In a diffraction pattern due to a single slit of width a, the first...

The condition for the first minimum in a single slit diffraction pattern is given by $a \sin \theta = \lambda$.

Substituting $\theta = 30^\circ$, we get $a \sin 30^\circ = \lambda$.

Since $\sin 30^\circ = \dfrac{1}{2}$, we have $a \left(\dfrac{1}{2}\right) = \lambda$, which gives $a = 2\lambda$.

The condition for the first secondary maximum is given by $a \sin \theta' = \dfrac{3\lambda}{2}$.

Substituting $a = 2\lambda$, we get $2\lambda \sin \theta' = \dfrac{3\lambda}{2}$.

Solving for $\sin \theta'$, we get $\sin \theta' = \dfrac{3}{4}$.

Therefore, $\theta' = \sin^{-1}\left(\dfrac{3}{4}\right)$.

Answer: $\sin^{-1}\left(\dfrac{3}{4}\right)