noi ddu Nham dap ung nhu cau on tap kien thuc va ren luyen ki nang lam bai cung nhu giup hpc sinh tu tin thi vao lop 10 chuyen hoac khong chuyen, chung toi bien soan bp sach on thi vao lop 10 On tap, cung co kien thifc Idp Bp sach gom c6 nam cuon : Toan, Ngu van, Tieng Anh, Vat li, Hoa hpc Pham vi kien thuc cua bp sach tap trung vao chuong trinh va chuan kien thuc, ki nang lop Bp Giao due va Dao tao ban hanh Cuon On tap, cung cokien thCfc Vat li gom c6 hai phan : Phan mot On tap va cung co kien thiifc A - Vat li 6, 7, B-Vatli9 Phan hai Gidi thieu mot so de thi tuyen sinh vao Idp 10 A - D e bai B - Huong dan giai Ngoai kien thuc trpng tarn va nhung bai tap de cung co kien thuc, cuon sach gioi thieu mot so de thi vao lop 10 kem voi huong dan each giai, qua khoi gpi su sang tao cua cae em on tap va lam bai Hi vpng cac em se su dung cuon sach mot each sang tao de dat dupe ket qua cao ki thi sap toi Mac du da rat co gkng qua trinh bien soan, nhung cung kho tranh khoi nhung so suat, ehung toi mong nhan dupe sy dong gop y kien tii phia ban dpe de Ian tai ban sau, sach dupe hoan ehinh hon Mpi y kien dong gop xin gui ve : Phong Khai thac - Thj tri/dng Cong ty co phan Oau tiTva Phat trien Giao due PhiTcfng Nam 231 Nguyin Van CiT, Quan 5, TP Ho Chi Minh hoac qua email: khaithacbanthao@yahoo.com TAC GIA ON T A P V A CUNG CO K I E N T H U G i A - ON T A P V A CCING C6 KIE'N T H U C V A T L( , , I - C d HOC Dan vi do ddi he thong lUdng hdp phap cua nU6c ta la met (m) Ngoai ra, ngUdi ta dung ddn v i k m , dm, cm, mm, Dan vi the tich thUdng dung la met khoi (m^), l i t (0- Ngoai ra, ngUdi ta dung ddn v i dm^, cm^, cc, 11 = dm^ ; Dan vi khoi litOng he thong lufdng hdp phap ciia nifdc ta la kilogam (kg), ta (ta), tan (t) Ngoai ra, ngifdi ta dung ddn v i g, lang, tan = 1000 kg ; m l = cm^ = cc ta = 100 kg ; lang = 100 g C a c loai lufc a) Trong lUc : P Trong lUc la lUc hut ciia Trai Dat, c6 phiTdng t h i n g diing va c6 chieu hadng ve phia Trai Dat b) Luc ddn hoi: F + Vat chim xuohg k h i : F^ hay hay d^ > d; d^ < d; + V a t Id \\ing t r o n g c h a t long k h i : P = FA T r o n g Ivldng v a k h o i hay = d; Ixidng H e t h i i c giijfa t r o n g liJdng va kho'i Ivfdng : P = lO.m (vdi m t i n h b a n g kg) Vi du : V a t c6 k h o i l i i d n g 100 g t h i t r o n g lUdng la N Chu y : K h o i l i f d n g m k h o n g t h a y doi theo v i t r i dat v a t , v i k h o i lUdng c h i l i f d n g c h a t c h i l a t r o n g v a t Con t r o n g l i i d n g l a lUc h u t ciia T r a i D a t len v a t n e n t r o n g l i l d n g ciia v a t p h u thuoc vao v i t r i cua v a t t r e n T r a i D a t K h o i Ixidng r i e n g : D Cong t h i i c t i n h k h o i l i i d n g r i e n g : D = Trong : D : k h o i lUdng r i e n g (kg/m^) ; m : kho'i l i i d n g (kg) ; V : t h e t i c h (m^) Trong lifoTng rieng : d Cong thiic t i n h t r o n g l i i d n g r i e n g : d = p Trong : d : t r o n g lUdng r i e n g (N/m ) ; P : t r o n g liTdng (N) ; V : the t i c h (m^) Cong thxic t i n h t r o n g lUdng r i e n g theo k h o i lUdng r i e n g : d = TO cong thijfc : d = l O D , t a suy r a : D = May lO.D ^ cor doTn g i a n a) Mat phang nghieng (Hinh 1.1) • r \, - Bo qua m a s a t : „ = - r Trong : F la lUc tac d u n g (N) ; Hinh 1.1 P la t r o n g lUdng v a t (N) ; h la cao cua m a t p h a n g n g h i e n g (m) ; I la chieu d a i cua m a t p h a n g n g h i e n g ( m ) - Co ma sat (hao p h i ) t h i h i e u suat H cua m a t ph&ng n g h i e n g la : Ph 100% b) Don bay (Hinh 1.2) : d i e m tUa ; O O O i , O2 : d i e m d a t lUc ; F i , F2 : cac lUc tac d u n g 0 = / i ; 0 = ^2 D i e u k i e n can