Question:
Consider a two-slit interference arrangements 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 λ such that the first minima on the screen falls at a distance D from the centre O.
Solution:
The minima will occur when ∆x = S2P – S1P = (2n-1/2)λ
S1P = √D2 + (D – x)2
S2P = √D2 + (D + x)2
T2P = D + x
T1P = D – x
[D2 + (D+x)2]-1/2 – [D2+(D-x)2]1/2 = λ/2
D = 0.404λ