Answer
We have shown that the FWHM ( full width at half-maximum ) is $\Delta \theta= 2 \arcsin \frac{0.442\lambda}{ a}$. A short answer is not appropriate, the steps in the proof are the solution. See step-by-step work for solution.
Work Step by Step
First we consider equation (36-6) when it's isolating for theta.
$\alpha = \frac{\pi a}{\lambda}\sin \theta$
$\theta = \arcsin \frac{\alpha \lambda}{\pi a}$
$\theta = \arcsin \frac{1.39\lambda}{\pi a}$
$\theta = \arcsin \frac{0.442\lambda}{ a}$
The difference in angle from one side of the pattern to the other is twice theta, so we have:
$\Delta\theta=2 \theta =2 \arcsin \frac{0.442\lambda}{ a}$
We have shown that the FWHM ( full width at half-maximum ) is $\Delta \theta= 2 \arcsin \frac{0.442\lambda}{ a}$
