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b. Pemantulan pada Cermin Cekung
                   5A=9> 35;D>7 141<18 35A=9> H1>7 =5=9<9;9 @5A=D;11> 35;D>7                             B
               5A=9> 35;D>7 25AB961C ;?>E5A75>  H19CD 25AB961C =5>7D=@D<;1> B9>1A
               1791> 21791> 35A=9> 35;D>7 C1=@1; B5@5AC9 @141  $0%$4               P        F             O
                  (0%(167.$1 %$:$1*$1 2/(+ &(40,1 &(.71*
                  %5C1; 41> B961C 21H1>71> H1>7 4925>CD; ?<58 35A=9> 35;D>7 25A71>CD>7            f
              @141 <5C1; 25>41  ,52D18 25>41 H1>7 49<5C1;;1> 49 45@1> B52D18 35A=9>
              35;D>7 1;1> =5=9<9;9 21H1>71> 45>71> B961C B961C C5AC5>CD  )5A81C9;1>          R
               $0%$4      B961C B961C 35A=9> 35;D>7 49 1>C1A1>H1                  f  = jarak fokus
              1  ,9>1A 41C1>7 H1>7 B5:1:1A BD=2D DC1=1 49@1>CD<;1> =5<1<D9 C9C9;  R = kelengkungan cermin
                  6?;DB DC1=1                                                     Gambar 6.8
                                                                                  Bagian-bagian cermin cekung
              2  ,9>1A  41C1>7  H1>7  =5<1<D9  C9C9;  6?;DB  DC1=1  49@1>CD<;1>  B5:1:1A
                  45>71> BD=2D DC1=1
              3  ,9>1A 41C1>7 H1>7 =5<1<D9 C9C9; @DB1C ;5<5>7;D>71> 35A=9> 1;1>           a)
                  49@1>CD<;1> ;5=21<9 =5<1<D9 <9>C1B1> H1>7 B1=1                  c)        b)    F
                   1H1>71> B52D18 25>41 ?<58 35A=9> 35;D>7 41@1C 49C5>CD;1> 45>71>       P                   O
              31A1 =5>771=21A;1>   41A9   B9>1A 9BC9=5F1 @141 35A=9> 35;D>7   1A1
              =5<D;9B 21H1>71> 25>41 ?<58 35A=9> 35;D>7 141<18 B521719 25A9;DC    Gambar 6.9
              )5A81C9;1>  $0%$4          +$( % -".0 *   35A=9> 35;D>7 4971=21A    Sinar istimewa pada cermin
              <5>7;1@ 45>71> 21791> 21791>>H1    +$( % ("!1   71A9B    4971=21A   cekung
              D>CD; =5<1=21>7;1> C9>779 25>41   41@D> 71A9B    C941; @5A<D C5A<1<D
              C9>779    +$( % ("0&$   71=21A;1> B9>1A 9BC9=5F1L     +$( % (""*- 0   B III     II     I
              71=21A;1>  B9>1A  9BC9=5F1L      +$( %  (")&*    C9C9;  H1>7  =5AD@1;1>
              @5A@?C?>71> 1>C1A1 B9>1A @1>CD<L  41> B9>1A @1>CD<L  4925A9 >1=1               A'
              -9C9; 9>9 =5AD@1;1> 21H1>71> 41A9 C9C9;    $5=D491>  71A9B C571; <DADB   A  P       F          O
              BD=2D DC1=1 49C1A9; 41A9 C9C9;    ;5 1A18 BD=2D DC1=1  -9C9; 9>9 4925A9          B'
              >1=1    H1>7 =5AD@1;1> 21H1>71> 41A9 C9C9;   B589>771      =5AD@1;1>
              21H1>71> 41A9 71A9B
                                                                                  Gambar 6.10
                III    II     I                  III    II      I                 Pembentukan bayangan di ruang III
                                                                                  AB pada cermin cekung
                     B                                                      B'
                                                                 B
                 A'                                                  O
                  P  A    F           O            P        F   A           A'
                                                                   S     S'

                B'

              Gambar 6.11                          Gambar 6.12
              Gambar bayangan jika letak bendanya di  Gambar bayangan jika bendanya di
              ruang II                             ruang I
                   5A41B1A;1>  $0%$4        $0%$4       41>  $0%$4      41@1C
              49B9=@D<;1> 218F1
               1  :9;1 25>41 @141 AD1>7 "  =1;1 21H1>71> H1>7 4981B9<;1> 141<18
                  =1H1  C571;  41> 49@5A25B1A
               2  :9;1 25>41 25A141 49 AD1>7 ""  =1;1 21H1>71> H1>7 4981B9<;1> 141<18
                  >H1C1  49 AD1>7 """   C5A21<9;  41> 49@5A25B1A
               3  :9;1 25>41 25A141 @141 AD1>7 """  =1;1 21H1>71> H1>7 4981B9<;1>
                  141<18 >H1C1  49 AD1>7 ""   C5A21<9; 41> 49@5A25B1A






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