Answer
$D=1.0\times10^{-6}m$
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.
$$\Delta \theta =51.0^{\circ}= 2 \theta_1$$
$$\theta_1=25.5^{\circ}$$
Use equation 24–3a to calculate the angular width from the middle of the central peak to the first minimum.
$$sin\theta_1=\frac{\lambda}{D}$$
$$D =\frac{\lambda}{ sin\theta_1}$$
$$D =\frac{440\times10^{-9}m }{sin25.5^{\circ}}$$
$$D=1.0\times10^{-6}m$$
