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f = 2 cm
s = 3 cm (ruang 2)
sifat bayangan __?
Benda yang terletak di ruang 2 lensa CEMBUNG, akan menghasilkan bayangan (di ruang 3) yang bersifat:
• nyata
• terbalik
• diperbesar
s = 3 cm
f = 2cm
ditanyakan: sifat bayangan?
jawab:
STEP 1: menentukan jarak bayangan
1/f = 1/s + 1/s'
1/2 = 1/3 + 1/s'
1/s' = 1/2 - 1/3
1/s' = 3-2/6
1/s' = 1/6
s' = 6/1 = 6 cm (nyata)
STEP 2: menentukan perbesaran
M = -s'/s = -6/3 = -2x perbesaran (terbalik, diperbesar)