghim th hai, nc trong cc un tr li nhit t10 ban u nh ln th nghim th nht. iu chnh bin tr am pe k c s ch I2 = 1,2 A. Lm tng t nh trn, o v ghi nhit ban u t10, nhit cui t20 ca nc cng vi thi gian un trong 7 pht. B7. Trong ln th nghim ln th 3, li nc trong cc un ngui tr li nhit t10 ban u nh ln th nghim ln th nht. iu chnh bin tr am pe k c s ch I3 = 1,8A. Lm tng t nh trn xc nh cc nhit u t10 v cui t20 ca nc v cng trong thi gian un l 7 pht. B8. Thc hin cc cng vic tip theo nh yu cu ca mu bo co.

C. Kt qu thc hnh:

Ln th nghimCng dng in I (A)Nhit ban u t01Nhit lc cui t02 tng nhit t0= t02- t01

1I1 = 0,6 t01 =

2I2 = 1,2 t02 =

3I3 = 1,8 t03 =

1. Tnh t s t02/ t01 v so snh vi t s I22/I21..

.... 2. Tnh t s t02/ t01 v so snh vi t s I32/I21...... D. Kt lun : T cc kt qu trn, hy pht biu mi quan h gia nhit lng Q to ra trn dy dn vi cng dng in I chy qua n.........BO CO TH NGHIM

Ngy thng nm 200

BI THC HNH S: 8.

TN TH NGHIM: XC NH T TNH CA NAM CHMTIT 23 - BI 21: NAM CHM VNH CUTng im (10)Chun b



(2)

I. MC CH YU CU:

- Bit c c im t tnh ca nam chm v cch xc nh cc t cc . - Bit tng tc gia cc t cc.

II. CHUN B DNG C: 1. Gio vin: Mt b th nghim nh ca nhm hc sinh.

2. Hc sinh: mi nhm: 2 thanh nam chm thng, mt thanh c bc kn bng giy khng nhn thy sn v tn cc t. 1 la bn. 1 chn .

1 kim nam chm t trn mi nhn thng ng.III. NI DUNG TH NGHIM:A. Cu hi chun b:Cu 1: Nam chm l vt c c tnh g?..

..

Cu 2: Hy xut mt th nghim pht hin xem mt thanh kim loi c phi l nam chm hay khng?

B. Cc bc tin hnh th nghim

1.Th nghim 1:C2. t kim nam chm trn gi thng ng nh hnh: 21.1(sgk).

- Khi ng cn bng, kim nam chm dc theo hng no?

- Xoay cho kim nam chm lch khi hng va xc nh, bung tay. Khi ng cn bng tr li, kim nam chm cn ch hng nh lc u na khng? Lm th nghim hai ln v cho nhn xt .

2.Th nghim 2: Tng tc gia hai nam chm C3 . a t cc ca hai nam chm li gn nhau (hnh 21.3sgk). Quan st hin tng xy ra, cho nhn xt .

C4. i u ca mt trong hai nam chm ri a li gn nhau, c hin tng g xy ra vi cc nam chm?C. Kt qu th nghim: - Th nghim 1: + Khi ng cn bng, kim nam chm dc theo hng:

+ Khi ng cn bng tr li, nam chm vn ch hng . nh c. - Th nghim 2:

C3 . Cc Bc ca kim nam chm b ht v pha cc .....ca thanh nam chm. C4. Cc cc cng tn ca hai thanh nam chm th .....nhau.

D. Kt lun:

- Bt k nam chm no cng c....cc. Khi t do, cc lun ch hng Bc gi l cc... cn cc lun ch hng Nam gi l cc . .

Quy c tn cc t: Mu : Cc Bc(N). Mu xanh: Cc Nam (S).

- Khi t hai nam chm gn nhau th chng ht nhau nu cc cc ......tn, y nhau nu cc cc..tn.BO CO TH NGHIM

Ngy thng nm 200

BI THC HNH S: 9.

TN TH NGHIM: TC DNG T CA DNG INTIT 24 - BI 22: TC DNG T CA DNG IN - T TRNGTng im (10)Chun b



(2)

I. MC CH YU CU:

- Kim tra xem dng in chy qua dy dn thng c tc dng t hay khng?II. CHUN B DNG C: 1. Gio vin: Mt b th nghim nh ca nhm hc sinh.

2. Hc sinh: mi nhm: - 1 dy dn thng lp trn bng, 1 la bn.

- 1 ngun in 1,5V

- 3 si dy dn

- 1 cng tcIII. NI DUNG TH NGHIM:

A. Cu hi chun b:Cu 1: Kim nam chm khi ng yn th ch theo hng no?..

..

Cu 2: Khi t mt kim nam chm li gn mt nam chm khc th c hin tng g xy ra?......

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

B. Cc bc tin hnh th nghim:

1. Lc t:.

B1: t gi lp si dy dn thng ln mt bn. B2: t la bn trn gi ca si dy dn, kim la bn nm ngay di si dy dn. B3: Quay gi lp si dy dn trn mt bn sao cho si dy dn song song vi kim la bn.

B4: ng mch in trong thi gian ngn, quan st v cho bit c hin tng g xy ra vi kim nam chm. Lc nm cn bng, kim nam chm cn song song vi dy dn na khng? B5: i chiu ngun in qua dy dn ri cng lm nh trn, quan st kim nam chm. 2. T trng: Mt kim nam chm c t t do trn trc thng ng, ang ch hng Nam - Bc. a n n cc v tr khc nhau xung quanh dy dn c dng in hoc xung quanh thanh nam chm. - Quan st hin tng xy ra i vi kim nam chm - mi v tr, sau khi nam chm ng yn, xoay cho n lch khi hng va xc nh, bung tay. Nhn xt hng ca kim nam chm sau khi tr li v tr cn bng.

