1.benda bergarak 2 cm di depan Cermin dengan Jari-jari kelengkungan Cermin 12 cm. Tentukan sifat bayangan dan Jarak bayangan yang terbentuk? 2. benda berjarak 30 cm dari cermin. Jika jarak OR 20 cm. Tentukan sifat bayangan dan perbesaran bayangan?

<p>ini tidak jelas jenis cerminnya apa. Jadi saya anggap ini cermin cekung saja.</p><p>&nbsp;</p><p>1. S = 2 cm</p><p>&nbsp; &nbsp; &nbsp;R = 12 cm</p><p>&nbsp; &nbsp; &nbsp;f = 6 cm</p><p>&nbsp; &nbsp; &nbsp;Benda berada di R-I karena S &lt; f</p><p>&nbsp; &nbsp; &nbsp;1/S' = 1/f - 1/S</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 1/6 - 1/2</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 1/6 - 3/6</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = -2/6</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = -1/3</p><p>&nbsp; &nbsp; S' = -3 cm &nbsp; (- berarti benda berada dibelakang cermin (maya))</p><p>&nbsp; &nbsp; Jarak bayangan adalah 3 cm dibelakang cermin</p><p>&nbsp; &nbsp; &nbsp;M = |S'/S| = |-3/2| = 1,5 kali &nbsp; (diperbesar)</p><p>&nbsp;</p><p>&nbsp; &nbsp; Sifat bayangan : Maya, tegak, diperbesar</p><p>&nbsp;</p><p>2. S = 30 cm</p><p>&nbsp; &nbsp; &nbsp;R =20 cm</p><p>&nbsp; &nbsp; &nbsp;f = 10 cm</p><p>&nbsp; &nbsp; &nbsp;1/S' = 1/f - 1/S</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 1/10 - 1/30</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 3/30 - 1/30</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 2/30</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 1/15</p><p>&nbsp; &nbsp; &nbsp; S' = 15 cm&nbsp;</p><p>&nbsp; &nbsp; &nbsp; Bayangan benda bersifat nyata karena didepan cermin</p><p>&nbsp;</p><p>&nbsp; &nbsp; &nbsp; M = |S'/S|&nbsp;</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = |15/30|</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 1/2 kali</p><p>&nbsp; &nbsp; &nbsp;Karena perbesarannya 1/2 kali maka bayangan benda diperkecil</p><p>&nbsp;</p><p>&nbsp; &nbsp; Sifat bayangannya : Nyata, terbalik, diperkecil</p><p>&nbsp;</p>