Answer
$\alpha = 0.118mm$
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;
$\alpha = m \lambda / \sin \theta$
$\alpha = 2* 441*10^{-9}m / \sin 0.430^{\circ}$
$\alpha = 1.18*10^{-4}m$
$\alpha = 0.118mm$
