Answer
48 km
Work Step by Step
Use equation 24–3a to calculate the angular width from the middle of the central peak to the first minimum.
$$sin\theta_1=\frac{\lambda}{D}$$
$$\theta_1=sin^{-1}(\frac{\lambda}{D})$$
$$= sin^{-1}(\frac{633\times10^{-9}m}{1.0\times10^{-2}m})=(3.627\times10^{-3})^{\circ}$$
Find the distance (on the screen) between the centerline and the edge of the central maximum, by using the distance to the screen $\mathcal{l}$ and the tangent of the angle.
$$x_1=(\mathcal{l})tan\theta_1=(380\times10^6m)tan(3.627\times10^{-3})^{\circ}=2.4\times10^{4}m$$
The width of the entire central peak is two times this distance.
$$\Delta x = 2 x_1=4.8\times10^{4}m $$
