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(B) Let the distance between the speakers be $d = 2.0 \ m$. The distance from the midpoint to the microphone is $D = 4.0 \ m$.Initially,the microphone

Vedclass · Accepted Answer

$0.4$

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(B) Let the distance between the speakers be $d = 2.0 \ m$. The distance from the midpoint to the microphone is $D = 4.0 \ m$.Initially,the microphone is on the perpendicular bisector,so the path difference $\Delta x = 0$. This corresponds to the central maximum $(n=0)$.When the box is rotated by $90^\circ$,the speakers are in line with the microphone. The distances from the speakers to the microphone are $D - d/2$ and $D + d/2$.The path difference at this final position is $\Delta x = (D + d/2) - (D - d/2) = d = 2.0 \ m$.For constructive interference (maxima),the path difference is $\Delta x = n\lambda$,where $n$ is an integer.In the process of rotating from $0^\circ$ to $90^\circ$,we observe $5$ maxima. The central maximum is at $0^\circ$ $(n=0)$. As we rotate,we pass through $n=1, 2, 3, 4$ and finally reach the $5^{th}$ maximum at $90^\circ$ where $n=5$.Thus,at the final position,$\Delta x = 5\lambda$.Given $\Delta x = 2.0 \ m$,we have $5\lambda = 2.0 \ m$.Therefore,$\lambda = \frac{2.0}{5} = 0.4 \ m$.

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