Wave Optics Part 1

Given, diameter of pinhole, $a=0.1\ \mu\mathrm{m}=0.1\times 10^{-6}\ \mathrm{m}$ $\because$ Path difference $(\Delta x)=a \sin \phi = n \lambda$ . . . (i) where, $\phi$ is the phase difference and $\lambda$ be the wavelength. As, $\;\; I=4 I_{0} \cos^{2} \phi$ and $\;\; \sin \phi = \; \frac{n\lambda}{a}$ [from Eq. (i)] If $a$ increases $\leftrightarrow \sin \phi$ or $\phi$ decreases As $\phi$ decreases $\leftrightarrow \cos \phi$ increases $\therefore$ Intensity increases. Hence, on decreasing diameter of pinhloe, the size of diffraction pattern decreases and intensity increases.