Answer
580 nm
Work Step by Step
The lines on the wing act like a reflection grating, discussed in section 24-6. Assume that the first order reflection is the one we see, m = 1. Use equation 24–4.
$$d sin \theta = m \lambda $$
Apply this to the first order fringe.
$$d = \frac{m \lambda}{ sin \theta }$$
$$d = \frac{1(480\times10^{-9}m)}{sin 56^{\circ}}$$
$$d = 5.8\times10^{-7}m $$