Comparing the waves y1=A*e^(i(k+b)x) y2=B*e^(i(kx+a)), the difference in wavelength for the two waves will depend on b right?
- marsLv 73 weeks ago
- Steve4PhysicsLv 73 weeks ago
The real coefficient1in front of the 'x' equals 2π/λ.
For example if we have e^(24ix) then
24 = 2π/λ
λ = 2π/24 = π/12
In e^(i(k+b)x) this coefficient is (k+b) giving:
k+b = 2π/λ
λ = 2π/(k+b)
In e^(i(kx+a)) this coefficient is k giving:
k = 2π/λ
λ = 2π/k
So the difference in wavelengths is:
Δλ = 2π/k – 2π/(k+b)
= 2π(1/k – 1/(k+b))