A single slit of width $a$ is illuminated by a monochromatic light of wavelength $600 \mathrm{~nm}$. The value of ' $a$ ' for which first minimum appears at $\theta=30^{\circ}$ on the screen will be :

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. <br/><br/> The condition for the first minimum in the diffraction pattern is given by : <br/><br/> $a\sin\theta = \lambda$ <br/><br/> where $a$ is the width of the slit, $\theta$ is the angle of diffraction, and $\lambda$ is the wavelength of the incident light. <br/><br/> 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. <br/><br/> Substituting the given values into the condition for the first minimum, we get : <br/><br/> $a\sin 30^\circ = 600\,\mathrm{nm}$ <br/><br/> Simplifying, we get : <br/><br/> $a\cdot \frac{1}{2} = 600\,\mathrm{nm}$ <br/><br/> Multiplying both sides by 2, we get : <br/><br/> $a = 1200\,\mathrm{nm} = \boxed{1.2\,\mu\mathrm{m}}$ <br/><br/> Therefore, the width of the slit for which the first minimum appears at $\theta = 30^\circ$ is $\boxed{1.2\,\mu\mathrm{m}}$.