bSng ciia don bay : = -y- Hinh 1.2 c) Rong roc : Rong roc la m o t b a n h xe q u a y difdc q u a n h m o t t r u e , v a n h b a n h xe c6 r a n h de d a t day keo LTng d u n g : Gin + Tae d u n g : D o i h i i d n g eiia lUe tae d u n g ; F = P + R o n g roe q u a y diJcJe q u a n h mot t r u e c6' d i n h + R o n g roe c6 d i n h ( H i n h 1.3) - t r e n d i n h eot ed de keo cd, cong n h a n xay dUng d u n g dua gach v i i a len eao, Rong roc q u a y diidc q u a n h + R o n g roc dong ( H i n h 1.4) - I Hinh 1.3 /////////// ® m o t t r u e d i dong, d i chuyen ciing vdi vat + Tae d u n g : T h a y doi Idn ciia Ivtc tae d u n g (giam lUc keo) F = | ; s - = 2h P a l a n g : G o m m o t hoac n h i e u cap r o n g roc D u n g p a l a n g cho phep g i a m Ixic keo, dong t h d i l a m doi h u d n g ciia lUc C i l d u n g m o t cap r o n g roc (mot r o n g roc eo' d i n h , m o t r o n g roc dong) t h i Idi Ian ve lUc ( H i n h 1.4a) F = P ; 2n s = n h (vdi n l a so' cap cua r o n g roc) Chuyen dong deu va chuyen dong khong deu a) Van toe chuyen dong deu s + Cong t h i i e t i n h v a n toe : v = — t Trong : v : v a n toe (km/h ; m/s) ; s : q u a n g dUdng d i ditde ( k m , m) ; b) Van t : t h d i g i a n d i het q u a n g d i f d n g (h, s) toe trung binh chuyen dong khong deu ^ X s - Cong t h i l c t i n h v a n toe t r u n g b i n h t r e n m o t q u a n g dUdng : v^^jj = — - Cong t h i i c t i n h v a n toe t r u n g b i n h t r e n ca q u a n g d i f d n g chuyen dong : _ + S2 + + Sn t l + t2 + + t„ 10 Ap s u a t A p s u a t c h a t l o n g B i n h t h o n g n h a u a) Ap suat F Cong t h i i c t i n h ap s u a t : p = Trong : p : ap suat (N/m^ ; Pa) ; F : ap l u c ( N ) ; S : dien t i c h m a t b i ep (m^) b) Ap suat chat long - Cong thijtc t i n h ap s u a t c h a t l o n g : p = d h Trong : p : ap s u a t chat l o n g (N/m^^; d : t r o n g lifdng r i e n g c h a t l o n g (N/m^) ; h : cao cot c h a t l o n g (m) (h diidc t i n h tvf d i e m t i n h ap suat den m a t t h o a n g c h a t long) c) Binh thong : T r o n g b i n h t h o n g n h a u chiia c i i n g m o t c h a t l o n g d i i n g yen, cac miic chat l o n g d cac n h a n h l u o n l u o n cl c u n g m o t cao d) Nguyen F tdc hoat dong cua may thuy lite : ^ *2 Trong - S ©2 do: F j la lUc tac d u n g l e n p i t t o n g c6 dien t i c h S j ; F2 la lUc tac d u n g l e n p i t t o n g c6 d i e n t i c h S2 11 C o n g ccf h o c C o n g s u a t a) Cong cd hoc - Cong t h i i c t i n h cong cd hoc : A = F s Trong : A : c o n g cd hoc ( J ) ; F : lUc t a c d u n g (N) ; s : q u a n g d U d n g v a t c h u y e n ddi (m) l J = l N l m = l N m Chii y : - Cong t h i i c t r e n c h i s\i d u n g k h i h i f d n g cua lUc tac d u n g trCing vdi h u d n g c h u y e n dong ciia v a t K h i hildng cua luc tac dung vuong goc vdi hudng chuyen dong t h i : A = - K h i hifdng cua lUc tac dung ngUdc vdi hifdng chuyen dong t h i : A = -F.s - AHieu suat cua may cd : H = -r^.100% A Trong : Ai : cong c6 ich (J) ; A : cong toan phan (J) Chii, y : Cong cd ich la cong can thiet de lam vat dich chuyen Cong toan phan la tdng cong cd ich va cong hao phi : A = A^ + Ajjp b) Cong suat - Cong thdc t i n h cong s u a t : 9°= — = F v Trong : v : van toe (m/s) ; t : thdi gian thiic hien cong (s) ; A : cong thvfc hien (J) ; P/': cong suat (W) ; F : luc tac dung (N) W = — = i J / s ; kW (kilooat) = 1000 W ; Is M W (megaoat) = 1000000 W - Cach t i n h cong cd hoc thong qua cong suat : Ttf cong thiJc : — ^ A = PAt (J ; W h ; kWh) BAITAP Mot vat xuat phat tiif A chuyen dong deu ve B each A 240 m vdi van toe 10 m/s Cving liie