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  4. Two coherent sources of sound, S1 and S2, produce sound waves of the same wavelength, λ=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 :

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Q. 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 :Physics Question Image

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Solution:

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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 $
$ \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 $
$\Rightarrow d =3 m$