To solve this problem, we need to understand the diffraction pattern due to a single slit.$\\$The condition for the first minimum in a single-slit diffraction pattern is given by:$\\ a \sin \theta = m \lambda \\$where $m$ is the order of the minimum ($m = 1$ for the first minimum), $\\ a$ is the slit width, $\theta$ is the angle of diffraction, and $\lambda$ is the wavelength of light.$\\$Given:$\\ \theta = 30^\circ, \lambda = 5000 \,$Å$= 5000 \times 10^{-10} \,$m$\\$For the first minimum ($m = 1$):$\\ a \sin 30^\circ = 1 \times 5000 \times 10^{-10} \\$Since $\sin 30^\circ = \frac{1}{2},$ we have:$\\ a \cdot \frac{1}{2} = 5000 \times 10^{-10} \\ a = 2 \times 5000 \times 10^{-10} \\ a = 10000 \times 10^{-10} \\ a = 10^{-6} \,$m$\\$The condition for the first secondary maximum is given by:$\\ a \sin \theta = \left(m + \frac{1}{2}\right) \lambda \\$For the first secondary maximum, $m = 1$ (since it occurs between the first and second minima):$\\ a \sin \theta = \left(1 + \frac{1}{2}\right) \lambda \\ a \sin \theta = \frac{3}{2} \lambda \\$Substitute the values of $a$ and $\lambda: \\ 10^{-6} \sin \theta = \frac{3}{2} \times 5000 \times 10^{-10} \\ \sin \theta = \frac{3 \times 5000 \times 10^{-10}}{2 \times 10^{-6}} \\ \sin \theta = \frac{15000 \times 10^{-10}}{2 \times 10^{-6}} \\ \sin \theta = \frac{15000}{20000} \\ \sin \theta = \frac{3}{4} \\$Therefore, the angle for the first secondary maximum is $\theta = \sin^{-1}\left(\frac{3}{4}\right). \\$This corresponds to Option 2.