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The resultant loudness at a point P is n...

The resultant loudness at a point P is `n` dB higher than the loudness of `S_1` which is one of the two identical sound sources `S_1` and `S_2` reaching at that point in phase. Find the value of n.

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The correct Answer is:
6
Loudness due to `S_(1)=J_(1)=ka^(2)` where a is the amplitude
Loudness due to `S_(1)` and `S_(2)`
`=I=k(2a)^(2)=4J_(1)`
`N=10log_(10)(4J_(1)//J_(1))`
`=10log_(10)(4)=10(0.6)=6`
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