The half angular width of the central bright maximum in Frauhoffer diffraction pattern, for a slit of width 1.2 × 10^-3 mm illuminated by monochromatic light of wavelength 600 nm is :

$\frac{\pi}{6}$

As we know that in franhaufer diffraction 

$sin\theta = \frac{\lambda}{a}$

where θ is the half angular width of the central maximum.

Given that $\lambda = 6\times 10^{-7}m , a = 1.2\times 10^{-6}m$

$\Rightarrow sin\theta = \frac{ 6\times 10^{-7}}{1.2 \times 10^{-6}} = 0.5 = 30^o$

$\theta = \frac{\pi}{6}$