In the double-concave lens ${\displaystyle r'}$ is negative,

⁠${\displaystyle {\frac {1}{F}}=(m-1)\left\{{\frac {1}{r}}+{\frac {1}{r'}}\right\}.}$

In the double-convex ${\displaystyle r}$ is negative,

⁠${\displaystyle {\frac {1}{F}}=-(m-1)\left\{{\frac {1}{r}}+{\frac {1}{r'}}\right\}.}$

In the plano-concave either ${\displaystyle r'}$ is infinite, or ${\displaystyle r}$ is infinite, and ${\displaystyle r'}$ negative; therefore putting ${\displaystyle r}$ for the single radius

⁠${\displaystyle {\frac {1}{F}}={\frac {m-1}{r}},\quad F={\frac {r}{m-1}}.}$

In the plane-convex,

⁠${\displaystyle {\frac {1}{F}}=-{\frac {m-1}{r}},\quad F={\frac {r}{m-1}}.}$

When in the double-concave, or double-convex lens the radii are equal,

⁠${\displaystyle {\frac {1}{F}}=\pm (m-1).{\frac {2}{r}},\quad \mathrm {or} \quad F=\pm {\frac {r}{2(m-1)}}.}$^[1]

88.⁠It appears from all this, that the place of the principal focus is the same, whichever side of a lens is turned towards the incident light, and that

The concavo-convex[2]	${\displaystyle \scriptstyle {\left.{\begin{matrix}\ \\\\\ \\\ \ \end{matrix}}\right\}\,}}$	make parallel rays diverge.
the double-concave
and the plano-concave
The meniscus	${\displaystyle \scriptstyle {\left.{\begin{matrix}\ \\\\\ \\\ \ \end{matrix}}\right\}\,}}$	make parallel rays converge.
the double-convex
and the plano-convex