Path different between waves
`y_(1) = A_(1) sin (omega t - (2pix)/(lamda)) and y _(2) = A_(2) cos (omega t - (2pi x)/(lamda) + phi)` at the point of superpositon is :

Options

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

Answer

B

Explanation

Here `y _(1) = A _(1) sin (omega t - (2pi x )/( lamda)) " "…(1)`
`y _(2) = A _(2) cos (omega t - (2pi x )/( lamda ) + pi ) `
`y _(2) = A _(2) sin ((pi)/(2) + omega t - (2pi x )/( lamda ) + phi) " "...(2)`
`{ because cos theta = sin ((pi)/(2) + theta ) }`
Now, phase difference, `Delta theta = (2pi)/(lamda ) xx Delta x`
`therefore ` Path difference `Delta x = (lamda )/(2pi ) xx (Delta theta )`
`therefore Delta x = (lamda )/(2pi ) (theta _(2) - theta _(1))`
`= (lamda )/( 2pi ) { ((pi )/(2) + omega t - (2pi x )/( lamda )+ phi )- (omega t - (2pi x )/( lamda )) }`
`therefore Delta x = (lamda )/( 2pi ) ((pi )/(2) + phi ) ` (From equation (1) and (2)