Question

Two waves of equal frequencies have their amplitudes in the ratio of 3:5. They are superimposed on each other. Calculate the ratio of maximum and minimum intensities of the resultant wave.

Answer 1

The ratio of maximum and minimum intensities of the resultant wave is 16 : 1

- Given :

Ratio of amplitude of two waves, A₁ / A₂ = 3 / 5

Maximum Intensity, I(max) = (A₁ + A₂)²

Minimum Intensity, I(min) = (A₁ - A₂)²

Thus ratio of maximum to minimum intensity

I(max) / I(min) = (A₁ + A₂)² / (A₁ - A₂)²

⇒ I(max) / I(min) =A₂² (A₁/A₂ + 1)² / A₂²(A₁/A₂ - 1)²

⇒ I(max) / I(min) = (3/5 + 1)² / (3/5 - 1)²

⇒ I(max) / I(min) = (8/5)² / (-2/5)²

⇒ I(max) / I(min) = 64/4 = 16/1

⇒ I(max) : I(min) =  16 : 1