Pemantulan Baur dan Pemantulan Teratur

111 Cahaya dan Optika A C B B D s s benda bayangan 5A41B1A;18D;D=5=1CD149C9C9;25A1;D 1 i 1 r 1 r J 1491A18B91A1CD?5835A=9=5=25CD;BD4DCJC5A814171A9B?A=1

a. Pemantulan pada Cermin Datar

1H1741981C14135A=941C1A;5C9;14125A49A949451 35A=941C1AC5AB52DC14135A=9C5A981C14121H17141 1719=11 B961C21H171H17C5A25CD;14135A=941C1A 5A81C9;104 -9C9;,=5AD1;121H17141A9C9C9;, 41 DA?B5B5=25CD;121H17114135A=941C1A14118B52171925A9;DC ,91A41C17, B91A:1CD814135A=941C1A4571BD4DC41C17 1 ;5=D491B91A99491CD;1 5A81C9;1:11B91A ,91A41C17, B91A :1CD814135A=94571BD4DC41C17 2 ;5=D491B91A99491CD;1 5A81C9;1:11B91A 5A1:171B91A1CD41B91A1CD 49251;1735A=9 49D;9B;1457171A9BC5ADCDBDCDB4125A?C?7149C9C9; 1495C1;21H171C9C9;14118H174925CD;41A95A?C?71 5A1:1714D1B91A1CD 57131A1H17B1=121H17125414D149=5B941C97149=5B9 411CC5A25CD;?5835A=941C1A A?B5B5=25CD;121H171H1B1=1 B55AC91412541C9C9; 1H17C5A5C9714118411=B5C91A?B5B 5=25CD;121H1718D;D=5=1CD1B51D25A1;D 5A81C9;104 25A9;DC .CD;2541H172D;125AD1C9C9;1C1D71A9B411;1=5411C ;1218F1D;DA121H17125415AB9BB1=14571D;DA12541H1 5414121H17181H125A2541411=811A18;9A941;11H1 5A81C9;1 04 1791 ;9A9 2541 =5:149 21791 ;11 21H17141B5219;H1 5A9BC9F19949B52DC5=219;1B9B90. +2.,+58;1A51141H15=219;1B9B999CD9B1H178541; 492131=51D935A=95D9B18DAD6H181ADB49219; 5A41B1A;121H171254114135A=941C1A411C49B9=D;1 218F1B961C21H1712541H174925CD;?5835A=941C1A14118 B52171925A9;DC K =1H1 K D;DA1B1=125B1A4571D;DA12541 K C571; K :1A1;2541C5A814135A=9B1=14571:1A1;21H171C5A8141 35A=9 Gambar 6.5 Ukuran benda persis sama dengan ukuran bayangan Gambar 6.4 Pembentukan bayangan pada cermin datar bayangan S S P 1 P 2 benda 1 2 Gambar 6.6 Ketika bercermin, posisi kiri benda menjadi posisi kanan bayangan mata pengamat 112 Mudah dan Aktif Belajar Fisika untuk Kelas X 58;1A51B91A1CDB1=14571 B91A41C17=1;1 149DCD;411C=5981C21H171 B5DAD8 CD2D8 495AD;1 1:17 35A=941C1A=99=1 41A9C9779 CD2D8B5B5?A17 :111:61:110: 1H171H1C11411821H171H17C5A25CD;41A95A?C?7117 BD7B91AB91A3181H1B5417;121H171=1H11411821H171H17 4981B9;141A95A?C?715A1:171B91AB91A3181H1 ,5217193?C?8 21H171H174981B9;114135A=941C1A=5AD1;121H171=1H1 ;1A51 21H171 C5AB52DC =5AD1;1 5A?C?71 17BD7 41A9 5A1:171B91AB91A3181H1 41D3?C?821H171H1C1C5A:149 14121H171H174981B9;1?58A?H5;C?A1411H1A 9CD71:17=99=D=BD1CD35A=9H17492DCD8;1171AB5?A1751A9141C 411C=5981CB5DAD8CD2D8H1H17C9779H1=5C5A 9 5A81C9;171=21A25A9;DC 70+:112+77+40,164 5A81C9;1 04 :9;1 B52D18 2541 25A141 49 451 4D1 2D1835A=9H17=5=25CD;BD4DCB1CDB1=1191;1C5A25CD; B5:D=18 21H171 D=18 21H171 25A71CD7 141 25A11 25B1A BD4DC 5A41B1A;181B95;B5A9=5:D=1821H171H174925CD; ?584D12D1835A=9H17=5=25CD;BD4DC 14118B52171925A9;DC ,52D182541C5A5C1;49451 2D1835A=941C1AH17=5=25CD;BD4DCJ 9CD7:D=1821H171H17C5A25CD; 9 9;5C18D9 J + 360 1 + J L J + L 149 :D=1821H171H17C5A25CD;2D18 J L + 5C5A171 + :D=1821H171 BD4DC1C1A14D12D1835A=9 L Contoh 6.3 G F H E A D C B cermin Gambar 6.7 Benda yang diletakkan di antara dua buah cermin datar dan membentuk sudut 90° akan menghasilkan 3 buah bayangan. cermin I cermin II bayangan bayangan bayangan benda Contoh 6.2