JEE Main 25 Feb 2021 Shift 2 Solved Paper

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.