Answer
0.54 cm
Work Step by Step
Use equation 24–3a to calculate the angular width from the middle of the central peak to the first minimum.
$$sin\theta_1=\frac{\lambda}{D}$$
$$\theta_1=sin^{-1}(\frac{\lambda}{D})$$
$$= sin^{-1}(\frac{450\times10^{-9}m}{1.0\times10^{-3}m})=0.02578^{\circ}$$
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.
$$x_1=(\mathcal{l})tan\theta_1=(6.0m)tan0.02578^{\circ}=2.70\times10^{-3}m$$
The width of the entire central peak is two times this distance.
$$\Delta x = 2 x_1=5.4\times10^{-3}m $$