Question

Sebuah lensa(n=1,5) mempunyai jarak titik fokus +10 cm diudara. jarak titik fokus lensa di air (n= 1,33) sebesar.. cm

Answer 1

OPTIK GEOMETRI
- Pemantulan

l = lensaa, u = udara, a = air

nl = 1,5 = 3/2
nu = 1, fu = 10 cm
na = 1,33 = 4/3, fa = __ ?

Rumus pembuat lensa
1/f = (nl/nmed - 1) (1/R1 + 1/R2)

di udara
1/fu = (nl/nu - 1) (1/R1 + 1/R2)

di air
1/fa = (nl/na - 1) (1/R1 + 1/R2)

perbandingan
(1/fu) / (1/fa) = [(nl/nu - 1) (1/R1 + 1/R2)] / [(nl/na - 1) (1/R1 + 1/R2)]
fa / fu = (nl/nu - 1) / (nl/na - 1)
fa / 10 = (1,5 / 1 - 1) / ((3/2) / (4/3) - 1)
fa / 10 = (1,5 - 1) / (9/8 - 1)
fa / 10 = (1/2) / (1/8)
fa / 10 = (1/2) × (8)
fa / 10 = 4
fa = 40 cm  ← jwb