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(B) For destructive interference,the path difference between two waves is given by $\Delta x = (2n + 1) \frac{\lambda}{2}$,where $n = 0, 1, 2, ...$. F

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$\frac{\lambda}{2}$

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(B) For destructive interference,the path difference between two waves is given by $\Delta x = (2n + 1) \frac{\lambda}{2}$,where $n = 0, 1, 2, ...$. For the first case of destructive interference $(n=0)$,the path difference is $\Delta x = \frac{\lambda}{2}$.To achieve constructive interference,the path difference must be an integer multiple of the wavelength,i.e.,$\Delta x' = n\lambda$. The condition for constructive interference is $\Delta x' = m\lambda$ (where $m$ is an integer).If we increase the path of one wave by $\frac{\lambda}{2}$,the new path difference becomes $\Delta x_{new} = \Delta x + \frac{\lambda}{2} = \frac{\lambda}{2} + \frac{\lambda}{2} = \lambda$. Since $\lambda$ is an integer multiple of the wavelength $(m=1)$,constructive interference occurs.

There is a destructive interference between two waves of wavelength $\lambda$ coming from two different paths at a point. To get maximum sound or constructive interference at that point,the path of one wave is to be increased by

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