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 :

Path different between waves y_(1) = A_(1) sin (omega t - (2pix)/(lam

1.	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 :
Answer» `(lamda)/(2pi) PHI` `(lamda)/(2pi) (phi + (pi)/(2))` `(2pi)/(lamda) (phi ~ (pi)/(2))` `(2pi)/(lamda) phi` SOLUTION :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)