Given, diameter of pinhole, a=0.1µm=0.1×10−6m ∵ Path difference (∆x)=asinφ=nλ . . . (i) where, φ is the phase difference and λ be the wavelength. As, I=4I0cos2φ and sinφ=
nλ
a
[from Eq. (i)] If a increases ↔sinφ or φ decreases As φ decreases ↔cosφ increases ∴ Intensity increases. Hence, on decreasing diameter of pinhloe, the size of diffraction pattern decreases and intensity increases.