 Solution

The independent, spatially homogeneous random-placement idealization gives a Poisson process of intersections along the sampling line. With line intensity $\mu$, an interval of length $s$ contains no ridge with probability $e^{-\mu s}$. The spacing $X$ thus has exponential distributionRandom orientation changes the intersection intensity, which has already been represented by the measured $\mu$. Merely saying that positions are random does not prove a Poisson process: independence and homogeneity are additional assumptions. Finite segments, clustering or excluded widths can invalidate them.

For a shifted lognormal distribution, write $Z=\log (X-\theta )\sim N(m,{\sigma }^2)$ with $\sigma >0$. Applying the change of variables formula, $dZ/dX=1/(X-\theta )$, givesThe threshold $\theta$ is a minimum observable separation. Finite keel widths impose geometric exclusion, and the sonar's footprint and ridge-identification criterion can suppress a shallow peak near a deeper peak. This sonar ridge shadowing or resolution effect means that $\theta$ need not be a universal physical distance between every pair of ridges; it can partly reflect how the profiles were processed.

A lognormal distribution is compatible with products of many positive factors: taking logarithms turns multiplicative changes into sums, which can approach a normal distribution. Repeated deformation, breakup, convergence and merging across scales are plausible contributors. The observed form therefore suggests correlated or multistage ridging rather than the simplest independent-intersection model. It does not identify a unique ridging mechanism. For example, directly generating $X=\theta +\exp Z$ with a normal distribution for $Z$ produces the same spacing law without specifying any particular mechanics. Distributional agreement must be supplemented by dynamical and spatial evidence.