Question

A slit is located 'at infinity' in front of a lens of focal length $1 \,m$ 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 $4 \,mm$. The width of the slit (in mm) is_______.

Answer 1

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
$\tan \theta \approx \theta=\frac{2 mm }{1 m }=0.002\, m$
Thus we have, $b=\frac{\lambda}{0.002}$
$=\frac{600 \times 10^{-9}}{0.002}=0.3\, mm$