Sebuah benda setinggi 8 cm diletakan didepan cermin cekung pada jarak 20 cm, jika jarak fokus cermin 12 cm tentukan

Jawaban:

a) s' = 30cm

b) M = 1,5

c) h' = 12cm

d) nyata, terbalik, diperbesar

Penjelasan:

h = 8cm

s = 20cm

f = 12cm

a) s' = ?

1/f = 1/s + 1/s'

1/s' = 1/f – 1/s

1/s' = 1/12 – 1/20 = 5/60 – 3/60 = 2/60 = 1/30

1/s' = 1 /30 (posisi dibalik)

s' = 30cm

b) M = ?

M = |s'/s| = |h'/h|

M = |s'/s|

M = |30/20| = |3/2|

M = 3/2 = 1,5

c) h' = ?

M = |h'/h| = |s'/s|

M = |h'/h|

1,5 = h'/8

h' = 12cm

d) jarak fokus (f) = 12cm

jarak fokus (2f) = 24cm

letak benda (s) = ruang 2 = 20cm

letak bayangan (s') = ruang 3 = 30cm

M > 1 = diperbesar

nyata, terbalik, diperbesar

semoga membantu!

maaf jika ada kesalahan