C. Kt qu th nghim:

1. Lc t: - Khi c dng in chy qua, kim nam chm....... (b lch khi v tr ban u, khng lch khi v tr ban u). Lc cn bng ......................... nm song song vi dy dn.

- i chiu dng in th kim nam chm.............( i, khng i ) chiu quay. 2. T trng: - Hin tng xy ra i vi kim nam chm.

(lch khi v tr cn bng, khng lch khi v tr cn bng) - Kim nam chm lun ch theo ....... ....( mt hng, nhiu hng).

D. Kt lun: - Dng in gy ra tc dng lc ln kim nam chm t gn n. Chng t dng in . (c tc dng t, khng c tc dng t). - Khng gian xung quanh nam chm, xung quanh dng in c kh nng tc dng lc t ln .... t gn n. Ta ni trong khng gian c ................BO CO TH NGHIM

Ngy thng nm 200

BI THC HNH S: 10.

TN TH NGHIM: S NHIM T CA ST, THP.TIT 27 - BI 25: S NHIM T CA ST, THP - NAM CHM IN.Tng im (10)Chun b



(2)

I. MC CH YU CU:

- Qua th nghim thy c s khc nhau v s nhim t ca st v thp. II. CHUN B DNG C: 1. Gio vin: Mt b th nghim nh ca nhm hc sinh.

2. Hc sinh: mi nhm: 1 bng lp in; 1 cun dy 400 vng (hoc thay bng cun dy c s vng khc); 1 li thp ch I; 1 li st non ch I; 1 kim nam chm t trn mi nhn; 1 am pe k; 1 bin tr; 1 cng tc; 1 ngun in 6V; 6 on dy ni.III. NI DUNG TH NGHIM:

A. Cu hi chun b: Nu cu to ca nam chm in?

B.Cc bc tin hnh th nghim:1. Th nghim 1: B1: B tr th nghim hnh 25.1. B2: Kim nam chm t trc cun dy, cch cun dy khong 10 cm.

B3: ng mch in, quan st gc lch ca kim nam chm so vi phng ban u. B4: t li st non vo trong lng cun dy, ng cng tc. Quan st v cho nhn xt v gc lch ca kim nam chm so vi trng hp ng dy khng c li st. B5: t li thp vo trong lng cun dy, ng cng tc. Quan st v cho nhn xt v gc lch ca kim nam chm so vi trng hp ng dy khng c li thp. 2. Th nghim 2: B1: B tr th nghim hnh 25.2. B2: Hy cho bit hin tng xy ra vi inh st trong cc trng hp sau: - ng dy c li st non ang ht inh st. Ngt cng tc K.

- ng dy c li thp ang ht inh st. Ngt cng tc K.

B3: Nhn xt v tc dng t ca ng dy c li st non v ng dy c li thp khi ngt dng in qua ng dy.C. Kt qu th nghim:

* Th nghim 1: - Khi c dng in chy qua cun dy, kim nam chm ........... - Khi t li st non vo trong lng cun dy th gc quay ca kim nam chm .. (nhiu hn, t hn) khi cun dy khng c li st.

- Khi t li thp vo trong lng cun dy th gc quay ca kim nam chm ................... nhiu hn, t hn) khi cun dy khng c thp. * Th nghim 2:Hin tng xy ra vi inh st l:

- ng dy c li st non ang ht inh st. Ngt cng tc K: - ng dy c li thp ang ht inh st. Ngt cng tc K: C1: Khi ngt dng in i qua ng dy, li ................ mt ht t tnh cn li thp ...... c t tnh . D. Kt lun: - Li st hoc li thp ................. tc dng t ca ng dy c dng in. - Khi ngt in, li st non........ t tnh cn li thp ....

BO CO TH NGHIM

Ngy thng nm 200

BI THC HNH S: 11.

TN TH NGHIM: TC DNG CA T TRNG LN DY DN C DNG INTIT 29 - BI 27: LC IN T.Tng im (10)Chun b



(2)

I. MC CH YU CU:

- Qua th nghim thy c si dy dn thng c dng in chy qua t vung gc vi t trng s chu mt lc tc dng. II. CHUN B DNG C: 1. Gio vin: Mt b th nghim nh ca nhm hc sinh.

2. Hc sinh: mi nhm: 1 nam chm hnh ch U; 1 ngun in 6 V; mt on dy dn thng AB bng ng, ng knh 2.5 mm, di 10cm; 1 bin tr 20 - 2A; 1 cng tc; 1 gi th nghim; 1 am pe k.III. NI DUNG TH NGHIM:

A. Cu hi chuan b: Trnh by th nghim - xtt?........................................

................................................................................................................................................................................................................................................B. Cc bc tin hnh th nghim:1. Th nghim 1: B1: t bng lp dng c th nghim trn mt bn nm ngang.

B2: t si ng ln hai thanh nm ngang v vung gc vi t trng ca nam chm ch U. B3: Ni mch in nh hnh 27.1, ngun in 1,5V. ng cng tc K, quan st chuyn ng ca si dy ng v tr li cu C1.2. Th nghim 2: Xc nh chiu ca lc in t ph thuc vo yu t no? B1: Lm li th nghim hnh 27.1, quan st chiu chuyn ng ca dy dn AB v cho bit, khi i chiu dng in qua AB hoc i chiu ng sc t th chiu ca lc in t tc dng ln dy dn AB c thay i hay khng? B2: B tr th nghim hnh 25.2.C. Kt qu th nghim:

* Th nghim 1: - Khi ng cng tc K, on dy ng hoc (b ht, b y, khng b ht, khng b y) vo trong lng nam chm ch U.

