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Sounds from two identical sources `S_(1)` and `S_(2)` reach a point `P`. When the sounds reach directly, and in the same phase, the intensity at `P` is `I_(0)`. The power of `S_(1)` is now reduced by `64%` and the phase difference between `S_(1)` and `S_(2)` is varied continuously. The maximum and minimum intensities recorded at `P` are now `I_("max")` and `I_("min")`

Text Solution

Verified by Experts

The correct Answer is:
`I_(max) // I_(min)` = 16
When waves are in phase resultant intensity is `I_(0) = 4I_(1)` where `I_(1)` is intensity by each source
If `I_(2)` = `0.36 I_(1)` Then we use
`I_(max)/(I_(min)) = ((sqrt(I_(1)) + sqrt(I_(2)))/(sqrt(I_(1) ) - sqrt(I_(2))))^(2)`
`((1.6)/(0.4))^(2)` = 16
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