Monochromatic light of wavelength ? is incident normally on a narrow slit of width 'a' to produce the diffraction pattern on the screen placed at a distance D from the set with the help of a relevant diagram deduce the condition for obtaining maxima and minima on the screen. Use these conditions to show that angular width of the central maximum is twice the angular width of the secondary maximum.

(a) When plane wavefront coming from distant source illuminate the slit of size (=d), each point within the slit becomes the source of secondary wavelets, and these wavelets superpose on each other to generate the maxima and minima on the screen; path difference between the rays, directing to the point P on the screen can be given as

Condition of Minima: If set AB is divided into the equal halves (or in even parts) each of size d/2, for every point in part AM, there is a point in part MB that contribute the secondary wavelets out of phase (i.e., 180º). So net contribution from two halves becomes zero and hence intensity falls to zero for path difference Δ = nλ.

∴ d sinθ = nλ
d.θ = nλ where n is integer except n = 0.

Condition of maxima:
If slit AB is divided into three equal parts (or in odd parts). First two thirds of the slit having a path difference λ/2 between them cancel each other, and only the remaining one third of the set contributes to the intensity at the point between two minima, so for path difference Δ =(n + 1/2)λ.
We have dsinθ =(n + 1/2)λ.
where n is integer except n = 0.

Regards