 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-338/3/b/ii/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

At the intermediate

d

1/ f_{i d} = ρ_{i} (n_{i d} - 1)

. Therefore the curvature-factor ratio from the previous part implies

\frac{f _{2 d}}{f _{1 d}} = \frac{ρ _{1} ( n _{1 d} - 1 )}{ρ _{2} ( n _{2 d} - 1 )},

and hence

\frac{f _{2 d}}{f _{1 d}} = - \frac{( n _{2 F} - n _{2 C} ) / ( n _{2 d} - 1 )}{( n _{1 F} - n _{1 C} ) / ( n _{1 d} - 1 )} = - \frac{V _{1 d}}{V _{2 d}} .

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