Cahaya kuning sebuah lampu natrium memiliki panjang gelombang 588nm (1nm = 10^-9) . jika satu berkas sinar ini masuk ke dalam air (indeks bias = 4/3) berapakah: a) cepat rambat? b) panjang gelombang? c) frekuensi cahaya dalam air?
FadlyHermaja
Λ1 = 588 nm = 5,88*10⁻⁷ m v1 = 3*10⁸ m/s n1 = 1 n2 = 4/3 a) v2=? v2/v1 = n1/n2 v2/(3*10⁸) = 1/(4/3) v2 = (3/4)*3*10⁸ = 2,25*10⁸ m/s b) λ2 = ? λ2/λ1 = n1/n2 λ2/5,88*10⁻⁷ = 1/(4/3) λ2 = (3/4)*5,88*10⁻⁷ = 4,41*10⁻⁷ m = 441 nm c) f2 =f1 f2 = v1/λ1 f2 = 3*10⁸/5,88*10⁻⁷ = 5,1*10¹⁴ Hz
v1 = 3*10⁸ m/s
n1 = 1
n2 = 4/3
a) v2=?
v2/v1 = n1/n2
v2/(3*10⁸) = 1/(4/3)
v2 = (3/4)*3*10⁸ = 2,25*10⁸ m/s
b) λ2 = ?
λ2/λ1 = n1/n2
λ2/5,88*10⁻⁷ = 1/(4/3)
λ2 = (3/4)*5,88*10⁻⁷ = 4,41*10⁻⁷ m = 441 nm
c) f2 =f1
f2 = v1/λ1
f2 = 3*10⁸/5,88*10⁻⁷ = 5,1*10¹⁴ Hz