Question

A slit is located 'at infinity' in front of a lens of focal length $1m$ and is illuminated normally with light of wavelength $6000\mathring{A}$. The first minima on either side of the central maximum of the diffraction pattern observed in the focal plane of the lens are separated by $4mm$. The width of the slit (in mm) is_______.

Answer 1

Answer: 0.30

First-order minima on either side of the central maxima are at an angular separation $\theta$ from the screen centre, which is given as $b\sin \theta =\lambda$ and for small $\theta$, we use $\sin \theta =\theta$, so we have $\theta =\frac{\lambda }{b}$ where $b$ is the slit width. The half angular spread of central maxima (up to first-order minima) is given as
$\tan \theta \approx \theta =\frac{2mm}{1m}=0.002m$
Thus we have, $b=\frac{\lambda }{0.002}$
$=\frac{600\times 10^{-9}}{0.002}=0.3mm$