C1. Hin tng trn chng t: on dy dn AB ........................ (chu tc dng, khng chu tc dng) ca mt lc no . * Th nghim 2: - Khi i chiu dng in, si dy dn AB chuyn ng......... (cng chiu, ngc chiu) vi trng hp trn. Vy chiu ca lc in t tc dng ln dy dn AB ....... (thay i, khng thay i). - Khi i chiu ng sc t (chiu dng in khng i), si dy dn AB chuyn ng(cng chiu, ngc chiu) vi trng hp trn. Vy chiu ca lc in t tc dng ln dy dn AB .................. (thay i, khng thay i).

D. Kt lun:

- T trng .......... (tc dng, khng tc dng) lc in t ln dy dn AB c dng in chy qua t trong t trng. Lc gi l lc in t.

- Chiu ca lc in t tc dng ln dy dn AB ph thuc vo ................ chy trong dy dn v chiu ............BO CO THC HNH

Ngy thng nm 200

BI THC HNH S: 12.

TIT 31 - BI 29: THC HNH: CH TO NAM CHM VNH CU, NGHIM LI T TNH CA NG DY C DNG IN.Tng im (10)Chun b



(2)

I. MC CH YU CU:

- Bit cch ch to nam chm vnh cu.

- Bit cch nhn bit mt vt c phi l nam chm hay khng?

- Bit cch xc nh t cc ca ng dy.II. CHUN B DNG C: 1. Gio vin: Mt b th nghim nh ca nhm hc sinh.

2. Hc sinh: mi nhm: 2 ngun in 3V v 6V; ng dy np t A khong 200 vng; ng dy B: 300 vng; 1 bt d; 1 on dy thp, 1 on dy ng; 1 cng tc; 1 la bn; Hai on chnh nilon mnh, mi on di 15cm.

III. NI DUNG THC HNH:

A. Cu hi chun b:Cu 1: Lm th no cho mt thanh thp nhim t?

Cu 2: C nhng cch no nhn bit chic kim bng thp b nhim t hay cha?

Cu 3: Nu cch xc nh tn t cc ca mt ng dy c dng in chy qua v chiu dng in trong cc vng dy bng mt kim nam chm.

B. Cc bc tin hnh thc hnh:1. Th nghim 1: Ch to nam chm vnh cu: a. Ni hai u ng dy A vi ngun in 3V. t ng thi cc on dy thp v ng dc trong lng ng dy trong khong ti gian t 1 n 3 pht.

b. Th nam chm: Ly cc on dy kim loi ra khi ng dy, ln lt treo cho mi on nm thng bng nh mt si ch khng xon, sau khi ng yn, n nm dc theo phng no?

Xoay cho on kim loi lch khi hng ban u, bung tay, sau khi cn bng tr li, on kim loi nm dc theo hng no? Lm nh vy 3 ln i vi mi on kim loi.

Ghi kt qu vo bng 1 ca mu bo co xc nh on kim loi no tr thnh nam chm vnh cu. c. Dng bt d nh du tn cc t ca nam chm va c ch to.

2. Th nghim 2: Nghim li t tnh ca ng dy c dng in chy qua. t ng dy B nm ngang, lun qua l trn trn ng dy B treo nam chm va c ch to vo trong lng ng dy. Xoay ng dy sao cho nam chm nm song song vi mt phng ca cc vng dy (hnh 29.2). C nh si ch treo nam chm vo gi th nghim. Mc ng dy vo mch in c ngun 6V. a. ng mch in. Quan st hin tng xy ra vi nam chm, cho nhn xt. Da vo chiu ca nam chm trong lng ng dy, xc nh tn cc t v chiu ca dng in chy qua ng dy. Kim tra li kt qu va thu c thng qua du cc cc ca ngun in, ghi vo bng 2 ca mu bo co. b. i cc ca ngun in i chiu dng in i vo cun dy. Lp li cng vic nh lm mc a, ghi kt qu vo mc 2 ca mu bo co.

C. Kt qu thc hnh:1. Kt qu ch to nam chm vnh cu.

Kt qu

Ln th

nghimthi gian nhim t (pht)Th nam chm. Sau khi ng cn bng, on dy dn nm theo phng no?

on dy no thnh nam chm vnh cu ?

Ln 1Ln 2Ln 3

Vi on dy ng

Vi on dy thp

2. Kt qu nghim li t tnh ca ng dy c dng in.

Nhn xtLn th nghimC hin tng g xy ra vi nam chm khi ng cng tc K?u no ca ng dy l t cc Bc?Dng mi tn cong k hiu chiu dng in chy trong cc vng dy mt u nht nh.

1

2(i cc ngun in)

BO CO TH NGHIM

Ngy thng nm 200

BI THC HNH S: 13.

TN TH NGHIM: DNG NAM CHM TO RA DNG IN.TIT 33 - BI 31: HIN TNG CM NG IN T.Tng im (10)Chun b



(2)

I. MC CH YU CU:

Lm th nghim dng nam chm to ra dng in.II. CHUN B DNG C:

1. Gio vin: Mt b th nghim nh ca nhm hc sinh.

2. Hc sinh: mi nhm: 1 cun dy 3000 vng, hai u c hai n LED mc song song v ngc chiu nhau; 1 nam chm in; Ngun in 3V; 1 li st ch I; 1 nam chm thng; 1 cng tc; 3 on dy ni; 1 trc quay nam chm lm th nghim 3.III. NI DUNG TH NGHIM:

A. Cu hi chun b:Nu cu to v hot ng ca inam xe p?

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

................................................................................................................................................................B. Cc bc tin hnh th nghim: 1. Th nghim 1: Dng nam chm vnh cu. B1: B tr th nghim hnh 31.2, hai u c hai n LED mc song song v ngc chiu nhau.