do, mot vat khae chuyen dong deu t i i B ve A Sau 15 s hai vat gap T i n h van toe ciia vat t h i i hai va vi t r i hai vat gap {DS • V = m/s ; s A C = 150 m) H a i xe chuyen dong deu tren cung mot diidng thang Neu di ngUdc chieu t h i sau 15 phut khoang each giiia hai xe giam 25 k m Neu di ciing chieu t h i sau 15 phiit, khoang each giiia hai xe chi giam k m Hay t i m van toe cua moi xe {DS ; v i = 60 km/h ; V = 40 km/h) H a i xe c h u y e n dong t h a n g deu tu: A den B each n h a u 120 k m Xe d i h e n tuc k h o n g n g h i v d i v a n toe V j = 15 k m / h Xe k h i h a n h sdm hdn xe l a h n h u n g doc dUdng p h a i n g h i 1,5 h H o i xe p h a i c6 v a n toe b a n g bao n h i e u de t i B eving m ot luc vdi xe ? (DS : V2 = 16 k m / h ) M o t cano chay xuoi dong song d a i 150 k m V a n toe cua cano k h i nxidc k h o n g chay l a 25 k m / h , v a n toe ciia dong nxidc chay l a k m / h T i n h thdi g i a n cano d i h e t doan song (DS • t = h) M o t chiee x u o n g may c h u y e n d o n g t r e n m ot d o n g song N e u x u o n g ehay x u o i dong tvf A den B t h i m a t h , eon neu x u o n g ehay ngUde dong txi B ve A t h i phai m a t h T i n h v a n toe eua x u o n g m a y k h i nUde y e n l a n g va v a n toe cua dong nxidc B i e t k h o a n g each gifla A va B l a 120 k m (DS : Vx = 40 k m / h ; Vn = 20 k m / h ) T r o n g m ot b i n h t h o n g n h a u chiia t h u y n g a n , ngUdi t a t h e m vao mot n h a n h a x i t s u n f u r i c v a n h a n h l a i t h e m nxidc K h i cot nifdc t r o n g n h a n h t h i i h a i cao 72 c m t h i t h a y mvte t h u y n g a n d h a i n h a n h n g a n g n h a u T i m cao ciia eot a x i t s u n f u r i c B i e t t r o n g liidng r i e n g eiia a x i t s u n f u r i c va nude I a n lUdt la d j = 18000 N / m ^ v a dg = 10000 N / m ^ (DS : hA = 40 em) M o t cue nxidc da eo t h e t i c h V = 360 cm"^ n o i t r e n m a t nxidc a) T i n h t h e t i c h V ciia p h a n 16 r a k h o i m a t nUdc B i e t k h o i liJdng r i e n g ciia nxidc da l a 0,92 g/cm^ (DS : V = 28,8 m^) b) So s a n h t h e t i c h cua cue nxidc da va p h a n t h e t i c h nUdc cue nxidc da t a n hoan toan M o t k h o i go h i n h hop c h i i n h a t c6 t i e t d i e n S = 40 em^, cao h = 10 c m , kho'i lifdng m = 160g a) T h a k h o i go vao nxidc T i m chieu cao cua p h a n go n o i t r e n m a t nxidc ? B i e t k h o i lUdng r i e n g ciia nxidc l a DQ = 1000 kg/m^ (DS • x = cm) b) Bay gid k h o i go difdc k h o e t m o t 16 h i n h t r u d gii3a eo t i e t d i e n AS = em", sau A h v a difdc l a p day c h i e6 kh6'i lu:dng r i e n g l a = 11300 k g / m ^ K h i t h a k h o i g6 vao nUde t h i ngUdi t a t h a y mxic nxidc b a n g v d i m a t t r e n cua kho'i go T i m sau A h ? (DS : A h = 5,5 em) 11 V a t sang d a t d m o i v i t r i trvfdc t h a u k i n h p h a n k i deu cho a n h ao, cung + Ddc diem anh cua mot vat tao bdi thau kinh phan - ki: c h i e u , nho h d n v a t va l u o n n i m t r o n g k h o a n g t i e u c\i cxia t h a u k i n h + V a t d a t r a t xa t h a u k i n h , a n h ao cua v a t c6 v i t r i each t h a u k i n h mot k h o a n g bSng t i e u c\i - Cdch dytng anh cua vat qua thau kinh phan DiCng anh tao bdi diem sang thau kinh phan S tao ki bdi ki : Txi S t a ve h a i t r o n g ba t i a dac b i e t den t h a u k i n h , ve h a i t i a 16 r a k h o i t h a u k i n h H a i t i a 16 k h o n g ckt n h a u t h a t sU m a c6 difdng keo d a i cua c h u n g c^t n h a u , Hint) 1.31 giao d i e m cat n h a u d6 e h i n h la a n h ao S' cua S qua t h a u k i n h ( H i n h 1.31 ) Diing thau anh kinh cua phan vat sang AB tao bdi ki : M u o n dUng a n h A ' B ' cua A B qua t h a u k i n h (AB v u o n g g6c v i t h a u k i n h , A nhm tren true c h i n h ) , t a c h i can d\ing a n h B' cua B b a n g h a i t r o n g ba t i a sang dac biet, sau tijf B' v u o n g goc xuo'ng t r u e Hinh 1.32 c h i n h t a c6 a n h A' cua A ( H i n h 1.32) S i i tao a n h t r e n p h i m t r o n g m a y a n h A n h cua v a t t r e n p h i m l u o n la a n h t h a t , ngUdc c h i e u va nho h d n vat + V a t k i n h cua m a y a n h la m o t t h a u k i n h h o i t u + d : k h o a n g each tH v a t sang den t h a u k i n h ; f : t i e u c i i ciia t h a u k i n h ; d' : k h o a n g each tit a n h den t h a u k i n h ; h : cao ciia v a t sang ; h ' : cao ciia a n h T a c6 : d' h' K i n h lup a) Kinh - lup Id gi ? K i n h l i i p l a m o t t h a u k i n h h o i t u eo t i e u eU ngSn NgUdi t a d i i n g k i n h l i i p de q u a n sat eac v a t nho 56 - Moi k i n h lup c6 mot so' boi giac (ki hieu G) dUdc ghi tren vanh k i n h bang cac so'nhu 2x, 3x, 5x, So'boi giac ciia k i n h lup cho biet k h i dung k i n h ta CO the thay dUdc mot anh Idn len gap bao nhieu I a n (tinh theo goc) so v6i k h i quan sat trUc tiep vat ma khong dung k i n h He thiic lien he giuta so' boi giac va tieu cU ciia thau k i n h : f(cm) b) Cdch quan G sat mot vat nho qua kinh lup K h i quan sat mot vat nho qua k i n h lup ta phai dat vat khoang tieu cU cua k i n h cho t h u dvtdc mot anh ao 16n hdn vat Mat nhin thay anh ao Mat Mat c a n v a m a t lao a) Mat - Can tao cua mat ve mat quang hoc : H a i bo phan quan nhat ciia m&t la the thuy t i n h la mang lufdi + T h e thuy t i n h la mot thau k i n h hoi t u bang mot chat suo't va mem No de dang phong len hay det xuo'ng k h i cd vong dd no bop lai hay dan lam cho tieu cU ciia no thay ddi + Mang liidi la mot mang d day mat, tai anh ciia vat ma ta nhin thay se hien len ro net - Si/ dieu tiet ciia mat: De nhin r5 nhiing vat d nhiing khoang each khac t h i anh ciia vat luon phai hien ro net tren mang Cd vong dd the thuy t i n h da phai co dan mot chut lam thay doi tieu cU ciia no, qua t r i n h goi la sU dieu tiet ciia mat S u dieu tiet xay hoan toan tU nhien - Diem cite can va diem cUc vien : + Diem xa mat nhat ma k h i co vat d do, mat khong dieu tiet co the nhin ro vat goi la diem cUc vien ( k i hieu C y ) + Diem gan mat nha't ma k h i co vat d do, mat co the nhin ro vat (khi dieu tiet toi da) goi la diem cUc can (ki hieu C c ) b) Mat can - Mat can la mat co the nhin ro nhiing vat cf gan, nhUng khong nhin ro difdc nhiJng vat d xa - De khac phuc tat can t h i , ngifdi can t h i phai deo k i n h de co the nhin ro nhiing vat d xa K i n h can la tha'u kinh phan k i K i n h can thich hdp co tieu diem F trtmg v6i diem cxic vien (Cy) ciia mat 57 c) Mat - Ido Mat lao l a m a t c6 t h e n h i n ro n h i i n g v a t d xa, n h i f n g k h o n g n h i n ro dUdc n h i i n g v a t gan - De k h a c phuc t a t lao t h i , ngUdi mSt lao p h a i deo k i n h de c6 t h e n h i n ro nhiJng v a t d gan n h u mSt ngvfdi b i n h t h i i d n g K i n h lao la t h a u k i n h hoi t u V BAITAP Bai a) So'boi giac ciia k i n h l i i p la lOx V a y t i e u cU cua k i n h l i i p la bao n h i e u ? b) M o t ngUdi d u n g k i n h l i i p c6 t i e u c i i cm de q u a n sat mot v a t nho V a t dat each k i n h c m D u n g a n h ciia v a t q u a k i n h l u p A n h cua v a t qua k i n h l u p la a n h t h a t hay - T i n h so'boi giac ciia k i n h l u p - a n h ao ? A n h 16n h d n hay nho h d n v a t bao n h i e u I a n ? Bai M o t ngifdi q u a n sat m o t cot dien cao m, each cho d i i n g 25 m Cho r a n g m a n