Answer
$$17.52\;cm$$
Work Step by Step
The path difference $(\Delta x)$ of the two interfering wave is
$(\Delta x)=(\pi r-2r)$
Now, the phase difference of the two interfering wave is
$\phi=\frac{2\pi}{\lambda}(\pi r-2r)$
The minimum value of the intensity at the detector arises when $\phi=\pi$
Let, when the intensity becomes minimum, $r=r_{min}$
Therefore,
$\frac{2\pi}{\lambda}(\pi r_{min}-2r_{min})=\pi$
or, $r_{min}=\frac{\lambda}{2(\pi -2)}$
Substituting the given values
$r_{min}=\frac{40}{2\times(\pi -2)}\;cm$
or, $\boxed{r_{min}=17.52\;cm}$
