Derive the relation a sin θ = λ for the first minimum of the diffraction pattern produced due to a single slit of width 'a' using light of wavelength λ.
As we can see from the figure, rays AP and BP reach to the screen from the two ends of a single slit. The ray BP travels an extra distance of BC as compared to ray AP.
Hence the path difference produced,Δx = BC ..........(1)
The ray reaching to screen at point P from point O makes the angle θ with the horizontal.
∵ ∠POQ = θ
∴∠BAC = θ
BC = a sin θ
Using equation (1), Δx = BC
⇒ Δx = a sin θ .............(2)
For diffraction pattern produced due to a single slit of width, 'a' using the light of wavelength, path difference for destructive interference is given by, Δx = mλ (where m = 1, -1, 2, -2, 3, -3, 4..............)
Using equation (2), we can write, a sin θ = mλ
For the first minimum,
m = 1,
∴ a sin θ = λ