Jawaban:

[Cahaya]

Penjelasan:

2. so=20 cm

f=10 cm

1/f=1/si+1/so

1/si = 1/f -1/so

1/si = 1/10 - 1/20

1/si = 2/20 - 1/20

1/si = 1/20

si = 20/1

si = 20 cm

3. so = 30 cm

nyata

m=2x

m = |si|

|so|

2 = |si|

|30|

si = 2 x 30

si = 60 cm

1/f = 1/si + 1/so

1/f = 1/60 + 1/30

1/f = 1/60 + 2/60

1/f = 3/60

f= 60/3

f= 20 cm

4. so = 50 cm

f = 100 cm

a. letak bayangan

lensa cembung

R<so< f

maka di ruang 2(II) antara jarak fokus dan jari jari pusat kelengkungan

b. m = |si|

|so|

= |-100|

|50|

= 2x

1/f= 1/si + 1/so

1/si = 1/f - 1/so

= 1/100 - 1/50

= 1/100 - 2/100

= -1/100

si = -100/1

si = -100 cm

c. p = 100/f

= 100/100 cm

= 1 dioptri

d. sifat bayangan

-nyata, terbalik, diperbesar

5. lensa cembung dengan kekuatan lensa 5 dioptri menghasilkan ukuran bayangan 1/2 x bendanya

P = 100/f(cm)

5 = 100/f

f = 100/5

f = 20 cm

M=|si|

|so|

1/2x = |si|

|so|

ukuran bayangan

nyata, terbalik, diperkecil

maka:

a. Jarak benda (so) = > lebih dari 2f/R (Ruang 3) => [so>2f/R]

b. Jarak bayangan (si) => antara f dan 2f (Ruang 2) => [f<si<2f]

6. Lensa cekung

So = 20 cm

maya

Si = 10 cm

1/f = 1/si + 1/so

1/f = 1/10 + 1/20

1/f = 2/20 + 1/20

1/f = 3/20

f = 20/3

f = 6 2/3 cm

maya, tegak, diperkecil

.

Detail Jawaban

Mapel Matematika

Kelas 8 ( VIII ) SMP

Materi Cahaya

Kode Kategorisasi -

Kata Kunci Cahaya

Demikian

Semoga membantu dan bermanfaat!