bài giảng Khúc xạ ánh sáng

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  • 1. BA NH LUT C BN CA QUANG HNH HC1) nh lut truyn thng nh2) nh lut phn x nh sngsngTrong mi trng trong sut v ng tnhnh sng truyn theo ng thng.3) nh lut khc x nh sng

2. S KHC X NH SNG 1)Hin tng khc x nh sngNi dung 2)nh lut khc x nh sng CHIT SUT CA MI TRNG 1)Chit sut t i 2)Chit sut tuyt i TNH THUN NGHCH CA S TRUYN NH SNG 3. 1) Hin tng khc x nh sng a) Th nghim : M t TN : Nhn xt : Cy bt ch ly b gy mt phn cch gia nc v khng kh b) nh ngha : Khc x nh sng l hin tng lch phng (gy ) ca tia sng khi truyn xin gc qua mt phn cch gia 2 mi trng trong sut khc nhau 4. SI : tia ti I : im ti2) nh lut khc x nh sngNIN : php tuyn mt phncch ti IN i: gc ti i : gc phn xS IR : tia khc x S iir: gc khc x1Kt lun 1 :2 Tia khc x nm trong mtIphng ti ( to bi tia ti vr php tuyn ) v bn kiaphp tuyn so vi tia ti N R 5. TH NGHIM M PHNGTH NGHIM 1 : Mi trng 1 : Khng kh . Mi trng 2 : NcTH NGHIM 2 : Mi trng 1 : Khng kh . Mi trng 2 : Thy tinhTH NGHIM 3 : Mi trng 1 : Nc. Mi trng 2 : Thy tinh 6. TH NGHIM 1: Mi trng 1 : khng kh , Mi trng 2 : nc Gc t i i00 200 300 400 500 600 700 800 Gc khc x 00 15.50 22.50 290 35.50 40.50 450 47.50 Sin i0.342 0.5 0.643 0.766 08660.940.985 Sin r0.267 0.383 0.485 0.581 0.645 0.707 0.737 7. TH NGHIM 2 : Mi trng 1 : Khng kh . Mi trng 2 : Thy tinhGc t i i 00 200 300 400 500 600700 800Gc khc x00 150 200 250 300 350380 400Sin i 0.342 0.5 0.643 0.766 0866 0.940.985Sin r 0.258 0.342 0.422 0.5 0.5730.615 0.642 8. TH NGHIM 3 : Mi trng 1 : Nc. Mi trng 2 : Thy tinh Gc t i i00 200 300 400 500 600 700 800 Gc khc x 00 170 250 330 410 500 560 600 Sin i0.342 0.5 0.643 0.766 08660.940.985 Sin r0.292 0.422 0.544 0.656 0.766 0.829 0.866 9. Kt lun 2:Vi 2 mi trng trong sut nht nh , t s giasin gc ti (sini ) v sin gc khc x (sinr) lunkhng i (26.1) 10. nh lut khc x nh sng : N Tia khc x nm trong mtphng ti (to bi tia ti vS php tuyn) v pha bnii kia php tuyn so vi tia ti Vi 2 mi trng trong sut1 nht nh , t s gia sin gc2 ti (sin i) v sin gc khc x I(sin r) lun khng i rNR (26.1) 11. 1) Chit sut t i(26.2)Theo thuyt nh sng v1 : tc nh sng i trong mi trng 1 (SI : m/s) v2 : tc nh sng i trong mi trng 2 (SI : m/s) 12. sini > sin r i >r sini < sin r i < r S S ii(1) II (1)(2)(2) rRr R mi trng (2) chit quang mi trng (2) chit quanghn mi trng (1) km hn mi trng (1) 13. 2) Chit sut tuyt inh ngha :Chit sut tuyt i (thnggi l chit sut ) ca mtmi trng l chit sut t ica mi trng i vichn khngTrong mi mi trng trong sutu c chit sut tuyt i ln hn 1Chit sut ca chn khng l 1Theo nh ngha chit sut n21tuyt i kt hp vi biuthc (*) (26.3)n1 :chit sut (tuyt i )ca mi trng 1n2 :chit sut (tuyt i )ca mi trng 2 14. Nhn xt :N (26.1)S ii (26.2)1(26.3)2Ir n1sini = n2sinr N R Cng thc ca nh lut khc x:n1: chit sut ca mi trng (1)n2 : chit sut ca mi trng (2)n1sini = n2sinri: gc ti ; r: gc khc x 15. Cng thc ca nh lut khc x: n1sini = n2sinr Ch : sin i i- Nu i v r nh hn 10 th:0 sin r rn1i = n2r- Trng hp i = 00 th r = 00 tia sng chiu vung gc mtphn cch th khng xy ra hin tng khc x.- Nu tia sng truyn qua n mi trng, khc x qua n mitrng, v cc mt phn cch song song nhau th:n1sini1 = n2sini2 = n3sini3 == nnsinin 16. I K n1n2 17. NNS S Riir=i 1 1 I 2 2 I i=r rNSRSNTnh thun nghch ca s truyn nh sng : nh sng truyn theong no th cng truyn ngc li theo ng Tnh thun nghch ny biu hin s truyn thng , s phn x v s khc x 18. Bi tp v dTia sng truyn t mt cht trong sut c chit sut n ti mtphn cch vi mi trng khng kh . Gc khc x trong khng khl 600 . Tia phn x mt phn cch c phng vung gc vi tiakhc x (H26.6) Tnh chit sut nNTm tt Rr=600r (IS,IR)= 900n=?ICng thc ca nh nlut khc x:iin1sini = n2sinrNS S 19. Theo bi i+i = 900 i + r = 900N R p dng nh lut khc x:rnsini = sinrIniiNSS