\(\mathrm{d} \theta=\frac{1.22 \lambda}{\mathrm{D}}(\mathrm{D}=\text { diameter })\)
\(\frac{\mathrm{x}}{\mathrm{d}}=\frac{1.22 \lambda}{\mathrm{D}}(\mathrm{d}=\text { distance between earth }\;and\; \text { moon })\)
\(\mathrm{x}=\frac{1.22 \times\left(5500 \times 10^{-10}\right) \times\left(4 \times 10^{8}\right)}{5}=53.68 \mathrm{m}\)
most appropriate is \(60 \mathrm{m}\)