\(S_{1} L=\sqrt{2^{2}+(3 / 2)^{2}}\)

\(S_{1} L=\frac{5}{2}=2.5 m\)

\(\Delta x =S_{1} L -S_{2} L =0.5 m\)

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. \(\Delta x=\lambda=S_{1} L-S_{2} L\)

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