= Solution
Use <Fresnel diffraction>, since the detector moves along the axis through successive near-field <interference> extrema. Write $z$ for its distance from the opening and $\rho$ for the integration coordinate within the opening. For a uniform incident <plane wave>, the on-axis <complex amplitude>, in the <paraxial approximation>, is proportional to
$$
\frac{e^{ikz}}{i\lambda z}\int_0^r e^{i\pi\rho^2/(\lambda z)}2\pi\rho\,d\rho
=e^{ikz}\left(1-e^{i\pi r^2/(\lambda z)}\right).
$$
Thus the <light intensity> relative to the incident value is
$$
\frac{I(z)}{I_{\rm inc}}=4\sin^2\left(\frac{\pi r^2}{2\lambda z}\right).
$$
The <Fresnel number> $F=r^2/(\lambda z)$ is an odd integer at a maximum and an even integer at a minimum. It decreases as the detector moves away. Adjacent extrema therefore differ by one:
$$
\frac{r^2}{\lambda}\left(\frac1{1.00\,\mathrm m}-\frac1{1.25\,\mathrm m}\right)=1.
$$
Consequently
$$
\boxed{r^2=2.25\times10^{-6}\,\mathrm{m^2},\qquad r=1.50\times10^{-3}\,\mathrm m=1.50\,\mathrm{mm}.}
$$
As a check, $F=5$ at $z=1\,\mathrm m$ and $F=4$ at $z=1.25\,\mathrm m$, so these are indeed a maximum followed by the next minimum. Also $r/z\simeq1.5\times10^{-3}$, consistent with the <paraxial approximation>.
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