g l u d i cua mkt each t h e t h u y t i n h cm C h i e u cao ciia cot dien t r o n g mSt la bao n h i e u ? Bai M o t ngiJdi c h i n h i n ro v a t each mat tii 15 c m den 50 cm a) Mat n g i i d i mac t a t gi ? NgUdi p h a i deo t h a u k i n h loai nao ? b) K i n h t h i c h hdp v d i ngUdi eo t i e u cxi OF bang bao n h i e u ? Liic ngifdi ay n h i n ro v a t xa n h a t each mat bao n h i e u ? c) G i a i t h i c h v i ngUdi p h a i deo l o a i k i n h ? d) N e u ba b i e n p h a p p h o n g chong t a t can t h i Bai M o t ngiJdi c h i n h i n ro cac v a t each mat i t nha't 50 cm a) Mat ngUdi eo b i t a t k h o n g ? V i ? b) N g U d i p h a i deo k i n h gi de c6 t h e n h i n dUde cac v a t d gan ? c) K h i d i dUdng, ngUdi eo can deo k i n h k h o n g ? V i ? 58 Bai5 M o t cay cao m, each may a n h m, a n h cua cay t r e n p h i m cao cm T i n h t i e u cU cua may a n h Ve h i n h Bale D a t v a t sang A B h i n h m u i t e n v u o n g goc V B t r u e c h i n h eiia t h a u k i n h p h a n k i c6 t r u e c h i n h A, q u a n g t a m O, cac t i e u d i e m F, A F' nh\i H i n h 1.33 : F o F (A) a) D u n g a n h A ' B ' cua A B b) A ' B ' la a n h t h a t hay a n h ao ? T a i ? Hinh 1.33 Bai T r e n H i n h 1.34 : A la t r u e c h i n h cua mot o S' t h a u k i n h h o i t u , S la v a t sang, S' la a n h eiia S qua t h a u k i n h B a n g each ve, hay xac d i n h q u a n g t a m O va cac t i e u d i e m (A)- F, F' eiia t h a u k i n h (neu each ve) Hinh 1.34 Bai M o t v a t sang A B d a t v u o n g goc v d i t r u e c h i n h (A nam t r e n t r u e c h i n h ) cua t h a u k i n h hoi t u c6 t i e u c^i 10 cm, v a t each t h a u k i n h 15 cm a) H a y ve va t r i n h bay each dUng a n h A^B^ eiia v a t A B ? (do eao cua v a t va a n h k h o n g can d u n g t i le) b) N e u t i n h chat cua a n h A ^ B j c) T i n h k h o a n g each t i f v a t den a n h Bai D a t v a t sang A B h i n h m i i i t e n trudc mot t h a u k i n h p h a n k i c6 t i e u c\i 10 cm V a t d a t t r e n t r u e c h i n h , each t h a u k i n h 15 cm V a t cao cm a) D i i n g a n h cua A B qua t h a u k i n h N e u t i n h c h a t cua a n h b) B a n g phvfdng p h a p h i n h hoe, t i n h k h o a n g each tii a n h den v a t va chieu cao cua a n h 59 B a i 10 V a t sang A B c6 d a n g mot m u i t e n , cao cm, v u o n g goc vdi t r u e c h i n h cua mot t h a u k i n h , d i e m A nam t r e n t r u e c h i n h , cho a n h A ' B ' cao 10 cm, ngitOc chieu v d i A B va each v a t A B mot k h o a n g 90 em a) T h a u k i n h da cho la t h a u k i n h g i ? T a i ? b) B a n g each ve, h a y xae d i n h q u a n g t a m O, va h a i t i e u d i e m F , F ' cua t h a u k i n h (Lay t i le cm t r e n thifdc i i n g v d i 10 em t r e n t r u e c h i n h va em t r e n true thang diing) c) T i n h k h o a n g each OA, OA' va O F cua tha'u k i n h B a i 11 M o t t h a u k i n h h o i t u c6 t i e u cU cm M o t v a t sang A B d a t t r e n true c h i n h , v u o n g goc v d i t r u e c h i n h va each t h a u k i n h 20 c m a) V e a n h A ' B ' ciia v a t sang A B b) K h i ddi v a t l a i gan tha'u k i n h mot doan cm H a y neu cae dac diem cua a n h tao bdi t h a u k i n h ? B a i 12 D a t m o t v a t A B eo d a n g m i i i t e n cao c m v u o n g goc vdi t r u e c h i n h ciia mot t h a u k i n h h o i t u , each tha'u k i n h 30 c m , tha'u k i n h eo t i e u cU 10 cm a) D i i n g a n h A ' B ' ciia v a t A B ( k h o n g can d u n g t i le) b) Xae d i n h v i t r i va c h i e u eao cua a n h c) Muo'n cho a n h eao gap I a n v a t t h i p h a i d i c h u y e n v a t t h e o c h i e u nao ? T a i ? B a i 13 M o t v a t sang A B cao c m (A t r e n t r u e c h i n h ) , dat v u o n g goc vdi t r u e c h i n h t r a d e tha'u k i n h h o i t u m o t k h o a n g d = 30 cm, cho m o t a n h t h a t A ' B ' = em a ; V e a n h A ' B ' cua A B qua tha'u k i n h T i m k h o a n g each tijf a n h den tha'u k i n h , va t i m t i e u cU cua tha'u k i n h b) Muo'n CO mot a n h ao cao cm t h i p h a i ddi v a t l a i gan h a y r a xa tha'u k i n h m o t d o a n l a bao n h i e u ? 