Question

Determine the angular separation between central maximum and first order maximum of the diffraction pattern due to a single slit of width 0.25 mm when light of wavelength 5890$A^0$ is incident on it normally?

Answer 1

$a\sin \theta =(2n+1)\frac{\lambda }{2}=(2\times 1+1)\frac{\lambda }{2}=\frac{3\lambda }{2}$
when $\theta$ s small, $\sin \theta =\theta$

$\Rightarrow a\theta =\frac{3\lambda }{2}$
$\Rightarrow \theta \frac{3\lambda }{2a}=\frac{3\times 5890\times {10}^{-10}}{2\times 0.25\times {10}^{-3}}$
$=3.534\times {10}^{-3}rad$
Total angular spread$=\pm 3.534\times {10}^{-3}rad$