For a single slit diffraction pattern, the condition for the $n^{\text{th}}$ minimum is given by $a \sin\theta = n\lambda$. For the first minimum ($n = 1$), $\theta = 30^\circ$: $a \sin 30^\circ = 1 \cdot \lambda$ $a \left(\frac{1}{2}\right) = \lambda \implies a = 2\lambda$ The condition for the $n^{\text{th}}$ secondary maximum is $a \sin\theta' = \left(n + \frac{1}{2}\right)\lambda$. For the first secondary maximum ($n = 1$): $a \sin\theta' = \frac{3\lambda}{2}$ Substitute $a = 2\lambda$ into the equation: $(2\lambda) \sin\theta' = \frac{3\lambda}{2}$ $\sin\theta' = \frac{3}{4}$ $\theta' = \sin^{-1}\left(\frac{3}{4}\right)$