Four light sources produce the following four waves: (i) y1 = a sin (ωt + ϕ1) (ii) y2 = a sin 2ωt (iii) y3 = a sin (ωt + ϕ2) (iv) y4 = a sin (3ωt + ϕ) Super position of which two waves give rise to interference?

Option 3 : i & iii

Concept:

When two waves of same frequency travelling in the same direction in a medium superpose with each other, their resultant intensity is maximum at some points and minimum at some other points. This phenomenon of superposition is called interference.

Let us consider two simple harmonic waves of same frequency travelling in the same direction. If a₁ and a₂ are the amplitudes of the waves and φ is the phase difference between them, then their instantaneous displacements are

y₁ = a₁sin ωt

y₂ = a₂sin (ωt + φ)

According to the principle of superposition, the resultant displacement is represented by

y = y₁ + y₂ = a₁ sin ωt + a₂ sin (ωt + φ) = (a₁ + a₂ cos φ) sin ωt + a₂ sin φ cos ωt

Put a₁ + a₂ cos φ = A cos θ, a₂ sin φ = A sin θ

Y = A sin (ωt + θ), This equation gives the resultant displacement with amplitude A.

Calculation:

Hence, y₁ = a sin (ωt + ϕ₁)

y₃ = a sin (ωt + ϕ₂)