Answer
$\Delta \theta= 52.5^{\circ}$.
Work Step by Step
We are given that the FWHM ( full width at half-maximum ) is $\Delta \theta= 2 \arcsin \frac{0.442\lambda}{ a}$.
For the given $a/\lambda=1.0$, we have:
$\Delta \theta= 2 \arcsin \frac{0.442\lambda}{ a}$
$\Delta \theta= 2 \arcsin0.442( \frac{ a}{\lambda})^{-1}$
$\Delta \theta= 2 \arcsin0.442( 1.0)^{-1}$
$\Delta \theta= 52.5^{\circ}$.
