Question

Light of wavelength 589.3 nm is incident normally on a slit of width 0.1 mm. The angular width of the central diffraction maximum at a distance of 1m from the slit is

• A. ${0.68}^{\circ }$
• B. ${0.34}^{\circ }$
• C. ${2.05}^{\circ }$
• D. None of these

Answer 1

The correct option is A ${0.68}^{\circ }$

Let D be the distance between source and screed d the distance between coherent source then for central diffraction maxima,
where $\lambda$ is wavelength
Given, $\lambda =589.3nm=589.3\times {10}^{-9}m$
$d=0.1\times {10}^{-3}m,D=1m$
$\therefore W=\frac{2\times 589.3\times {10}^{-9}}{0.1\times {10}^{-3}}\times 1$
Angular width
$\theta =\frac{W}{D}=\frac{2\times 589.3\times {10}^{-9}}{0.1\times {10}^{-3}}rad={0.68}^{\circ }$