Given that b=1 cm, a=2.286 cm and f=9 GHz

We know that

In case of dominant mode \(f_c=\frac{C}{2a}=\frac{3\times{10}^{10}}{2\times2.286}=6.56\) GHz

\(\lambda_g=\frac{\lambda}{\sqrt{1-\left(f_c/f\right)^2}}\)

\(\lambda_g=\frac{\frac{C}{f}}{\sqrt{1-\left(f_c/f\right)^2}}\)

\(\lambda_g=\frac{\frac{3\times{10}^{10}}{9\times{10}^9}}{\sqrt{1-\left(6.56\times{10}^9/9\times{10}^9\right)^2}}\)

\(\lambda_g=\frac{3.33}{\sqrt{1-0.532}}=4.87\) cm

Therefore the distance between two holes in a two hole coupler is \(\frac{\lambda_g}{4}=\frac{4.87}{4}=1.22\) cm