`Delta x_(min) = (2n-1) (lambda)/(2)` The farthest minima has path difference `lambda//2` while 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)xx2xx(3)/(2)lambda xxD`
`implies 3D = 4lambda- (9lambda)/(4) = (7lambda)/(4) implies D = (7)/(12) lambda`
For the farthest minima, `S_(1)P-S_(2)P = (lambda)/(2)`
`implies sqrt(4lambda^(2)+D^(2))-D = (lambda)/(2)`
`implies 4 lambda^(2)+D^(2)= (lambda^(2))/(4)+D^(2)+Dlambda implies D= 4lambda-lambda//4 = (15lambda)/(4)`