B2: Di chuyn nhanh u cc nam chm n u cun dy, quan st hai n LED B3: t nam chm ng yn trc cun dy, quan st hai n LED

B4: t nam chm nm yn trong cun dy, quan st hai n LED

B5: Di chuyn nhanh nam chm ra xa cun dy, quan st hai n LED

2. Th nghim 2: Dng nam chm in. t mt nam chm in nm yn trc mt cun dy dn c mc hai n LED song song v ngc chiu. Hy lm th nghim xc nh trong nhng trng hp no di y xut hin dng in cun dy c mc n LED: + Trong khi ng mch in ca nam chm in.

+ Trong khi dng in n nh.

+ Trong khi ngt mch in ca nam chm in.

+ Sau khi ngt mch in.

C. Kt qu th nghim: * Th nghim 1: - Khi nam chm ang di chuyn n u cc ca cun dy th.................... (n LED 1 sng, khng c n no sng). - t nam chm ng yn trc cun dy........... (n LED 2 sng, khng n LED no sng).

- t nam chm nm yn trong cun dy:................................ (n LED 1 sng, khng c n LED no sng, n LED 2 sng). - Di chuyn nhanh nam chm ra xa cun dy:.............................. (n LED 1 sng, khng c n LED no sng, n LED 2 sng). * Th nghim 2: - Khi ng mch in............ (n LED 1 sng, khng n LED no sng)

- Khi mch in ng, dng in n nh.......................... (n LED 1 sng, khng n LED no sng, n LED 2 sng). - Khi ngt mch in................................................ (n LED 1 sng, khng n LED no sng, n LED 2 sng).

- Sau khi ngt mch in................................................. (n LED 1 sng, khng n LED no sng, n LED 2 sng).

D. Rt ra nhn xt v kt lun: - Nhn xt 1: Dng in xut hin trong cun dy dn kn khi ta a mt cc ca nam chm............. hay .................. mt u cun dy hoc ngc li.

- Nhn xt: Dng in xut hin cun dy dn kn trong thi gian ........ mch ca nam chm in, ngha l trong thi gian dng in ca nam chm in ............... BO CO TH NGHIM

Ngy thng nm 200

BI THC HNH S: 14.

TN TH NGHIM: TO RA DNG IN XOAY CHIUTRONG CUN DY DN KNTIT 37 - BI 33: DNG IN XOAY CHIU.Tng im (10)Chun b



(2)

I. MC CH YU CU:

Bit th no l dng in xoay chiu v cch to ra dng in xoay chiu.

II. CHUN B DNG C:


2. Hc sinh: mi nhm: 1 cun dy 3000 vng, hai u c hai n LED (1 n mu , 1 n mu vng) mc song song v ngc chiu nhau; 1 nam chm vnh cu c th quay quanh mt trc thng ng; 1 my pht in xoay chiu c c gp.


A. Cu hi chun b: Hy trnh by cc cch to ra dng in cm ng m em hc?

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

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

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

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

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

B. Cc bc tin hnh th nghim: 1. Th nghim 1: Chiu ca dng in cm ng.B1: a thanh nam chm vo trong cun dy, quan st 2 n LED.

B2: Ko thanh nam chm ra khi cun dy, quan st 2 n LED.2. Th nghim 2: Cch to ra dng in xoay chiu. - Cho nam chm quay trc cun dy dn kn

B1: t cun dy 3000 vng nh th nghim 33.2 trn mt bn nm ngang, trc quay ca nam chm thng ng, nam chm nm trong mt phng qua trc ca cun dy.

B2: Quay nam chm quanh trc thng ng, quan st 2 n LED cun dy.

- Cho cun dy quay trong t trng. Dng my pht in xoay chiu, hai thanh qut tip xc vi 2 c gp ca khung. Quay khung dy trong t trng ca nam chm, quan st hai n LED.

C. Kt qu th nghim:

1. Th nghim 1: - Khi a nam chm vo th...., lc ko nam chm ra th n............ Hai n LED thay nhau sng, chng t trong hai trng hp dng in trong cun dy c chiu

2. Th nghim 2: - Cho nam chm quay trc cun dy dn kn:

Khi quay nam chm trc cc cun dy th 2 n LED ............, chng t dng in trong trong cun dy lun phin i chiu. Nu quay nhanh s thy 2 n LED sng

- Cho cun dy quay trong t trng: Khi khung dy quay trong t trng th 2 n LED sng nhp nhy, chng t dng in trong khung dy lun lun.

D. Nhn xt:

- Cho nam chm quay trc cun dy dn kn:C2. Khi cc N ca nam chm li gn cun dy th s ng sc t xuyn qua tit din S ca cun dy .(tng, gim). Khi cc N ra xa cun dy th s ng sc t qua S

..... (tng, gim). Khi nam chm quay lin tc th s ng sc t xuyn qua S lun phin ...............Vy dng in cm ng xut hin trong cun dy l dng in ......

- Cho cun dy quay trong t trng:C3. Khi cun dy quay t v tr 1 sang v tr 2 th s ng sc t xuyn qua tit din S ca cun dy.. (tng, gim). Khi cun dy t v tr 2 quay tip th s ng sc t xuyn qua tit din S ca cun dy.. (tng, gim). Nu cun dy quay lin tc th s ng sc t xuyn qua tit din S lun phin ...............Vy dng in cm ng xut hin trong cun dy l dng in E. Kt lun: - Khi s ng sc t xuyn qua tit din S ca cun dy tng th dng in cm ng trong cun dy c chiu ............. vi chiu dng in cm ng khi s ng sc t xuyn qua tit din .............

