When light passes through a narrow slit, it diffracts and produces a diffraction pattern on a screen. The pattern consists of a central bright maximum flanked by a series of alternating bright and dark fringes.

The condition for the first minimum in the diffraction pattern is given by :

$$ a\sin\theta = \lambda $$

where $a$ is the width of the slit, $\theta$ is the angle of diffraction, and $\lambda$ is the wavelength of the incident light.

In this problem, we are given that the wavelength of the incident light is $\lambda = 600\,\mathrm{nm}$, and the angle of diffraction for the first minimum is $\theta = 30^\circ$. We want to find the width of the slit, $a$, for which this occurs.

Substituting the given values into the condition for the first minimum, we get :

$$ a\sin 30^\circ = 600\,\mathrm{nm} $$

Simplifying, we get :

$$ a\cdot \frac{1}{2} = 600\,\mathrm{nm} $$

Multiplying both sides by 2, we get :

$$ a = 1200\,\mathrm{nm} = \boxed{1.2\,\mu\mathrm{m}} $$

Therefore, the width of the slit for which the first minimum appears at $\theta = 30^\circ$ is $\boxed{1.2\,\mu\mathrm{m}}$.