Chapter 28 - Physical Optics: Interference and Diffraction - Problems and Conceptual Exercises - Page 1004: 5

$0.18KHz$

We can find the required frequency as follows: $d=\sqrt{(1.5)^2+(3.0)^2}$ $d=3.3m$ We know that $\lambda=3.3m-1.5m=1.8m$ We know that $f=\frac{c}{\lambda}$ We plug in the known values to obtain: $f=\frac{340m/s}{1.8m}$ $f=0.18KHz$