- Trong cun dy dn kn, dng in cm ng xoay chiu xut hin khi cho nam chm quay trc ............. hay cho cun dy .......

BO CO TH NGHIM

Ngy thng nm 200

BI THC HNH S: 15.

TN TH NGHIM: CC TC DNG CA DNG IN XOAY CHIUTIT 39 - BI 35: CC TC DNG CA DNG IN XOAY CHIU -O CNG V HIU IN TH XOAY CHIU

Tng im (10)Chun b



(2)

I. MC CH YU CU:

- Qua th nghim thy c cc tc dng ca dng in xoay chiu.

II. CHUN B DNG C:


2. Hc sinh: mi nhm: - 1 nam chm in - 1 nam chm vnh cu.

- 1 ngun in mt chiu 3V-6V.

- 1 ngun in xoay chiu 3V-6V.


A. Cu hi chun b: Hy m t hin tng xy ra trong mi th nghim hnh 35.1 v cho bit hin tng no chng t dng in xoay chiu c tc dng nhit, tc dng quang, tc dng t.............................

B. Cc bc tin hnh th nghim: B1: B tr th nghim hnh 35.2. B2: ng mch in, quan st hin tng xy ra. i chiu ngun in ri ng mch in, quan st hin tng xy ra. B3: Vn mch in nh trn nhng thay bng ngun in xoay chiu 6V. ng mch in, hin tng xy ra vi thanh nam chm c g khc so vi trng hp dng ngun in 1 chiu? Gii thch v sao?C. Kt qu th nghim: - Trng hp s dng dng in 1 chiu, nu lc u cc N ca thanh nam chm b ht th khi i chiu dng in n s b ...................v ngc li. - Khi dng in xoay chiu chy qua ng dy th cc N ca thanh nam chm ln lt b .............. Nguyn nhn l do dng in......D. Kt lun: Khi dng in i chiu th lc t ca dng in tc dng ln nam chm .. .......................

BO CO THC HNH

Ngy thng nm 200

BI THC HNH S: 16.

TIT 42 - BI 38: THC HNH: VN HNH MY PHT IN V MY BIN THTng im (10)Chun b



(2)

I. MC CH YU CU: Quan st my pht in n gin bit cu to v nhn bit hiu qu tc dng ca dng in do my pht ra khng ph thuc vo chiu quay

- Cng quay nhanh th hiu in th hai u cun dy ca my cng cao.

- Nghim li cng thc ca my bin th: =

II. CHUN B DNG C:


2. Hc sinh: mi nhm:1 my pht in xoay chiu loi nh; 1 bng n 3V c ; 1 my bin th nh, cc cun dy c ghi s vng dy, li st c th tho lp c; 1 ngun in xoay chiu 3V - 6V; 1 vn k xoay chiu; 6 on dy ni.III. NI DUNG THC HNH:

A. Cu hi chun b:Cu 1: Hy nu b phn chnh v nguyn tc hot ng ca my pht in xoay chiu?Cu 2: Hy nu cu to v nguyn tc hot ng ca my bin th?B. Cc bc tin hnh thc hnh:1. Vn hnh my pht in xoay chiu n gin B1: Mc bng n vo hai u ly in ra ca my pht in (Hnh 38.1). Mc vn k xoay chiu song song vi bng n. B2: iu khin tay quay cun dy ca my pht in quay u n, quan st ng thi sng ca bng n v s ch ca vn k.

Ghi cc kt qu o vo bo co th nghim.

2. Vn hnh my bin th: B1: Dng cun dy 200 vng lm cun s cp v 400 vng lm cun th cp ca my bin th. Mc 2 u cun s cp vo ngun in xoay chiu 6V. Dng vn k xoay chiu o hiu in th U1 2 u cun s cp v U2 2 u cun th cp. Ghi kt qu o vo bng 1. B2: Sau dng cun 500 vng ri 1000 vng lm cun s cp, cun 500 vng lm cun th cp. o hiu in th 2 u cun s cp v th cp. Ghi kt qu vo bng 1.

C. Kt qu th nghim: 1. Vn hnh my pht in n gin V s th nghim hnh 38.1

C1. Khi my quay cng nhanh th hiu in th hai u dy ra ca my cng............ Hiu in th ln nht t c l..............C2. i chiu quay ca cun dy n c sng khng? Vn k c hot ng khng?

2. Vn hnh my bin th:- V s th nghim hnh 38.2

Bng 1: Kt qu o

Lnth nghimn1(vng)n2(vng)U1(V)U2(V)

1

2

3

C3. Quan h gia s o hiu in th 2 u cun dy ca my bin th v s vng ca cc cun dy?

..

..

..

BO CO TH NGHIM

Ngy thng nm 200

BI THC HNH S: 17.

TN TH NGHIM: QUAN H GIA GC TI V GC KHC XTIT 45 - BI 41: QUAN H GIA GC TI V GC KHC XTng im (10)Chun b



(2)

I. MC CH YU CU:

Lm th nghim tm c mi quan h gia gc ti v gc khc x khi chiu tia sng t mi trng trong sut ny sang mi trng trong sut khc.