60 Bai 14 a) Ke ten nguon phat anh sang trSng va nguon phat anh sang mau ? b) Muo'n tao anh sang t i m ta lam each nao ? c) Tam loc mau xanh hap t h u tot nhiing anh sang nao ? d) Chieu ehum sang qua tam loc mau xanh, ta thu dUde anh sang mau gi ? V i sad ? Bai 15 a) Neu cac each phan tich ehum sang trang nhflng chum sang mau b) Ke vai hien tUdng ddi song c6 lien quan den noi dung tren Bai 16 - K h i tron hai chiim anh sang mau vdi ta dUde anh sang mau gi ? - Co the tao anh sang tr&ng bang each nao ? Bai 17 Mot vat sang A B c6 dang mui ten dUdc dat trUde mot thau k i n h , vuong gdc vdi true ehinh eua thau k i n h , diem A n&m tren true chinh eua thau k i n h Mot man M dat vuong goc vdi true ehinh phia sau thau k i n h K h i vat d v i t r i A j B i , ta t i m dUdc mot v i t r i cua man M cd anh A'^B'^ eiia A^B^ qua thau k i n h hien ro tren man va A\B\o gap l^n AB Ddi vat xa thau k i n h them mot doan den v i t r i A2B2 K h i de cd anh r5 tren man, ta phai ddi man M lai gan thau k i n h them mdt doan x = 30 cm va nhan thay anh mdi A2B2 tren man cao bang niia AB a) Hoi thau k i n h la hoi t u hay phan k i ? Giai thieh v i ? b) Ve hinh minh hoa cho hai trifdng hdp tao anh neu tren va dUa vao hinh ve dving eae phep tinh hinh hoc de t i n h tieu c\l cua thau k i n h Bai 18 Hai gUdng phang va G2 dat song song, doi dien nhau, mat phan xa quay vao Khoang each giula hai giidng la h = AC = 20 cm, ehieu dai la d = A B = CD = 85 em Mdt bdng den nho S dat each deu hai ngang vdi eae mep A va C cua hai hai gvfdng gUdng gUdng Mot ngUdi dat mat t a i O d mdi gUdng each deu va each S doan / = SO = 100 cm nhif H i n h 1.35 61 a) Hay ve va neu each ve dUdng di cua tia sang tic S den va phan xa tren gifdng G j hai Ian, tren gifcJng G2 mot Ian, roi di den mat Tinh chieu dai ditOng di ciia tia sang b) NgUdi nhin vao mot giJdng se thay diidc toi da bao nhieu anh cua S gifdng ? HUONG DAN A //////////////// B - -o O G2 Hmh 1.35 GIAI Bai a) Tieu c\i cua k i n h lup : „ 25 „ 25 25 ^^ G = - ^ = > f = - ^ = - = 2,5em b) So boi giae cua k i n h lup : G = f = f =3,125 Hinh 1.36 * DUng anh cua vat qua k i n h lup (Hinh 1.36) * A n h cua vat qua k i n h liip la anh ao, eCing chieu va Idn hdn vat A B B ' I 00 A O B F (g-g) : BB' OB' OB ^ OB' " AOAB GO AOA'B'(g-g) : BI OB- OB'-OB OB' OA OF OB OB' 3^1 ~ OA A B _ OB OA' " A B' " OB' " Vay anh Idn gap Ian vat Bai Chieu cao cua cot dien m S t : h' h d' d hd' d 800.2 2500 62 Bai3 a) Mat ngiidi m&c tat can t h i NgUdi phai deo thau k i n h phan k i b) K i n h c^n thich hdp v6i ngiidi c6 tieu cU OF bang 50 cm K h i deo k i n h nay, ngUdi nhin ro vat ci rat xa (v6 ciJc) c) NgUdi phai deo loai k i n h de nhin ro nhflng vat d xa mSt d) Khong doc sach qua gan mSt, khong doc sach k h i dang di t a u xe ; khong ngoi qua hai gid trUdc man h i n h v i t i n h ; hoc tap va lam viec d ndi d i i anh sang Bai4 a) Mat ngUdi hi tat lao t h i , v i chi nhin ro cac vat d xa b) NgUdi phai deo k i n h lao la thau k i n h hoi t u c) K h i di dUdng, ngUdi khong can deo kinh, v i c6 the nhin ro cac vat d xa Bai h = 6m = 600 cm ; d = 3m = 300 cm ; h' = cm Tren H i n h 1.37 : A