Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 338 / 1 / c / ii

Table of contents
- Solution ii

Solution

At the

d

line, the focal lengths obey

1/ f_{i d} = (n_{i d} - 1) ρ_{i}

. Thus

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

Rearranging gives

\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}} .

The last equality uses the Abbe number definition. The minus sign signifies a converging and a diverging component, with the normal optical glass convention and nonzero finite focal lengths.

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