A standing wave is produced due to a superposition of the incident wave and the wave reflected from a boundary. It is observed that the amplitude at antinode is 9 times that at node. The percentage of the incident intensity reflected from the boundary is

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36%

81%

64%

72%

answer is B.

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Detailed Solution

Let $A_{i}$ and $A_{r}$ be the amplitudes of the incident and reflected waves. Then
Amplitude at antinode $= A_{i} + A_{r}$
Amplitude at node $= A_{i} - A_{r}$
Given $\frac{A_{i} + A_{r}}{A_{i} - A_{r}} = 9 \Rightarrow \frac{A_{r}}{A_{i}} = 0.8$
Intensity is proportional to the square of the amplitude.
Hence
$\frac{I_{r}}{I_{i}} = {(\frac{A_{r}}{A_{i}})}^{2} = {(0.8)}^{2} = 0.64 = 64 %$

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