Answer
$4636\;m$
Work Step by Step
The angular location of the edge of the central diffraction disk is given by
$\sin\theta=\frac{1.22\lambda}{d}$ ....................................$(1)$
where, $d$ is the diameter of the beam aperture.
Let, $R$ is the radius of the central diffraction disk on the Moon’s surface and $D$ the distance between the earth and the Moon. Therefore, we can write
$\tan\theta=\frac{R}{D}$
As $\theta$ is very small, $\tan\theta\approx\sin\theta$
therefore,
$\sin\theta=\frac{R}{D}$ ....................................$(2)$
Equating Eq. $1$ and Eq. $2$, we obtain
$\frac{1.22\lambda}{d}=\frac{R}{D}$
$R=\frac{1.22\lambda D}{d}$
Substituting the given values, we obtain
$R=\frac{1.22\times600\times 10^{-9}\times3.8\times10^{5}\times10^{3}}{0.12}\;m=2318\;m$
Therefore, the diameter of the central diffraction disk on the Moon’s surface is $(2\times2318)\;m=4636\;m$
