A narrow tube is bent in the form of a circle of radius $R,$ as shown in the figure. Two small holes $S$ and $D$ are made in the tube at the positions right angle to each other. A source placed at $S$ generated a wave of intensity $I_0$ which is equally divided into two parts : One part travels along the longer path, while the other travels along the shorter path. Both the part waves meet at the point $D$ where a detector is placed The maximum value of $\lambda$ to produce a minima at $D$ is given by
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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 $\lambda_{\max }=2 \pi R$
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