Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2018 / S19 / 1 / b / Solution

If the quoted coupling is $k={\epsilon }_y/{\epsilon }_x$, then ${\epsilon }_y=k{\epsilon }_x$. Combining the horizontal factor of eight with $k$ falling from $0.1$ to $0.02$ givesThis is the usual leading estimate treating the horizontal emittance in part a as unchanged by small coupling.

If one instead accounts for exact redistribution of a fixed natural-emittance sum, ${\epsilon }_x={\epsilon }_0/(1+k)$ and ${\epsilon }_y=k{\epsilon }_0/(1+k)$. Then the horizontal ratio is $(1/8)(1.1/1.02)\simeq 0.1348$ and the vertical ratio is $0.02696$, an improvement by about $37$. These are consistent refinements of the stated approximate factor forty, not a second independent reduction to apply afterward.