B B ' I co AOB'F' (g-g) : BE' BI OB + OB' OA OB OA +1= OB' OF ^ O B ' OF' OB' OB OB' OA f 300 -1 f (1) oo AOA'B' (g-g) : OA OA' 0A' = TC ,1) (2) : M AB A'B' OA _ = 200 OB OB' 30 600 Hinh 1.37 (2) = 15 cm ^ = 201 => f = |22 , 1,5 cm Bale a) Dung anh A'B' cua vat AB (Hinh 1.38) b) A n h A'B' ciia vat A B la anh ao v i vat A B dat trUdc thau k i n h phan k i luon cho anh ao, ciing chieu va nho hdn vat Hinh 1.38 63 Bai7 X e m H i n h 1.39 Bai a) Ve a n h ( H i n h 1.40) * Trinh bay each ve : V e t i a t d i BI//(A) cho t i a 16 (1) d i qua t i e u d i e m F j , t i a B O qua q u a n g t a m O cua t h a u k i n h cho t i a 16 (2) d i t h S n g H a i t i a 16 cSt n h a u t a i B j B^ l a a n h cua B Tu: B i h a dUdng v u o n g goc x u o n g (A), cat (A) t a i A^ A^ l a a n h cua A A j E j l a a n h cua v a t A B cAn ve b) A n h A j B i : A n h t h a t , ngUdc c h i e u v a 16n h d n vat c) C h i i n g m i n h : AA^ = 45 c m A B B i I c/5 A O B i F ' (g-g) : OF' OBi BI BB^ OB +0B, OBi OA OF OB + = OBi 15 10 OB = ^ - = OBi 2 A O A B CO A O A ' B ' (g-g) : OA OAi OA OA^ AB A^Bi OB OB ^ = 1^ OKJDi Bi 2 O A i = A = 2.15 = c m K h o a n g each tii v a t den a n h : A A i = O A + O A i = 15 + 30 = 45 cm 64 Bai9 Giai Tom tat Thau kinh phan k i a) - Dung anh A'B' cua A B qua thau kinh : AB = cm ; V f = 10 cm ; A = 15 cm a) Diing anh A'B' ciia AB qua thau kinh Neu tinh chat anh b) Bang phvfdng phap hinh hoc : AA' va A'B' = ? (cm) - Tinh chat anh A'B' : Anh A'B' cua vat A B la anh ao, cung chieu va nho hdn vat b) ABB'I CO AOB'F (g-g) : BB' OB' BI OF OB OB' OB - O B ' OB' OA OF OB OB' ^1^^ ~ 10 " - = 2" AOAB oo AOA'B' (g-g) : OA _ AB _ OB _ OA' ~ A ' B ' " OB' " AB ^ A ^ 2.AB 2.5 ^ = - - " = '^"^ = ^ V a ^ = OA' ^ - A ' = ^ ^ c m OB 5 Khoang each t i i vat den anh : AA' = OA - OA' = 15 - = cm Do cao cua anh la cm Bai 10 Giai Tom tat AB = cm ; A A' = 90 cm ; a) Anh A'B' ngUdc chieu va Idn hdn vat, nen thau k i n h da cho la thau k i n h hoi t u A'B' = 10 cm b) a) Thau kinh da cho la thau kinh gi ? Tai ? - Ve vat AB vuong goc vdi true chinh (A) tai A - Xac dinh AA' = 90 cm 65 - N o i B B ' cat (A) t a i c h i n h la q u a n g t a m cua b) B a n g each ve h a y xac Giai Tom tat dinh : quang t a m 0, h a i t i e u d i e m F, va F'cua thau kinh ? (Lay t i le cm thau kinh => Ve t h a u k i n h h o i t u - Ve t i a t d i BI//(A) cat t h a u k i n h t a i I N o i I B ' cat (A) tren thitdc l i n g v d i 10 cm t r e n t r u e c h i n h va cm tai F => t i e u d i e m F t r e n true thSng diing) \ !i c) OA, OA', O F = ? (cm) A ^ ' ^ ^ F ^ \ F ' OA c) A O A B c n A O A ' B ' i ^ - ^ ) •• ^ = A' AR = (A) ^ = OA' = A m a A A ' = O A + O A = A AA' 90 O A = - — = — = 30 em o V a OA' = 30 = 60 em A O I F ' CO A A ' B ' F ' (g-g) : 01 OF' A'B' _ OI.A'F' " A'F' A'B' 10 A'B' 5(60-OF') AB.(OA'-OF') lO.OF' = 300 - F ' ^ F ' = 300 => O F ' = 20 em, hay O F = 20 cm B a i 11 a) Ve a n h A ' B ' cua v a t sang A B ( H i n h 1.41) Hinh 1.41 66 b) B K h i d6i v a t l a i gan t h a u ki'nh 1I , m ot V' doan cm ( H i n h 1.42) i Cac dac d i e m cua a n h tao bcii t h a u F ^ O A' k i n h : a n h t h a t ngifcic c h i e u v a I d n hdn vat B' V Hinh 1.42 Bai 12 Tom tat Giai Thau k f n h hoi t u a) - Ve v a t A B v u o n g goc vdi t r u e c h i n h (A) t a i A A B = cm ; - Ve t i a t d i BI//(A) cho t i a 16 (1) d i qua t i e u d i e m F' f = 10 cm ; cua t h a u k i n h OA = 30 cm - Ve t i a t d i B O qua q u a n g t a m ciia t h a u k i n h cho a) DUng anh A'B' ciia t i a 16 (2) t i e p tuc d i t h S n g theo phUdng cua t i a t d i vat A B - T i a 16 (1) v a (2) cat n h a u t a i B' B' l a a n