`Delta x_(min) = (2n - 1) (lambda)/(2)`
The farthest mimima has path difference `lambda//2` while the nearest minima has path difference `(3//2)lambda`. For the nearest minima,
`S_(1) P - S_(2) P = (3)/(2) lambda` [as maximum path difference is `2lambda`]
`implies sqrt((2lambda)^(2) + D^(2)) - D = (3)/(2) lambda`
`implies (2lambda)^(2) + D^(2) = ((3)/(2) lambda + D)^(2)`
`implies 4lambda^(2) + D^(2) = (9)/(4) lambda^(2) + D^(2) xx 2 xx (3)/(2) lambda xx D`
`implies 3D = 4lambda - (9lambda)/(4) = (7lambda)/(4) implies D = (7)/(12) lambda`
For the farthest minima,
`S_(1)P - S_(2)P = (lambda)/(2)`
`impliessqrt (4 lambda^(2) + D^(2)) - D = (lambda)/(2)`
`4lambda^(2) + D^(2) = (lambda^2)/(4) + D^(2) + D lambda implies D = 4 lambda - lambda//4 = (15 lambda)/(4)`.