Answer
$ \theta =2.2 * 10^{–4}rad$
Work Step by Step
From part A, we know the slit width, $\alpha$, is 2.5mm, so we use equation (36-3) and solve for theta;
$\alpha \sin \theta = m \lambda $ where $m=1,2,3,....$ (minima)
$ \theta =\arcsin( m \lambda /\alpha )$
$ \theta =\arcsin( 1 *550*10^{-9}m /(2.5*10^{-3}m ))$
$ \theta =2.2 * 10^{–4}rad$
