b
Net path difference between the waves travelling through path I and II, $x =$
$\frac{3}{4} 2 \pi R-\frac{1}{4} 2 \pi R$
$\Rightarrow x=\pi R$
Let $\lambda$ be the wavelength of the waves.
Now for minima to occur at D, path difference $=\left( n +\frac{1}{2}\right) \lambda \quad$ where $n =$
$0,1,2, \ldots \ldots \ldots$
Thus $\frac{(2 n+1) \lambda}{2}=\pi R \Rightarrow \lambda=\frac{2 \pi R}{2 n+1}$
Now wavelength is maximum for $n =0$
Thus $\lambda_{\max }=2 \pi R$