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The path difference between the two wave...

The path difference between the two waves
`y_(1)=a_(1) sin(omega t-(2pi x)/(lambda)) and y(2)=a_(2) cos(omega t-(2pi x)/(lambda)+phi)` is

A

`(lambda)/(2pi) phi`

B

`(lambda)/(2pi) ( phi+(pi)/(2))`

C

`(2pi)/(lambda) (phi-(pi)/(2))`

D

`(2pi)/(lambda) phi`

Text Solution

Verified by Experts

The correct Answer is:
B
Phase diff. `=((pi)/(2)+ phi)`, patth diff ?
Path diff. `(lambda)/(2pi)` Phase diff. `(lambda)/(2pi) (( pi)/(2)+ phi)`
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