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JEE Mains · Physics · STD 12 - 10. Wave optics

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

  1. A \(12\)
  2. B \(3\)
  3. C \(5\)
  4. D \(2\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(3\)

Step-by-step Solution

Detailed explanation

Initially \(S _{2} L =2 m\) \(S_{1} L=\sqrt{2^{2}+(3 / 2)^{2}}\) \(S_{1} L=\frac{5}{2}=2.5 m\) \(\Delta x =S_{1} L -S_{2} L =0.5 m\) So since \(\lambda=1 m \quad \therefore \Delta x =\frac{\lambda}{2}\) So while listener moves away from \(S_{1}\) Then,…
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