Two coherent sources of sound, S₁ and S₂, produce sound waves of the same wavelength, λ = 1 m, in phase. S₁ and S₂ are placed 1.5 m apart (see fig). A listener, located at L, directly in front of S₂ finds that the intensity is at a minimum
when he is 2 m away from S₂. The listener moves away from S₁, keeping his distance from S₂ fixed. The adjacent maximum of intensity is observed when the listener is at a distance d from S₁. Then, d is :

Solution

Physics sir · 4 steps

At L the path difference is half a wavelength, so the next maximum needs a full wavelength.

1. Path difference at L. S₂L = 2 m, and the angle at S₂ is a right angle, so S₁L = √(1.5² + 2²) = 2.5 m. The path difference is 2.5 - 2 = 0.5 m = λ/2. That is why L is a minimum.
2. Which way it changes. The listener stays 2 m from S₂ and moves away from S₁. So the distance from S₁ grows beyond 2.5 m while the distance from S₂ stays 2 m. The path difference increases.

   Δ x = d - 2 = λ = 1 m

3. Next maximum. After λ/2, the next path difference that gives a maximum is one whole wavelength, Δ x = 1 m. This needs d = 3 m, which is reachable because d can be at most 1.5 + 2 = 3.5 m.
4. So. The adjacent maximum is at d = 3 m, option (c).

The idea to keep. Track the path difference, not the positions. A neighbouring minimum and maximum are λ/2 apart in path difference, and the direction of motion tells you whether it goes up or down.

Answer from the NTA final answer key. Solution by PrepWiser.