For a convergent doublet of separated lens, we have

$${1 \over f} = {1 \over {{f_1}}} + {1 \over {{f_2}}} - {d \over {{f_1}{f_2}}}$$ ...... (1)

where d is separation between two lens, f₁ and f₂ are focal lengths of component lenses, f is resultant focal length. Therefore, Eq. (1) becomes

$${1 \over {10}} = {1 \over {{f_1}}} + {1 \over {{f_2}}} - {2 \over {{f_1}{f_2}}} \Rightarrow {1 \over {10}} = \left( {{{{f_2} + {f_1} - 2} \over {{f_1}{f_2}}}} \right)$$

$$ \Rightarrow {f_1}{f_2} = 10{f_2} + 10{f_1} - 20$$

$$ \Rightarrow 10{f_1} + 10{f_2} - {f_1}{f_2} = + 20$$

For f₁ = 18 cm and f₂ = 20 cm, the above equation satisfies.