Past exam of the particle physics course of the University of Oxford / 2022 / 4 / b / ii

Let $\ell =\ln ({\theta }_{\max }/{\theta }_{\min })$. The supplied second moment gives the single-scatter root mean square angleFor independent scatters with uniform azimuth, the small angular-deflection vectors have zero mean. Their cross terms average to zero, so their variances add. After $N$ scatters,This is the root mean square magnitude of the two-dimensional angular deflection. Its projection into one specified plane has a variance smaller by a factor of two, hence a root mean square smaller by $\sqrt{2}$.