II. CHUN B DNG C:1. Gio vin: V to hnh 41.1; bng 1 SGK.

2. Hc sinh: (mi nhm) 1 ming thu tinh trong sut hnh bn nguyt, 1 ming g phng, 1 thc o , 3 chic inh ghim.


A. Cu hi chun b:

Cu 1: Khi chiu tia sng t mi trng trong sut ny sang mi trng trong sut khc th ng i ca tia sng s nh th no? Hin tng gi l hin tng g?

Tr li:

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

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

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

Cu 2: So snh gc khc x v gc ti khi chiu tia sng t khng kh vo mi trng nc, thu tinh?

Tr li:

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

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

B. Cc bc tin hnh:

B tr th nghim nh hnh 41.1 SGK.

Khi gc ti bng 600 :

B1: Cm mt inh ghim ti im A vi gc NA = 600.

B2: t pha mt cnh cong ca ming thu tinh sao cho nhn qua khe I thy A. a inh ghim A ti v tr sao cho n che khut ng thi c khe I v inh ghim A.

B3: Chng minh ng ni cc v tr A, I, A l ng truyn ca tia sng t inh ghim ti mt l ng truyn ca tia sng t inh ghim A ti mt.

B4: Ch ra tia ti, tia khc x, gc ti, gc khc x. o ln gc khc x v ghi vo bng bo co.

Tin hnh th nghim theo cc bc tng t nh trn i vi gc ti bng 450, 300, 00. V ng truyn ca tia sng t inh ghim n mt trong tng trng hp, o cc gc khc x tng ng v ghi vo bng bo co.C. Kt qu thc hnh:

Kt qu o

Ln oGc ti iGc khc x r

1600

2450

3300

400

D. Nhn xt kt qu v rt ra kt lun:

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

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

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

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

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

E. Cu hi vn dng:

1. Khi chiu tia sng t nc sang khng kh, vi gc ti i = 500 th c hin tng g xy ra?................................................................................................................................................

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

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

2. Khi chiu tia sng t khng kh sang cc mi trng trong sut rn, lng khc nhau th gc khc x nh th no so vi gc ti?................................................................................................................................................

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

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

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

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

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

BO CO TH NGHIM

Ngy thng nm 200

BI THC HNH S: 18.

TN TH NGHIM: C IM CA NH CA MT VT TO BITHU KNH HI TTIT 47 - BI 43: NH CA MT VT TO BI THU KNH HI T Tng im (10)Chun b



(2)

I. MC CH YU CU:

HS tin hnh th nghim, nu c trong trng hp no thu knh hi t cho nh tht v cho nh o ca mt vt v ch ra c c im ca cc nh ny.II. CHUN B DNG C:1. Gio vin: V to bng 1, b th nghim nh hnh 43.2 SGK.

2. Hc sinh: (mi nhm) 1 thu knh hi t c tiu c khong 12cm, 1 gi quang hc, 1 cy nn cao khong 5cm, 1 mn hng nh, 1 bt la.III. NI DUNG TH NGHIM:

A. Cu hi chun b:

Cu 1: Nu c im ca thu knh hi t?

Tr li:

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

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

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

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

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

Cu 2: Tiu im ca thu knh l g? Mi thu knh c my tiu im? V tr ca chng c c im g?

Tr li:

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

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

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

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

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

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

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

B. Cc bc tin hnh:

B tr th nghim nh hnh 43.2 SGK. C vt v mn c t vung gc vi trc chnh ca thu knh.

B1: t vt ngoi khong tiu c : t vt xa thu knh v mn st thu knh. T t dch chuyn mn ra xa thu knh cho n khi xut hin nh r nt ca vt trn mn, quan st nh tht cng chiu hay ngc chiu so vi vt.

B2: Dch vt vo gn thu knh hn. Tin hnh nh trn, quan st v nu c im ca nh. Ghi kt qu quan st vo bo co.

B3: t vt trong khong tiu c. Chng t rng khng hng c nh ca vt trn mn. Quan st v nu c im ca nh trong trng hp ny. Ghi kt qu quan st vo bo co.C. Kt qu thc hnh: Kt qu QS

Ln TNKhong cch t vt n thu knh (d)c im ca nh

Tht hay o?Cng chiu hay ngc chiu so vi vt?Ln hn hay nh hn vt?

1Vt rt xathu knh

2d > 2f

3f < d < 2f

4d < f

Ch : Mt im sng nm ngay trn trc chnh, rt xa thu knh, cho nh ti tiu im ca thu knh.

Vt t vung gc vi trc chnh ca thu knh cho nh cng vung gc vi trc chnh.


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

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

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

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

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

BO CO THC HNH

Ngy thng nm 200

BI THC HNH S: 19.

TIT 50 - BI 46: THC HNH: O TIU C CA THU KNH HI TTng im (10)Chun b



(2)

I. MC CH YU CU:

- HS nm c phng php o tiu c ca thu knh hi t.

- Tin hnh o tiu c ca thu knh hi t theo phng php trn.II. CHUN B DNG C:1. Gio vin: V to hnh 46.1 SGK.

2. Hc sinh: (mi nhm) 1 thu knh hi t; mt vt sng phng c dng ch F, khot trn mt mn chn sng; 1 mn nh; 1 gi quang hc; 1 thc thng c chia n mm.


A. Cu hi chun b:

Cu 1: Dng nh ca mt vt t cch thu knh hi t mt khong bng 2f.................................................................................................................................................................................................................................................................................................

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

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

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

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

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

Cu 2: Da vo hnh v chng minh rng trong trng hp ny th khong cch t vt v t nh n thu knh l bng nhau.

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

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

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

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

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

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

Cu 3: nh ny c kch thc nh th no so vi vt?........................Cu 4: Lp cng thc tnh tiu c ca thu knh trong trng hp ny?................................................................................................................................................

