Derive the relation asinθ=λ for the first minimum of the diffraction pattern produced due to a single slit of width 'a' using light of wavelength λ.

A point 'P' is on the screen. At point P: The path difference = Δx → A → x = (BP - AP) = BC
In ΔABC
sinθ = BC/AB = Δx/a
Δx = asinθ ———(1)
From experimental observations, we know that intensity at central maxima at θ=0 and secondary maxima at θ=(n+1/2)λ/a and has minima at θ=nλ/a
So, sinθ = nλ/a (for minima)
For 1st minima, n=1
sinθ = λ/a
→ a sinθ = λ
Hence, proved