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 If the minima is formed at the detector then, the magnitude of wavelength $\lambda$ of the wave produced is given by
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Net path difference between the waves traveling through path $I$ and $I I$, $\Delta x=\frac{3}{4} 2 \pi R-\frac{1}{4} 2 \pi R$
$\Delta x=\pi R$
Let $\lambda$ be the wavelength of the waves. Now for minima to occur at $\mathrm{D}$, path difference $=\left(n+\frac{1}{2}\right) \lambda$ where $n=0,1,2, \ldots \ldots \ldots$ Thus $\frac{(2 n+1) \lambda}{2}=\pi R \Longrightarrow \lambda=\frac{2 \pi R}{2 n+1}$
Now $n=0,1,$ and $2 \Longrightarrow \lambda=2 \pi R, \frac{2 \pi R}{3}$ and $\frac{2 \pi R}{5}$ respectively.
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