Answer
$d = 2.35\times10^{-6}m $; $48.7^{\circ}$
Work Step by Step
Solve Eq. 24–4 for the slit separation d.
$$d sin \theta = m \lambda $$
$$d = \frac{m \lambda}{ sin \theta }$$
$$d = \frac{1(589\times10^{-9}m)}{sin 14.5^{\circ}}$$
$$d = 2.35\times10^{-6}m $$
Now set m = 3 and find the angle of the third-order maximum.
$$d sin \theta = m \lambda $$
$$ \theta = sin^{-1}(\frac{m \lambda} {d})$$
$$ \theta = sin^{-1}(\frac{3(589\times10^{-9}m)} {2.352\times10^{-6}m})=48.7^{\circ}$$