= Solution
Use SI units throughout: $d=1.0\times10^{-3}\,\mathrm m$, $\lambda=1.0\times10^{-6}\,\mathrm m$, and $p=q=1.0\,\mathrm m$. The effective propagation distance is $R_{\rm eff}=0.50\,\mathrm m$, giving
$$
w=\frac{10^{-3}}2\sqrt{\frac{2}{(10^{-6})(0.50)}}=1.
$$
In the supplied <Cornu spiral>, the endpoint at $w=1$ is approximately $(C,S)=(0.78,0.44)$. Hence
$$
\boxed{\frac{I_P}{I_0}=4(0.78^2+0.44^2)^2\simeq2.6.}
$$
More accurate <Fresnel integrals>, $C(1)=0.7798934$ and $S(1)=0.4382591$, give $I_P/I_0=2.562$. The central <light intensity> is increased to about $2.6$ times the unobscured value, an increase of roughly $160\%$.
There is no violation of <conservation of energy>: the aperture reduces the total transmitted <power>, while removing contributions that partly cancel the central field redistributes the transmitted <power> across the observation plane. In particular, the square aperture does not give the same central field as a circular aperture of a superficially similar size.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2021/paper-2-b7-square-fresnel-intensity.png]
{title=Calculated central intensity and an off-axis section for the square aperture in the Fresnel approximation}
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