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(D) The circumference of the circular tube is $2\pi R$. The source $S$ and detector $D$ are at positions at a right angle to each other.The shorter pa

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All of the above

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(D) The circumference of the circular tube is $2\pi R$. The source $S$ and detector $D$ are at positions at a right angle to each other.The shorter path length is $L_1 = \frac{1}{4} (2\pi R) = \frac{\pi R}{2}$.The longer path length is $L_2 = \frac{3}{4} (2\pi R) = \frac{3\pi R}{2}$.The path difference between the waves traveling through the two paths is $\Delta x = L_2 - L_1 = \frac{3\pi R}{2} - \frac{\pi R}{2} = \pi R$.For a minimum to occur at the detector,the path difference must be an odd multiple of half-wavelengths:$\Delta x = (n + \frac{1}{2}) \lambda$,where $n = 0, 1, 2, \dots$$\pi R = \frac{(2n + 1) \lambda}{2}$$\lambda = \frac{2\pi R}{2n + 1}$For $n = 0$,$\lambda = 2\pi R$.For $n = 1$,$\lambda = \frac{2\pi R}{3}$.For $n = 2$,$\lambda = \frac{2\pi R}{5}$.Thus,all the given options are possible values for the wavelength $\lambda$.

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