Answer
629 nm
Work Step by Step
The width of the entire central peak is two times this angular width of the central peak to the first minimum.
$$\Delta \theta =28.0^{\circ}= 2 \theta_1$$
$$\theta_1=14.0^{\circ}$$
Use equation 24–3a to calculate the angular width from the middle of the central peak to the first minimum.
$$sin\theta_1=\frac{\lambda}{D}$$
$$\lambda = D sin\theta_1=(2.60\times10^{-6}m)sin14^{\circ}$$
$$\lambda = 6.29\times10^{-7}m=629nm$$