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(D) The length of the shorter path is $L_1 = \frac{1}{4}(2\pi R) = \frac{\pi R}{2}$.The length of the longer path is $L_2 = \frac{3}{4}(2\pi R) = \fra

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All of the above

Vedclass · Answer

(D) The length of the shorter path is $L_1 = \frac{1}{4}(2\pi R) = \frac{\pi R}{2}$.The length of the longer path is $L_2 = \frac{3}{4}(2\pi R) = \frac{3\pi R}{2}$.The path difference between the two waves is $\Delta = L_2 - L_1 = \frac{3\pi R}{2} - \frac{\pi R}{2} = \pi R$.For constructive interference (maxima) at the detector,the path difference must be an integer multiple of the wavelength:$\Delta = n\lambda$,where $n = 1, 2, 3, \dots$Substituting $\Delta = \pi R$,we get $\pi R = n\lambda$.Therefore,$\lambda = \frac{\pi R}{n}$.For $n=1$,$\lambda = \pi R$.For $n=2$,$\lambda = \frac{\pi R}{2}$.For $n=3$,$\lambda = \frac{\pi R}{3}$.Since the options include $\pi R$,$\frac{\pi R}{2}$,and $\frac{\pi R}{4}$ (which corresponds to $n=4$),and all these are possible values for different integers $n$,the correct option is 'All of the above'.

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