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

Options

A. `(lamda)/(2pi) (phi + (pi)/(2))`
B. `(2pi)/(lamda)phi`
C. `(lamda)/(2pi) phi`
D. `(2pi)/(lamda) (phi - (pi)/(2))`

Answer

A

Explanation

The two waves are `y_(1) = a_(1) sin (omega t - (2pi x)/(lamda))`
and `y_(2) = a_(2) cos (omega t - (2pi x)/(lamda) + phi)`
`= a_(2) sin (omega t - (2pi x)/(lamda) + phi + (pi)/(2))`
`:.` The phase difference `= phi + (pi)/(2) = (2pi)/(lamda)` (path difference)
`:.` The path difference `= (lamda (phi + (pi)/(2)))/(2pi)`