Answer
$\alpha = 60.4\mu m$
Work Step by Step
Since we know the angle, $\theta$, from part 6a, we can use the given order and wavelength in equation (36-3) to solve for the slit width, $\alpha$.
$\alpha \sin \theta = m \lambda $ where $m=1,2,3,....$ (minima)
Therefore;
$\alpha = m \lambda / \sin \theta$
Because the angle given was from minima to minima and not from the center to a minima, we must use half the given angle, $1.20^{\circ}$, to obtain $\theta=1.20^{\circ}/2=0.60^{\circ}$
$\alpha = 1* 633*10^{-9}m / \sin 0.60^{\circ}$
$\alpha = 6.04*10^{-5}m$
$\alpha = 60.4\mu m$
