Answer
$\theta =15.2^{\circ}$
Work Step by Step
We know the slit width from part 12a. We can use the given values and set $m=1$ in equation (36-3) and solve for the angle, $\theta$;
$\alpha \sin \theta = m \lambda $ where $m=1,2,3,....$ (minima)
Therefore;
$\theta = \arcsin( m \lambda / \alpha)$
$\theta = \arcsin( 1* 610nm /2330nm )$
$\theta =15.2^{\circ}$
