In a single-slit Fraunhofer diffraction experiment, the first...

For a single-slit Fraunhofer diffraction, the condition for the $n$-th secondary minimum is given by:

$a \sin \theta = n\lambda$

For the first secondary minimum ($n=1$) at \theta = 30^{}^{\circ}:

a \sin 30^{}^{\circ} = \lambda

$a \left(\dfrac{1}{2}\right) = \lambda \Rightarrow a = 2\lambda$

The condition for the $n$-th secondary maximum is given by:

$a \sin \theta' = \left(n + \dfrac{1}{2}\right)\lambda$

For the first secondary maximum ($n=1$):

$a \sin \theta' = \dfrac{3\lambda}{2}$

Substituting $a = 2\lambda$ into the equation:

$(2\lambda) \sin \theta' = \dfrac{3\lambda}{2}$

$\sin \theta' = \dfrac{3}{4}$

$\theta' = \sin^{-1} \left(\dfrac{3}{4}\right)<br><br>Answer: \sin^{-1}(3/4)$