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

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

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

B. Cc bc tin hnh:

B1: B tr th nghim nh hnh 46.1. t thu knh gia gi quang hc, ri t vt v mn nh st gn v cch u thu knh.

B2: Dch vt v mn xa dn thu knh nhng khong bng nhau cho ti khi thu c nh ca vt r nt trn mn v nh c kch thc bng vt.

B3: o khong cch L t vt ti mn v tnh tiu c:

f = =

C. Kt qu o:

Kt qu o

Ln oKhong cch t vt n mn nh (mm)Chiu cao ca vt (mm)Chiu cao ca nh (mm)Tiu c ca thu knh (mm)

1

2

3

4

Gi tr trung bnh ca tiu c thu knh o c l: ........................................ (mm).


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

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

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

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

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

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

E. Cu hi vn dng:

1. nh ca mt vt to bi cc thu knh cng chiu vi vt th nh l nh g?.......................

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

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

2. Nu mt vi phng php khc o tiu c ca thu knh hi t........................

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

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

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

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

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

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

BO CO TH NGHIM

Ngy thng nm 200

BI THC HNH S: 20.

TN TH NGHIM: TO RA NH SNG MU BNG TM LC MUTIT 58 - BI 52: NH SNG TRNG V NH SNG MUTng im (10)Chun b



(2)

I. MC CH YU CU:

Quan st nh sng pha sau tm lc mu v nu nhn xt.

Gii thch c s to ra nh sng mu bng tm lc mu trong mt s ng dng thc t.

II. CHUN B DNG C:1. Gio vin: V to hnh 52.1 SGK.

2. Hc sinh: (mi nhm) 1 n pht nh sng trng, 1 n pht nh sng , 1 n pht nh sng xanh, 1 b cc tm lc mu , lc, lam, ...


A. Cu hi chun b:

Cu 1: Nu mt s ngun sng trng v ngun sng mu m em bit.

Tr li:

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Cu 2: Nu mt vi tm lc mu m em bit.

Tr li:

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B. Cc bc tin hnh:

B1: Chiu mt chm sng trng qua mt tm lc mu .

B2: Chiu mt chm sng mu qua mt tm lc mu .

B3: Chiu mt chm sng mu qua tm lc mu xanh.

Lm th nghim tng t vi cc tm lc mu v nh sng mu khc v cho bit mu ca nh sng sau cc tm lc mu.

C. Kt qu thc hnh:Chiu mt chm sng muQua tm lc muCho nh sng mu

Trng

Xanh


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

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E. Cu hi vn dng:

Hy gii thch kt qu th nghim trn? .......................

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

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

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

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

BO CO TH NGHIM

Ngy thng nm 200

BI THC HNH S: 21.

TN TH NGHIM: TRN HAI V BA NH SNG MU VI NHAUTIT 60 - BI 54: S TRN CC NH SNG MUTng im (10)Chun b



(2)

I. MC CH YU CU:

HS lm th nghim trn hai hay ba nh sng mu vi nhau, quan st c nh sng mu thu c sau khi trn.

II. CHUN B DNG C:1. Gio vin: Dng c th nghim nh hnh 54.1 SGK.

2. Hc sinh: (mi nhm) mt n chiu c ba ca s v hai gng phng, 1 b cc tm lc mu (, lc, lam) v mt tm chn sng, 1 mn nh, 1 gi quang hc.


A. Cu hi chun b:

Cu 1: Nu mt vi ngun nh sng mu m em bit. C th to ra nh sng mu bng cch no m em bit?

Tr li: ............

...

Cu 2: Th no l trn cc nh sng mu vi nhau?

Tr li: ............

...

...

B. Cc bc tin hnh:

1. Trn hai nh sng mu vi nhau:

B1. B tr th nghim nh hnh 54.1 SGK

B2. Chn hai ca s bng hai tm lc mu (tm lc mu v tm lc mu lc). Chn ca s cn li bng tm chn sng.

B3. t mn nh vo ch hai chm sng mu giao nhau v nhn xt v mu m ta thu c trn mn nh.

Lm tng t nh trn vi tm lc mu vi tm lc mu lam; ri tm lc mu lc vi tm lc mu lam, quan st nhn xt v mu m ta thu c trn mn nh. Ghi kt qu quan st c vo bo co.

2. Trn ba nh sng mu vi nhau c nh sng trng:

B1. B tr th nghim nh hnh 54.1 SGK.

B2. Chn ba ca s bng ba tm lc mu , lc v lam.

B3. iu chnh ba chm sng gp nhau v nhn xt v mu m ta thu c ch . Ghi kt qu quan st c vo bo co.

Nu c iu kin cho HS lm thm v trn ba nh sng mu khc.

C. Kt qu thc hnh:

Bng 1. Trn hai nh sng mu

Cc nh sng mu c trnMu nh sng thu c

- Lc

- Lam

Lc - Lam

Bng 2. Trn ba nh sng muCc nh sng mu c trnMu nh sng thu c

- Lc - Lam


BO CO TH NGHIM

Ngy thng nm 200

BI THC HNH S: 22.

TN TH NGHIM: KH NNG TN X NH SNG MU CA CC VTTIT 61 - BI 55: MU SC CC VT DI NH SNG TRNG V DI NH SNG MUTng im (10)Chun b



(2)

I. MC CH YU CU:

Qua th nghim HS gii thch c hin tng: khi t cc vt di nh sng th ch cc mu mi gi nguyn c mu, cn cc vt c mu khc th mu sc s b thay i.

