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\(......m\)

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Answer & Solution

Correct Answer

(B) \(3\)

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, \(\Delta x\left(=S_{1} L-S_{2} L\right)\) increases

and hence, at \(\Delta x=\lambda\) first maxima will appear. \(\Delta x=\lambda=S_{1} L-S_{2} L\)

\(1=d-2 \Rightarrow d=3 m\)

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