A single slit of width 0.1 mm is illuminated by a parallel beam of light of wavelength  $6000\stackrel{o}{A}$ and diffraction bands are observed on a screen 0.5 m from the slit. The distance of the third dark band from the central bright band is 

The angular separation of $n^{th}$  minima from screen centre (central maxima) is given as  $b\sin {\theta }_n=n\lambda$
From 3rd dark band $(n=3)$ we have
${\theta }_3\simeq \sin {\theta }_3=\frac{3\lambda }{b}\left[as\theta isverysmall\right]$
Distance of 3rd minima from O is
$x_3=D{\theta }_3=0.5\times \frac{3\times 6000\times {10}^{-10}}{0.1\times {10}^{-3}}$
$\Rightarrow x_3=9\times {10}^{-3}m=9mm$