II. CHUN B:1.Gio vin: Chun b mt vi tm nh phong cnh c mu xanh da tri.

2. Hc sinh: (mi nhm) mt hp kn c mt ca s c th chn bng cc tm lc mu hoc lc; cc vt mu trng, , lc v en, t trong hp, mt tm lc mu v mt tm lc mu lc.III. NI DUNG TH NGHIM:

A. Cu hi chun b:

Cu 1: Khi chiu nh sng trng n cc vt mu , xanh lc v trng th cc vt c mu g?

Tr li:

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Cu 2: Ta thy cc vt mu en nh vo u?

Tr li:

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B. Cc bc tin hnh:

B1. Chiu nh sng ln cc vt mu , xanh lc, en v trng.

B2. Chiu nh sng xanh lc vo cc vt mu , xanh lc, en v trng.

B3. Chiu nh sng trng vo cc vt mu , xanh lc, en v trng. C. Kt qu thc hnh:Chiu nh sngCc vt muC mu g?Tn x ttKhng tn xTn x km

Xanh lc

en

Trng

Xanh lc

Xanh lc

en

Trng

Trng

Xanh lc

en

Trng


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

BO CO TH NGHIM

Ngy thng nm 200

BI THC HNH S: 23.

TN TH NGHIM: NGHIN CU TC DNG NHIT CA NH SNG TRN VT MU TRNG V VT MU ENTIT 62 - BI 56: CC TC DNG CA NH SNGTng im (10)Chun b



(2)

I. MC CH YU CU:

So snh s hp th nng lng nh sng ca cc vt mu en v mu trng.

II. CHUN B DNG C:1.Gio vin: V to bng 1 SGK.

2. Hc sinh: (mi nhm) 2 tm kim loi ging nhau, mt sn trng, mt sn en; 2 nhit k; 1 bng n khong 25W; ngun in 12V ; 1 chic ng h.III. NI DUNG TH NGHIM:

A. Cu hi chun b:

Cu 1: Hy nu mt s hin tng chng t nh sng c tc dng nhit, ng dng ca n.

Tr li:

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Cu 2: Ti sao v ma ng nn mc qun o mu ti, cn v ma h nn mc qun o mu sng?

Tr li:

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........................B. Cc bc tin hnh:

B1: t hai tm kim loi ging nhau, mt sn trng, mt sn en cch u hai bn dy tc bng n. Cp hai nhit k du vo hai tm kim loi, c nhit ca hai nhit k lc u cha bt n.

B2: Bt cng tc, n sng, tnh thi gian (sau 1 pht, 2 pht, 3 pht) c v ghi s ch ca hai nhit k vo bng bo co.

C. Kt qu thc hnh:

Nhit

Ln TNLc uSau 1 phtSau 2 phtSau 3 pht

Vi vt mu trng

Vi vt mu en


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E. Cu hi vn dng:

Cu 1: Hy gii thch ti sao bnh cha xng, du trn cc xe t hay cc toa tu ch du phi sn cc mu sng nh mu nh bc, mu trng, mu vng

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Cu 2: Chiu nh sng vo mt t giy trng ta s thy t giy c mu g? Nu thay bng t giy xanh, ta s thy t giy c mu g?

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BO CO TH NGHIM

Ngy thng nm 200

BI THC HNH S: 24.

TIT 63 - BI 57: THC HNH: NHN BIT NH SNG N SC

V NH SNG KHNG N SC BNG A CD.Tng im (10)Chun b



(2)

I. MC CH YU CU:

- HS nm c khi nim nh sng n sc v nh sng khng n sc.

- HS bit cch dng a CD nhn bit c nh sng n sc v nh sng khng n sc.

II. CHUN B DNG C:1. Gio vin: Mt b th nghim nh ca nhm hc sinh.2. Hc sinh: (mi nhm) 1 n pht nh sng trng, cc tm lc mu vng, lc, lam, 1 a CD, mt s ngun sng n sc nh n laze, n LED, ngun in, mu bo co thc hnh.


A. Cu hi chun b:Cu 1: Th no l nh sng n sc? V d?........................

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

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

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

Cu 2: Th no l nh sng khng n sc? V d?........................

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Cu 3: Nu vi hin tng thc t v s phn tch nh sng trng?........................

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

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

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

B. Cc bc tin hnh: B1: Ln lt chn cc tm lc mu trc n ri a a CD vo chm nh sng l ra. Phi cm a trong tay sao cho c th thay i nghing ca a mt cch d dng.

B2: Quan st nh sng phn x c th rt ra nhng nhn xt cn thit.

Lu : nhn bit mt chm nh sng mu c phi l n sc hay khng, ta chiu chm sng vo mt ghi ca mt a CD v quan st chm sng phn x. Nu thy chm sng phn x ch c mt mu nht nh th nh sng chu ti a l nh sng n sc. Nu thy trong chm phn x c nhiu nh sng mu th nh sng chiu ti a l nh sng khng n sc.C. Kt qu thc hnh:

Mu ca cc nh sng c phn tch ra t cc nh sng mu to ra nh c tm lc mu khc nhau.

Kt qu QS

Ln TNCc mu ca nh sng c phn tch ranh sng mu c to ra nh tm lc mu l n sc hay khng n sc.

Vi tm lc mu

Vi tm lc mu vng

Vi tm lc mu lc

Vi tm lc mu lam


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E. Cu hi vn dng:

1. Trong nh sng phn x khi chiu chm nh sng trng n a CD c nhng mu no?.......................

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2. Gii thch hin tng cu vng.

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

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.......................A

V

V

A

A

V

X

A

l1

K

A B

A

3V

V

K

V

R2

R1

A B

A

K

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_1267448983.unknown

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_1173378584.unknown

_1102911908.unknown

_1102885715.unknown

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