Chapter 28 - Physical Optics: Interference and Diffraction - Problems and Conceptual Exercises - Page 1004: 8

$203Hz; 608.8Hz$

We can find the required frequencies as follows: $\Delta l=\frac{d}{2}+\frac{d}{2}=d=0.845m$ Now $f_1=\frac{v}{\lambda_1}$ We plug in the known values to obtain: $f_1=\frac{343m/s}{2(0.845m)}=203Hz$ and $f_2=\frac{v}{\lambda_2}$ We plug in the known values to obtain: $f_2=\frac{2(343m/s)}{2(0.845m)}=608.8Hz$