h cua B b) OA', A ' B ' = ? (cm) Tvf B' h a d i i d n g v u o n g g6c v d i (A) t a i A ' A ' l a a n h c) M u o n cho a n h cao gap ba I a n v a t t h i p h a i d i chuyen vat theo ciia A - A ' B ' la a n h cua v a t A B chieu nao ? T a i ? b) A B B ' I cn> A OB ' F' (g-g) : BB' BI OB + OB' OB' OF' OB' OB OA OB' 10 OA - = F OB OB' , OA ^ ^ = 10- 30 -1 = ^ - = ^ 10 A O A B c o A O A ' B ' (g-g) : OA AB ^ OB OA' " A ' B ' " OB' 0A' = OA VaA'B' = ^ 30 = - 15 c m = § = lcm 67 Tom tat Giai c) Muo'n anh cao gap ba Ian vat t h i phai dcii vat lai gan thau k i n h de OF < OA < 2.OF / B I \F' A F O \ f Bai 13 Tom t^t Thau k i n h hoi t u Giai a) A B = cm ; B OA = d = 30 cm ; A A' A n h that A'B' = cm ^ a) Dang anh A'B' ciia vat A B F B' OA', f = ? ( c m ) b) De A'B' = cm la anh ao t h i phai ddi vat lai gan hay xa thau k i n h ? OA? ABB'Ic/5 AOB'F' (g-g) : BB' B I _ OB + OB' OA _ OB OB' " OF' " OB' " OF OB' ' OB OB' 30 " f (1) AOABco AOA'B' (g-g) : OA AB OB OA' " A ' B ' " OB' " " ^ ^ (2) => OA' = 2.OA = 2.30 cm = 60 cm Tvi (1), (2) : ^ - 1= i f=20cm = 2=^ 68 GiSi Tom t i t B' b) F A' A O >f ABB'Ic/D A O B ' F (g-g) : BE' _ BI OB'-OB OA OB' " OF' OB' OF OB OA OB' - = ^ 20 OB OA OB' (1) AOABo-5 A O A ' B ' (g-g) : OA AB OB OA' A'B' " OB' Tiif (1), (2) : 0A = 20 " OA _ OA 20 20 (2) = 10 cm V a y de A ' B ' = c m l a a n h ao t h i p h a i ddi v a t l a i g a n t h a u k i n h va each t h a u k i n h 10 cm B a i 14 a) - N g u o n p h a t a n h sang t r a n g : M a t T r d i , den p i n , l i i a t r a i , den p h a oto, den day toe, - N g u o n p h a t a n h sang m a u : D e n L E D , den bao r e , den q u a n g cao, den laze, den giao t h o n g , b) C h i e u a n h sang t r S n g qua t a m loc m a u t i m t a c6 a n h s a n g t i m ( t h i n g h i e m diidc x e m n h u t h i n g h i e m ve s\i p h a n t i c h a n h sang t r S n g n e u I a n l i f d t d i i n g n h i e u t a m loc m a u ) c) T a m loc m a u x a n h h a p t h u t o t cac a n h sang k h o n g p h a i m a u x a n h d) C h i e u c h u m sang qua t a m loc m a u x a n h , t a k h o n g t h u diJdc a n h sang nao v i t a m loc m a u x a n h h a p t h u t o t a n h sang B a i 15 a) Co t h e p h a n t i c h c h i i m sang t r a n g t h a n h n h i i n g c h u m sang m a u b a n g each cho c h u m sang t r a n g d i qua m o t l a n g k i n h hoac cho no p h a n xa t r e n m a t g h i cua d i a C D (c6 t h e cho c h i i m sang t r S n g I a n lUdt d i qua cac t a m loc m a u ) 69 f , b) Hien titcing cau vong, vang dkn, bong bong xa phong, hit nhieu anh sang mau den mat ta k h i c6 anh sang trSng chieu vao Bai 16 - ^ K h i tron hai chum sang mau vdi ta dvfcic anh sang c6 mau khac v6i hai anh sang ban dau - Co the tao anh sang trSng bang each chieu ba chiim sang do, lam, luc vao cung mot cho tren man mau trang Bai 17 a) A n h hien ro tren man : anh that Vat A B cho anh that : thau k i n h hoi t u b) Ve hinh hai triidng hdp tao anh cua thau k i n h A F ' A ' ^ B j va AF'A'2B'2 dong dang, do ta c6 : AoBg F'B, F'A'i - X ^ = ^ = — — i = — => F A , = 40 cm AiB'i F'B'i F'A'i AF'A'^B'i va A F ' O j I i dong dang, do ta c6 : Bai 18 S\i tao anh : S qua Gx -> qua G2 -> S2 qua G^ -> S3 Caeh ve : Ve Sj doi xiing vdi S qua G^, S2 do'i xiing vdi Sj qua G2, S3 doi xiing vdi S2 qua G j S2I3 cat Gg t a i I2, S^ I2 cat Gj tai I i , SI1I2I3O la difdng di ciia tia sang can ve Chieu dai cua tia sang bang : OS3 = ^/oS^Tssf = 116,6 cm Noi OB cat AC t a i K Dung tam giac dong dang tinh dUdc : SK = 67 cm Khoang each ttf anh Sn den S la : SS^ = n.h = 20n em Dieu kien : SS^ < SK => n < mat thay tol da ba anh cua S mot giidng 70