Consider a two slit interference arrangements (figure) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of D in terms of `lambda` such that the first minima on the screen falls at a distance D from the centre O.

Explanation

Using figure,
`T_1P = OP - OT_1 = OP - CS_1 = x - D`
Similarly,
`T_2 P = OT_2 + OP = CS_2 + OP = D + x `
`:. S_1 P = sqrt((S_1T_1)^2 + (PT_1)^(2)) = [D^2 + (D = x)^(2)]^(1//2)`
and `S_2P = sqrt((S_2T_2)^(2) + (PT_2)^2)`
`= [D^2 + (D - x)^2]^(1//2)`
Path difference between the two waves reaching point P from slits `S_1 and S_2` is
`Deltax = S_2P - S_1P`
`= [D^2+ (D + x)^(2)]^(1//2) - [D^2 + (D - x)^2]^(1//2)`
Minima will occur when path difference
`Dleta x = (n lambda)/2`
For `n = 1`,
`Delta x = lambda/2`
`implies [D^2 + 4D^2]^(1//2) - [D^2 + (D - x)^2]^(1//2) = lambda/2`
When `x = D`
`Dleta x = [D^2 + 4D^2]^(1//2) - D = lambda/2`
or `D(sqrt(5) - 1) = lambda/2`
`implies D = lambda/(2(sqrt(5) - 1)) = lambda/(2.472)`.