In standard YDSE slits were moved apart symmetrically with relative velocity v, calculate the rate at which fringes pass a point at a distance x from the centre of the fringe system formed on a screen at 'y' distance away from the double slits if wavelength of light is `lambda` , and distance between the slits is 'b' . Assuming `y gtgt b & b gtgt lambda`

Answer

`(x)/(lambday)v`

Explanation

`x=nbeta`
differetiating the equation with respect to time
`0=(ndbeta)/(dt)+(betadn)/(dt) rArr (dn)/(dt)=-(n)/(beta).(dbeta)/(dt)`
`(dn)/(dt)=-(n)/(beta).(d)/(dt)((lambday)/(d)) rArr (dn)/(dt)=-(n)/(beta).lambday(d)/(dt)((1)/(d))`
`(dn)/(dt)=-(n)/(beta).-((lambdayv)/(d^(2))) rArr (dn)/(dt)=(n)/(beta)(lambday)/d^(2)v`
`(dn)/(dt)=(nbeta)/(beta^(2))(lambday)/(d^(2))v rArr (dn)/(dt)=(x)/(beta^(2))(lambda^(2)y^(2))/(d^(2)lambday)V`
`(dn)/(dt)=(xv)/(lambday)`